measur11.miz



    begin

    theorem :: MEASUR11:1

    

     Th72: for F be disjoint_valued FinSequence, n,m be Nat st n < m holds ( union ( rng (F | n))) misses (F . m)

    proof

      let F be disjoint_valued FinSequence, n,m be Nat;

      assume

       A1: n < m;

      per cases ;

        suppose n >= ( len F);

        then m > ( len F) by A1, XXREAL_0: 2;

        then not m in ( dom F) by FINSEQ_3: 25;

        then (F . m) = {} by FUNCT_1:def 2;

        hence ( union ( rng (F | n))) misses (F . m);

      end;

        suppose

         A2: n < ( len F);

        for A be set st A in ( rng (F | n)) holds A misses (F . m)

        proof

          let A be set;

          assume A in ( rng (F | n));

          then

          consider k be object such that

           A3: k in ( dom (F | n)) & A = ((F | n) . k) by FUNCT_1:def 3;

          reconsider k as Element of NAT by A3;

          1 <= k <= ( len (F | n)) by A3, FINSEQ_3: 25;

          then

           A4: k <= n by A2, FINSEQ_1: 59;

          then A = (F . k) by A3, FINSEQ_3: 112;

          hence A misses (F . m) by A1, A4, PROB_2:def 2;

        end;

        hence ( union ( rng (F | n))) misses (F . m) by ZFMISC_1: 80;

      end;

    end;

    theorem :: MEASUR11:2

    

     Th73: for F be FinSequence, m,n be Nat st m <= n holds ( len (F | m)) <= ( len (F | n))

    proof

      let F be FinSequence, m,n be Nat;

      assume m <= n;

      then (F | m) = ((F | n) | m) by FINSEQ_1: 82;

      hence ( len (F | m)) <= ( len (F | n)) by FINSEQ_1: 79;

    end;

    theorem :: MEASUR11:3

    

     Th74: for F be FinSequence, n be Nat holds (( union ( rng (F | n))) \/ (F . (n + 1))) = ( union ( rng (F | (n + 1))))

    proof

      let F be FinSequence, n be Nat;

      now

        let x be set;

        assume x in (( union ( rng (F | n))) \/ (F . (n + 1)));

        per cases by XBOOLE_0:def 3;

          suppose x in ( union ( rng (F | n)));

          then

          consider A be set such that

           A2: x in A & A in ( rng (F | n)) by TARSKI:def 4;

          consider k be object such that

           A3: k in ( dom (F | n)) & A = ((F | n) . k) by A2, FUNCT_1:def 3;

          reconsider k as Element of NAT by A3;

          

           A4: 1 <= k <= ( len (F | n)) by A3, FINSEQ_3: 25;

          ( len (F | n)) <= n by FINSEQ_1: 86;

          then

           A5: k <= n & A = (F . k) by A4, A3, FINSEQ_3: 112, XXREAL_0: 2;

          n <= (n + 1) by NAT_1: 11;

          then

           A6: A = ((F | (n + 1)) . k) by A5, XXREAL_0: 2, FINSEQ_3: 112;

          ( len (F | n)) <= ( len (F | (n + 1))) by NAT_1: 11, Th73;

          then k <= ( len (F | (n + 1))) by A4, XXREAL_0: 2;

          then k in ( dom (F | (n + 1))) by A4, FINSEQ_3: 25;

          then A in ( rng (F | (n + 1))) by A6, FUNCT_1: 3;

          hence x in ( union ( rng (F | (n + 1)))) by A2, TARSKI:def 4;

        end;

          suppose x in (F . (n + 1));

          then

           A7: x in ((F | (n + 1)) . (n + 1)) by FINSEQ_3: 112;

          then (n + 1) in ( dom (F | (n + 1))) by FUNCT_1:def 2;

          then ((F | (n + 1)) . (n + 1)) in ( rng (F | (n + 1))) by FUNCT_1: 3;

          hence x in ( union ( rng (F | (n + 1)))) by A7, TARSKI:def 4;

        end;

      end;

      hence (( union ( rng (F | n))) \/ (F . (n + 1))) c= ( union ( rng (F | (n + 1))));

      let x be object;

      assume x in ( union ( rng (F | (n + 1))));

      then

      consider A be set such that

       A9: x in A & A in ( rng (F | (n + 1))) by TARSKI:def 4;

      consider k be object such that

       A10: k in ( dom (F | (n + 1))) & A = ((F | (n + 1)) . k) by A9, FUNCT_1:def 3;

      reconsider k as Element of NAT by A10;

      1 <= k <= ( len (F | (n + 1))) <= (n + 1) by A10, FINSEQ_1: 86, FINSEQ_3: 25;

      then

       A11: k <= (n + 1) & ((F | (n + 1)) . k) = (F . k) by XXREAL_0: 2, FINSEQ_3: 112;

      per cases ;

        suppose k = (n + 1);

        hence x in (( union ( rng (F | n))) \/ (F . (n + 1))) by A9, A10, A11, XBOOLE_0:def 3;

      end;

        suppose k <> (n + 1);

        then k < (n + 1) by A11, XXREAL_0: 1;

        then k <= n by NAT_1: 13;

        then

         A12: ((F | n) . k) = (F . k) by FINSEQ_3: 112;

        then k in ( dom (F | n)) by A11, A10, A9, FUNCT_1:def 2;

        then A in ( rng (F | n)) by A12, A11, A10, FUNCT_1: 3;

        then x in ( union ( rng (F | n))) by A9, TARSKI:def 4;

        hence x in (( union ( rng (F | n))) \/ (F . (n + 1))) by XBOOLE_0:def 3;

      end;

    end;

    theorem :: MEASUR11:4

    

     Th101: for F be disjoint_valued FinSequence, n be Nat holds ( Union (F | n)) misses (F . (n + 1))

    proof

      let F be disjoint_valued FinSequence, n be Nat;

      assume ( Union (F | n)) meets (F . (n + 1));

      then

      consider x be object such that

       A1: x in ( Union (F | n)) & x in (F . (n + 1)) by XBOOLE_0: 3;

      x in ( union ( rng (F | n))) by A1, CARD_3:def 4;

      then

      consider A be set such that

       A2: x in A & A in ( rng (F | n)) by TARSKI:def 4;

      consider m be object such that

       A3: m in ( dom (F | n)) & A = ((F | n) . m) by A2, FUNCT_1:def 3;

      reconsider m as Element of NAT by A3;

      m <= ( len (F | n)) & ( len (F | n)) <= n by A3, FINSEQ_3: 25, FINSEQ_1: 86;

      then m <> (n + 1) by NAT_1: 13;

      then (F . m) misses (F . (n + 1)) by PROB_2:def 2;

      then (((F | n) . m) /\ (F . (n + 1))) = {} by A3, FUNCT_1: 47;

      hence contradiction by A1, A2, A3, XBOOLE_0:def 4;

    end;

    theorem :: MEASUR11:5

    

     Th41: for P be set, F be FinSequence st P is cup-closed & {} in P & (for n be Nat st n in ( dom F) holds (F . n) in P) holds ( Union F) in P

    proof

      let P be set, F be FinSequence;

      assume that

       A0: P is cup-closed and

       A1: {} in P and

       A2: for n be Nat st n in ( dom F) holds (F . n) in P;

      defpred P[ Nat] means ( union ( rng (F | $1))) in P;

      

       A3: P[ 0 ] by A1, ZFMISC_1: 2;

      

       A4: for k be Nat st P[k] holds P[(k + 1)]

      proof

        let k be Nat;

        assume

         A5: P[k];

        

         A6: k <= (k + 1) by NAT_1: 13;

        per cases ;

          suppose

           A7: ( len F) >= (k + 1);

          then ( len (F | (k + 1))) = (k + 1) by FINSEQ_1: 59;

          

          then (F | (k + 1)) = (((F | (k + 1)) | k) ^ <*((F | (k + 1)) . (k + 1))*>) by FINSEQ_3: 55

          .= ((F | k) ^ <*((F | (k + 1)) . (k + 1))*>) by A6, FINSEQ_1: 82

          .= ((F | k) ^ <*(F . (k + 1))*>) by FINSEQ_3: 112;

          

          then ( rng (F | (k + 1))) = (( rng (F | k)) \/ ( rng <*(F . (k + 1))*>)) by FINSEQ_1: 31

          .= (( rng (F | k)) \/ {(F . (k + 1))}) by FINSEQ_1: 38;

          

          then

           A8: ( union ( rng (F | (k + 1)))) = (( union ( rng (F | k))) \/ ( union {(F . (k + 1))})) by ZFMISC_1: 78

          .= (( union ( rng (F | k))) \/ (F . (k + 1))) by ZFMISC_1: 25;

          1 <= (k + 1) by NAT_1: 11;

          then (F . (k + 1)) in P by A2, A7, FINSEQ_3: 25;

          hence P[(k + 1)] by A0, A5, A8, FINSUB_1:def 1;

        end;

          suppose ( len F) < (k + 1);

          then (F | (k + 1)) = F & (F | k) = F by FINSEQ_3: 49, NAT_1: 13;

          hence P[(k + 1)] by A5;

        end;

      end;

      for k be Nat holds P[k] from NAT_1:sch 2( A3, A4);

      then ( union ( rng (F | ( len F)))) in P;

      then ( union ( rng F)) in P by FINSEQ_3: 49;

      hence thesis by CARD_3:def 4;

    end;

    definition

      let A,X be set;

      :: original: chi

      redefine

      func chi (A,X) -> Function of X, ExtREAL ;

      coherence

      proof

        ( dom ( chi (A,X))) = X by FUNCT_3:def 3;

        hence thesis by FUNCT_2:def 1;

      end;

    end

    definition

      let X be non empty set, S be SigmaField of X, F be FinSequence of S;

      :: original: Union

      redefine

      func Union F -> Element of S ;

      coherence by PROB_3: 57;

    end

    definition

      let X be non empty set, S be SigmaField of X, F be sequence of S;

      :: original: Union

      redefine

      func Union F -> Element of S ;

      coherence

      proof

        ( union ( rng F)) is Element of S;

        hence thesis by CARD_3:def 4;

      end;

    end

    definition

      let X be non empty set;

      let F be FinSequence of ( PFuncs (X, ExtREAL ));

      let x be Element of X;

      :: MEASUR11:def1

      func F # x -> FinSequence of ExtREAL means

      : DEF5: ( dom it ) = ( dom F) & (for n be Element of NAT st n in ( dom it ) holds (it . n) = ((F . n) . x));

      existence

      proof

        defpred P[ Nat, set] means $2 = ((F . $1) . x);

        

         A1: for n be Nat st n in ( Seg ( len F)) holds ex z be Element of ExtREAL st P[n, z]

        proof

          let n be Nat;

          assume n in ( Seg ( len F));

          then n in ( dom F) by FINSEQ_1:def 3;

          then

          reconsider G = (F . n) as Element of ( PFuncs (X, ExtREAL )) by FINSEQ_2: 11;

          

           X1: G is PartFunc of X, ExtREAL by PARTFUN1: 46;

           X2:

          now

            per cases ;

              suppose x in ( dom G);

              hence (G . x) is Element of ExtREAL by X1, PARTFUN1: 4;

            end;

              suppose not x in ( dom G);

              then (G . x) = 0 by FUNCT_1:def 2;

              hence (G . x) is Element of ExtREAL by NUMBERS: 31, XREAL_0:def 1;

            end;

          end;

          take (G . x);

          thus thesis by X2;

        end;

        consider p be FinSequence of ExtREAL such that

         A2: ( dom p) = ( Seg ( len F)) and

         A3: for n be Nat st n in ( Seg ( len F)) holds P[n, (p . n)] from FINSEQ_1:sch 5( A1);

        take p;

        thus ( dom p) = ( dom F) by A2, FINSEQ_1:def 3;

        thus thesis by A2, A3;

      end;

      uniqueness

      proof

        let p1,p2 be FinSequence of ExtREAL ;

        assume that

         A4: ( dom p1) = ( dom F) and

         A5: for n be Element of NAT st n in ( dom p1) holds (p1 . n) = ((F . n) . x) and

         A6: ( dom p2) = ( dom F) and

         A7: for n be Element of NAT st n in ( dom p2) holds (p2 . n) = ((F . n) . x);

        

         B1: ( len p1) = ( len p2) by A4, A6, FINSEQ_3: 29;

        now

          let n be Nat;

          assume

           A10: n in ( dom p1);

          then (p1 . n) = ((F . n) . x) by A5;

          hence (p1 . n) = (p2 . n) by A4, A6, A7, A10;

        end;

        hence thesis by B1, FINSEQ_2: 9;

      end;

    end

    theorem :: MEASUR11:6

    for X be non empty set, S be non empty Subset-Family of X, f be FinSequence of S, F be FinSequence of ( PFuncs (X, ExtREAL )) st ( dom f) = ( dom F) & f is disjoint_valued & (for n be Nat st n in ( dom F) holds (F . n) = ( chi ((f . n),X))) holds (for x be Element of X holds (( chi (( Union f),X)) . x) = ( Sum (F # x)))

    proof

      let X be non empty set, S be non empty Subset-Family of X, f be FinSequence of S, F be FinSequence of ( PFuncs (X, ExtREAL ));

      assume that

       A0: ( dom f) = ( dom F) and

       A1: f is disjoint_valued and

       A2: for n be Nat st n in ( dom F) holds (F . n) = ( chi ((f . n),X));

      let x be Element of X;

      reconsider x1 = x as Element of X;

      consider Sf be sequence of ExtREAL such that

       B1: ( Sum (F # x)) = (Sf . ( len (F # x))) & (Sf . 0 ) = 0 & for i be Nat st i < ( len (F # x)) holds (Sf . (i + 1)) = ((Sf . i) + ((F # x) . (i + 1))) by EXTREAL1:def 2;

      per cases ;

        suppose

         A8: x in ( Union f);

        then x in ( union ( rng f)) by CARD_3:def 4;

        then

        consider fn be set such that

         A9: x in fn & fn in ( rng f) by TARSKI:def 4;

        consider n be Element of NAT such that

         A10: n in ( dom f) & fn = (f . n) by A9, PARTFUN1: 3;

        

         A11: for m be Nat holds (m = n implies ((F # x) . m) = 1) & (m <> n implies ((F # x) . m) = 0 )

        proof

          let m be Nat;

          hereby

            assume

             A12: m = n;

            then m in ( dom (F # x)) by A0, A10, DEF5;

            then ((F # x) . m) = ((F . m) . x) by DEF5;

            then ((F # x) . m) = (( chi ((f . m),X)) . x) by A2, A12, A10, A0;

            hence ((F # x) . m) = 1 by A9, A10, A12, FUNCT_3:def 3;

          end;

          assume m <> n;

          then

           A13: not x in (f . m) by A9, A10, A1, PROB_2:def 2, XBOOLE_0: 3;

          per cases ;

            suppose m in ( dom (F # x));

            then m in ( dom F) & ((F # x) . m) = ((F . m) . x) by DEF5;

            then ((F # x) . m) = (( chi ((f . m),X)) . x) by A2;

            hence ((F # x) . m) = 0 by A13, FUNCT_3:def 3;

          end;

            suppose not m in ( dom (F # x));

            hence ((F # x) . m) = 0 by FUNCT_1:def 2;

          end;

        end;

        defpred P1[ Nat] means $1 < n implies (Sf . $1) = 0 ;

        

         A14: P1[ 0 ] by B1;

        

         A15: for m be Nat st P1[m] holds P1[(m + 1)]

        proof

          let m be Nat;

          assume

           A16: P1[m];

          assume

           A17: (m + 1) < n;

          then

           A18: m < n by NAT_1: 13;

          

           A20: ((F # x) . (m + 1)) = 0 by A17, A11;

          n in ( dom (F # x)) by A0, A10, DEF5;

          then 1 <= n <= ( len (F # x)) by FINSEQ_3: 25;

          then m < ( len (F # x)) by A18, XXREAL_0: 2;

          

          then (Sf . (m + 1)) = ((Sf . m) + ((F # x) . (m + 1))) by B1

          .= ( 0 + 0 ) by A20, A16, A17, NAT_1: 13;

          hence (Sf . (m + 1)) = 0 ;

        end;

        

         A21: for m be Nat holds P1[m] from NAT_1:sch 2( A14, A15);

        defpred P2[ Nat] means n <= $1 <= ( len (F # x)) implies (Sf . $1) = 1;

        

         A23: P2[ 0 ] by A10, FINSEQ_3: 25;

        

         A24: for m be Nat st P2[m] holds P2[(m + 1)]

        proof

          let m be Nat;

          assume

           A25: P2[m];

          assume

           A26: n <= (m + 1) <= ( len (F # x));

          then

           A27: (Sf . (m + 1)) = ((Sf . m) + ((F # x) . (m + 1))) by B1, NAT_1: 13;

          per cases by A26, XXREAL_0: 1;

            suppose

             A28: n = (m + 1);

            then m < n by NAT_1: 13;

            then (Sf . m) = 0 & ((F # x) . (m + 1)) = 1 by A21, A28, A11;

            hence (Sf . (m + 1)) = 1 by A27, XXREAL_3: 4;

          end;

            suppose n < (m + 1);

            then (Sf . m) = 1 & ((F # x) . (m + 1)) = 0 by A25, A11, A26, NAT_1: 13;

            hence (Sf . (m + 1)) = 1 by A27, XXREAL_3: 4;

          end;

        end;

        

         A30: for m be Nat holds P2[m] from NAT_1:sch 2( A23, A24);

        n in ( dom (F # x)) by A10, A0, DEF5;

        then n <= ( len (F # x)) by FINSEQ_3: 25;

        then (Sf . ( len (F # x))) = 1 by A30;

        hence (( chi (( Union f),X)) . x) = ( Sum (F # x)) by A8, B1, FUNCT_3:def 3;

      end;

        suppose

         A31: not x in ( Union f);

        then not x in ( union ( rng f)) by CARD_3:def 4;

        then

         A32: for V be set st V in ( rng f) holds not x in V by TARSKI:def 4;

        defpred P3[ Nat] means $1 <= ( len (F # x)) implies (Sf . $1) = 0 ;

        

         A33: P3[ 0 ] by B1;

        

         A34: for m be Nat st P3[m] holds P3[(m + 1)]

        proof

          let m be Nat;

          assume

           A35: P3[m];

          assume

           A37: (m + 1) <= ( len (F # x));

          then

           A38: (m + 1) in ( dom (F # x)) by NAT_1: 11, FINSEQ_3: 25;

          then

           C2: (m + 1) in ( dom f) by A0, DEF5;

          then

           A39: not x in (f . (m + 1)) by A32, FUNCT_1: 3;

          ((F # x) . (m + 1)) = ((F . (m + 1)) . x) by A38, DEF5

          .= (( chi ((f . (m + 1)),X)) . x) by A2, C2, A0;

          then ((F # x) . (m + 1)) = 0 by A39, FUNCT_3:def 3;

          then ((Sf . m) + ((F # x) . (m + 1))) = 0 by A35, A37, NAT_1: 13;

          hence (Sf . (m + 1)) = 0 by A37, B1, NAT_1: 13;

        end;

        for m be Nat holds P3[m] from NAT_1:sch 2( A33, A34);

        then ( Sum (F # x)) = 0 by B1;

        hence (( chi (( Union f),X)) . x) = ( Sum (F # x)) by A31, FUNCT_3:def 3;

      end;

    end;

    begin

    theorem :: MEASUR11:7

    

     Th1: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2 holds ( sigma ( DisUnion ( measurable_rectangles (S1,S2)))) = ( sigma ( measurable_rectangles (S1,S2)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2;

      ( Field_generated_by ( measurable_rectangles (S1,S2))) = ( DisUnion ( measurable_rectangles (S1,S2))) by SRINGS_3: 22;

      hence thesis by SRINGS_3: 23;

    end;

    definition

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2;

      :: MEASUR11:def2

      func product_Measure (M1,M2) -> induced_Measure of ( measurable_rectangles (S1,S2)), ( product-pre-Measure (M1,M2)) means for E be set st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) holds for F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st E = ( Union F) holds (it . E) = ( Sum (( product-pre-Measure (M1,M2)) * F));

      existence

      proof

        consider IT be Measure of ( Field_generated_by ( measurable_rectangles (S1,S2))) such that

         A1: for E be set st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) holds for F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st E = ( Union F) holds (IT . E) = ( Sum (( product-pre-Measure (M1,M2)) * F)) by MEASURE9: 55;

        reconsider IT as induced_Measure of ( measurable_rectangles (S1,S2)), ( product-pre-Measure (M1,M2)) by A1, MEASURE9:def 8;

        take IT;

        thus thesis by A1;

      end;

      uniqueness

      proof

        let f1,f2 be induced_Measure of ( measurable_rectangles (S1,S2)), ( product-pre-Measure (M1,M2));

        assume that

         A1: for E be set st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) holds for F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st E = ( Union F) holds (f1 . E) = ( Sum (( product-pre-Measure (M1,M2)) * F)) and

         A2: for E be set st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) holds for F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st E = ( Union F) holds (f2 . E) = ( Sum (( product-pre-Measure (M1,M2)) * F));

        now

          let E be Element of ( Field_generated_by ( measurable_rectangles (S1,S2)));

          ( Field_generated_by ( measurable_rectangles (S1,S2))) = ( DisUnion ( measurable_rectangles (S1,S2))) by SRINGS_3: 22;

          then E in ( DisUnion ( measurable_rectangles (S1,S2)));

          then E in { A where A be Subset of [:X1, X2:] : ex F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st A = ( Union F) } by SRINGS_3:def 3;

          then

          consider A be Subset of [:X1, X2:] such that

           A3: E = A & ex F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st A = ( Union F);

          consider F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) such that

           A4: E = ( Union F) by A3;

          (f1 . E) = ( Sum (( product-pre-Measure (M1,M2)) * F)) by A1, A4;

          hence (f1 . E) = (f2 . E) by A2, A4;

        end;

        hence f1 = f2 by FUNCT_2: 63;

      end;

    end

    definition

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2;

      :: MEASUR11:def3

      func product_sigma_Measure (M1,M2) -> induced_sigma_Measure of ( measurable_rectangles (S1,S2)), ( product_Measure (M1,M2)) equals (( sigma_Meas ( C_Meas ( product_Measure (M1,M2)))) | ( sigma ( measurable_rectangles (S1,S2))));

      correctness

      proof

        ( Field_generated_by ( measurable_rectangles (S1,S2))) = ( DisUnion ( measurable_rectangles (S1,S2))) by SRINGS_3: 22;

        then

         A1: ( sigma ( Field_generated_by ( measurable_rectangles (S1,S2)))) = ( sigma ( measurable_rectangles (S1,S2))) by Th1;

        (( sigma_Meas ( C_Meas ( product_Measure (M1,M2)))) | ( sigma ( measurable_rectangles (S1,S2)))) is sigma_Measure of ( sigma ( measurable_rectangles (S1,S2))) by A1, MEASURE9: 61;

        hence thesis by A1, MEASURE9:def 9;

      end;

    end

    theorem :: MEASUR11:8

    

     Th2: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2 holds ( product_sigma_Measure (M1,M2)) is sigma_Measure of ( sigma ( measurable_rectangles (S1,S2)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2;

      ( Field_generated_by ( measurable_rectangles (S1,S2))) = ( DisUnion ( measurable_rectangles (S1,S2))) by SRINGS_3: 22;

      then ( sigma ( Field_generated_by ( measurable_rectangles (S1,S2)))) = ( sigma ( measurable_rectangles (S1,S2))) by Th1;

      hence thesis;

    end;

    theorem :: MEASUR11:9

    

     Th3: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, F1 be Set_Sequence of S1, F2 be Set_Sequence of S2, n be Nat holds [:(F1 . n), (F2 . n):] is Element of ( sigma ( measurable_rectangles (S1,S2)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, F1 be Set_Sequence of S1, F2 be Set_Sequence of S2, n be Nat;

      set S = ( measurable_rectangles (S1,S2));

      (F1 . n) in S1 & (F2 . n) in S2 by MEASURE8:def 2;

      then [:(F1 . n), (F2 . n):] in the set of all [:A, B:] where A be Element of S1, B be Element of S2;

      then

       A1: [:(F1 . n), (F2 . n):] in S by MEASUR10:def 5;

      

       A2: S c= ( DisUnion S) by SRINGS_3: 12;

      ( DisUnion S) c= ( sigma ( DisUnion S)) by PROB_1:def 9;

      then [:(F1 . n), (F2 . n):] is Element of ( sigma ( DisUnion ( measurable_rectangles (S1,S2)))) by A1, A2;

      hence thesis by Th1;

    end;

    theorem :: MEASUR11:10

    

     Th4: for X1,X2 be set, F1 be SetSequence of X1, F2 be SetSequence of X2, n be Nat st F1 is non-descending & F2 is non-descending holds [:(F1 . n), (F2 . n):] c= [:(F1 . (n + 1)), (F2 . (n + 1)):]

    proof

      let X1,X2 be set, F1 be SetSequence of X1, F2 be SetSequence of X2, n be Nat;

      assume F1 is non-descending & F2 is non-descending;

      then (F1 . n) c= (F1 . (n + 1)) & (F2 . n) c= (F2 . (n + 1)) by PROB_1:def 5, NAT_1: 11;

      hence [:(F1 . n), (F2 . n):] c= [:(F1 . (n + 1)), (F2 . (n + 1)):] by ZFMISC_1: 96;

    end;

    theorem :: MEASUR11:11

    

     Th5: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, A be Element of S1, B be Element of S2 holds (( product_Measure (M1,M2)) . [:A, B:]) = ((M1 . A) * (M2 . B))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, A be Element of S1, B be Element of S2;

      set S = ( measurable_rectangles (S1,S2));

      set P = ( product-pre-Measure (M1,M2));

      set m = ( product_Measure (M1,M2));

      

       A1: ( DisUnion S) = ( Field_generated_by S) by SRINGS_3: 22;

       [:A, B:] in the set of all [:A, B:] where A be Element of S1, B be Element of S2;

      then

       A2: [:A, B:] in S by MEASUR10:def 5;

      then

      reconsider F = <* [:A, B:]*> as disjoint_valued FinSequence of S by FINSEQ_1: 74;

      

       A3: S c= ( DisUnion S) by SRINGS_3: 12;

      consider SumPF be sequence of ExtREAL such that

       A4: ( Sum (P * F)) = (SumPF . ( len (P * F))) & (SumPF . 0 ) = 0. & (for n be Nat st n < ( len (P * F)) holds (SumPF . (n + 1)) = ((SumPF . n) + ((P * F) . (n + 1)))) by EXTREAL1:def 2;

      

       A5: ( len F) = 1 by FINSEQ_1: 39;

      then

       A6: 1 in ( dom F) by FINSEQ_3: 25;

      ( len (P * F)) = 1 by A5, FINSEQ_3: 120;

      then ( Sum (P * F)) = ((SumPF . 0 ) + ((P * F) . ( 0 + 1))) by A4;

      then ( Sum (P * F)) = ((P * F) . 1) by A4, XXREAL_3: 4;

      then ( Sum (P * F)) = (P . (F . 1)) by A6, FUNCT_1: 13;

      then ( Sum (P * F)) = (P . [:A, B:]) by FINSEQ_1: 40;

      then

       A7: ( Sum (P * F)) = ((M1 . A) * (M2 . B)) by MEASUR10: 22;

      ( rng <* [:A, B:]*>) = { [:A, B:]} by FINSEQ_1: 39;

      then ( union ( rng <* [:A, B:]*>)) = [:A, B:] by ZFMISC_1: 25;

      then [:A, B:] = ( Union <* [:A, B:]*>) by CARD_3:def 4;

      hence (m . [:A, B:]) = ((M1 . A) * (M2 . B)) by A1, A2, A3, A7, MEASURE9:def 8;

    end;

    theorem :: MEASUR11:12

    

     Th6: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F1 be Set_Sequence of S1, F2 be Set_Sequence of S2, n be Nat holds (( product_Measure (M1,M2)) . [:(F1 . n), (F2 . n):]) = ((M1 . (F1 . n)) * (M2 . (F2 . n)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F1 be Set_Sequence of S1, F2 be Set_Sequence of S2, n be Nat;

      (F1 . n) in S1 & (F2 . n) in S2 by MEASURE8:def 2;

      hence (( product_Measure (M1,M2)) . [:(F1 . n), (F2 . n):]) = ((M1 . (F1 . n)) * (M2 . (F2 . n))) by Th5;

    end;

    theorem :: MEASUR11:13

    for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F1 be FinSequence of S1, F2 be FinSequence of S2, n be Nat st n in ( dom F1) & n in ( dom F2) holds (( product_Measure (M1,M2)) . [:(F1 . n), (F2 . n):]) = ((M1 . (F1 . n)) * (M2 . (F2 . n))) by Th5;

    theorem :: MEASUR11:14

    for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Subset of [:X1, X2:] holds (( C_Meas ( product_Measure (M1,M2))) . E) = ( inf ( Svc (( product_Measure (M1,M2)),E))) by MEASURE8:def 8;

    theorem :: MEASUR11:15

    

     Th9: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2 holds ( sigma ( measurable_rectangles (S1,S2))) c= ( sigma_Field ( C_Meas ( product_Measure (M1,M2))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2;

      set C = ( C_Meas ( product_Measure (M1,M2)));

      set F = ( Field_generated_by ( measurable_rectangles (S1,S2)));

      F c= ( sigma_Field ( C_Meas ( product_Measure (M1,M2)))) by MEASURE8: 20;

      then ( sigma F) c= ( sigma_Field ( C_Meas ( product_Measure (M1,M2)))) by PROB_1:def 9;

      then ( sigma ( DisUnion ( measurable_rectangles (S1,S2)))) c= ( sigma_Field ( C_Meas ( product_Measure (M1,M2)))) by SRINGS_3: 22;

      hence thesis by Th1;

    end;

    theorem :: MEASUR11:16

    

     Th10: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2 st E = [:A, B:] holds (( product_sigma_Measure (M1,M2)) . E) = ((M1 . A) * (M2 . B))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2;

      assume

       A1: E = [:A, B:];

      then

       A2: (( product_sigma_Measure (M1,M2)) . [:A, B:]) = (( sigma_Meas ( C_Meas ( product_Measure (M1,M2)))) . [:A, B:]) by FUNCT_1: 49;

      

       A3: ( measurable_rectangles (S1,S2)) c= ( Field_generated_by ( measurable_rectangles (S1,S2))) by SRINGS_3: 21;

       [:A, B:] in the set of all [:A, B:] where A be Element of S1, B be Element of S2;

      then

       A4: [:A, B:] in ( measurable_rectangles (S1,S2)) by MEASUR10:def 5;

      ( product_Measure (M1,M2)) is completely-additive by MEASURE9: 60;

      then

       A5: (( product_Measure (M1,M2)) . [:A, B:]) = (( C_Meas ( product_Measure (M1,M2))) . [:A, B:]) by A3, A4, MEASURE8: 18;

      ( sigma ( measurable_rectangles (S1,S2))) c= ( sigma_Field ( C_Meas ( product_Measure (M1,M2)))) by Th9;

      then (( product_sigma_Measure (M1,M2)) . [:A, B:]) = (( product_Measure (M1,M2)) . [:A, B:]) by A1, A2, A5, MEASURE4:def 3;

      hence thesis by A1, Th5;

    end;

    theorem :: MEASUR11:17

    for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F1 be Set_Sequence of S1, F2 be Set_Sequence of S2, n be Nat holds (( product_sigma_Measure (M1,M2)) . [:(F1 . n), (F2 . n):]) = ((M1 . (F1 . n)) * (M2 . (F2 . n)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F1 be Set_Sequence of S1, F2 be Set_Sequence of S2, n be Nat;

      

       A1: [:(F1 . n), (F2 . n):] is Element of ( sigma ( measurable_rectangles (S1,S2))) by Th3;

      then

       A2: (( product_sigma_Measure (M1,M2)) . [:(F1 . n), (F2 . n):]) = (( sigma_Meas ( C_Meas ( product_Measure (M1,M2)))) . [:(F1 . n), (F2 . n):]) by FUNCT_1: 49;

      

       A3: ( measurable_rectangles (S1,S2)) c= ( Field_generated_by ( measurable_rectangles (S1,S2))) by SRINGS_3: 21;

      (F1 . n) in S1 & (F2 . n) in S2 by MEASURE8:def 2;

      then [:(F1 . n), (F2 . n):] in the set of all [:A, B:] where A be Element of S1, B be Element of S2;

      then

       A4: [:(F1 . n), (F2 . n):] in ( measurable_rectangles (S1,S2)) by MEASUR10:def 5;

      ( product_Measure (M1,M2)) is completely-additive by MEASURE9: 60;

      then

       A5: (( product_Measure (M1,M2)) . [:(F1 . n), (F2 . n):]) = (( C_Meas ( product_Measure (M1,M2))) . [:(F1 . n), (F2 . n):]) by A3, A4, MEASURE8: 18;

      ( sigma ( measurable_rectangles (S1,S2))) c= ( sigma_Field ( C_Meas ( product_Measure (M1,M2)))) by Th9;

      then (( product_sigma_Measure (M1,M2)) . [:(F1 . n), (F2 . n):]) = (( product_Measure (M1,M2)) . [:(F1 . n), (F2 . n):]) by A1, A2, A5, MEASURE4:def 3;

      hence thesis by Th6;

    end;

    theorem :: MEASUR11:18

    

     Th12: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E1,E2 be Element of ( sigma ( measurable_rectangles (S1,S2))) st E1 misses E2 holds (( product_sigma_Measure (M1,M2)) . (E1 \/ E2)) = ((( product_sigma_Measure (M1,M2)) . E1) + (( product_sigma_Measure (M1,M2)) . E2))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E1,E2 be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume

       A1: E1 misses E2;

      ( product_sigma_Measure (M1,M2)) is sigma_Measure of ( sigma ( measurable_rectangles (S1,S2))) by Th2;

      hence thesis by A1, MEASURE1: 30;

    end;

    theorem :: MEASUR11:19

    for X1,X2,A,B be set, F1 be SetSequence of X1, F2 be SetSequence of X2, F be SetSequence of [:X1, X2:] st F1 is non-descending & ( lim F1) = A & F2 is non-descending & ( lim F2) = B & (for n be Nat holds (F . n) = [:(F1 . n), (F2 . n):]) holds ( lim F) = [:A, B:]

    proof

      let X1,X2,A,B be set, F1 be SetSequence of X1, F2 be SetSequence of X2, F be SetSequence of [:X1, X2:];

      assume that

       A1: F1 is non-descending and

       A2: ( lim F1) = A and

       A3: F2 is non-descending and

       A4: ( lim F2) = B and

       A5: for n be Nat holds (F . n) = [:(F1 . n), (F2 . n):];

      now

        let n be Nat;

        (F . n) = [:(F1 . n), (F2 . n):] & (F . (n + 1)) = [:(F1 . (n + 1)), (F2 . (n + 1)):] by A5;

        hence (F . n) c= (F . (n + 1)) by A1, A3, Th4;

      end;

      then F is non-descending by PROB_2: 7;

      then

       A6: ( lim F) = ( Union F) by SETLIM_1: 63;

      ( Union F1) = A & ( Union F2) = B by A1, A2, A3, A4, SETLIM_1: 63;

      then

       A8: ( union ( rng F1)) = A & ( union ( rng F2)) = B by CARD_3:def 4;

      then

       A7: [:A, B:] = ( union { [:P, Q:] where P be Element of ( rng F1), Q be Element of ( rng F2) : P in ( rng F1) & Q in ( rng F2) }) by LATTICE5: 2;

      now

        let z be object;

        assume z in [:A, B:];

        then

        consider Z be set such that

         X1: z in Z & Z in { [:A, B:] where A be Element of ( rng F1), B be Element of ( rng F2) : A in ( rng F1) & B in ( rng F2) } by A7, TARSKI:def 4;

        consider A be Element of ( rng F1), B be Element of ( rng F2) such that

         X2: Z = [:A, B:] & A in ( rng F1) & B in ( rng F2) by X1;

        consider n1 be Element of NAT such that

         X3: n1 in ( dom F1) & A = (F1 . n1) by PARTFUN1: 3;

        consider n2 be Element of NAT such that

         X4: n2 in ( dom F2) & B = (F2 . n2) by PARTFUN1: 3;

        set n = ( max (n1,n2));

        A c= (F1 . n) & B c= (F2 . n) by A1, A3, X3, X4, PROB_1:def 5, XXREAL_0: 25;

        then

         X5: Z c= [:(F1 . n), (F2 . n):] by X2, ZFMISC_1: 96;

        n in NAT ;

        then n in ( dom F) by FUNCT_2:def 1;

        then (F . n) in ( rng F) by FUNCT_1: 3;

        then [:(F1 . n), (F2 . n):] in ( rng F) by A5;

        hence z in ( union ( rng F)) by X1, X5, TARSKI:def 4;

      end;

      then

       X6: [:A, B:] c= ( union ( rng F));

      now

        let z be object;

        assume z in ( union ( rng F));

        then

        consider Z be set such that

         Y1: z in Z & Z in ( rng F) by TARSKI:def 4;

        consider n be Element of NAT such that

         Y2: n in ( dom F) & Z = (F . n) by Y1, PARTFUN1: 3;

        

         Y3: Z = [:(F1 . n), (F2 . n):] by A5, Y2;

        ( dom F1) = NAT & ( dom F2) = NAT by FUNCT_2:def 1;

        then (F1 . n) c= ( union ( rng F1)) & (F2 . n) c= ( union ( rng F2)) by FUNCT_1: 3, ZFMISC_1: 74;

        then Z c= [:A, B:] by A8, Y3, ZFMISC_1: 96;

        hence z in [:A, B:] by Y1;

      end;

      then ( union ( rng F)) c= [:A, B:];

      hence ( lim F) = [:A, B:] by A6, X6, CARD_3:def 4;

    end;

    begin

    definition

      let X be set, Y be non empty set, E be Subset of [:X, Y:], x be set;

      :: MEASUR11:def4

      func X-section (E,x) -> Subset of Y equals { y where y be Element of Y : [x, y] in E };

      correctness

      proof

        now

          let y be set;

          assume y in { y where y be Element of Y : [x, y] in E };

          then ex y1 be Element of Y st y = y1 & [x, y1] in E;

          hence y in Y;

        end;

        then { y where y be Element of Y : [x, y] in E } c= Y;

        hence thesis;

      end;

    end

    definition

      let X be non empty set, Y be set, E be Subset of [:X, Y:], y be set;

      :: MEASUR11:def5

      func Y-section (E,y) -> Subset of X equals { x where x be Element of X : [x, y] in E };

      correctness

      proof

        now

          let x be set;

          assume x in { x where x be Element of X : [x, y] in E };

          then ex x1 be Element of X st x = x1 & [x1, y] in E;

          hence x in X;

        end;

        then { x where x be Element of X : [x, y] in E } c= X;

        hence thesis;

      end;

    end

    theorem :: MEASUR11:20

    

     Th14: for X be set, Y be non empty set, E1,E2 be Subset of [:X, Y:], p be set st E1 c= E2 holds ( X-section (E1,p)) c= ( X-section (E2,p))

    proof

      let X be set, Y be non empty set, E1,E2 be Subset of [:X, Y:], p be set;

      assume

       A1: E1 c= E2;

      now

        let y be set;

        assume y in ( X-section (E1,p));

        then ex y1 be Element of Y st y = y1 & [p, y1] in E1;

        hence y in ( X-section (E2,p)) by A1;

      end;

      hence ( X-section (E1,p)) c= ( X-section (E2,p));

    end;

    theorem :: MEASUR11:21

    

     Th15: for X be non empty set, Y be set, E1,E2 be Subset of [:X, Y:], p be set st E1 c= E2 holds ( Y-section (E1,p)) c= ( Y-section (E2,p))

    proof

      let X be non empty set, Y be set, E1,E2 be Subset of [:X, Y:], p be set;

      assume

       A1: E1 c= E2;

      let y be object;

      assume y in ( Y-section (E1,p));

      then ex y1 be Element of X st y = y1 & [y1, p] in E1;

      hence y in ( Y-section (E2,p)) by A1;

    end;

    theorem :: MEASUR11:22

    

     Th16: for X,Y be non empty set, A be Subset of X, B be Subset of Y, p be set holds (p in A implies ( X-section ( [:A, B:],p)) = B) & ( not p in A implies ( X-section ( [:A, B:],p)) = {} ) & (p in B implies ( Y-section ( [:A, B:],p)) = A) & ( not p in B implies ( Y-section ( [:A, B:],p)) = {} )

    proof

      let X,Y be non empty set, A be Subset of X, B be Subset of Y, p be set;

      set E = [:A, B:];

      hereby

        assume

         A2: p in A;

        now

          let y be set;

          assume y in ( X-section ( [:A, B:],p));

          then ex y1 be Element of Y st y = y1 & [p, y1] in E;

          hence y in B by ZFMISC_1: 87;

        end;

        then

         A3: ( X-section ( [:A, B:],p)) c= B;

        now

          let y be set;

          assume

           A4: y in B;

          then [p, y] in [:A, B:] by A2, ZFMISC_1: 87;

          hence y in ( X-section ( [:A, B:],p)) by A4;

        end;

        then B c= ( X-section ( [:A, B:],p));

        hence ( X-section ( [:A, B:],p)) = B by A3;

      end;

      hereby

        assume

         A5: not p in A;

        now

          let y be set;

          assume y in ( X-section ( [:A, B:],p));

          then ex y1 be Element of Y st y = y1 & [p, y1] in E;

          hence contradiction by A5, ZFMISC_1: 87;

        end;

        then ( X-section ( [:A, B:],p)) is empty;

        hence ( X-section ( [:A, B:],p)) = {} ;

      end;

      hereby

        assume

         A4: p in B;

        now

          let x be set;

          assume x in ( Y-section ( [:A, B:],p));

          then ex x1 be Element of X st x = x1 & [x1, p] in E;

          hence x in A by ZFMISC_1: 87;

        end;

        then

         A5: ( Y-section ( [:A, B:],p)) c= A;

        now

          let x be set;

          assume

           A6: x in A;

          then [x, p] in [:A, B:] by A4, ZFMISC_1: 87;

          hence x in ( Y-section ( [:A, B:],p)) by A6;

        end;

        then A c= ( Y-section ( [:A, B:],p));

        hence ( Y-section ( [:A, B:],p)) = A by A5;

      end;

      assume

       A7: not p in B;

      now

        let x be set;

        assume x in ( Y-section ( [:A, B:],p));

        then ex x1 be Element of X st x = x1 & [x1, p] in E;

        hence contradiction by A7, ZFMISC_1: 87;

      end;

      then ( Y-section ( [:A, B:],p)) is empty;

      hence ( Y-section ( [:A, B:],p)) = {} ;

    end;

    theorem :: MEASUR11:23

    

     Th17: for X,Y be non empty set, E be Subset of [:X, Y:], p be set holds ( not p in X implies ( X-section (E,p)) = {} ) & ( not p in Y implies ( Y-section (E,p)) = {} )

    proof

      let X,Y be non empty set, E be Subset of [:X, Y:], p be set;

      hereby

        assume

         A1: not p in X;

        now

          let y be set;

          assume y in ( X-section (E,p));

          then ex y1 be Element of Y st y = y1 & [p, y1] in E;

          hence contradiction by A1, ZFMISC_1: 87;

        end;

        then ( X-section (E,p)) is empty;

        hence ( X-section (E,p)) = {} ;

      end;

      assume

       A7: not p in Y;

      now

        let y be set;

        assume y in ( Y-section (E,p));

        then ex y1 be Element of X st y = y1 & [y1, p] in E;

        hence contradiction by A7, ZFMISC_1: 87;

      end;

      then ( Y-section (E,p)) is empty;

      hence ( Y-section (E,p)) = {} ;

    end;

    theorem :: MEASUR11:24

    

     Th18: for X,Y be non empty set, p be set holds ( X-section (( {} [:X, Y:]),p)) = {} & ( Y-section (( {} [:X, Y:]),p)) = {} & (p in X implies ( X-section (( [#] [:X, Y:]),p)) = Y) & (p in Y implies ( Y-section (( [#] [:X, Y:]),p)) = X)

    proof

      let X,Y be non empty set, p be set;

      now

        let q be set;

        assume q in ( X-section (( {} [:X, Y:]),p));

        then ex y1 be Element of Y st q = y1 & [p, y1] in ( {} [:X, Y:]);

        hence contradiction;

      end;

      then ( X-section (( {} [:X, Y:]),p)) is empty;

      hence ( X-section (( {} [:X, Y:]),p)) = {} ;

      now

        let q be set;

        assume q in ( Y-section (( {} [:X, Y:]),p));

        then ex x1 be Element of X st q = x1 & [x1, p] in ( {} [:X, Y:]);

        hence contradiction;

      end;

      then ( Y-section (( {} [:X, Y:]),p)) is empty;

      hence ( Y-section (( {} [:X, Y:]),p)) = {} ;

      

       A3: ( [#] X) = X & ( [#] Y) = Y by SUBSET_1:def 3;

      then

       A4: ( [#] [:X, Y:]) = [:( [#] X), ( [#] Y):] by SUBSET_1:def 3;

      hence p in X implies ( X-section (( [#] [:X, Y:]),p)) = Y by A3, Th16;

      assume p in Y;

      hence ( Y-section (( [#] [:X, Y:]),p)) = X by A3, A4, Th16;

    end;

    theorem :: MEASUR11:25

    

     Th19: for X,Y be non empty set, E be Subset of [:X, Y:], p be set holds (p in X implies ( X-section (( [:X, Y:] \ E),p)) = (Y \ ( X-section (E,p)))) & (p in Y implies ( Y-section (( [:X, Y:] \ E),p)) = (X \ ( Y-section (E,p))))

    proof

      let X,Y be non empty set, E be Subset of [:X, Y:], p be set;

      hereby

        assume

         A1: p in X;

        now

          let y be set;

          assume

           A2: y in ( X-section (( [:X, Y:] \ E),p));

          then

           A3: ex y1 be Element of Y st y = y1 & [p, y1] in ( [:X, Y:] \ E);

          now

            assume y in ( X-section (E,p));

            then ex y2 be Element of Y st y = y2 & [p, y2] in E;

            hence contradiction by A3, XBOOLE_0:def 5;

          end;

          hence y in (Y \ ( X-section (E,p))) by A2, XBOOLE_0:def 5;

        end;

        then

         A4: ( X-section (( [:X, Y:] \ E),p)) c= (Y \ ( X-section (E,p)));

        now

          let y be set;

          assume

           A5: y in (Y \ ( X-section (E,p)));

          then y in Y & not y in ( X-section (E,p)) by XBOOLE_0:def 5;

          then

           A6: not [p, y] in E;

           [p, y] in [:X, Y:] by A1, A5, ZFMISC_1:def 2;

          then [p, y] in ( [:X, Y:] \ E) by A6, XBOOLE_0:def 5;

          hence y in ( X-section (( [:X, Y:] \ E),p)) by A5;

        end;

        then (Y \ ( X-section (E,p))) c= ( X-section (( [:X, Y:] \ E),p));

        hence ( X-section (( [:X, Y:] \ E),p)) = (Y \ ( X-section (E,p))) by A4;

      end;

      assume

       A7: p in Y;

      now

        let y be set;

        assume

         A8: y in ( Y-section (( [:X, Y:] \ E),p));

        then

         A9: ex y1 be Element of X st y = y1 & [y1, p] in ( [:X, Y:] \ E);

        now

          assume y in ( Y-section (E,p));

          then ex y2 be Element of X st y = y2 & [y2, p] in E;

          hence contradiction by A9, XBOOLE_0:def 5;

        end;

        hence y in (X \ ( Y-section (E,p))) by A8, XBOOLE_0:def 5;

      end;

      then

       A10: ( Y-section (( [:X, Y:] \ E),p)) c= (X \ ( Y-section (E,p)));

      now

        let y be set;

        assume

         A11: y in (X \ ( Y-section (E,p)));

        then y in X & not y in ( Y-section (E,p)) by XBOOLE_0:def 5;

        then

         A12: not [y, p] in E;

         [y, p] in [:X, Y:] by A7, A11, ZFMISC_1:def 2;

        then [y, p] in ( [:X, Y:] \ E) by A12, XBOOLE_0:def 5;

        hence y in ( Y-section (( [:X, Y:] \ E),p)) by A11;

      end;

      then (X \ ( Y-section (E,p))) c= ( Y-section (( [:X, Y:] \ E),p));

      hence ( Y-section (( [:X, Y:] \ E),p)) = (X \ ( Y-section (E,p))) by A10;

    end;

    theorem :: MEASUR11:26

    

     Th20: for X,Y be non empty set, E1,E2 be Subset of [:X, Y:], p be set holds ( X-section ((E1 \/ E2),p)) = (( X-section (E1,p)) \/ ( X-section (E2,p))) & ( Y-section ((E1 \/ E2),p)) = (( Y-section (E1,p)) \/ ( Y-section (E2,p)))

    proof

      let X,Y be non empty set, E1,E2 be Subset of [:X, Y:], p be set;

      now

        let q be set;

        assume q in ( X-section ((E1 \/ E2),p));

        then

        consider y1 be Element of Y such that

         A2: q = y1 & [p, y1] in (E1 \/ E2);

         [p, y1] in E1 or [p, y1] in E2 by A2, XBOOLE_0:def 3;

        then q in ( X-section (E1,p)) or q in ( X-section (E2,p)) by A2;

        hence q in (( X-section (E1,p)) \/ ( X-section (E2,p))) by XBOOLE_0:def 3;

      end;

      then

       A3: ( X-section ((E1 \/ E2),p)) c= (( X-section (E1,p)) \/ ( X-section (E2,p)));

      now

        let q be set;

        assume

         A4: q in (( X-section (E1,p)) \/ ( X-section (E2,p)));

        per cases by A4, XBOOLE_0:def 3;

          suppose q in ( X-section (E1,p));

          then

          consider y1 be Element of Y such that

           A5: q = y1 & [p, y1] in E1;

           [p, y1] in (E1 \/ E2) by A5, XBOOLE_0:def 3;

          hence q in ( X-section ((E1 \/ E2),p)) by A5;

        end;

          suppose q in ( X-section (E2,p));

          then

          consider y1 be Element of Y such that

           A6: q = y1 & [p, y1] in E2;

           [p, y1] in (E1 \/ E2) by A6, XBOOLE_0:def 3;

          hence q in ( X-section ((E1 \/ E2),p)) by A6;

        end;

      end;

      then (( X-section (E1,p)) \/ ( X-section (E2,p))) c= ( X-section ((E1 \/ E2),p));

      hence ( X-section ((E1 \/ E2),p)) = (( X-section (E1,p)) \/ ( X-section (E2,p))) by A3;

      now

        let q be set;

        assume q in ( Y-section ((E1 \/ E2),p));

        then

        consider x1 be Element of X such that

         A2: q = x1 & [x1, p] in (E1 \/ E2);

         [x1, p] in E1 or [x1, p] in E2 by A2, XBOOLE_0:def 3;

        then q in ( Y-section (E1,p)) or q in ( Y-section (E2,p)) by A2;

        hence q in (( Y-section (E1,p)) \/ ( Y-section (E2,p))) by XBOOLE_0:def 3;

      end;

      then

       A3: ( Y-section ((E1 \/ E2),p)) c= (( Y-section (E1,p)) \/ ( Y-section (E2,p)));

      now

        let q be set;

        assume

         A4: q in (( Y-section (E1,p)) \/ ( Y-section (E2,p)));

        per cases by A4, XBOOLE_0:def 3;

          suppose q in ( Y-section (E1,p));

          then

          consider x1 be Element of X such that

           A5: q = x1 & [x1, p] in E1;

           [x1, p] in (E1 \/ E2) by A5, XBOOLE_0:def 3;

          hence q in ( Y-section ((E1 \/ E2),p)) by A5;

        end;

          suppose q in ( Y-section (E2,p));

          then

          consider x1 be Element of X such that

           A6: q = x1 & [x1, p] in E2;

           [x1, p] in (E1 \/ E2) by A6, XBOOLE_0:def 3;

          hence q in ( Y-section ((E1 \/ E2),p)) by A6;

        end;

      end;

      then (( Y-section (E1,p)) \/ ( Y-section (E2,p))) c= ( Y-section ((E1 \/ E2),p));

      hence ( Y-section ((E1 \/ E2),p)) = (( Y-section (E1,p)) \/ ( Y-section (E2,p))) by A3;

    end;

    theorem :: MEASUR11:27

    

     Th21: for X,Y be non empty set, E1,E2 be Subset of [:X, Y:], p be set holds ( X-section ((E1 /\ E2),p)) = (( X-section (E1,p)) /\ ( X-section (E2,p))) & ( Y-section ((E1 /\ E2),p)) = (( Y-section (E1,p)) /\ ( Y-section (E2,p)))

    proof

      let X,Y be non empty set, E1,E2 be Subset of [:X, Y:], p be set;

      now

        let q be set;

        assume q in ( X-section ((E1 /\ E2),p));

        then

        consider y1 be Element of Y such that

         A2: q = y1 & [p, y1] in (E1 /\ E2);

         [p, y1] in E1 & [p, y1] in E2 by A2, XBOOLE_0:def 4;

        then q in ( X-section (E1,p)) & q in ( X-section (E2,p)) by A2;

        hence q in (( X-section (E1,p)) /\ ( X-section (E2,p))) by XBOOLE_0:def 4;

      end;

      then

       A3: ( X-section ((E1 /\ E2),p)) c= (( X-section (E1,p)) /\ ( X-section (E2,p)));

      now

        let q be set;

        assume q in (( X-section (E1,p)) /\ ( X-section (E2,p)));

        then

         A4: q in ( X-section (E1,p)) & q in ( X-section (E2,p)) by XBOOLE_0:def 4;

        then

        consider y1 be Element of Y such that

         A5: q = y1 & [p, y1] in E1;

        consider y2 be Element of Y such that

         A6: q = y2 & [p, y2] in E2 by A4;

         [p, q] in (E1 /\ E2) by A5, A6, XBOOLE_0:def 4;

        hence q in ( X-section ((E1 /\ E2),p)) by A5;

      end;

      then (( X-section (E1,p)) /\ ( X-section (E2,p))) c= ( X-section ((E1 /\ E2),p));

      hence ( X-section ((E1 /\ E2),p)) = (( X-section (E1,p)) /\ ( X-section (E2,p))) by A3;

      now

        let q be set;

        assume q in ( Y-section ((E1 /\ E2),p));

        then

        consider x1 be Element of X such that

         A2: q = x1 & [x1, p] in (E1 /\ E2);

         [x1, p] in E1 & [x1, p] in E2 by A2, XBOOLE_0:def 4;

        then q in ( Y-section (E1,p)) & q in ( Y-section (E2,p)) by A2;

        hence q in (( Y-section (E1,p)) /\ ( Y-section (E2,p))) by XBOOLE_0:def 4;

      end;

      then

       A3: ( Y-section ((E1 /\ E2),p)) c= (( Y-section (E1,p)) /\ ( Y-section (E2,p)));

      now

        let q be set;

        assume q in (( Y-section (E1,p)) /\ ( Y-section (E2,p)));

        then

         A4: q in ( Y-section (E1,p)) & q in ( Y-section (E2,p)) by XBOOLE_0:def 4;

        then

        consider x1 be Element of X such that

         A5: q = x1 & [x1, p] in E1;

        consider x2 be Element of X such that

         A6: q = x2 & [x2, p] in E2 by A4;

         [x1, p] in (E1 /\ E2) by A5, A6, XBOOLE_0:def 4;

        hence q in ( Y-section ((E1 /\ E2),p)) by A5;

      end;

      then (( Y-section (E1,p)) /\ ( Y-section (E2,p))) c= ( Y-section ((E1 /\ E2),p));

      hence ( Y-section ((E1 /\ E2),p)) = (( Y-section (E1,p)) /\ ( Y-section (E2,p))) by A3;

    end;

    theorem :: MEASUR11:28

    

     Th22: for X be set, Y be non empty set, F be FinSequence of ( bool [:X, Y:]), Fy be FinSequence of ( bool Y), p be set st ( dom F) = ( dom Fy) & (for n be Nat st n in ( dom Fy) holds (Fy . n) = ( X-section ((F . n),p))) holds ( X-section (( union ( rng F)),p)) = ( union ( rng Fy))

    proof

      let X be set, Y be non empty set, F be FinSequence of ( bool [:X, Y:]), Fy be FinSequence of ( bool Y), p be set;

      assume that

       A1: ( dom F) = ( dom Fy) and

       A2: for n be Nat st n in ( dom Fy) holds (Fy . n) = ( X-section ((F . n),p));

      now

        let q be set;

        assume q in ( X-section (( union ( rng F)),p));

        then

        consider q1 be Element of Y such that

         A3: q = q1 & [p, q1] in ( union ( rng F));

        consider T be set such that

         A4: [p, q1] in T & T in ( rng F) by A3, TARSKI:def 4;

        consider m be Element of NAT such that

         A5: m in ( dom F) & T = (F . m) by A4, PARTFUN1: 3;

        (Fy . m) = ( X-section ((F . m),p)) by A1, A2, A5;

        then q in (Fy . m) & (Fy . m) in ( rng Fy) by A1, A3, A4, A5, FUNCT_1: 3;

        hence q in ( union ( rng Fy)) by TARSKI:def 4;

      end;

      then

       A6: ( X-section (( union ( rng F)),p)) c= ( union ( rng Fy));

      now

        let q be set;

        assume q in ( union ( rng Fy));

        then

        consider T be set such that

         A7: q in T & T in ( rng Fy) by TARSKI:def 4;

        consider m be Element of NAT such that

         A8: m in ( dom Fy) & T = (Fy . m) by A7, PARTFUN1: 3;

        q in ( X-section ((F . m),p)) by A2, A7, A8;

        then

        consider q1 be Element of Y such that

         A9: q = q1 & [p, q1] in (F . m);

        (F . m) in ( rng F) by A1, A8, FUNCT_1: 3;

        then [p, q1] in ( union ( rng F)) by A9, TARSKI:def 4;

        hence q in ( X-section (( union ( rng F)),p)) by A9;

      end;

      then ( union ( rng Fy)) c= ( X-section (( union ( rng F)),p));

      hence ( X-section (( union ( rng F)),p)) = ( union ( rng Fy)) by A6;

    end;

    theorem :: MEASUR11:29

    

     Th23: for X be non empty set, Y be set, F be FinSequence of ( bool [:X, Y:]), Fx be FinSequence of ( bool X), p be set st ( dom F) = ( dom Fx) & (for n be Nat st n in ( dom Fx) holds (Fx . n) = ( Y-section ((F . n),p))) holds ( Y-section (( union ( rng F)),p)) = ( union ( rng Fx))

    proof

      let X be non empty set, Y be set, F be FinSequence of ( bool [:X, Y:]), Fx be FinSequence of ( bool X), p be set;

      assume that

       A1: ( dom F) = ( dom Fx) and

       A2: for n be Nat st n in ( dom Fx) holds (Fx . n) = ( Y-section ((F . n),p));

      now

        let q be set;

        assume q in ( Y-section (( union ( rng F)),p));

        then

        consider q1 be Element of X such that

         A3: q = q1 & [q1, p] in ( union ( rng F));

        consider T be set such that

         A4: [q1, p] in T & T in ( rng F) by A3, TARSKI:def 4;

        consider m be Element of NAT such that

         A5: m in ( dom F) & T = (F . m) by A4, PARTFUN1: 3;

        (Fx . m) = ( Y-section ((F . m),p)) by A1, A2, A5;

        then q in (Fx . m) & (Fx . m) in ( rng Fx) by A1, A3, A4, A5, FUNCT_1: 3;

        hence q in ( union ( rng Fx)) by TARSKI:def 4;

      end;

      then

       A6: ( Y-section (( union ( rng F)),p)) c= ( union ( rng Fx));

      now

        let q be set;

        assume q in ( union ( rng Fx));

        then

        consider T be set such that

         A7: q in T & T in ( rng Fx) by TARSKI:def 4;

        consider m be Element of NAT such that

         A8: m in ( dom Fx) & T = (Fx . m) by A7, PARTFUN1: 3;

        q in ( Y-section ((F . m),p)) by A2, A7, A8;

        then

        consider q1 be Element of X such that

         A9: q = q1 & [q1, p] in (F . m);

        (F . m) in ( rng F) by A1, A8, FUNCT_1: 3;

        then [q1, p] in ( union ( rng F)) by A9, TARSKI:def 4;

        hence q in ( Y-section (( union ( rng F)),p)) by A9;

      end;

      then ( union ( rng Fx)) c= ( Y-section (( union ( rng F)),p));

      hence ( Y-section (( union ( rng F)),p)) = ( union ( rng Fx)) by A6;

    end;

    theorem :: MEASUR11:30

    

     Th24: for X be set, Y be non empty set, p be set, F be SetSequence of [:X, Y:], Fy be SetSequence of Y st (for n be Nat holds (Fy . n) = ( X-section ((F . n),p))) holds ( X-section (( union ( rng F)),p)) = ( union ( rng Fy))

    proof

      let X be set, Y be non empty set, p be set, F be SetSequence of [:X, Y:], Fy be SetSequence of Y;

      assume

       A2: for n be Nat holds (Fy . n) = ( X-section ((F . n),p));

      now

        let q be set;

        assume q in ( X-section (( union ( rng F)),p));

        then

        consider y1 be Element of Y such that

         A3: q = y1 & [p, y1] in ( union ( rng F));

        consider T be set such that

         A4: [p, y1] in T & T in ( rng F) by A3, TARSKI:def 4;

        consider m be Element of NAT such that

         A5: T = (F . m) by A4, FUNCT_2: 113;

        (Fy . m) = ( X-section ((F . m),p)) by A2;

        then q in (Fy . m) & (Fy . m) in ( rng Fy) by A3, A4, A5, FUNCT_2: 112;

        hence q in ( union ( rng Fy)) by TARSKI:def 4;

      end;

      then

       A7: ( X-section (( union ( rng F)),p)) c= ( union ( rng Fy));

      now

        let q be set;

        assume q in ( union ( rng Fy));

        then

        consider T be set such that

         A8: q in T & T in ( rng Fy) by TARSKI:def 4;

        consider m be Element of NAT such that

         A9: T = (Fy . m) by A8, FUNCT_2: 113;

        q in ( X-section ((F . m),p)) by A2, A8, A9;

        then

        consider y1 be Element of Y such that

         A10: q = y1 & [p, y1] in (F . m);

        (F . m) in ( rng F) by FUNCT_2: 112;

        then [p, y1] in ( union ( rng F)) by A10, TARSKI:def 4;

        hence q in ( X-section (( union ( rng F)),p)) by A10;

      end;

      then ( union ( rng Fy)) c= ( X-section (( union ( rng F)),p));

      hence ( X-section (( union ( rng F)),p)) = ( union ( rng Fy)) by A7;

    end;

    theorem :: MEASUR11:31

    

     Th25: for X be set, Y be non empty set, p be set, F be SetSequence of [:X, Y:], Fy be SetSequence of Y st (for n be Nat holds (Fy . n) = ( X-section ((F . n),p))) holds ( X-section (( meet ( rng F)),p)) = ( meet ( rng Fy))

    proof

      let X be set, Y be non empty set, p be set, F be SetSequence of [:X, Y:], Fy be SetSequence of Y;

      assume

       A2: for n be Nat holds (Fy . n) = ( X-section ((F . n),p));

      now

        let q be set;

        assume q in ( X-section (( meet ( rng F)),p));

        then

        consider y1 be Element of Y such that

         A3: q = y1 & [p, y1] in ( meet ( rng F));

        for T be set st T in ( rng Fy) holds q in T

        proof

          let T be set;

          assume T in ( rng Fy);

          then

          consider n be object such that

           B1: n in ( dom Fy) & T = (Fy . n) by FUNCT_1:def 3;

          reconsider n as Element of NAT by B1;

          ( dom F) = NAT by FUNCT_2:def 1;

          then (F . n) in ( rng F) by FUNCT_1: 3;

          then [p, q] in (F . n) by A3, SETFAM_1:def 1;

          then q in ( X-section ((F . n),p)) by A3;

          hence q in T by B1, A2;

        end;

        hence q in ( meet ( rng Fy)) by SETFAM_1:def 1;

      end;

      then

       A7: ( X-section (( meet ( rng F)),p)) c= ( meet ( rng Fy));

      now

        let q be set;

        assume

         B0: q in ( meet ( rng Fy));

        now

          let T be set;

          assume T in ( rng F);

          then

          consider n be object such that

           B2: n in ( dom F) & T = (F . n) by FUNCT_1:def 3;

          reconsider n as Nat by B2;

          ( dom Fy) = NAT by FUNCT_2:def 1;

          then (Fy . n) in ( rng Fy) by B2, FUNCT_1: 3;

          then q in (Fy . n) by B0, SETFAM_1:def 1;

          then q in ( X-section ((F . n),p)) by A2;

          then ex y be Element of Y st q = y & [p, y] in (F . n);

          hence [p, q] in T by B2;

        end;

        then [p, q] in ( meet ( rng F)) by SETFAM_1:def 1;

        hence q in ( X-section (( meet ( rng F)),p)) by B0;

      end;

      then ( meet ( rng Fy)) c= ( X-section (( meet ( rng F)),p));

      hence ( X-section (( meet ( rng F)),p)) = ( meet ( rng Fy)) by A7;

    end;

    theorem :: MEASUR11:32

    

     Th26: for X be non empty set, Y be set, p be set, F be SetSequence of [:X, Y:], Fx be SetSequence of X st (for n be Nat holds (Fx . n) = ( Y-section ((F . n),p))) holds ( Y-section (( union ( rng F)),p)) = ( union ( rng Fx))

    proof

      let X be non empty set, Y be set, p be set, F be SetSequence of [:X, Y:], Fx be SetSequence of X;

      assume

       A2: for n be Nat holds (Fx . n) = ( Y-section ((F . n),p));

      now

        let q be set;

        assume q in ( Y-section (( union ( rng F)),p));

        then

        consider x1 be Element of X such that

         A3: q = x1 & [x1, p] in ( union ( rng F));

        consider T be set such that

         A4: [x1, p] in T & T in ( rng F) by A3, TARSKI:def 4;

        consider m be Element of NAT such that

         A5: T = (F . m) by A4, FUNCT_2: 113;

        (Fx . m) = ( Y-section ((F . m),p)) by A2;

        then q in (Fx . m) & (Fx . m) in ( rng Fx) by A3, A4, A5, FUNCT_2: 112;

        hence q in ( union ( rng Fx)) by TARSKI:def 4;

      end;

      then

       A7: ( Y-section (( union ( rng F)),p)) c= ( union ( rng Fx));

      now

        let q be set;

        assume q in ( union ( rng Fx));

        then

        consider T be set such that

         A8: q in T & T in ( rng Fx) by TARSKI:def 4;

        consider m be Element of NAT such that

         A9: T = (Fx . m) by A8, FUNCT_2: 113;

        q in ( Y-section ((F . m),p)) by A2, A8, A9;

        then

        consider x1 be Element of X such that

         A10: q = x1 & [x1, p] in (F . m);

        (F . m) in ( rng F) by FUNCT_2: 112;

        then [x1, p] in ( union ( rng F)) by A10, TARSKI:def 4;

        hence q in ( Y-section (( union ( rng F)),p)) by A10;

      end;

      then ( union ( rng Fx)) c= ( Y-section (( union ( rng F)),p));

      hence ( Y-section (( union ( rng F)),p)) = ( union ( rng Fx)) by A7;

    end;

    theorem :: MEASUR11:33

    

     Th27: for X be non empty set, Y be set, p be set, F be SetSequence of [:X, Y:], Fx be SetSequence of X st (for n be Nat holds (Fx . n) = ( Y-section ((F . n),p))) holds ( Y-section (( meet ( rng F)),p)) = ( meet ( rng Fx))

    proof

      let X be non empty set, Y be set, p be set, F be SetSequence of [:X, Y:], Fx be SetSequence of X;

      assume

       A2: for n be Nat holds (Fx . n) = ( Y-section ((F . n),p));

      now

        let q be set;

        assume q in ( Y-section (( meet ( rng F)),p));

        then

        consider y1 be Element of X such that

         A3: q = y1 & [y1, p] in ( meet ( rng F));

        for T be set st T in ( rng Fx) holds q in T

        proof

          let T be set;

          assume T in ( rng Fx);

          then

          consider n be object such that

           B1: n in ( dom Fx) & T = (Fx . n) by FUNCT_1:def 3;

          reconsider n as Element of NAT by B1;

          ( dom F) = NAT by FUNCT_2:def 1;

          then (F . n) in ( rng F) by FUNCT_1: 3;

          then [q, p] in (F . n) by A3, SETFAM_1:def 1;

          then q in ( Y-section ((F . n),p)) by A3;

          hence q in T by B1, A2;

        end;

        hence q in ( meet ( rng Fx)) by SETFAM_1:def 1;

      end;

      then

       A7: ( Y-section (( meet ( rng F)),p)) c= ( meet ( rng Fx));

      now

        let q be set;

        assume

         B0: q in ( meet ( rng Fx));

        now

          let T be set;

          assume T in ( rng F);

          then

          consider n be object such that

           B2: n in ( dom F) & T = (F . n) by FUNCT_1:def 3;

          reconsider n as Nat by B2;

          ( dom Fx) = NAT by FUNCT_2:def 1;

          then (Fx . n) in ( rng Fx) by B2, FUNCT_1: 3;

          then q in (Fx . n) by B0, SETFAM_1:def 1;

          then q in ( Y-section ((F . n),p)) by A2;

          then ex y be Element of X st q = y & [y, p] in (F . n);

          hence [q, p] in T by B2;

        end;

        then [q, p] in ( meet ( rng F)) by SETFAM_1:def 1;

        hence q in ( Y-section (( meet ( rng F)),p)) by B0;

      end;

      then ( meet ( rng Fx)) c= ( Y-section (( meet ( rng F)),p));

      hence ( Y-section (( meet ( rng F)),p)) = ( meet ( rng Fx)) by A7;

    end;

    theorem :: MEASUR11:34

    

     Th28: for X,Y be non empty set, x,y be set, E be Subset of [:X, Y:] holds (( chi (E, [:X, Y:])) . (x,y)) = (( chi (( X-section (E,x)),Y)) . y) & (( chi (E, [:X, Y:])) . (x,y)) = (( chi (( Y-section (E,y)),X)) . x)

    proof

      let X,Y be non empty set, x,y be set, E be Subset of [:X, Y:];

      set z = [x, y];

      per cases ;

        suppose

         A1: [x, y] in E;

        then

        consider x1,y1 be object such that

         A2: x1 in X & y1 in Y & [x, y] = [x1, y1] by ZFMISC_1: 84;

        x = x1 & y = y1 by A2, XTUPLE_0: 1;

        then

         A3: y in ( X-section (E,x)) & x in ( Y-section (E,y)) by A1, A2;

        (( chi (E, [:X, Y:])) . z) = 1 by A1, RFUNCT_1: 63;

        hence (( chi (E, [:X, Y:])) . (x,y)) = (( chi (( X-section (E,x)),Y)) . y) & (( chi (E, [:X, Y:])) . (x,y)) = (( chi (( Y-section (E,y)),X)) . x) by A3, RFUNCT_1: 63;

      end;

        suppose

         A4: not [x, y] in E;

        

         A5: (( chi (E, [:X, Y:])) . (x,y)) = 0

        proof

          per cases ;

            suppose [x, y] in [:X, Y:];

            hence (( chi (E, [:X, Y:])) . (x,y)) = 0 by A4, FUNCT_3:def 3;

          end;

            suppose not [x, y] in [:X, Y:];

            then not [x, y] in ( dom ( chi (E, [:X, Y:])));

            hence (( chi (E, [:X, Y:])) . (x,y)) = 0 by FUNCT_1:def 2;

          end;

        end;

         A6:

        now

          assume y in ( X-section (E,x));

          then ex y1 be Element of Y st y = y1 & [x, y1] in E;

          hence contradiction by A4;

        end;

        

         A7: (( chi (( X-section (E,x)),Y)) . y) = 0

        proof

          per cases ;

            suppose y in Y;

            hence thesis by A6, FUNCT_3:def 3;

          end;

            suppose not y in Y;

            then not y in ( dom ( chi (( X-section (E,x)),Y)));

            hence thesis by FUNCT_1:def 2;

          end;

        end;

         A8:

        now

          assume x in ( Y-section (E,y));

          then ex x1 be Element of X st x = x1 & [x1, y] in E;

          hence contradiction by A4;

        end;

        (( chi (( Y-section (E,y)),X)) . x) = 0

        proof

          per cases ;

            suppose x in X;

            hence thesis by A8, FUNCT_3:def 3;

          end;

            suppose not x in X;

            then not x in ( dom ( chi (( Y-section (E,y)),X)));

            hence thesis by FUNCT_1:def 2;

          end;

        end;

        hence (( chi (E, [:X, Y:])) . (x,y)) = (( chi (( X-section (E,x)),Y)) . y) & (( chi (E, [:X, Y:])) . (x,y)) = (( chi (( Y-section (E,y)),X)) . x) by A5, A7;

      end;

    end;

    theorem :: MEASUR11:35

    

     Th29: for X,Y be non empty set, E1,E2 be Subset of [:X, Y:], p be set st E1 misses E2 holds ( X-section (E1,p)) misses ( X-section (E2,p)) & ( Y-section (E1,p)) misses ( Y-section (E2,p))

    proof

      let X,Y be non empty set, E1,E2 be Subset of [:X, Y:], p be set;

      assume

       A1: E1 misses E2;

      now

        let q be set;

        assume q in (( X-section (E1,p)) /\ ( X-section (E2,p)));

        then

         A2: q in ( X-section (E1,p)) & q in ( X-section (E2,p)) by XBOOLE_0:def 4;

        then

         A3: ex y1 be Element of Y st q = y1 & [p, y1] in E1;

        ex y2 be Element of Y st q = y2 & [p, y2] in E2 by A2;

        hence contradiction by A1, A3, XBOOLE_0:def 4;

      end;

      then (( X-section (E1,p)) /\ ( X-section (E2,p))) is empty;

      hence ( X-section (E1,p)) misses ( X-section (E2,p));

      now

        let q be set;

        assume q in (( Y-section (E1,p)) /\ ( Y-section (E2,p)));

        then

         A4: q in ( Y-section (E1,p)) & q in ( Y-section (E2,p)) by XBOOLE_0:def 4;

        then

         A5: ex x1 be Element of X st q = x1 & [x1, p] in E1;

        ex x2 be Element of X st q = x2 & [x2, p] in E2 by A4;

        hence contradiction by A1, A5, XBOOLE_0:def 4;

      end;

      then (( Y-section (E1,p)) /\ ( Y-section (E2,p))) is empty;

      hence ( Y-section (E1,p)) misses ( Y-section (E2,p));

    end;

    theorem :: MEASUR11:36

    for X,Y be non empty set, F be disjoint_valued FinSequence of ( bool [:X, Y:]), p be set holds (ex Fy be disjoint_valued FinSequence of ( bool X) st (( dom F) = ( dom Fy) & for n be Nat st n in ( dom Fy) holds (Fy . n) = ( Y-section ((F . n),p)))) & (ex Fx be disjoint_valued FinSequence of ( bool Y) st (( dom F) = ( dom Fx) & for n be Nat st n in ( dom Fx) holds (Fx . n) = ( X-section ((F . n),p))))

    proof

      let X,Y be non empty set, F be disjoint_valued FinSequence of ( bool [:X, Y:]);

      let p be set;

      deffunc f1( Nat) = ( Y-section ((F . $1),p));

      deffunc f2( Nat) = ( X-section ((F . $1),p));

      thus ex Fy be disjoint_valued FinSequence of ( bool X) st (( dom F) = ( dom Fy) & for n be Nat st n in ( dom Fy) holds (Fy . n) = ( Y-section ((F . n),p)))

      proof

        consider Fy be FinSequence of ( bool X) such that

         A3: ( len Fy) = ( len F) & (for j be Nat st j in ( dom Fy) holds (Fy . j) = f1(j)) from FINSEQ_2:sch 1;

        reconsider Fy as FinSequence of ( bool X);

        now

          let n,m be object;

          assume n <> m;

          then

           A4: (F . n) misses (F . m) by PROB_2:def 2;

          per cases ;

            suppose

             A5: n in ( dom Fy) & m in ( dom Fy);

            then

            reconsider n1 = n, m1 = m as Nat;

            (Fy . n) = ( Y-section ((F . n1),p)) & (Fy . m) = ( Y-section ((F . m1),p)) by A3, A5;

            hence (Fy . n) misses (Fy . m) by A4, Th29;

          end;

            suppose not n in ( dom Fy) or not m in ( dom Fy);

            then (Fy . n) = {} or (Fy . m) = {} by FUNCT_1:def 2;

            hence (Fy . n) misses (Fy . m);

          end;

        end;

        then

        reconsider Fy as disjoint_valued FinSequence of ( bool X) by PROB_2:def 2;

        take Fy;

        thus ( dom F) = ( dom Fy) & for n be Nat st n in ( dom Fy) holds (Fy . n) = ( Y-section ((F . n),p)) by A3, FINSEQ_3: 29;

      end;

      thus ex Fx be disjoint_valued FinSequence of ( bool Y) st (( dom F) = ( dom Fx) & for n be Nat st n in ( dom Fx) holds (Fx . n) = ( X-section ((F . n),p)))

      proof

        consider Fx be FinSequence of ( bool Y) such that

         A3: ( len Fx) = ( len F) & (for j be Nat st j in ( dom Fx) holds (Fx . j) = f2(j)) from FINSEQ_2:sch 1;

        reconsider Fx as FinSequence of ( bool Y);

        now

          let n,m be object;

          assume n <> m;

          then

           A4: (F . n) misses (F . m) by PROB_2:def 2;

          per cases ;

            suppose

             A5: n in ( dom Fx) & m in ( dom Fx);

            then

            reconsider n1 = n, m1 = m as Nat;

            (Fx . n) = ( X-section ((F . n1),p)) & (Fx . m) = ( X-section ((F . m1),p)) by A3, A5;

            hence (Fx . n) misses (Fx . m) by A4, Th29;

          end;

            suppose not n in ( dom Fx) or not m in ( dom Fx);

            then (Fx . n) = {} or (Fx . m) = {} by FUNCT_1:def 2;

            hence (Fx . n) misses (Fx . m);

          end;

        end;

        then

        reconsider Fx as disjoint_valued FinSequence of ( bool Y) by PROB_2:def 2;

        take Fx;

        thus ( dom F) = ( dom Fx) & for n be Nat st n in ( dom Fx) holds (Fx . n) = ( X-section ((F . n),p)) by A3, FINSEQ_3: 29;

      end;

    end;

    theorem :: MEASUR11:37

    for X,Y be non empty set, F be disjoint_valued SetSequence of [:X, Y:], p be set holds (ex Fy be disjoint_valued SetSequence of X st (for n be Nat holds (Fy . n) = ( Y-section ((F . n),p)))) & (ex Fx be disjoint_valued SetSequence of Y st (for n be Nat holds (Fx . n) = ( X-section ((F . n),p))))

    proof

      let X,Y be non empty set, F be disjoint_valued SetSequence of [:X, Y:], p be set;

      thus ex Fy be disjoint_valued SetSequence of X st (for n be Nat holds (Fy . n) = ( Y-section ((F . n),p)))

      proof

        deffunc f( Nat) = ( Y-section ((F . $1),p));

        consider Fy be SetSequence of X such that

         A1: for n be Element of NAT holds (Fy . n) = f(n) from FUNCT_2:sch 4;

        now

          let n,m be Nat;

          

           A2: n is Element of NAT & m is Element of NAT by ORDINAL1:def 12;

          assume n <> m;

          then (F . n) misses (F . m) by PROB_3:def 4;

          then ( Y-section ((F . n),p)) misses ( Y-section ((F . m),p)) by Th29;

          then (Fy . n) misses ( Y-section ((F . m),p)) by A1, A2;

          hence (Fy . n) misses (Fy . m) by A1, A2;

        end;

        then

        reconsider Fy as disjoint_valued SetSequence of X by PROB_3:def 4;

        take Fy;

        let n be Nat;

        n is Element of NAT by ORDINAL1:def 12;

        hence (Fy . n) = ( Y-section ((F . n),p)) by A1;

      end;

      deffunc f( Nat) = ( X-section ((F . $1),p));

      consider Fx be SetSequence of Y such that

       A3: for n be Element of NAT holds (Fx . n) = f(n) from FUNCT_2:sch 4;

      now

        let n,m be Nat;

        

         A4: n is Element of NAT & m is Element of NAT by ORDINAL1:def 12;

        assume n <> m;

        then (F . n) misses (F . m) by PROB_3:def 4;

        then ( X-section ((F . n),p)) misses ( X-section ((F . m),p)) by Th29;

        then (Fx . n) misses ( X-section ((F . m),p)) by A3, A4;

        hence (Fx . n) misses (Fx . m) by A3, A4;

      end;

      then

      reconsider Fx as disjoint_valued SetSequence of Y by PROB_3:def 4;

      take Fx;

      let n be Nat;

      n is Element of NAT by ORDINAL1:def 12;

      hence (Fx . n) = ( X-section ((F . n),p)) by A3;

    end;

    theorem :: MEASUR11:38

    for X,Y be non empty set, x,y be set, E1,E2 be Subset of [:X, Y:] st E1 misses E2 holds (( chi ((E1 \/ E2), [:X, Y:])) . (x,y)) = ((( chi (( X-section (E1,x)),Y)) . y) + (( chi (( X-section (E2,x)),Y)) . y)) & (( chi ((E1 \/ E2), [:X, Y:])) . (x,y)) = ((( chi (( Y-section (E1,y)),X)) . x) + (( chi (( Y-section (E2,y)),X)) . x))

    proof

      let X,Y be non empty set, x,y be set, E1,E2 be Subset of [:X, Y:];

      assume E1 misses E2;

      then

       A1: ( X-section (E1,x)) misses ( X-section (E2,x)) & ( Y-section (E1,y)) misses ( Y-section (E2,y)) by Th29;

      

       A2: (( chi ((E1 \/ E2), [:X, Y:])) . (x,y)) = (( chi (( X-section ((E1 \/ E2),x)),Y)) . y) by Th28

      .= (( chi ((( X-section (E1,x)) \/ ( X-section (E2,x))),Y)) . y) by Th20;

      thus (( chi ((E1 \/ E2), [:X, Y:])) . (x,y)) = ((( chi (( X-section (E1,x)),Y)) . y) + (( chi (( X-section (E2,x)),Y)) . y))

      proof

        per cases ;

          suppose

           B1: not y in Y;

          ( dom ( chi ((( X-section (E1,x)) \/ ( X-section (E2,x))),Y))) = Y & ( dom ( chi (( X-section (E1,x)),Y))) = Y & ( dom ( chi (( X-section (E2,x)),Y))) = Y by FUNCT_3:def 3;

          then (( chi ((E1 \/ E2), [:X, Y:])) . (x,y)) = 0 & (( chi (( X-section (E1,x)),Y)) . y) = 0 & (( chi (( X-section (E2,x)),Y)) . y) = 0 by A2, B1, FUNCT_1:def 2;

          hence thesis;

        end;

          suppose

           A3: y in Y & y in (( X-section (E1,x)) \/ ( X-section (E2,x)));

          then

           A4: (( chi ((E1 \/ E2), [:X, Y:])) . (x,y)) = 1 by A2, FUNCT_3:def 3;

          per cases by A1, A3, XBOOLE_0: 5;

            suppose y in ( X-section (E1,x)) & not y in ( X-section (E2,x));

            then (( chi (( X-section (E1,x)),Y)) . y) = 1 & (( chi (( X-section (E2,x)),Y)) . y) = 0 by FUNCT_3:def 3;

            hence thesis by A4, XXREAL_3: 4;

          end;

            suppose not y in ( X-section (E1,x)) & y in ( X-section (E2,x));

            then (( chi (( X-section (E1,x)),Y)) . y) = 0 & (( chi (( X-section (E2,x)),Y)) . y) = 1 by FUNCT_3:def 3;

            hence thesis by A4, XXREAL_3: 4;

          end;

        end;

          suppose

           A5: y in Y & not y in (( X-section (E1,x)) \/ ( X-section (E2,x)));

          then

           A6: (( chi ((E1 \/ E2), [:X, Y:])) . (x,y)) = 0 by A2, FUNCT_3:def 3;

           not y in ( X-section (E1,x)) & not y in ( X-section (E2,x)) by A5, XBOOLE_0:def 3;

          then (( chi (( X-section (E1,x)),Y)) . y) = 0 & (( chi (( X-section (E2,x)),Y)) . y) = 0 by A5, FUNCT_3:def 3;

          hence thesis by A6;

        end;

      end;

      

       C2: (( chi ((E1 \/ E2), [:X, Y:])) . (x,y)) = (( chi (( Y-section ((E1 \/ E2),y)),X)) . x) by Th28

      .= (( chi ((( Y-section (E1,y)) \/ ( Y-section (E2,y))),X)) . x) by Th20;

      per cases ;

        suppose

         B1: not x in X;

        ( dom ( chi ((( Y-section (E1,y)) \/ ( Y-section (E2,y))),X))) = X & ( dom ( chi (( Y-section (E1,y)),X))) = X & ( dom ( chi (( Y-section (E2,y)),X))) = X by FUNCT_3:def 3;

        then (( chi ((E1 \/ E2), [:X, Y:])) . (x,y)) = 0 & (( chi (( Y-section (E1,y)),X)) . x) = 0 & (( chi (( Y-section (E2,y)),X)) . x) = 0 by C2, B1, FUNCT_1:def 2;

        hence thesis;

      end;

        suppose

         C3: x in X & x in (( Y-section (E1,y)) \/ ( Y-section (E2,y)));

        then

         C4: (( chi ((E1 \/ E2), [:X, Y:])) . (x,y)) = 1 by C2, FUNCT_3:def 3;

        per cases by A1, C3, XBOOLE_0: 5;

          suppose x in ( Y-section (E1,y)) & not x in ( Y-section (E2,y));

          then (( chi (( Y-section (E1,y)),X)) . x) = 1 & (( chi (( Y-section (E2,y)),X)) . x) = 0 by FUNCT_3:def 3;

          hence thesis by C4, XXREAL_3: 4;

        end;

          suppose not x in ( Y-section (E1,y)) & x in ( Y-section (E2,y));

          then (( chi (( Y-section (E1,y)),X)) . x) = 0 & (( chi (( Y-section (E2,y)),X)) . x) = 1 by FUNCT_3:def 3;

          hence thesis by C4, XXREAL_3: 4;

        end;

      end;

        suppose

         C5: x in X & not x in (( Y-section (E1,y)) \/ ( Y-section (E2,y)));

        then

         C6: (( chi ((E1 \/ E2), [:X, Y:])) . (x,y)) = 0 by C2, FUNCT_3:def 3;

         not x in ( Y-section (E1,y)) & not x in ( Y-section (E2,y)) by C5, XBOOLE_0:def 3;

        then (( chi (( Y-section (E1,y)),X)) . x) = 0 & (( chi (( Y-section (E2,y)),X)) . x) = 0 by C5, FUNCT_3:def 3;

        hence thesis by C6;

      end;

    end;

    theorem :: MEASUR11:39

    

     Th33: for X be set, Y be non empty set, x be set, E be SetSequence of [:X, Y:], G be SetSequence of Y st E is non-descending & (for n be Nat holds (G . n) = ( X-section ((E . n),x))) holds G is non-descending

    proof

      let X be set, Y be non empty set, x be set, E be SetSequence of [:X, Y:], G be SetSequence of Y;

      assume that

       A1: E is non-descending and

       A2: for n be Nat holds (G . n) = ( X-section ((E . n),x));

      for n be Nat holds (G . n) c= (G . (n + 1))

      proof

        let n be Nat;

        ( X-section ((E . n),x)) c= ( X-section ((E . (n + 1)),x)) by Th14, A1, KURATO_0:def 4;

        then (G . n) c= ( X-section ((E . (n + 1)),x)) by A2;

        hence (G . n) c= (G . (n + 1)) by A2;

      end;

      hence G is non-descending by KURATO_0:def 4;

    end;

    theorem :: MEASUR11:40

    

     Th34: for X be non empty set, Y be set, x be set, E be SetSequence of [:X, Y:], G be SetSequence of X st E is non-descending & (for n be Nat holds (G . n) = ( Y-section ((E . n),x))) holds G is non-descending

    proof

      let X be non empty set, Y be set, x be set, E be SetSequence of [:X, Y:], G be SetSequence of X;

      assume that

       A1: E is non-descending and

       A2: for n be Nat holds (G . n) = ( Y-section ((E . n),x));

      for n be Nat holds (G . n) c= (G . (n + 1))

      proof

        let n be Nat;

        ( Y-section ((E . n),x)) c= ( Y-section ((E . (n + 1)),x)) by Th15, A1, KURATO_0:def 4;

        then (G . n) c= ( Y-section ((E . (n + 1)),x)) by A2;

        hence (G . n) c= (G . (n + 1)) by A2;

      end;

      hence G is non-descending by KURATO_0:def 4;

    end;

    theorem :: MEASUR11:41

    

     Th35: for X be set, Y be non empty set, x be set, E be SetSequence of [:X, Y:], G be SetSequence of Y st E is non-ascending & (for n be Nat holds (G . n) = ( X-section ((E . n),x))) holds G is non-ascending

    proof

      let X be set, Y be non empty set, x be set, E be SetSequence of [:X, Y:], G be SetSequence of Y;

      assume that

       A1: E is non-ascending and

       A2: for n be Nat holds (G . n) = ( X-section ((E . n),x));

      for n be Nat holds (G . (n + 1)) c= (G . n)

      proof

        let n be Nat;

        ( X-section ((E . (n + 1)),x)) c= ( X-section ((E . n),x)) by Th14, A1, KURATO_0:def 3;

        then (G . (n + 1)) c= ( X-section ((E . n),x)) by A2;

        hence (G . (n + 1)) c= (G . n) by A2;

      end;

      hence G is non-ascending by KURATO_0:def 3;

    end;

    theorem :: MEASUR11:42

    

     Th36: for X be non empty set, Y be set, x be set, E be SetSequence of [:X, Y:], G be SetSequence of X st E is non-ascending & (for n be Nat holds (G . n) = ( Y-section ((E . n),x))) holds G is non-ascending

    proof

      let X be non empty set, Y be set, x be set, E be SetSequence of [:X, Y:], G be SetSequence of X;

      assume that

       A1: E is non-ascending and

       A2: for n be Nat holds (G . n) = ( Y-section ((E . n),x));

      for n be Nat holds (G . (n + 1)) c= (G . n)

      proof

        let n be Nat;

        ( Y-section ((E . (n + 1)),x)) c= ( Y-section ((E . n),x)) by Th15, A1, KURATO_0:def 3;

        then (G . (n + 1)) c= ( Y-section ((E . n),x)) by A2;

        hence (G . (n + 1)) c= (G . n) by A2;

      end;

      hence G is non-ascending by KURATO_0:def 3;

    end;

    theorem :: MEASUR11:43

    

     Th37: for X be set, Y be non empty set, E be SetSequence of [:X, Y:], x be set st E is non-descending holds ex G be SetSequence of Y st G is non-descending & (for n be Nat holds (G . n) = ( X-section ((E . n),x)))

    proof

      let X be set, Y be non empty set, E be SetSequence of [:X, Y:], x be set;

      assume

       A1: E is non-descending;

      deffunc F( Nat) = ( X-section ((E . $1),x));

      consider G be Function of NAT , ( bool Y) such that

       A2: for n be Element of NAT holds (G . n) = F(n) from FUNCT_2:sch 4;

      reconsider G as SetSequence of Y;

      

       A3: for n be Nat holds (G . n) = ( X-section ((E . n),x))

      proof

        let n be Nat;

        n is Element of NAT by ORDINAL1:def 12;

        hence (G . n) = ( X-section ((E . n),x)) by A2;

      end;

      take G;

      thus G is non-descending by A1, A3, Th33;

      thus thesis by A3;

    end;

    theorem :: MEASUR11:44

    

     Th38: for X be non empty set, Y be set, E be SetSequence of [:X, Y:], x be set st E is non-descending holds ex G be SetSequence of X st G is non-descending & (for n be Nat holds (G . n) = ( Y-section ((E . n),x)))

    proof

      let X be non empty set, Y be set, E be SetSequence of [:X, Y:], x be set;

      assume

       A1: E is non-descending;

      deffunc F( Nat) = ( Y-section ((E . $1),x));

      consider G be Function of NAT , ( bool X) such that

       A2: for n be Element of NAT holds (G . n) = F(n) from FUNCT_2:sch 4;

      reconsider G as SetSequence of X;

      

       A3: for n be Nat holds (G . n) = ( Y-section ((E . n),x))

      proof

        let n be Nat;

        n is Element of NAT by ORDINAL1:def 12;

        hence (G . n) = ( Y-section ((E . n),x)) by A2;

      end;

      take G;

      thus G is non-descending by A1, A3, Th34;

      thus thesis by A3;

    end;

    theorem :: MEASUR11:45

    

     Th39: for X be set, Y be non empty set, E be SetSequence of [:X, Y:], x be set st E is non-ascending holds ex G be SetSequence of Y st G is non-ascending & (for n be Nat holds (G . n) = ( X-section ((E . n),x)))

    proof

      let X be set, Y be non empty set, E be SetSequence of [:X, Y:], x be set;

      assume

       A1: E is non-ascending;

      deffunc F( Nat) = ( X-section ((E . $1),x));

      consider G be Function of NAT , ( bool Y) such that

       A2: for n be Element of NAT holds (G . n) = F(n) from FUNCT_2:sch 4;

      reconsider G as SetSequence of Y;

      

       A3: for n be Nat holds (G . n) = ( X-section ((E . n),x))

      proof

        let n be Nat;

        n is Element of NAT by ORDINAL1:def 12;

        hence (G . n) = ( X-section ((E . n),x)) by A2;

      end;

      take G;

      thus G is non-ascending by A1, A3, Th35;

      thus thesis by A3;

    end;

    theorem :: MEASUR11:46

    

     Th40: for X be non empty set, Y be set, E be SetSequence of [:X, Y:], x be set st E is non-ascending holds ex G be SetSequence of X st G is non-ascending & (for n be Nat holds (G . n) = ( Y-section ((E . n),x)))

    proof

      let X be non empty set, Y be set, E be SetSequence of [:X, Y:], x be set;

      assume

       A1: E is non-ascending;

      deffunc F( Nat) = ( Y-section ((E . $1),x));

      consider G be Function of NAT , ( bool X) such that

       A2: for n be Element of NAT holds (G . n) = F(n) from FUNCT_2:sch 4;

      reconsider G as SetSequence of X;

      

       A3: for n be Nat holds (G . n) = ( Y-section ((E . n),x))

      proof

        let n be Nat;

        n is Element of NAT by ORDINAL1:def 12;

        hence (G . n) = ( Y-section ((E . n),x)) by A2;

      end;

      take G;

      thus G is non-ascending by A1, A3, Th36;

      thus thesis by A3;

    end;

    begin

    theorem :: MEASUR11:47

    

     Th42: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), K be set st K = { C where C be Subset of [:X1, X2:] : for p be set holds ( X-section (C,p)) in S2 } holds ( Field_generated_by ( measurable_rectangles (S1,S2))) c= K & K is SigmaField of [:X1, X2:]

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), K be set;

      assume

       AS: K = { C where C be Subset of [:X1, X2:] : for x be set holds ( X-section (C,x)) in S2 };

       A1:

      now

        let C1,C2 be set;

        assume

         A2: C1 in K & C2 in K;

        then

        consider SC1 be Subset of [:X1, X2:] such that

         A3: C1 = SC1 & for x be set holds ( X-section (SC1,x)) in S2 by AS;

        consider SC2 be Subset of [:X1, X2:] such that

         A4: C2 = SC2 & for x be set holds ( X-section (SC2,x)) in S2 by AS, A2;

        now

          let x be set;

          

           A5: ( X-section (SC1,x)) in S2 & ( X-section (SC2,x)) in S2 by A3, A4;

          ( X-section ((SC1 \/ SC2),x)) = (( X-section (SC1,x)) \/ ( X-section (SC2,x))) by Th20;

          hence ( X-section ((SC1 \/ SC2),x)) in S2 by A5, PROB_1: 3;

        end;

        hence (C1 \/ C2) in K by AS, A3, A4;

      end;

      then

       A6: K is cup-closed by FINSUB_1:def 1;

      for x be set holds ( X-section (( {} [:X1, X2:]),x)) in S2

      proof

        let x be set;

        ( X-section (( {} [:X1, X2:]),x)) = {} by Th18;

        hence thesis by MEASURE1: 7;

      end;

      then

       A7: {} in K by AS;

      now

        let C be set;

        assume C in ( DisUnion ( measurable_rectangles (S1,S2)));

        then C in { A where A be Subset of [:X1, X2:] : ex F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st A = ( Union F) } by SRINGS_3:def 3;

        then

        consider C1 be Subset of [:X1, X2:] such that

         A8: C = C1 & ex F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st C1 = ( Union F);

        consider F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) such that

         A9: C1 = ( Union F) by A8;

        for n be Nat st n in ( dom F) holds (F . n) in K

        proof

          let n be Nat;

          assume n in ( dom F);

          then (F . n) in ( measurable_rectangles (S1,S2)) by PARTFUN1: 4;

          then (F . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

          then

          consider A be Element of S1, B be Element of S2 such that

           A10: (F . n) = [:A, B:];

          now

            let x be set;

            ( X-section ( [:A, B:],x)) = B or ( X-section ( [:A, B:],x)) = {} by Th16;

            hence ( X-section ( [:A, B:],x)) in S2 by MEASURE1: 7;

          end;

          hence (F . n) in K by AS, A10;

        end;

        hence C in K by A1, A7, A8, A9, Th41, FINSUB_1:def 1;

      end;

      then ( DisUnion ( measurable_rectangles (S1,S2))) c= K;

      hence ( Field_generated_by ( measurable_rectangles (S1,S2))) c= K by SRINGS_3: 22;

      now

        let A be set;

        assume A in K;

        then ex A1 be Subset of [:X1, X2:] st A = A1 & for x be set holds ( X-section (A1,x)) in S2 by AS;

        hence A in ( bool [:X1, X2:]);

      end;

      then K c= ( bool [:X1, X2:]);

      then

      reconsider K as Subset-Family of [:X1, X2:];

      for C be Subset of [:X1, X2:] st C in K holds (C ` ) in K

      proof

        let C be Subset of [:X1, X2:];

        assume C in K;

        then

        consider C1 be Subset of [:X1, X2:] such that

         A11: C = C1 & for x be set holds ( X-section (C1,x)) in S2 by AS;

        now

          let x be set;

          per cases ;

            suppose

             A12: x in X1;

            

             A13: ( X-section (C1,x)) in S2 by A11;

            X2 in S2 by PROB_1: 5;

            then (X2 \ ( X-section (C1,x))) in S2 by A13, PROB_1: 6;

            hence ( X-section (( [:X1, X2:] \ C1),x)) in S2 by A12, Th19;

          end;

            suppose not x in X1;

            then ( X-section (( [:X1, X2:] \ C1),x)) = {} by Th17;

            hence ( X-section (( [:X1, X2:] \ C1),x)) in S2 by MEASURE1: 7;

          end;

        end;

        then ( [:X1, X2:] \ C) in K by AS, A11;

        hence (C ` ) in K by SUBSET_1:def 4;

      end;

      then K is compl-closed by PROB_1:def 1;

      then

      reconsider K as Field_Subset of [:X1, X2:] by A7, A6;

      now

        let M be N_Sub_set_fam of [:X1, X2:];

        assume

         A15: M c= K;

        consider E be SetSequence of [:X1, X2:] such that

         A16: ( rng E) = M by SUPINF_2:def 8;

        now

          let x be set;

          defpred P[ Nat, object] means $2 = { y where y be Element of X2 : [x, y] in (E . $1) };

          

           A18: for n be Element of NAT holds ex d be Element of ( bool X2) st P[n, d]

          proof

            let n be Element of NAT ;

            set d = { y where y be Element of X2 : [x, y] in (E . n) };

            now

              let y be set;

              assume y in d;

              then ex y1 be Element of X2 st y = y1 & [x, y1] in (E . n);

              hence y in X2;

            end;

            then d c= X2;

            then

            reconsider d as Element of ( bool X2);

            take d;

            thus P[n, d];

          end;

          consider D be Function of NAT , ( bool X2) such that

           A19: for n be Element of NAT holds P[n, (D . n)] from FUNCT_2:sch 3( A18);

          reconsider D as SetSequence of X2;

          

           A20: for n be Nat holds (D . n) = ( X-section ((E . n),x))

          proof

            let n be Nat;

            n is Element of NAT by ORDINAL1:def 12;

            hence thesis by A19;

          end;

          

           A21: ( dom D) = NAT by FUNCT_2:def 1;

          now

            let D0 be set;

            assume D0 in ( rng D);

            then

            consider n0 be Element of NAT such that

             A22: D0 = (D . n0) by FUNCT_2: 113;

            

             A23: D0 = ( X-section ((E . n0),x)) by A20, A22;

            (E . n0) in K by A15, A16, FUNCT_2: 112;

            then ex C0 be Subset of [:X1, X2:] st (E . n0) = C0 & for x be set holds ( X-section (C0,x)) in S2 by AS;

            hence D0 in S2 by A23;

          end;

          then ( rng D) c= S2;

          then D is sequence of S2 by A21, FUNCT_2: 2;

          then

           A24: ( union ( rng D)) is Element of S2 by MEASURE1: 24;

          ( X-section (( union ( rng E)),x)) = ( union ( rng D)) by A20, Th24;

          hence ( X-section (( union ( rng E)),x)) in S2 by A24;

        end;

        hence ( union M) in K by AS, A16;

      end;

      then K is sigma-additive by MEASURE1:def 5;

      hence thesis;

    end;

    theorem :: MEASUR11:48

    

     Th43: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), K be set st K = { C where C be Subset of [:X1, X2:] : for p be set holds ( Y-section (C,p)) in S1 } holds ( Field_generated_by ( measurable_rectangles (S1,S2))) c= K & K is SigmaField of [:X1, X2:]

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), K be set;

      assume

       AS: K = { C where C be Subset of [:X1, X2:] : for p be set holds ( Y-section (C,p)) in S1 };

       A1:

      now

        let C1,C2 be set;

        assume

         A2: C1 in K & C2 in K;

        then

        consider SC1 be Subset of [:X1, X2:] such that

         A3: C1 = SC1 & for x be set holds ( Y-section (SC1,x)) in S1 by AS;

        consider SC2 be Subset of [:X1, X2:] such that

         A4: C2 = SC2 & for x be set holds ( Y-section (SC2,x)) in S1 by AS, A2;

        now

          let x be set;

          

           A5: ( Y-section (SC1,x)) in S1 & ( Y-section (SC2,x)) in S1 by A3, A4;

          ( Y-section ((SC1 \/ SC2),x)) = (( Y-section (SC1,x)) \/ ( Y-section (SC2,x))) by Th20;

          hence ( Y-section ((SC1 \/ SC2),x)) in S1 by A5, PROB_1: 3;

        end;

        hence (C1 \/ C2) in K by AS, A3, A4;

      end;

      then

       A6: K is cup-closed by FINSUB_1:def 1;

      for y be set holds ( Y-section (( {} [:X1, X2:]),y)) in S1

      proof

        let y be set;

        ( Y-section (( {} [:X1, X2:]),y)) = {} by Th18;

        hence thesis by MEASURE1: 7;

      end;

      then

       A7: {} in K by AS;

      now

        let C be set;

        assume C in ( DisUnion ( measurable_rectangles (S1,S2)));

        then C in { A where A be Subset of [:X1, X2:] : ex F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st A = ( Union F) } by SRINGS_3:def 3;

        then

        consider C1 be Subset of [:X1, X2:] such that

         A8: C = C1 & ex F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st C1 = ( Union F);

        consider F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) such that

         A9: C1 = ( Union F) by A8;

        for n be Nat st n in ( dom F) holds (F . n) in K

        proof

          let n be Nat;

          assume n in ( dom F);

          then (F . n) in ( measurable_rectangles (S1,S2)) by PARTFUN1: 4;

          then (F . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

          then

          consider A be Element of S1, B be Element of S2 such that

           A10: (F . n) = [:A, B:];

          now

            let x be set;

            ( Y-section ( [:A, B:],x)) = A or ( Y-section ( [:A, B:],x)) = {} by Th16;

            hence ( Y-section ( [:A, B:],x)) in S1 by MEASURE1: 7;

          end;

          hence (F . n) in K by AS, A10;

        end;

        hence C in K by A1, A7, A8, A9, Th41, FINSUB_1:def 1;

      end;

      then ( DisUnion ( measurable_rectangles (S1,S2))) c= K;

      hence ( Field_generated_by ( measurable_rectangles (S1,S2))) c= K by SRINGS_3: 22;

      now

        let A be set;

        assume A in K;

        then ex A1 be Subset of [:X1, X2:] st A = A1 & for x be set holds ( Y-section (A1,x)) in S1 by AS;

        hence A in ( bool [:X1, X2:]);

      end;

      then K c= ( bool [:X1, X2:]);

      then

      reconsider K as Subset-Family of [:X1, X2:];

      for C be Subset of [:X1, X2:] st C in K holds (C ` ) in K

      proof

        let C be Subset of [:X1, X2:];

        assume C in K;

        then

        consider C1 be Subset of [:X1, X2:] such that

         A11: C = C1 & for x be set holds ( Y-section (C1,x)) in S1 by AS;

        now

          let x be set;

          per cases ;

            suppose

             A12: x in X2;

            

             A13: ( Y-section (C1,x)) in S1 by A11;

            X1 in S1 by PROB_1: 5;

            then (X1 \ ( Y-section (C1,x))) in S1 by A13, PROB_1: 6;

            hence ( Y-section (( [:X1, X2:] \ C1),x)) in S1 by A12, Th19;

          end;

            suppose not x in X2;

            then ( Y-section (( [:X1, X2:] \ C1),x)) = {} by Th17;

            hence ( Y-section (( [:X1, X2:] \ C1),x)) in S1 by MEASURE1: 7;

          end;

        end;

        then ( [:X1, X2:] \ C) in K by AS, A11;

        hence (C ` ) in K by SUBSET_1:def 4;

      end;

      then

       A14: K is compl-closed by PROB_1:def 1;

      now

        let p be set;

        ( Y-section (( {} [:X1, X2:]),p)) = {} by Th18;

        hence ( Y-section (( {} [:X1, X2:]),p)) in S1 by SETFAM_1:def 8;

      end;

      then {} in { C where C be Subset of [:X1, X2:] : for p be set holds ( Y-section (C,p)) in S1 };

      then

      reconsider K as Field_Subset of [:X1, X2:] by A14, AS, A6;

      now

        let M be N_Sub_set_fam of [:X1, X2:];

        assume

         A15: M c= K;

        consider E be SetSequence of [:X1, X2:] such that

         A16: ( rng E) = M by SUPINF_2:def 8;

        now

          let x be set;

          defpred P[ Nat, object] means $2 = { y where y be Element of X1 : [y, x] in (E . $1) };

          

           A18: for n be Element of NAT holds ex d be Element of ( bool X1) st P[n, d]

          proof

            let n be Element of NAT ;

            set d = { y where y be Element of X1 : [y, x] in (E . n) };

            now

              let y be set;

              assume y in d;

              then ex y1 be Element of X1 st y = y1 & [y1, x] in (E . n);

              hence y in X1;

            end;

            then d c= X1;

            then

            reconsider d as Element of ( bool X1);

            take d;

            thus P[n, d];

          end;

          consider D be Function of NAT , ( bool X1) such that

           A19: for n be Element of NAT holds P[n, (D . n)] from FUNCT_2:sch 3( A18);

          reconsider D as SetSequence of X1;

          

           A20: for n be Nat holds (D . n) = ( Y-section ((E . n),x))

          proof

            let n be Nat;

            n is Element of NAT by ORDINAL1:def 12;

            hence thesis by A19;

          end;

          

           A21: ( dom D) = NAT by FUNCT_2:def 1;

          now

            let D0 be set;

            assume D0 in ( rng D);

            then

            consider n0 be Element of NAT such that

             A22: D0 = (D . n0) by FUNCT_2: 113;

            

             A23: D0 = ( Y-section ((E . n0),x)) by A20, A22;

            (E . n0) in K by A15, A16, FUNCT_2: 112;

            then ex C0 be Subset of [:X1, X2:] st (E . n0) = C0 & for x be set holds ( Y-section (C0,x)) in S1 by AS;

            hence D0 in S1 by A23;

          end;

          then ( rng D) c= S1;

          then D is sequence of S1 by A21, FUNCT_2: 2;

          then

           A24: ( union ( rng D)) is Element of S1 by MEASURE1: 24;

          ( Y-section (( union ( rng E)),x)) = ( union ( rng D)) by A20, Th26;

          hence ( Y-section (( union ( rng E)),x)) in S1 by A24;

        end;

        hence ( union M) in K by AS, A16;

      end;

      then K is sigma-additive by MEASURE1:def 5;

      hence thesis;

    end;

    theorem :: MEASUR11:49

    

     Th44: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2))) holds (for p be set holds ( X-section (E,p)) in S2) & (for p be set holds ( Y-section (E,p)) in S1)

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      set K = { C where C be Subset of [:X1, X2:] : for x be set holds ( X-section (C,x)) in S2 };

      reconsider K as SigmaField of [:X1, X2:] by Th42;

      

       A1: ( measurable_rectangles (S1,S2)) c= ( Field_generated_by ( measurable_rectangles (S1,S2))) & ( Field_generated_by ( measurable_rectangles (S1,S2))) c= K by Th42, SRINGS_3: 21;

      then ( measurable_rectangles (S1,S2)) c= K;

      then ( sigma ( measurable_rectangles (S1,S2))) c= K by PROB_1:def 9;

      then E in K;

      then ex C be Subset of [:X1, X2:] st E = C & for x be set holds ( X-section (C,x)) in S2;

      hence for x be set holds ( X-section (E,x)) in S2;

      set K2 = { C where C be Subset of [:X1, X2:] : for x be set holds ( Y-section (C,x)) in S1 };

      reconsider K2 as SigmaField of [:X1, X2:] by Th43;

      ( Field_generated_by ( measurable_rectangles (S1,S2))) c= K2 by Th43;

      then ( measurable_rectangles (S1,S2)) c= K2 by A1;

      then ( sigma ( measurable_rectangles (S1,S2))) c= K2 by PROB_1:def 9;

      then E in K2;

      then ex C be Subset of [:X1, X2:] st E = C & for x be set holds ( Y-section (C,x)) in S1;

      hence for x be set holds ( Y-section (E,x)) in S1;

    end;

    definition

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), x be set;

      :: MEASUR11:def6

      func Measurable-X-section (E,x) -> Element of S2 equals ( X-section (E,x));

      correctness by Th44;

    end

    definition

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), y be set;

      :: MEASUR11:def7

      func Measurable-Y-section (E,y) -> Element of S1 equals ( Y-section (E,y));

      correctness by Th44;

    end

    theorem :: MEASUR11:50

    for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, F be FinSequence of ( sigma ( measurable_rectangles (S1,S2))), Fy be FinSequence of S2, p be set st ( dom F) = ( dom Fy) & (for n be Nat st n in ( dom Fy) holds (Fy . n) = ( Measurable-X-section ((F . n),p))) holds ( Measurable-X-section (( Union F),p)) = ( Union Fy)

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, F be FinSequence of ( sigma ( measurable_rectangles (S1,S2))), Fy be FinSequence of S2, p be set;

      assume that

       A1: ( dom F) = ( dom Fy) and

       A2: for n be Nat st n in ( dom Fy) holds (Fy . n) = ( Measurable-X-section ((F . n),p));

      

       A3: ( union ( rng F)) = ( Union F) by CARD_3:def 4;

      reconsider F1 = F as FinSequence of ( bool [:X1, X2:]) by FINSEQ_2: 24;

      reconsider F1y = Fy as FinSequence of ( bool X2) by FINSEQ_2: 24;

      for n be Nat st n in ( dom F1y) holds (F1y . n) = ( X-section ((F1 . n),p))

      proof

        let n be Nat;

        assume n in ( dom F1y);

        then (Fy . n) = ( Measurable-X-section ((F . n),p)) by A2;

        hence thesis;

      end;

      then ( X-section (( union ( rng F1)),p)) = ( union ( rng F1y)) by A1, Th22;

      hence thesis by A3, CARD_3:def 4;

    end;

    theorem :: MEASUR11:51

    for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, F be FinSequence of ( sigma ( measurable_rectangles (S1,S2))), Fx be FinSequence of S1, p be set st ( dom F) = ( dom Fx) & (for n be Nat st n in ( dom Fx) holds (Fx . n) = ( Measurable-Y-section ((F . n),p))) holds ( Measurable-Y-section (( Union F),p)) = ( Union Fx)

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, F be FinSequence of ( sigma ( measurable_rectangles (S1,S2))), Fx be FinSequence of S1, p be set;

      assume that

       A1: ( dom F) = ( dom Fx) and

       A2: for n be Nat st n in ( dom Fx) holds (Fx . n) = ( Measurable-Y-section ((F . n),p));

      

       A3: ( union ( rng F)) = ( Union F) by CARD_3:def 4;

      reconsider F1 = F as FinSequence of ( bool [:X1, X2:]) by FINSEQ_2: 24;

      reconsider F1x = Fx as FinSequence of ( bool X1) by FINSEQ_2: 24;

      for n be Nat st n in ( dom F1x) holds (F1x . n) = ( Y-section ((F1 . n),p))

      proof

        let n be Nat;

        assume n in ( dom F1x);

        then (Fx . n) = ( Measurable-Y-section ((F . n),p)) by A2;

        hence thesis;

      end;

      then ( Y-section (( union ( rng F1)),p)) = ( union ( rng F1x)) by A1, Th23;

      hence thesis by A3, CARD_3:def 4;

    end;

    theorem :: MEASUR11:52

    

     Th47: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, A be Element of S1, B be Element of S2, x be Element of X1 holds ((M2 . B) * (( chi (A,X1)) . x)) = ( Integral (M2,( ProjMap1 (( chi ( [:A, B:], [:X1, X2:])),x))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, A be Element of S1, B be Element of S2, x be Element of X1;

      

       A1: for y be Element of X2 holds (( ProjMap1 (( chi ( [:A, B:], [:X1, X2:])),x)) . y) = ((( chi (A,X1)) . x) * (( chi (B,X2)) . y))

      proof

        let y be Element of X2;

        (( ProjMap1 (( chi ( [:A, B:], [:X1, X2:])),x)) . y) = (( chi ( [:A, B:], [:X1, X2:])) . (x,y)) by MESFUNC9:def 6;

        hence thesis by MEASUR10: 2;

      end;

      set CAB = ( chi ( [:A, B:], [:X1, X2:]));

      per cases ;

        suppose x in A;

        then

         A2: (( chi (A,X1)) . x) = 1 by FUNCT_3:def 3;

        then

         A3: ((M2 . B) * (( chi (A,X1)) . x)) = (M2 . B) by XXREAL_3: 81;

        

         A4: ( dom ( ProjMap1 (( chi ( [:A, B:], [:X1, X2:])),x))) = X2 by FUNCT_2:def 1

        .= ( dom ( chi (B,X2))) by FUNCT_3:def 3;

        for y be Element of X2 st y in ( dom ( ProjMap1 (CAB,x))) holds (( ProjMap1 (CAB,x)) . y) = (( chi (B,X2)) . y)

        proof

          let y be Element of X2;

          assume y in ( dom ( ProjMap1 (CAB,x)));

          (( ProjMap1 (CAB,x)) . y) = ((( chi (A,X1)) . x) * (( chi (B,X2)) . y)) by A1;

          hence (( ProjMap1 (CAB,x)) . y) = (( chi (B,X2)) . y) by A2, XXREAL_3: 81;

        end;

        then ( ProjMap1 (CAB,x)) = ( chi (B,X2)) by A4, PARTFUN1: 5;

        hence ((M2 . B) * (( chi (A,X1)) . x)) = ( Integral (M2,( ProjMap1 (CAB,x)))) by A3, MESFUNC9: 14;

      end;

        suppose not x in A;

        then

         A5: (( chi (A,X1)) . x) = 0 by FUNCT_3:def 3;

        then

         A6: ((M2 . B) * (( chi (A,X1)) . x)) = 0 ;

        

         A7: {} is Element of S2 by PROB_1: 4;

        

         A8: ( dom ( ProjMap1 (CAB,x))) = X2 by FUNCT_2:def 1

        .= ( dom ( chi ( {} ,X2))) by FUNCT_3:def 3;

        for y be Element of X2 st y in ( dom ( ProjMap1 (CAB,x))) holds (( ProjMap1 (CAB,x)) . y) = (( chi ( {} ,X2)) . y)

        proof

          let y be Element of X2;

          assume y in ( dom ( ProjMap1 (CAB,x)));

          (( ProjMap1 (CAB,x)) . y) = ((( chi (A,X1)) . x) * (( chi (B,X2)) . y)) by A1;

          then (( ProjMap1 (CAB,x)) . y) = 0 by A5;

          hence (( ProjMap1 (CAB,x)) . y) = (( chi ( {} ,X2)) . y) by FUNCT_3:def 3;

        end;

        then ( ProjMap1 (CAB,x)) = ( chi ( {} ,X2)) by A8, PARTFUN1: 5;

        then ( Integral (M2,( ProjMap1 (CAB,x)))) = (M2 . {} ) by A7, MESFUNC9: 14;

        hence ((M2 . B) * (( chi (A,X1)) . x)) = ( Integral (M2,( ProjMap1 (CAB,x)))) by A6, VALUED_0:def 19;

      end;

    end;

    theorem :: MEASUR11:53

    

     Th48: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2, x be Element of X1 st E = [:A, B:] holds (M2 . ( Measurable-X-section (E,x))) = ((M2 . B) * (( chi (A,X1)) . x))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2, x be Element of X1;

      assume

       A1: E = [:A, B:];

      per cases ;

        suppose

         A4: x in A;

        then

         A2: (M2 . ( Measurable-X-section (E,x))) = (M2 . B) by A1, Th16;

        (( chi (A,X1)) . x) = 1 by A4, FUNCT_3:def 3;

        hence (M2 . ( Measurable-X-section (E,x))) = ((M2 . B) * (( chi (A,X1)) . x)) by A2, XXREAL_3: 81;

      end;

        suppose

         A5: not x in A;

        then ( Measurable-X-section (E,x)) = {} by A1, Th16;

        then

         A3: (M2 . ( Measurable-X-section (E,x))) = 0 by VALUED_0:def 19;

        (( chi (A,X1)) . x) = 0 by A5, FUNCT_3:def 3;

        hence (M2 . ( Measurable-X-section (E,x))) = ((M2 . B) * (( chi (A,X1)) . x)) by A3;

      end;

    end;

    theorem :: MEASUR11:54

    

     Th49: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, A be Element of S1, B be Element of S2, y be Element of X2 holds ((M1 . A) * (( chi (B,X2)) . y)) = ( Integral (M1,( ProjMap2 (( chi ( [:A, B:], [:X1, X2:])),y))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, A be Element of S1, B be Element of S2, y be Element of X2;

      

       A1: for x be Element of X1 holds (( ProjMap2 (( chi ( [:A, B:], [:X1, X2:])),y)) . x) = ((( chi (A,X1)) . x) * (( chi (B,X2)) . y))

      proof

        let x be Element of X1;

        (( ProjMap2 (( chi ( [:A, B:], [:X1, X2:])),y)) . x) = (( chi ( [:A, B:], [:X1, X2:])) . (x,y)) by MESFUNC9:def 7;

        hence thesis by MEASUR10: 2;

      end;

      set CAB = ( chi ( [:A, B:], [:X1, X2:]));

      per cases ;

        suppose y in B;

        then

         A2: (( chi (B,X2)) . y) = 1 by FUNCT_3:def 3;

        then

         A3: ((M1 . A) * (( chi (B,X2)) . y)) = (M1 . A) by XXREAL_3: 81;

        

         A4: ( dom ( ProjMap2 (( chi ( [:A, B:], [:X1, X2:])),y))) = X1 by FUNCT_2:def 1

        .= ( dom ( chi (A,X1))) by FUNCT_3:def 3;

        for x be Element of X1 st x in ( dom ( ProjMap2 (CAB,y))) holds (( ProjMap2 (CAB,y)) . x) = (( chi (A,X1)) . x)

        proof

          let x be Element of X1;

          assume x in ( dom ( ProjMap2 (CAB,y)));

          (( ProjMap2 (CAB,y)) . x) = ((( chi (A,X1)) . x) * (( chi (B,X2)) . y)) by A1;

          hence (( ProjMap2 (CAB,y)) . x) = (( chi (A,X1)) . x) by A2, XXREAL_3: 81;

        end;

        then ( ProjMap2 (CAB,y)) = ( chi (A,X1)) by A4, PARTFUN1: 5;

        hence ((M1 . A) * (( chi (B,X2)) . y)) = ( Integral (M1,( ProjMap2 (CAB,y)))) by A3, MESFUNC9: 14;

      end;

        suppose not y in B;

        then

         A5: (( chi (B,X2)) . y) = 0 by FUNCT_3:def 3;

        then

         A6: ((M1 . A) * (( chi (B,X2)) . y)) = 0 ;

        

         A7: {} is Element of S1 by PROB_1: 4;

        

         A8: ( dom ( ProjMap2 (CAB,y))) = X1 by FUNCT_2:def 1

        .= ( dom ( chi ( {} ,X1))) by FUNCT_3:def 3;

        for x be Element of X1 st x in ( dom ( ProjMap2 (CAB,y))) holds (( ProjMap2 (CAB,y)) . x) = (( chi ( {} ,X1)) . x)

        proof

          let x be Element of X1;

          assume x in ( dom ( ProjMap2 (CAB,y)));

          (( ProjMap2 (CAB,y)) . x) = ((( chi (A,X1)) . x) * (( chi (B,X2)) . y)) by A1;

          then (( ProjMap2 (CAB,y)) . x) = 0 by A5;

          hence (( ProjMap2 (CAB,y)) . x) = (( chi ( {} ,X1)) . x) by FUNCT_3:def 3;

        end;

        then ( ProjMap2 (CAB,y)) = ( chi ( {} ,X1)) by A8, PARTFUN1: 5;

        then ( Integral (M1,( ProjMap2 (CAB,y)))) = (M1 . {} ) by A7, MESFUNC9: 14;

        hence ((M1 . A) * (( chi (B,X2)) . y)) = ( Integral (M1,( ProjMap2 (CAB,y)))) by A6, VALUED_0:def 19;

      end;

    end;

    theorem :: MEASUR11:55

    

     Th50: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2, y be Element of X2 st E = [:A, B:] holds (M1 . ( Measurable-Y-section (E,y))) = ((M1 . A) * (( chi (B,X2)) . y))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2, y be Element of X2;

      assume

       A1: E = [:A, B:];

      per cases ;

        suppose

         A4: y in B;

        then

         A2: (M1 . ( Measurable-Y-section (E,y))) = (M1 . A) by A1, Th16;

        (( chi (B,X2)) . y) = 1 by A4, FUNCT_3:def 3;

        hence (M1 . ( Measurable-Y-section (E,y))) = ((M1 . A) * (( chi (B,X2)) . y)) by A2, XXREAL_3: 81;

      end;

        suppose

         A5: not y in B;

        then ( Measurable-Y-section (E,y)) = {} by A1, Th16;

        then

         A3: (M1 . ( Measurable-Y-section (E,y))) = 0 by VALUED_0:def 19;

        (( chi (B,X2)) . y) = 0 by A5, FUNCT_3:def 3;

        hence (M1 . ( Measurable-Y-section (E,y))) = ((M1 . A) * (( chi (B,X2)) . y)) by A3;

      end;

    end;

    begin

    definition

      let X,Y be non empty set, G be FUNCTION_DOMAIN of X, Y, F be FinSequence of G, n be Nat;

      :: original: /.

      redefine

      func F /. n -> Element of G ;

      correctness ;

    end

    definition

      let X be set;

      let F be FinSequence of ( Funcs (X, ExtREAL ));

      :: MEASUR11:def8

      attr F is without_+infty-valued means

      : DEF10: for n be Nat st n in ( dom F) holds (F . n) is without+infty;

      :: MEASUR11:def9

      attr F is without_-infty-valued means

      : DEF11: for n be Nat st n in ( dom F) holds (F . n) is without-infty;

    end

    theorem :: MEASUR11:56

    

     Th52: for X be non empty set holds <*(X --> 0 )*> is FinSequence of ( Funcs (X, ExtREAL )) & (for n be Nat st n in ( dom <*(X --> 0 )*>) holds ( <*(X --> 0 )*> . n) is without+infty) & (for n be Nat st n in ( dom <*(X --> 0 )*>) holds ( <*(X --> 0 )*> . n) is without-infty)

    proof

      let X be non empty set;

      (X --> 0 ) is Function of X, ExtREAL by XXREAL_0:def 1, FUNCOP_1: 45;

      then

      reconsider f = (X --> 0 ) as Element of ( Funcs (X, ExtREAL )) by FUNCT_2: 8;

       <*f*> is FinSequence of ( Funcs (X, ExtREAL ));

      hence <*(X --> 0 )*> is FinSequence of ( Funcs (X, ExtREAL ));

      hereby

        let n be Nat;

        assume n in ( dom <*(X --> 0 )*>);

        then n in ( Seg 1) by FINSEQ_1: 38;

        then n = 1 by FINSEQ_1: 2, TARSKI:def 1;

        then

         A1: ( <*(X --> 0 )*> . n) = (X --> 0 ) by FINSEQ_1: 40;

         not +infty in ( rng (X --> 0 ));

        hence ( <*(X --> 0 )*> . n) is without+infty by A1, MESFUNC5:def 4;

      end;

      let n be Nat;

      assume n in ( dom <*(X --> 0 )*>);

      then n in ( Seg 1) by FINSEQ_1: 38;

      then n = 1 by FINSEQ_1: 2, TARSKI:def 1;

      then

       A2: ( <*(X --> 0 )*> . n) = (X --> 0 ) by FINSEQ_1: 40;

       not -infty in ( rng (X --> 0 ));

      hence ( <*(X --> 0 )*> . n) is without-infty by A2, MESFUNC5:def 3;

    end;

    registration

      let X be non empty set;

      cluster without_+infty-valued without_-infty-valued for FinSequence of ( Funcs (X, ExtREAL ));

      existence

      proof

        reconsider F = <*(X --> 0 )*> as FinSequence of ( Funcs (X, ExtREAL )) by Th52;

        take F;

        thus thesis by Th52;

      end;

    end

    theorem :: MEASUR11:57

    

     Th53: for X be non empty set, F be without_+infty-valued FinSequence of ( Funcs (X, ExtREAL )), n be Nat st n in ( dom F) holds ((F /. n) " { +infty }) = {}

    proof

      let X be non empty set, F be without_+infty-valued FinSequence of ( Funcs (X, ExtREAL )), n be Nat;

      assume

       A1: n in ( dom F);

      then (F . n) is without+infty by DEF10;

      then not +infty in ( rng (F . n)) by MESFUNC5:def 4;

      then not +infty in ( rng (F /. n)) by A1, PARTFUN1:def 6;

      hence ((F /. n) " { +infty }) = {} by FUNCT_1: 72;

    end;

    theorem :: MEASUR11:58

    

     Th54: for X be non empty set, F be without_-infty-valued FinSequence of ( Funcs (X, ExtREAL )), n be Nat st n in ( dom F) holds ((F /. n) " { -infty }) = {}

    proof

      let X be non empty set, F be without_-infty-valued FinSequence of ( Funcs (X, ExtREAL )), n be Nat;

      assume

       A1: n in ( dom F);

      then (F . n) is without-infty by DEF11;

      then not -infty in ( rng (F . n)) by MESFUNC5:def 3;

      then not -infty in ( rng (F /. n)) by A1, PARTFUN1:def 6;

      hence ((F /. n) " { -infty }) = {} by FUNCT_1: 72;

    end;

    theorem :: MEASUR11:59

    for X be non empty set, F be FinSequence of ( Funcs (X, ExtREAL )) st F is without_+infty-valued or F is without_-infty-valued holds for n,m be Nat st n in ( dom F) & m in ( dom F) holds ( dom ((F /. n) + (F /. m))) = X

    proof

      let X be non empty set, F be FinSequence of ( Funcs (X, ExtREAL ));

      assume

       A1: F is without_+infty-valued or F is without_-infty-valued;

      per cases by A1;

        suppose

         A2: F is without_+infty-valued;

        let n,m be Nat;

        assume n in ( dom F) & m in ( dom F);

        then ((F /. n) " { +infty }) = {} & ((F /. m) " { +infty }) = {} by A2, Th53;

        then

         A4: ((( dom (F /. n)) /\ ( dom (F /. m))) \ ((((F /. n) " { -infty }) /\ ((F /. m) " { +infty })) \/ (((F /. n) " { +infty }) /\ ((F /. m) " { -infty })))) = (( dom (F /. n)) /\ ( dom (F /. m)));

        ( dom (F /. n)) = X & ( dom (F /. m)) = X by FUNCT_2:def 1;

        hence ( dom ((F /. n) + (F /. m))) = X by A4, MESFUNC1:def 3;

      end;

        suppose

         A5: F is without_-infty-valued;

        let n,m be Nat;

        assume n in ( dom F) & m in ( dom F);

        then ((F /. n) " { -infty }) = {} & ((F /. m) " { -infty }) = {} by A5, Th54;

        then

         A7: ((( dom (F /. n)) /\ ( dom (F /. m))) \ ((((F /. n) " { -infty }) /\ ((F /. m) " { +infty })) \/ (((F /. n) " { +infty }) /\ ((F /. m) " { -infty })))) = (( dom (F /. n)) /\ ( dom (F /. m)));

        ( dom (F /. n)) = X & ( dom (F /. m)) = X by FUNCT_2:def 1;

        hence ( dom ((F /. n) + (F /. m))) = X by A7, MESFUNC1:def 3;

      end;

    end;

    definition

      let X be non empty set;

      let F be FinSequence of ( Funcs (X, ExtREAL ));

      :: MEASUR11:def10

      attr F is summable means

      : DEF12: F is without_+infty-valued or F is without_-infty-valued;

    end

    registration

      let X be non empty set;

      cluster summable for FinSequence of ( Funcs (X, ExtREAL ));

      existence

      proof

        reconsider F = <*(X --> 0 )*> as FinSequence of ( Funcs (X, ExtREAL )) by Th52;

        take F;

        for n be Nat st n in ( dom F) holds (F . n) is without+infty by Th52;

        hence F is summable by DEF10;

      end;

    end

    definition

      let X be non empty set;

      let F be summable FinSequence of ( Funcs (X, ExtREAL ));

      :: MEASUR11:def11

      func Partial_Sums F -> FinSequence of ( Funcs (X, ExtREAL )) means

      : DEF13: ( len F) = ( len it ) & (F . 1) = (it . 1) & (for n be Nat st 1 <= n < ( len F) holds (it . (n + 1)) = ((it /. n) + (F /. (n + 1))));

      existence

      proof

        set G = ( Funcs (X, ExtREAL ));

        per cases by DEF12;

          suppose

           a1: F is without_+infty-valued;

          per cases ;

            suppose ( len F) > 0 ;

            then

             a2: ( 0 + 1) <= ( len F) by NAT_1: 13;

            then

             a3: 1 in ( dom F) by FINSEQ_3: 25;

            now

              let n be Nat;

              assume n in ( dom <*(F /. 1)*>);

              then n in ( Seg 1) by FINSEQ_1: 38;

              then n = 1 by FINSEQ_1: 2, TARSKI:def 1;

              then ( <*(F /. 1)*> . n) = (F /. 1) by FINSEQ_1: 40;

              then ( <*(F /. 1)*> . n) = (F . 1) by a3, PARTFUN1:def 6;

              hence ( <*(F /. 1)*> . n) is without+infty by a1, a2, FINSEQ_3: 25;

            end;

            then

            reconsider q = <*(F /. 1)*> as without_+infty-valued FinSequence of G by DEF10;

            (F /. 1) = (F . 1) by a2, FINSEQ_4: 15;

            then

             A3: (q . 1) = (F . 1) by FINSEQ_1: 40;

            defpred S1[ Nat] means (($1 + 1) <= ( len F) implies ex g be without_+infty-valued FinSequence of G st (($1 + 1) = ( len g) & (F . 1) = (g . 1) & (for i be Nat st 1 <= i < ($1 + 1) holds (g . (i + 1)) = ((g /. i) + (F /. (i + 1))))));

            

             A4: for i be Nat st 1 <= i < ( 0 + 1) holds (q . (i + 1)) = ((q /. i) + (F /. (i + 1)));

            

             A5: for k be Nat st S1[k] holds S1[(k + 1)]

            proof

              let k be Nat;

              assume

               A6: S1[k];

              per cases ;

                suppose

                 A7: ((k + 1) + 1) <= ( len F);

                (k + 1) < ((k + 1) + 1) by XREAL_1: 29;

                then

                consider g be without_+infty-valued FinSequence of G such that

                 A8: (k + 1) = ( len g) and

                 A9: (F . 1) = (g . 1) and

                 A10: for i be Nat st 1 <= i & i < (k + 1) holds (g . (i + 1)) = ((g /. i) + (F /. (i + 1))) by A6, A7, XXREAL_0: 2;

                

                 A11: 1 <= ((k + 1) + 1) by NAT_1: 12;

                then

                 A12: ((F /. ((k + 1) + 1)) " { +infty }) = {} by a1, A7, Th53, FINSEQ_3: 25;

                1 <= (k + 1) by NAT_1: 12;

                then

                 A13: (k + 1) in ( dom g) by A8, FINSEQ_3: 25;

                then ((g /. (k + 1)) " { +infty }) = {} by Th53;

                then ((( dom (g /. (k + 1))) /\ ( dom (F /. ((k + 1) + 1)))) \ ((((g /. (k + 1)) " { -infty }) /\ ((F /. ((k + 1) + 1)) " { +infty })) \/ (((g /. (k + 1)) " { +infty }) /\ ((F /. ((k + 1) + 1)) " { -infty })))) = (( dom (g /. (k + 1))) /\ ( dom (F /. ((k + 1) + 1)))) by A12;

                then

                 A14: ( dom ((g /. (k + 1)) + (F /. ((k + 1) + 1)))) = (( dom (g /. (k + 1))) /\ ( dom (F /. ((k + 1) + 1)))) by MESFUNC1:def 3;

                ( dom (g /. (k + 1))) = X & ( dom (F /. ((k + 1) + 1))) = X by FUNCT_2:def 1;

                then ((g /. (k + 1)) + (F /. ((k + 1) + 1))) is Function of X, ExtREAL by A14, FUNCT_2:def 1;

                then ((g /. (k + 1)) + (F /. ((k + 1) + 1))) is Element of G by FUNCT_2: 8;

                then <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*> is FinSequence of G by FINSEQ_1: 74;

                then

                reconsider g2 = (g ^ <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*>) as FinSequence of G by FINSEQ_1: 75;

                now

                  let n be Nat;

                  assume n in ( dom g2);

                  then

                   A15: 1 <= n <= ( len g2) by FINSEQ_3: 25;

                  then

                   A16: 1 <= n <= (( len g) + 1) by FINSEQ_2: 16;

                  per cases ;

                    suppose

                     A17: n = (( len g) + 1);

                    ( len <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*>) = 1 by FINSEQ_1: 40;

                    then 1 in ( dom <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*>) by FINSEQ_3: 25;

                    then

                     A18: (g2 . n) = ( <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*> . 1) by A17, FINSEQ_1:def 7;

                    

                     A19: ((k + 1) + 1) in ( dom F) by A7, NAT_1: 12, FINSEQ_3: 25;

                    (g . (k + 1)) is without+infty & (F . ((k + 1) + 1)) is without+infty by a1, A13, A11, A7, DEF10, FINSEQ_3: 25;

                    then

                    reconsider p = (g /. (k + 1)), q = (F /. ((k + 1) + 1)) as without+infty Function of X, ExtREAL by A13, A19, PARTFUN1:def 6;

                    (p + q) is without+infty Function of X, ExtREAL ;

                    hence (g2 . n) is without+infty by A18, FINSEQ_1: 40;

                  end;

                    suppose n <> (( len g) + 1);

                    then n < (( len g) + 1) by A16, XXREAL_0: 1;

                    then n <= ( len g) by NAT_1: 13;

                    then

                     A20: n in ( dom g) by A15, FINSEQ_3: 25;

                    then (g2 . n) = (g . n) by FINSEQ_1:def 7;

                    hence (g2 . n) is without+infty by A20, DEF10;

                  end;

                end;

                then

                reconsider g2 as without_+infty-valued FinSequence of G by DEF10;

                

                 A21: ( Seg ( len g)) = ( dom g) by FINSEQ_1:def 3;

                

                 A22: ( len g2) = (( len g) + ( len <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*>)) by FINSEQ_1: 22

                .= ((k + 1) + 1) by A8, FINSEQ_1: 40;

                

                 A23: for i be Nat st 1 <= i < ((k + 1) + 1) holds (g2 . (i + 1)) = ((g2 /. i) + (F /. (i + 1)))

                proof

                  let i be Nat;

                  

                   A28: 1 <= (i + 1) by NAT_1: 12;

                  assume that

                   A24: 1 <= i and

                   A25: i < ((k + 1) + 1);

                  

                   A26: i <= (k + 1) by A25, NAT_1: 13;

                  per cases by A26, XXREAL_0: 1;

                    suppose

                     A27: i < (k + 1);

                    then (i + 1) <= (k + 1) by NAT_1: 13;

                    then (i + 1) in ( Seg ( len g)) by A8, A28;

                    then

                     A29: (g2 . (i + 1)) = (g . (i + 1)) by A21, FINSEQ_1:def 7;

                    i in ( Seg ( len g)) by A8, A24, A26;

                    then

                     A30: (g2 . i) = (g . i) by A21, FINSEQ_1:def 7;

                    

                     A31: (g /. i) = (g . i) by A8, A24, A27, FINSEQ_4: 15;

                    (g2 /. i) = (g2 . i) by A22, A24, A25, FINSEQ_4: 15;

                    hence thesis by A10, A24, A27, A29, A30, A31;

                  end;

                    suppose

                     A32: i = (k + 1);

                    

                     A33: (g2 /. i) = (g2 . i) by A22, A24, A25, FINSEQ_4: 15;

                    i in ( Seg ( len g)) by A8, A24, A26;

                    then

                     A34: (g . i) = (g2 . i) by A21, FINSEQ_1:def 7;

                    (g /. i) = (g . i) by A8, A24, A26, FINSEQ_4: 15;

                    hence thesis by A8, A32, A34, A33, FINSEQ_1: 42;

                  end;

                end;

                1 <= (k + 1) by NAT_1: 11;

                then 1 in ( Seg ( len g)) by A8;

                then (g2 . 1) = (F . 1) by A9, A21, FINSEQ_1:def 7;

                hence S1[(k + 1)] by A22, A23;

              end;

                suppose ((k + 1) + 1) > ( len F);

                hence S1[(k + 1)];

              end;

            end;

            

             A35: (( len F) -' 1) = (( len F) - 1) by a2, XREAL_1: 233;

            ( len q) = 1 by FINSEQ_1: 40;

            then

             A36: S1[ 0 ] by A3, A4;

            for k be Nat holds S1[k] from NAT_1:sch 2( A36, A5);

            then ex IT be without_+infty-valued FinSequence of G st ((( len F) -' 1) + 1) = ( len IT) & (F . 1) = (IT . 1) & for i be Nat st 1 <= i < ((( len F) -' 1) + 1) holds (IT . (i + 1)) = ((IT /. i) + (F /. (i + 1))) by A35;

            hence ex IT be FinSequence of G st ( len F) = ( len IT) & (F . 1) = (IT . 1) & (for n be Nat st 1 <= n < ( len F) holds (IT . (n + 1)) = ((IT /. n) + (F /. (n + 1)))) by A35;

          end;

            suppose

             A37: ( len F) <= 0 ;

            take F;

            thus ( len F) = ( len F) & (F . 1) = (F . 1);

            let n be Nat;

            thus 1 <= n & n < ( len F) implies (F . (n + 1)) = ((F /. n) + (F /. (n + 1))) by A37;

          end;

        end;

          suppose

           A38: F is without_-infty-valued;

          per cases ;

            suppose

             A39: ( len F) > 0 ;

            then

             A40: ( 0 + 1) <= ( len F) by NAT_1: 13;

            then

             A41: 1 in ( dom F) by FINSEQ_3: 25;

            now

              let n be Nat;

              assume n in ( dom <*(F /. 1)*>);

              then n in ( Seg 1) by FINSEQ_1: 38;

              then n = 1 by FINSEQ_1: 2, TARSKI:def 1;

              then ( <*(F /. 1)*> . n) = (F /. 1) by FINSEQ_1: 40;

              then ( <*(F /. 1)*> . n) = (F . 1) by A41, PARTFUN1:def 6;

              hence ( <*(F /. 1)*> . n) is without-infty by A38, A40, FINSEQ_3: 25;

            end;

            then

            reconsider q = <*(F /. 1)*> as without_-infty-valued FinSequence of G by DEF11;

            

             A42: ( 0 + 1) <= ( len F) by A39, NAT_1: 13;

            then (F /. 1) = (F . 1) by FINSEQ_4: 15;

            then

             A43: (q . 1) = (F . 1) by FINSEQ_1: 40;

            defpred S1[ Nat] means ($1 + 1) <= ( len F) implies ex g be without_-infty-valued FinSequence of G st (($1 + 1) = ( len g) & (F . 1) = (g . 1) & (for i be Nat st 1 <= i < ($1 + 1) holds (g . (i + 1)) = ((g /. i) + (F /. (i + 1)))));

            

             A44: for i be Nat st 1 <= i < ( 0 + 1) holds (q . (i + 1)) = ((q /. i) + (F /. (i + 1)));

            

             A45: for k be Nat st S1[k] holds S1[(k + 1)]

            proof

              let k be Nat;

              assume

               A46: S1[k];

              per cases ;

                suppose

                 A47: ((k + 1) + 1) <= ( len F);

                (k + 1) < ((k + 1) + 1) by XREAL_1: 29;

                then

                consider g be without_-infty-valued FinSequence of G such that

                 A48: (k + 1) = ( len g) and

                 A49: (F . 1) = (g . 1) and

                 A50: for i be Nat st 1 <= i & i < (k + 1) holds (g . (i + 1)) = ((g /. i) + (F /. (i + 1))) by A46, A47, XXREAL_0: 2;

                

                 A51: ((k + 1) + 1) in ( dom F) by A47, NAT_1: 12, FINSEQ_3: 25;

                then

                 A52: ((F /. ((k + 1) + 1)) " { -infty }) = {} by A38, Th54;

                1 <= (k + 1) by NAT_1: 12;

                then

                 A53: (k + 1) in ( dom g) by A48, FINSEQ_3: 25;

                then ((g /. (k + 1)) " { -infty }) = {} by Th54;

                then ((( dom (g /. (k + 1))) /\ ( dom (F /. ((k + 1) + 1)))) \ ((((g /. (k + 1)) " { -infty }) /\ ((F /. ((k + 1) + 1)) " { +infty })) \/ (((g /. (k + 1)) " { +infty }) /\ ((F /. ((k + 1) + 1)) " { -infty })))) = (( dom (g /. (k + 1))) /\ ( dom (F /. ((k + 1) + 1)))) by A52;

                then

                 A54: ( dom ((g /. (k + 1)) + (F /. ((k + 1) + 1)))) = (( dom (g /. (k + 1))) /\ ( dom (F /. ((k + 1) + 1)))) by MESFUNC1:def 3;

                ( dom (g /. (k + 1))) = X & ( dom (F /. ((k + 1) + 1))) = X by FUNCT_2:def 1;

                then ((g /. (k + 1)) + (F /. ((k + 1) + 1))) is Function of X, ExtREAL by A54, FUNCT_2:def 1;

                then ((g /. (k + 1)) + (F /. ((k + 1) + 1))) is Element of G by FUNCT_2: 8;

                then <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*> is FinSequence of G by FINSEQ_1: 74;

                then

                reconsider g2 = (g ^ <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*>) as FinSequence of G by FINSEQ_1: 75;

                now

                  let n be Nat;

                  assume n in ( dom g2);

                  then

                   A55: 1 <= n <= ( len g2) by FINSEQ_3: 25;

                  then

                   A56: 1 <= n <= (( len g) + 1) by FINSEQ_2: 16;

                  per cases ;

                    suppose

                     A57: n = (( len g) + 1);

                    ( len <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*>) = 1 by FINSEQ_1: 40;

                    then 1 in ( dom <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*>) by FINSEQ_3: 25;

                    then

                     A58: (g2 . n) = ( <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*> . 1) by A57, FINSEQ_1:def 7;

                    (g . (k + 1)) is without-infty & (F . ((k + 1) + 1)) is without-infty by A38, A51, A53, DEF11;

                    then

                    reconsider p = (g /. (k + 1)), q = (F /. ((k + 1) + 1)) as without-infty Function of X, ExtREAL by A51, A53, PARTFUN1:def 6;

                    (p + q) is without-infty Function of X, ExtREAL ;

                    hence (g2 . n) is without-infty by A58, FINSEQ_1: 40;

                  end;

                    suppose n <> (( len g) + 1);

                    then n < (( len g) + 1) by A56, XXREAL_0: 1;

                    then n <= ( len g) by NAT_1: 13;

                    then

                     A59: n in ( dom g) by A55, FINSEQ_3: 25;

                    then (g2 . n) = (g . n) by FINSEQ_1:def 7;

                    hence (g2 . n) is without-infty by A59, DEF11;

                  end;

                end;

                then

                reconsider g2 as without_-infty-valued FinSequence of G by DEF11;

                

                 A60: ( Seg ( len g)) = ( dom g) by FINSEQ_1:def 3;

                

                 A61: ( len g2) = (( len g) + ( len <*((g /. (k + 1)) + (F /. ((k + 1) + 1)))*>)) by FINSEQ_1: 22

                .= ((k + 1) + 1) by A48, FINSEQ_1: 40;

                

                 A62: for i be Nat st 1 <= i < ((k + 1) + 1) holds (g2 . (i + 1)) = ((g2 /. i) + (F /. (i + 1)))

                proof

                  let i be Nat;

                  assume

                   A63: 1 <= i & i < ((k + 1) + 1);

                  then

                   A65: i <= (k + 1) by NAT_1: 13;

                  per cases by A65, XXREAL_0: 1;

                    suppose

                     A66: i < (k + 1);

                    

                     A67: 1 <= (i + 1) by NAT_1: 12;

                    (i + 1) <= (k + 1) by A66, NAT_1: 13;

                    then (i + 1) in ( Seg ( len g)) by A48, A67;

                    then

                     A68: (g2 . (i + 1)) = (g . (i + 1)) by A60, FINSEQ_1:def 7;

                    i in ( Seg ( len g)) by A48, A63, A65;

                    then

                     A69: (g2 . i) = (g . i) by A60, FINSEQ_1:def 7;

                    

                     A70: (g /. i) = (g . i) by A48, A63, A66, FINSEQ_4: 15;

                    (g2 /. i) = (g2 . i) by A61, A63, FINSEQ_4: 15;

                    hence thesis by A50, A63, A66, A68, A69, A70;

                  end;

                    suppose

                     A71: i = (k + 1);

                    

                     A72: (g2 /. i) = (g2 . i) by A61, A63, FINSEQ_4: 15;

                    i in ( Seg ( len g)) by A48, A63, A65;

                    then

                     A73: (g . i) = (g2 . i) by A60, FINSEQ_1:def 7;

                    (g /. i) = (g . i) by A48, A63, A65, FINSEQ_4: 15;

                    hence thesis by A48, A71, A72, A73, FINSEQ_1: 42;

                  end;

                end;

                1 <= (k + 1) by NAT_1: 11;

                then 1 in ( Seg ( len g)) by A48;

                then (g2 . 1) = (F . 1) by A49, A60, FINSEQ_1:def 7;

                hence S1[(k + 1)] by A61, A62;

              end;

                suppose ((k + 1) + 1) > ( len F);

                hence S1[(k + 1)];

              end;

            end;

            

             A74: (( len F) -' 1) = (( len F) - 1) by A42, XREAL_1: 233;

            ( len q) = 1 by FINSEQ_1: 40;

            then

             A75: S1[ 0 ] by A43, A44;

            for k be Nat holds S1[k] from NAT_1:sch 2( A75, A45);

            then ex IT be without_-infty-valued FinSequence of G st ((( len F) -' 1) + 1) = ( len IT) & (F . 1) = (IT . 1) & for i be Nat st 1 <= i < ((( len F) -' 1) + 1) holds (IT . (i + 1)) = ((IT /. i) + (F /. (i + 1))) by A74;

            hence ex IT be FinSequence of G st ( len F) = ( len IT) & (F . 1) = (IT . 1) & (for n be Nat st 1 <= n < ( len F) holds (IT . (n + 1)) = ((IT /. n) + (F /. (n + 1)))) by A74;

          end;

            suppose

             A76: ( len F) <= 0 ;

            take F;

            thus ( len F) = ( len F) & (F . 1) = (F . 1);

            let n be Nat;

            thus 1 <= n & n < ( len F) implies (F . (n + 1)) = ((F /. n) + (F /. (n + 1))) by A76;

          end;

        end;

      end;

      uniqueness

      proof

        let g1,g2 be FinSequence of ( Funcs (X, ExtREAL ));

        assume that

         A28: ( len F) = ( len g1) and

         A29: (F . 1) = (g1 . 1) and

         A30: for i be Nat st 1 <= i & i < ( len F) holds (g1 . (i + 1)) = ((g1 /. i) + (F /. (i + 1)));

        defpred P[ Nat] means 1 <= $1 & $1 <= ( len F) implies (g1 . $1) = (g2 . $1);

        assume that

         A31: ( len F) = ( len g2) and

         A32: (F . 1) = (g2 . 1) and

         A33: for i be Nat st 1 <= i & i < ( len F) holds (g2 . (i + 1)) = ((g2 /. i) + (F /. (i + 1)));

        

         A34: for k be Nat st P[k] holds P[(k + 1)]

        proof

          let k be Nat;

          assume

           A35: P[k];

          1 <= (k + 1) & (k + 1) <= ( len F) implies (g1 . (k + 1)) = (g2 . (k + 1))

          proof

            assume that 1 <= (k + 1) and

             A36: (k + 1) <= ( len F);

            

             A37: k < (k + 1) by XREAL_1: 29;

            then

             A38: k < ( len F) by A36, XXREAL_0: 2;

            per cases ;

              suppose

               A39: 1 <= k;

              then

               A40: (g2 . (k + 1)) = ((g2 /. k) + (F /. (k + 1))) by A33, A38;

              

               A41: k <= ( len g2) by A31, A36, A37, XXREAL_0: 2;

              

               A42: (g1 /. k) = (g1 . k) by A28, A38, A39, FINSEQ_4: 15;

              (g1 . (k + 1)) = ((g1 /. k) + (F /. (k + 1))) by A30, A38, A39;

              hence thesis by A35, A36, A37, A39, A40, A42, A41, FINSEQ_4: 15, XXREAL_0: 2;

            end;

              suppose 1 > k;

              then ( 0 + 1) > k;

              then k = 0 by NAT_1: 13;

              hence (g1 . (k + 1)) = (g2 . (k + 1)) by A29, A32;

            end;

          end;

          hence P[(k + 1)];

        end;

        

         A43: P[ 0 ];

        for k be Nat holds P[k] from NAT_1:sch 2( A43, A34);

        hence g1 = g2 by A28, A31, FINSEQ_1: 14;

      end;

    end

    registration

      let X be non empty set;

      cluster without_+infty-valued -> summable for FinSequence of ( Funcs (X, ExtREAL ));

      correctness ;

      cluster without_-infty-valued -> summable for FinSequence of ( Funcs (X, ExtREAL ));

      correctness ;

    end

    theorem :: MEASUR11:60

    

     Th56: for X be non empty set, F be without_+infty-valued FinSequence of ( Funcs (X, ExtREAL )) holds ( Partial_Sums F) is without_+infty-valued

    proof

      let X be non empty set, F be without_+infty-valued FinSequence of ( Funcs (X, ExtREAL ));

      defpred P[ Nat] means $1 in ( dom ( Partial_Sums F)) implies (( Partial_Sums F) . $1) is without+infty;

      

       A1: P[ 0 ] by FINSEQ_3: 24;

      

       A2: for n be Nat st P[n] holds P[(n + 1)]

      proof

        let n be Nat;

        assume

         A3: P[n];

        

         B1: ( len F) = ( len ( Partial_Sums F)) by DEF13;

        then

         A4: ( dom F) = ( dom ( Partial_Sums F)) by FINSEQ_3: 29;

        assume

         A5: (n + 1) in ( dom ( Partial_Sums F));

        per cases ;

          suppose

           A6: n = 0 ;

          then (F . 1) is without+infty by A4, A5, DEF10;

          hence (( Partial_Sums F) . (n + 1)) is without+infty by A6, DEF13;

        end;

          suppose

           A7: n <> 0 ;

          then

           A8: n >= 1 by NAT_1: 14;

          (n + 1) <= ( len F) by A5, B1, FINSEQ_3: 25;

          then

           A9: n < ( len F) by NAT_1: 13;

          (F . (n + 1)) is without+infty by A4, A5, DEF10;

          then

          reconsider p = (( Partial_Sums F) /. n), q = (F /. (n + 1)) as without+infty Function of X, ExtREAL by A3, A4, A5, A8, A9, FINSEQ_3: 25, PARTFUN1:def 6;

          (p + q) is without+infty Function of X, ExtREAL ;

          hence (( Partial_Sums F) . (n + 1)) is without+infty by A7, A9, DEF13, NAT_1: 14;

        end;

      end;

      for n be Nat holds P[n] from NAT_1:sch 2( A1, A2);

      hence thesis;

    end;

    theorem :: MEASUR11:61

    

     Th57: for X be non empty set, F be without_-infty-valued FinSequence of ( Funcs (X, ExtREAL )) holds ( Partial_Sums F) is without_-infty-valued

    proof

      let X be non empty set, F be without_-infty-valued FinSequence of ( Funcs (X, ExtREAL ));

      defpred P[ Nat] means $1 in ( dom ( Partial_Sums F)) implies (( Partial_Sums F) . $1) is without-infty;

      

       A1: P[ 0 ] by FINSEQ_3: 24;

      

       A2: for n be Nat st P[n] holds P[(n + 1)]

      proof

        let n be Nat;

        assume

         A3: P[n];

        

         B4: ( len F) = ( len ( Partial_Sums F)) by DEF13;

        then

         A4: ( dom F) = ( dom ( Partial_Sums F)) by FINSEQ_3: 29;

        assume

         A5: (n + 1) in ( dom ( Partial_Sums F));

        per cases ;

          suppose

           A6: n = 0 ;

          then (F . 1) is without-infty by A4, A5, DEF11;

          hence (( Partial_Sums F) . (n + 1)) is without-infty by A6, DEF13;

        end;

          suppose n <> 0 ;

          then

           A8: n >= 1 by NAT_1: 14;

          

           A7: (n + 1) <= ( len F) by A5, B4, FINSEQ_3: 25;

          then

           A9: n < ( len F) by NAT_1: 13;

          (F . (n + 1)) is without-infty by A4, A5, DEF11;

          then

          reconsider p = (( Partial_Sums F) /. n), q = (F /. (n + 1)) as without-infty Function of X, ExtREAL by A9, A3, A4, A5, A8, FINSEQ_3: 25, PARTFUN1:def 6;

          (p + q) is without-infty Function of X, ExtREAL ;

          hence (( Partial_Sums F) . (n + 1)) is without-infty by A8, A7, DEF13, NAT_1: 13;

        end;

      end;

      for n be Nat holds P[n] from NAT_1:sch 2( A1, A2);

      hence thesis;

    end;

    theorem :: MEASUR11:62

    for X be non empty set, A be set, er be ExtReal, f be Function of X, ExtREAL st (for x be Element of X holds (f . x) = (er * (( chi (A,X)) . x))) holds (er = +infty implies f = ( Xchi (A,X))) & (er = -infty implies f = ( - ( Xchi (A,X)))) & (er <> +infty & er <> -infty implies ex r be Real st r = er & f = (r (#) ( chi (A,X))))

    proof

      let X be non empty set, A be set, er be ExtReal, f be Function of X, ExtREAL ;

      assume

       A1: for x be Element of X holds (f . x) = (er * (( chi (A,X)) . x));

      hereby

        assume

         A2: er = +infty ;

        for x be Element of X holds (f . x) = (( Xchi (A,X)) . x)

        proof

          let x be Element of X;

          per cases ;

            suppose

             A3: x in A;

            then

             A4: (( Xchi (A,X)) . x) = +infty by MEASUR10:def 7;

            (( chi (A,X)) . x) = 1 by A3, FUNCT_3:def 3;

            then (f . x) = (er * 1) by A1;

            hence (f . x) = (( Xchi (A,X)) . x) by A2, A4, XXREAL_3: 81;

          end;

            suppose

             A5: not x in A;

            then (( chi (A,X)) . x) = 0 by FUNCT_3:def 3;

            then (f . x) = (er * 0 ) by A1;

            hence (f . x) = (( Xchi (A,X)) . x) by A5, MEASUR10:def 7;

          end;

        end;

        hence f = ( Xchi (A,X)) by FUNCT_2:def 8;

      end;

      hereby

        assume

         A6: er = -infty ;

        for x be Element of X holds (f . x) = (( - ( Xchi (A,X))) . x)

        proof

          let x be Element of X;

          ( dom ( Xchi (A,X))) = X by FUNCT_2:def 1;

          then x in ( dom ( Xchi (A,X)));

          then

           A7: x in ( dom ( - ( Xchi (A,X)))) by MESFUNC1:def 7;

          per cases ;

            suppose

             A8: x in A;

            then (( Xchi (A,X)) . x) = +infty by MEASUR10:def 7;

            then

             A9: (( - ( Xchi (A,X))) . x) = ( - +infty ) by A7, MESFUNC1:def 7;

            (( chi (A,X)) . x) = 1 by A8, FUNCT_3:def 3;

            then (f . x) = (er * 1) by A1;

            hence (f . x) = (( - ( Xchi (A,X))) . x) by A6, A9, XXREAL_3: 6, XXREAL_3: 81;

          end;

            suppose

             A10: not x in A;

            then

             A11: ( - (( Xchi (A,X)) . x)) = ( - 0 ) by MEASUR10:def 7;

            (( chi (A,X)) . x) = 0 by A10, FUNCT_3:def 3;

            then (f . x) = (er * 0 ) by A1;

            hence (f . x) = (( - ( Xchi (A,X))) . x) by A7, A11, MESFUNC1:def 7;

          end;

        end;

        hence f = ( - ( Xchi (A,X))) by FUNCT_2:def 8;

      end;

      assume er <> +infty & er <> -infty ;

      then er in REAL by XXREAL_0: 14;

      then

      reconsider r = er as Real;

      ( dom f) = X & ( dom ( chi (A,X))) = X by FUNCT_2:def 1;

      then

       A12: ( dom f) = ( dom (r (#) ( chi (A,X)))) by MESFUNC1:def 6;

      for x be Element of X st x in ( dom f) holds (f . x) = ((r (#) ( chi (A,X))) . x)

      proof

        let x be Element of X;

        assume x in ( dom f);

        then ((r (#) ( chi (A,X))) . x) = (r * (( chi (A,X)) . x)) by A12, MESFUNC1:def 6;

        hence (f . x) = ((r (#) ( chi (A,X))) . x) by A1;

      end;

      hence ex r be Real st r = er & f = (r (#) ( chi (A,X))) by A12, PARTFUN1: 5;

    end;

    theorem :: MEASUR11:63

    

     Th59: for X be non empty set, S be SigmaField of X, f be PartFunc of X, ExtREAL , A be Element of S st f is A -measurable & A c= ( dom f) holds ( - f) is A -measurable

    proof

      let X be non empty set, S be SigmaField of X, f be PartFunc of X, ExtREAL , A be Element of S;

      assume that

       A1: f is A -measurable and

       A2: A c= ( dom f);

      ( - f) = (( - 1) (#) f) by MESFUNC2: 9;

      hence thesis by A1, A2, MESFUNC1: 37;

    end;

    registration

      let X be non empty set, f be without-infty PartFunc of X, ExtREAL ;

      cluster ( - f) -> without+infty;

      correctness

      proof

        now

          let x be set;

          assume

           A1: x in ( dom ( - f));

          then x in ( dom f) by MESFUNC1:def 7;

          then ((f . x) * ( - 1)) < ( -infty * ( - 1)) by MESFUNC5: 10, XXREAL_3: 102;

          then ((f . x) * ( - 1)) < ( - -infty ) by XXREAL_3: 91;

          then ( - (f . x)) < +infty by XXREAL_3: 5, XXREAL_3: 91;

          hence (( - f) . x) < +infty by A1, MESFUNC1:def 7;

        end;

        hence ( - f) is without+infty by MESFUNC5: 11;

      end;

    end

    registration

      let X be non empty set, f be without+infty PartFunc of X, ExtREAL ;

      cluster ( - f) -> without-infty;

      correctness

      proof

        now

          let x be set;

          assume

           A1: x in ( dom ( - f));

          then x in ( dom f) by MESFUNC1:def 7;

          then (( - 1) * +infty ) < (( - 1) * (f . x)) by MESFUNC5: 11, XXREAL_3: 102;

          then ( - +infty ) < (( - 1) * (f . x)) by XXREAL_3: 91;

          then -infty < ( - (f . x)) by XXREAL_3: 6, XXREAL_3: 91;

          hence -infty < (( - f) . x) by A1, MESFUNC1:def 7;

        end;

        hence ( - f) is without-infty by MESFUNC5: 10;

      end;

    end

    definition

      let X be non empty set;

      let f1,f2 be without+infty PartFunc of X, ExtREAL ;

      :: original: +

      redefine

      func f1 + f2 -> without+infty PartFunc of X, ExtREAL ;

      correctness by MESFUNC9: 4;

    end

    definition

      let X be non empty set;

      let f1,f2 be without-infty PartFunc of X, ExtREAL ;

      :: original: +

      redefine

      func f1 + f2 -> without-infty PartFunc of X, ExtREAL ;

      correctness by MESFUNC9: 3;

    end

    definition

      let X be non empty set;

      let f1 be without+infty PartFunc of X, ExtREAL ;

      let f2 be without-infty PartFunc of X, ExtREAL ;

      :: original: -

      redefine

      func f1 - f2 -> without+infty PartFunc of X, ExtREAL ;

      correctness by MESFUNC9: 6;

    end

    definition

      let X be non empty set;

      let f1 be without-infty PartFunc of X, ExtREAL ;

      let f2 be without+infty PartFunc of X, ExtREAL ;

      :: original: -

      redefine

      func f1 - f2 -> without-infty PartFunc of X, ExtREAL ;

      correctness by MESFUNC9: 5;

    end

    

     LEM10: for X be non empty set, f be PartFunc of X, ExtREAL holds (f " { +infty }) = (( - f) " { -infty }) & (f " { -infty }) = (( - f) " { +infty })

    proof

      let X be non empty set, f be PartFunc of X, ExtREAL ;

      now

        let x be set;

        assume x in (f " { +infty });

        then

         A1: x in ( dom f) & (f . x) in { +infty } by FUNCT_1:def 7;

        then

         A2: x in ( dom ( - f)) by MESFUNC1:def 7;

        (f . x) = +infty by A1, TARSKI:def 1;

        then (( - f) . x) = ( - +infty ) by A2, MESFUNC1:def 7;

        then (( - f) . x) in { -infty } by XXREAL_3: 6, TARSKI:def 1;

        hence x in (( - f) " { -infty }) by A2, FUNCT_1:def 7;

      end;

      then

       A3: (f " { +infty }) c= (( - f) " { -infty });

      now

        let x be set;

        assume x in (( - f) " { -infty });

        then

         A4: x in ( dom ( - f)) & (( - f) . x) in { -infty } by FUNCT_1:def 7;

        then

         A5: x in ( dom f) by MESFUNC1:def 7;

        (( - f) . x) = -infty by A4, TARSKI:def 1;

        then ( - (f . x)) = -infty by A4, MESFUNC1:def 7;

        then (f . x) in { +infty } by XXREAL_3: 5, TARSKI:def 1;

        hence x in (f " { +infty }) by A5, FUNCT_1:def 7;

      end;

      then (( - f) " { -infty }) c= (f " { +infty });

      hence (f " { +infty }) = (( - f) " { -infty }) by A3;

      now

        let x be set;

        assume x in (f " { -infty });

        then

         A7: x in ( dom f) & (f . x) in { -infty } by FUNCT_1:def 7;

        then

         A8: x in ( dom ( - f)) by MESFUNC1:def 7;

        (f . x) = -infty by A7, TARSKI:def 1;

        then (( - f) . x) = ( - -infty ) by A8, MESFUNC1:def 7;

        then (( - f) . x) in { +infty } by XXREAL_3: 5, TARSKI:def 1;

        hence x in (( - f) " { +infty }) by A8, FUNCT_1:def 7;

      end;

      then

       A9: (f " { -infty }) c= (( - f) " { +infty });

      now

        let x be set;

        assume x in (( - f) " { +infty });

        then

         A10: x in ( dom ( - f)) & (( - f) . x) in { +infty } by FUNCT_1:def 7;

        then

         A11: x in ( dom f) by MESFUNC1:def 7;

        (( - f) . x) = +infty by A10, TARSKI:def 1;

        then ( - (f . x)) = +infty by A10, MESFUNC1:def 7;

        then (f . x) in { -infty } by XXREAL_3: 6, TARSKI:def 1;

        hence x in (f " { -infty }) by A11, FUNCT_1:def 7;

      end;

      then (( - f) " { +infty }) c= (f " { -infty });

      hence (f " { -infty }) = (( - f) " { +infty }) by A9;

    end;

    theorem :: MEASUR11:64

    

     Th60: for X be non empty set, f,g be PartFunc of X, ExtREAL holds ( - (f + g)) = (( - f) + ( - g)) & ( - (f - g)) = (( - f) + g) & ( - (f - g)) = (g - f) & ( - (( - f) + g)) = (f - g) & ( - (( - f) + g)) = (f + ( - g))

    proof

      let X be non empty set, f,g be PartFunc of X, ExtREAL ;

      

       A1: (f " { -infty }) = (( - f) " { +infty }) & (f " { +infty }) = (( - f) " { -infty }) & (g " { -infty }) = (( - g) " { +infty }) & (g " { +infty }) = (( - g) " { -infty }) by LEM10;

      

       A2: ( dom f) = ( dom ( - f)) & ( dom g) = ( dom ( - g)) by MESFUNC1:def 7;

      

       A3: ( dom ( - (f + g))) = ( dom (f + g)) & ( dom ( - f)) = ( dom f) & ( dom ( - g)) = ( dom g) & ( dom ( - (f - g))) = ( dom (f - g)) by MESFUNC1:def 7;

      then

       A4: ( dom ( - (f + g))) = ((( dom f) /\ ( dom g)) \ (((f " { -infty }) /\ (g " { +infty })) \/ ((f " { +infty }) /\ (g " { -infty })))) by MESFUNC1:def 3;

      then

       A5: ( dom ( - (f + g))) = ( dom (( - f) + ( - g))) by A1, A2, MESFUNC1:def 3;

      

       A6: ( dom ( - (f - g))) = ((( dom f) /\ ( dom g)) \ (((f " { +infty }) /\ (g " { +infty })) \/ ((f " { -infty }) /\ (g " { -infty })))) by A3, MESFUNC1:def 4;

      then

       A7: ( dom ( - (f - g))) = ( dom (( - f) + g)) by A1, A2, MESFUNC1:def 3;

      then

       C1: ( dom ( - (( - f) + g))) = ( dom ( - (f - g))) by MESFUNC1:def 7;

      then

       A10: ( dom ( - (( - f) + g))) = ( dom (f - g)) by MESFUNC1:def 7;

      then ( dom ( - (( - f) + g))) = ((( dom f) /\ ( dom g)) \ (((f " { +infty }) /\ (g " { +infty })) \/ ((f " { -infty }) /\ (g " { -infty })))) by MESFUNC1:def 4;

      then

       A12: ( dom ( - (( - f) + g))) = ( dom (f + ( - g))) by A1, A2, MESFUNC1:def 3;

      

       A8: ( dom ( - (f - g))) = ( dom (g - f)) by A6, MESFUNC1:def 4;

      

       A9: ( dom ( - (( - f) + g))) = ( dom (( - f) + g)) by MESFUNC1:def 7;

      

       B3: ( dom ( - (f + g))) c= ( dom f) & ( dom ( - (f + g))) c= ( dom g) by A4, XBOOLE_1: 18, XBOOLE_1: 36;

      

       B4: ( dom ( - (f - g))) c= ( dom f) & ( dom ( - (f - g))) c= ( dom g) by A6, XBOOLE_1: 18, XBOOLE_1: 36;

      

       B5: ( dom ( - (( - f) + g))) c= ( dom ( - f)) & ( dom ( - (( - f) + g))) c= ( dom g) by C1, A3, A6, XBOOLE_1: 18, XBOOLE_1: 36;

      now

        let x be Element of X;

        assume

         B2: x in ( dom ( - (f + g)));

        

        then (( - (f + g)) . x) = ( - ((f + g) . x)) by MESFUNC1:def 7

        .= ( - ((f . x) + (g . x))) by A3, B2, MESFUNC1:def 3

        .= (( - (f . x)) + ( - (g . x))) by XXREAL_3: 9

        .= ((( - f) . x) + ( - (g . x))) by A2, B2, B3, MESFUNC1:def 7

        .= ((( - f) . x) + (( - g) . x)) by A2, B2, B3, MESFUNC1:def 7;

        hence (( - (f + g)) . x) = ((( - f) + ( - g)) . x) by B2, A5, MESFUNC1:def 3;

      end;

      hence ( - (f + g)) = (( - f) + ( - g)) by A5, PARTFUN1: 5;

      now

        let x be Element of X;

        assume

         B2: x in ( dom ( - (f - g)));

        

        then (( - (f - g)) . x) = ( - ((f - g) . x)) by MESFUNC1:def 7

        .= ( - ((f . x) - (g . x))) by A3, B2, MESFUNC1:def 4

        .= (( - (f . x)) + (g . x)) by XXREAL_3: 26

        .= ((( - f) . x) + (g . x)) by A2, B4, B2, MESFUNC1:def 7;

        hence (( - (f - g)) . x) = ((( - f) + g) . x) by B2, A7, MESFUNC1:def 3;

      end;

      hence ( - (f - g)) = (( - f) + g) by A7, PARTFUN1: 5;

      now

        let x be Element of X;

        assume

         B2: x in ( dom ( - (f - g)));

        

        then (( - (f - g)) . x) = ( - ((f - g) . x)) by MESFUNC1:def 7

        .= ( - ((f . x) - (g . x))) by A3, B2, MESFUNC1:def 4

        .= ((g . x) - (f . x)) by XXREAL_3: 26;

        hence (( - (f - g)) . x) = ((g - f) . x) by B2, A8, MESFUNC1:def 4;

      end;

      hence ( - (f - g)) = (g - f) by A8, PARTFUN1: 5;

      now

        let x be Element of X;

        assume

         B2: x in ( dom ( - (( - f) + g)));

        

        then (( - (( - f) + g)) . x) = ( - ((( - f) + g) . x)) by MESFUNC1:def 7

        .= ( - ((( - f) . x) + (g . x))) by A9, B2, MESFUNC1:def 3

        .= ( - (( - (f . x)) + (g . x))) by B5, B2, MESFUNC1:def 7

        .= ((f . x) - (g . x)) by XXREAL_3: 27;

        hence (( - (( - f) + g)) . x) = ((f - g) . x) by B2, A10, MESFUNC1:def 4;

      end;

      hence ( - (( - f) + g)) = (f - g) by A10, PARTFUN1: 5;

      now

        let x be Element of X;

        assume

         B2: x in ( dom ( - (( - f) + g)));

        

        then (( - (( - f) + g)) . x) = ( - ((( - f) + g) . x)) by MESFUNC1:def 7

        .= ( - ((( - f) . x) + (g . x))) by A9, B2, MESFUNC1:def 3

        .= ( - (( - (f . x)) + (g . x))) by B5, B2, MESFUNC1:def 7

        .= ((f . x) + ( - (g . x))) by XXREAL_3: 27

        .= ((f . x) + (( - g) . x)) by B2, B5, A3, MESFUNC1:def 7;

        hence (( - (( - f) + g)) . x) = ((f + ( - g)) . x) by B2, A12, MESFUNC1:def 3;

      end;

      hence ( - (( - f) + g)) = (f + ( - g)) by A12, PARTFUN1: 5;

    end;

    theorem :: MEASUR11:65

    

     Th61: for X be non empty set, S be SigmaField of X, f,g be without+infty PartFunc of X, ExtREAL , A be Element of S st f is A -measurable & g is A -measurable & A c= ( dom (f + g)) holds (f + g) is A -measurable

    proof

      let X be non empty set, S be SigmaField of X, f,g be without+infty PartFunc of X, ExtREAL , A be Element of S;

      assume that

       A3: f is A -measurable and

       A4: g is A -measurable and

       A5: A c= ( dom (f + g));

      

       A6: ( dom (f + g)) = (( dom f) /\ ( dom g)) by MESFUNC9: 1;

      (( dom f) /\ ( dom g)) c= ( dom f) & (( dom f) /\ ( dom g)) c= ( dom g) by XBOOLE_1: 17;

      then A c= ( dom f) & A c= ( dom g) by A5, A6;

      then ( - f) is A -measurable & ( - g) is A -measurable by A3, A4, Th59;

      then

       A7: (( - f) + ( - g)) is A -measurable by MESFUNC5: 31;

      ( dom f) = ( dom ( - f)) & ( dom g) = ( dom ( - g)) by MESFUNC1:def 7;

      

      then ( dom (( - f) + ( - g))) = (( dom f) /\ ( dom g)) by MESFUNC5: 16

      .= ( dom (f + g)) by MESFUNC9: 1;

      then

       A8: ( - (( - f) + ( - g))) is A -measurable by A5, A7, Th59;

      (( - f) + ( - g)) = ( - (f + g)) by Th60;

      hence thesis by A8, DBLSEQ_3: 2;

    end;

    theorem :: MEASUR11:66

    for X be non empty set, S be SigmaField of X, A be Element of S, f be without+infty PartFunc of X, ExtREAL , g be without-infty PartFunc of X, ExtREAL st f is A -measurable & g is A -measurable & A c= ( dom (f - g)) holds (f - g) is A -measurable

    proof

      let X be non empty set, S be SigmaField of X, A be Element of S, f be without+infty PartFunc of X, ExtREAL , g be without-infty PartFunc of X, ExtREAL ;

      assume that

       A1: f is A -measurable and

       A2: g is A -measurable and

       A3: A c= ( dom (f - g));

      

       A4: ( dom (f - g)) = (( dom f) /\ ( dom g)) by MESFUNC9: 2;

      (( dom f) /\ ( dom g)) c= ( dom f) & (( dom f) /\ ( dom g)) c= ( dom g) by XBOOLE_1: 17;

      then A c= ( dom f) & A c= ( dom g) by A3, A4;

      then ( - f) is A -measurable by A1, Th59;

      then

       A5: (( - f) + g) is A -measurable by A2, MESFUNC5: 31;

      ( dom f) = ( dom ( - f)) & ( dom g) = ( dom ( - g)) by MESFUNC1:def 7;

      then ( dom (( - f) + g)) = (( dom f) /\ ( dom g)) by MESFUNC5: 16;

      then ( dom (( - f) + g)) = ( dom (f - g)) by MESFUNC9: 2;

      then ( - (( - f) + g)) is A -measurable by A3, A5, Th59;

      hence (f - g) is A -measurable by Th60;

    end;

    theorem :: MEASUR11:67

    for X be non empty set, S be SigmaField of X, A be Element of S, f be without-infty PartFunc of X, ExtREAL , g be without+infty PartFunc of X, ExtREAL st f is A -measurable & g is A -measurable & A c= ( dom (f - g)) holds (f - g) is A -measurable

    proof

      let X be non empty set, S be SigmaField of X, A be Element of S, f be without-infty PartFunc of X, ExtREAL , g be without+infty PartFunc of X, ExtREAL ;

      assume that

       A1: f is A -measurable and

       A2: g is A -measurable and

       A3: A c= ( dom (f - g));

      

       A4: ( dom (f - g)) = (( dom f) /\ ( dom g)) by MESFUNC5: 17;

      ( dom (( - f) + g)) = ( dom ( - (f - g))) by Th60;

      then

       A5: ( dom (( - f) + g)) = ( dom (f - g)) by MESFUNC1:def 7;

      (( dom f) /\ ( dom g)) c= ( dom f) & (( dom f) /\ ( dom g)) c= ( dom g) by XBOOLE_1: 17;

      then A c= ( dom f) & A c= ( dom g) by A3, A4;

      then ( - f) is A -measurable by A1, Th59;

      then

       A6: (( - f) + g) is A -measurable by A2, A3, A5, Th61;

      ( dom f) = ( dom ( - f)) & ( dom g) = ( dom ( - g)) by MESFUNC1:def 7;

      then ( dom (( - f) + g)) = (( dom f) /\ ( dom g)) by MESFUNC9: 1;

      then ( dom (( - f) + g)) = ( dom (f - g)) by MESFUNC5: 17;

      then ( - (( - f) + g)) is A -measurable by A3, A6, Th59;

      hence (f - g) is A -measurable by Th60;

    end;

    theorem :: MEASUR11:68

    

     Th64: for X be non empty set, S be SigmaField of X, P be Element of S, F be summable FinSequence of ( Funcs (X, ExtREAL )) st (for n be Nat st n in ( dom F) holds (F /. n) is P -measurable) holds for n be Nat st n in ( dom F) holds (( Partial_Sums F) /. n) is P -measurable

    proof

      let X be non empty set, S be SigmaField of X, P be Element of S, F be summable FinSequence of ( Funcs (X, ExtREAL ));

      assume

       A1: for n be Nat st n in ( dom F) holds (F /. n) is P -measurable;

      

       A2: P c= X;

      

       A3: ( len F) = ( len ( Partial_Sums F)) by DEF13;

      then

       A4: ( dom F) = ( dom ( Partial_Sums F)) by FINSEQ_3: 29;

      defpred P[ Nat] means $1 in ( dom F) implies (( Partial_Sums F) /. $1) is P -measurable;

      per cases by DEF12;

        suppose

         A5: F is without_+infty-valued;

        

         A6: P[ 0 ] by FINSEQ_3: 24;

        

         A7: for n be Nat st P[n] holds P[(n + 1)]

        proof

          let n be Nat;

          assume

           A8: P[n];

          assume

           A9: (n + 1) in ( dom F);

          per cases ;

            suppose

             A10: n = 0 ;

            

            then (( Partial_Sums F) /. (n + 1)) = (( Partial_Sums F) . 1) by A4, A9, PARTFUN1:def 6

            .= (F . 1) by DEF13

            .= (F /. 1) by A9, A10, PARTFUN1:def 6;

            hence (( Partial_Sums F) /. (n + 1)) is P -measurable by A1, A9, A10;

          end;

            suppose

             A11: n <> 0 ;

            then

             A12: n >= 1 by NAT_1: 14;

            (n + 1) <= ( len F) by A9, FINSEQ_3: 25;

            then

             A13: n < ( len F) by NAT_1: 13;

            then

             A15: (F /. (n + 1)) = (F . (n + 1)) & (( Partial_Sums F) /. n) = (( Partial_Sums F) . n) & (( Partial_Sums F) /. (n + 1)) = (( Partial_Sums F) . (n + 1)) by A4, A9, A12, FINSEQ_3: 25, PARTFUN1:def 6;

            then

             A16: (( Partial_Sums F) /. (n + 1)) = ((( Partial_Sums F) /. n) + (F /. (n + 1))) by A11, A13, NAT_1: 14, DEF13;

            ( Partial_Sums F) is without_+infty-valued by A5, Th56;

            then

             A17: (F /. (n + 1)) is without+infty & (( Partial_Sums F) /. n) is without+infty by A5, A9, A12, A15, A13, A3, FINSEQ_3: 25;

            then

             A19: ( dom ((( Partial_Sums F) /. n) + (F /. (n + 1)))) = (( dom (( Partial_Sums F) /. n)) /\ ( dom (F /. (n + 1)))) by MESFUNC9: 1;

            

             A18: P c= ( dom (( Partial_Sums F) /. n)) & P c= ( dom (F /. (n + 1))) by A2, FUNCT_2:def 1;

            (F /. (n + 1)) is P -measurable by A9, A1;

            hence (( Partial_Sums F) /. (n + 1)) is P -measurable by A8, A12, A13, A16, A17, A18, A19, Th61, FINSEQ_3: 25, XBOOLE_1: 19;

          end;

        end;

        for n be Nat holds P[n] from NAT_1:sch 2( A6, A7);

        hence for n be Nat st n in ( dom F) holds (( Partial_Sums F) /. n) is P -measurable;

      end;

        suppose

         A19: F is without_-infty-valued;

        

         A20: P[ 0 ] by FINSEQ_3: 24;

        

         A21: for n be Nat st P[n] holds P[(n + 1)]

        proof

          let n be Nat;

          assume

           A24: P[n];

          assume

           A25: (n + 1) in ( dom F);

          per cases ;

            suppose

             A26: n = 0 ;

            

            then (( Partial_Sums F) /. (n + 1)) = (( Partial_Sums F) . 1) by A25, A4, PARTFUN1:def 6

            .= (F . 1) by DEF13

            .= (F /. 1) by A25, A26, PARTFUN1:def 6;

            hence (( Partial_Sums F) /. (n + 1)) is P -measurable by A1, A25, A26;

          end;

            suppose

             A27: n <> 0 ;

            then

             A28: n >= 1 by NAT_1: 14;

            (n + 1) <= ( len F) by A25, FINSEQ_3: 25;

            then

             A29: n < ( len F) by NAT_1: 13;

            then

             A30: (F /. (n + 1)) = (F . (n + 1)) & (( Partial_Sums F) /. n) = (( Partial_Sums F) . n) & (( Partial_Sums F) /. (n + 1)) = (( Partial_Sums F) . (n + 1)) by A4, A25, A28, FINSEQ_3: 25, PARTFUN1:def 6;

            then

             A31: (( Partial_Sums F) /. (n + 1)) = ((( Partial_Sums F) /. n) + (F /. (n + 1))) by A27, A29, DEF13, NAT_1: 14;

            ( Partial_Sums F) is without_-infty-valued by A19, Th57;

            then

             A32: (F /. (n + 1)) is without-infty & (( Partial_Sums F) /. n) is without-infty by A19, A25, A29, A3, A28, A30, FINSEQ_3: 25;

            (F /. (n + 1)) is P -measurable by A25, A1;

            hence (( Partial_Sums F) /. (n + 1)) is P -measurable by A31, A32, A29, A24, A28, FINSEQ_3: 25, MESFUNC5: 31;

          end;

        end;

        for n be Nat holds P[n] from NAT_1:sch 2( A20, A21);

        hence for n be Nat st n in ( dom F) holds (( Partial_Sums F) /. n) is P -measurable;

      end;

    end;

    begin

    theorem :: MEASUR11:69

    

     Th65: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2, x be Element of X1, y be Element of X2 st E = [:A, B:] holds ( Integral (M2,( ProjMap1 (( chi (E, [:X1, X2:])),x)))) = ((M2 . ( Measurable-X-section (E,x))) * (( chi (A,X1)) . x)) & ( Integral (M1,( ProjMap2 (( chi (E, [:X1, X2:])),y)))) = ((M1 . ( Measurable-Y-section (E,y))) * (( chi (B,X2)) . y))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2, x be Element of X1, y be Element of X2;

      assume

       A1: E = [:A, B:];

      then

       A2: ( Integral (M2,( ProjMap1 (( chi (E, [:X1, X2:])),x)))) = ((M2 . B) * (( chi (A,X1)) . x)) by Th47;

      

       A3: ((M2 . B) * (( chi (A,X1)) . x)) = (M2 . ( Measurable-X-section (E,x))) by A1, Th48;

      

       A4: ( Integral (M1,( ProjMap2 (( chi (E, [:X1, X2:])),y)))) = ((M1 . A) * (( chi (B,X2)) . y)) by A1, Th49;

      

       A5: ((M1 . A) * (( chi (B,X2)) . y)) = (M1 . ( Measurable-Y-section (E,y))) by A1, Th50;

      thus ( Integral (M2,( ProjMap1 (( chi (E, [:X1, X2:])),x)))) = ((M2 . ( Measurable-X-section (E,x))) * (( chi (A,X1)) . x))

      proof

        per cases ;

          suppose x in A;

          then (( chi (A,X1)) . x) = 1 by FUNCT_3:def 3;

          hence thesis by A2, A3, XXREAL_3: 81;

        end;

          suppose not x in A;

          then (( chi (A,X1)) . x) = 0 by FUNCT_3:def 3;

          hence thesis by A2;

        end;

      end;

      per cases ;

        suppose y in B;

        then (( chi (B,X2)) . y) = 1 by FUNCT_3:def 3;

        hence thesis by A4, A5, XXREAL_3: 81;

      end;

        suppose not y in B;

        then (( chi (B,X2)) . y) = 0 by FUNCT_3:def 3;

        hence thesis by A4;

      end;

    end;

    theorem :: MEASUR11:70

    

     Th66: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2))) st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) holds ex f be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)), A be FinSequence of S1, B be FinSequence of S2 st ( len f) = ( len A) & ( len f) = ( len B) & E = ( Union f) & (for n be Nat st n in ( dom f) holds ( proj1 (f . n)) = (A . n) & ( proj2 (f . n)) = (B . n)) & (for n be Nat, x,y be set st n in ( dom f) & x in X1 & y in X2 holds (( chi ((f . n), [:X1, X2:])) . (x,y)) = ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume E in ( Field_generated_by ( measurable_rectangles (S1,S2)));

      then E in ( DisUnion ( measurable_rectangles (S1,S2))) by SRINGS_3: 22;

      then E in { E1 where E1 be Subset of [:X1, X2:] : ex f be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st E1 = ( Union f) } by SRINGS_3:def 3;

      then

      consider E1 be Subset of [:X1, X2:] such that

       A1: E = E1 & ex f be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st E1 = ( Union f);

      consider f be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) such that

       A2: E1 = ( Union f) by A1;

      defpred P1[ Nat, object] means $2 = ( proj1 (f . $1));

      

       A3: for i be Nat st i in ( Seg ( len f)) holds ex Ai be Element of S1 st P1[i, Ai]

      proof

        let i be Nat;

        assume i in ( Seg ( len f));

        then i in ( dom f) by FINSEQ_1:def 3;

        then (f . i) in ( measurable_rectangles (S1,S2)) by PARTFUN1: 4;

        then (f . i) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

        then

        consider Ai be Element of S1, Bi be Element of S2 such that

         A4: (f . i) = [:Ai, Bi:];

        per cases ;

          suppose

           A5: Bi <> {} ;

          take Ai;

          thus ( proj1 (f . i)) = Ai by A4, A5, FUNCT_5: 9;

        end;

          suppose

           A6: Bi = {} ;

          reconsider Ai = {} as Element of S1 by MEASURE1: 7;

          take Ai;

          thus ( proj1 (f . i)) = Ai by A4, A6;

        end;

      end;

      consider A be FinSequence of S1 such that

       A7: ( dom A) = ( Seg ( len f)) & for i be Nat st i in ( Seg ( len f)) holds P1[i, (A . i)] from FINSEQ_1:sch 5( A3);

      defpred P2[ Nat, object] means $2 = ( proj2 (f . $1));

      

       A8: for i be Nat st i in ( Seg ( len f)) holds ex Bi be Element of S2 st P2[i, Bi]

      proof

        let i be Nat;

        assume i in ( Seg ( len f));

        then i in ( dom f) by FINSEQ_1:def 3;

        then (f . i) in ( measurable_rectangles (S1,S2)) by PARTFUN1: 4;

        then (f . i) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

        then

        consider Ai be Element of S1, Bi be Element of S2 such that

         A9: (f . i) = [:Ai, Bi:];

        per cases ;

          suppose

           A10: Ai <> {} ;

          take Bi;

          thus ( proj2 (f . i)) = Bi by A9, A10, FUNCT_5: 9;

        end;

          suppose

           A11: Ai = {} ;

          reconsider Bi = {} as Element of S2 by MEASURE1: 7;

          take Bi;

          thus ( proj2 (f . i)) = Bi by A9, A11;

        end;

      end;

      consider B be FinSequence of S2 such that

       A12: ( dom B) = ( Seg ( len f)) & for i be Nat st i in ( Seg ( len f)) holds P2[i, (B . i)] from FINSEQ_1:sch 5( A8);

      take f, A, B;

      thus ( len f) = ( len A) by A7, FINSEQ_1:def 3;

      thus ( len f) = ( len B) by A12, FINSEQ_1:def 3;

      thus E = ( Union f) by A1, A2;

      thus

       A13: for n be Nat st n in ( dom f) holds ( proj1 (f . n)) = (A . n) & ( proj2 (f . n)) = (B . n)

      proof

        let n be Nat;

        assume n in ( dom f);

        then n in ( Seg ( len f)) by FINSEQ_1:def 3;

        hence (A . n) = ( proj1 (f . n)) & (B . n) = ( proj2 (f . n)) by A7, A12;

      end;

      let n be Nat, x,y be set;

      assume

       A14: n in ( dom f) & x in X1 & y in X2;

      then

       A15: (A . n) = ( proj1 (f . n)) & (B . n) = ( proj2 (f . n)) by A13;

      (f . n) in ( measurable_rectangles (S1,S2)) by A14, PARTFUN1: 4;

      then (f . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

      then

      consider An be Element of S1, Bn be Element of S2 such that

       A16: (f . n) = [:An, Bn:];

      

       A17: [x, y] in [:X1, X2:] by A14, ZFMISC_1: 87;

      per cases ;

        suppose (f . n) = {} ;

        then (( chi ((f . n), [:X1, X2:])) . (x,y)) = 0 & (( chi ((A . n),X1)) . x) = 0 & (( chi ((B . n),X2)) . y) = 0 by A14, A15, A17, FUNCT_3:def 3;

        hence (( chi ((f . n), [:X1, X2:])) . (x,y)) = ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y));

      end;

        suppose (f . n) <> {} ;

        then

         A18: (A . n) = An & (B . n) = Bn by A15, A16, FUNCT_5: 9;

        per cases ;

          suppose

           A19: x in (A . n) & y in (B . n);

          then (( chi ((A . n),X1)) . x) = 1 & (( chi ((B . n),X2)) . y) = 1 by FUNCT_3:def 3;

          then

           A20: ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y)) = 1 by XXREAL_3: 81;

          ( proj1 (f . n)) c= An & ( proj2 (f . n)) c= Bn by A16, FUNCT_5: 10;

          then [x, y] in (f . n) & [x, y] in [:X1, X2:] by A19, A15, A16, ZFMISC_1:def 2;

          hence (( chi ((f . n), [:X1, X2:])) . (x,y)) = ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y)) by A20, FUNCT_3:def 3;

        end;

          suppose

           A21: not x in (A . n) or not y in (B . n);

          then (( chi ((A . n),X1)) . x) = 0 or (( chi ((B . n),X2)) . y) = 0 by A14, FUNCT_3:def 3;

          then

           A22: ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y)) = 0 ;

           not [x, y] in (f . n) by A18, A21, A16, ZFMISC_1: 87;

          hence (( chi ((f . n), [:X1, X2:])) . (x,y)) = ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y)) by A17, A22, FUNCT_3:def 3;

        end;

      end;

    end;

    theorem :: MEASUR11:71

    

     Th67: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), x be Element of X1, y be Element of X2, U be Element of S1, V be Element of S2 holds (M1 . (( Measurable-Y-section (E,y)) /\ U)) = ( Integral (M1,( ProjMap2 (( chi ((E /\ [:U, X2:]), [:X1, X2:])),y)))) & (M2 . (( Measurable-X-section (E,x)) /\ V)) = ( Integral (M2,( ProjMap1 (( chi ((E /\ [:X1, V:]), [:X1, X2:])),x))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), x be Element of X1, y be Element of X2, U be Element of S1, V be Element of S2;

      for x be Element of X1 holds (( ProjMap2 (( chi ((E /\ [:U, X2:]), [:X1, X2:])),y)) . x) = (( chi ((( Measurable-Y-section (E,y)) /\ U),X1)) . x)

      proof

        let x be Element of X1;

        

         A1: X2 = ( [#] X2) by SUBSET_1:def 3;

        (( ProjMap2 (( chi ((E /\ [:U, X2:]), [:X1, X2:])),y)) . x) = (( chi ((E /\ [:U, X2:]), [:X1, X2:])) . (x,y)) by MESFUNC9:def 7

        .= (( chi (( Y-section ((E /\ [:U, X2:]),y)),X1)) . x) by Th28

        .= (( chi ((( Y-section (E,y)) /\ ( Y-section ( [:U, ( [#] X2):],y))),X1)) . x) by A1, Th21;

        hence (( ProjMap2 (( chi ((E /\ [:U, X2:]), [:X1, X2:])),y)) . x) = (( chi ((( Measurable-Y-section (E,y)) /\ U),X1)) . x) by A1, Th16;

      end;

      then ( ProjMap2 (( chi ((E /\ [:U, X2:]), [:X1, X2:])),y)) = ( chi ((( Measurable-Y-section (E,y)) /\ U),X1)) by FUNCT_2:def 8;

      hence (M1 . (( Measurable-Y-section (E,y)) /\ U)) = ( Integral (M1,( ProjMap2 (( chi ((E /\ [:U, X2:]), [:X1, X2:])),y)))) by MESFUNC9: 14;

      for y be Element of X2 holds (( ProjMap1 (( chi ((E /\ [:X1, V:]), [:X1, X2:])),x)) . y) = (( chi ((( Measurable-X-section (E,x)) /\ V),X2)) . y)

      proof

        let y be Element of X2;

        

         A3: X1 = ( [#] X1) by SUBSET_1:def 3;

        (( ProjMap1 (( chi ((E /\ [:X1, V:]), [:X1, X2:])),x)) . y) = (( chi ((E /\ [:X1, V:]), [:X1, X2:])) . (x,y)) by MESFUNC9:def 6

        .= (( chi (( X-section ((E /\ [:X1, V:]),x)),X2)) . y) by Th28

        .= (( chi ((( X-section (E,x)) /\ ( X-section ( [:( [#] X1), V:],x))),X2)) . y) by A3, Th21;

        hence (( ProjMap1 (( chi ((E /\ [:X1, V:]), [:X1, X2:])),x)) . y) = (( chi ((( Measurable-X-section (E,x)) /\ V),X2)) . y) by A3, Th16;

      end;

      then ( ProjMap1 (( chi ((E /\ [:X1, V:]), [:X1, X2:])),x)) = ( chi ((( Measurable-X-section (E,x)) /\ V),X2)) by FUNCT_2:def 8;

      hence (M2 . (( Measurable-X-section (E,x)) /\ V)) = ( Integral (M2,( ProjMap1 (( chi ((E /\ [:X1, V:]), [:X1, X2:])),x)))) by MESFUNC9: 14;

    end;

    theorem :: MEASUR11:72

    

     Th68: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), x be Element of X1, y be Element of X2 holds (M1 . ( Measurable-Y-section (E,y))) = ( Integral (M1,( ProjMap2 (( chi (E, [:X1, X2:])),y)))) & (M2 . ( Measurable-X-section (E,x))) = ( Integral (M2,( ProjMap1 (( chi (E, [:X1, X2:])),x))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), x be Element of X1, y be Element of X2;

      

       A1: X1 in S1 & X2 in S2 by MEASURE1: 7;

      (E /\ [:X1, X2:]) = E & (( Measurable-Y-section (E,y)) /\ X1) = ( Measurable-Y-section (E,y)) & (( Measurable-X-section (E,x)) /\ X2) = ( Measurable-X-section (E,x)) by XBOOLE_1: 28;

      hence thesis by Th67, A1;

    end;

    theorem :: MEASUR11:73

    for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, f be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)), x be Element of X1, n be Nat, En be Element of ( sigma ( measurable_rectangles (S1,S2))), An be Element of S1, Bn be Element of S2 st n in ( dom f) & (f . n) = En & En = [:An, Bn:] holds ( Integral (M2,( ProjMap1 (( chi ((f . n), [:X1, X2:])),x)))) = ((M2 . ( Measurable-X-section (En,x))) * (( chi (An,X1)) . x)) by Th65;

    theorem :: MEASUR11:74

    

     Th70: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2))) st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) & E <> {} holds ex f be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)), A be FinSequence of S1, B be FinSequence of S2, Xf be summable FinSequence of ( Funcs ( [:X1, X2:], ExtREAL )) st E = ( Union f) & ( len f) in ( dom f) & ( len f) = ( len A) & ( len f) = ( len B) & ( len f) = ( len Xf) & (for n be Nat st n in ( dom f) holds (f . n) = [:(A . n), (B . n):]) & (for n be Nat st n in ( dom Xf) holds (Xf . n) = ( chi ((f . n), [:X1, X2:]))) & (( Partial_Sums Xf) . ( len Xf)) = ( chi (E, [:X1, X2:])) & (for n be Nat, x,y be set st n in ( dom Xf) & x in X1 & y in X2 holds ((Xf . n) . (x,y)) = ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y))) & (for x be Element of X1 holds ( ProjMap1 (( chi (E, [:X1, X2:])),x)) = ( ProjMap1 ((( Partial_Sums Xf) /. ( len Xf)),x))) & (for y be Element of X2 holds ( ProjMap2 (( chi (E, [:X1, X2:])),y)) = ( ProjMap2 ((( Partial_Sums Xf) /. ( len Xf)),y)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume

       A1: E in ( Field_generated_by ( measurable_rectangles (S1,S2))) & E <> {} ;

      consider f be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)), A be FinSequence of S1, B be FinSequence of S2 such that

       A2: ( len f) = ( len A) & ( len f) = ( len B) & E = ( Union f) & (for n be Nat st n in ( dom f) holds ( proj1 (f . n)) = (A . n) & ( proj2 (f . n)) = (B . n)) & (for n be Nat, x,y be set st n in ( dom f) & x in X1 & y in X2 holds (( chi ((f . n), [:X1, X2:])) . (x,y)) = ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y))) by A1, Th66;

      deffunc F( set) = ( chi ((f . $1), [:X1, X2:]));

      consider Xf be FinSequence such that

       A3: ( len Xf) = ( len f) & for n be Nat st n in ( dom Xf) holds (Xf . n) = F(n) from FINSEQ_1:sch 2;

      now

        let z be set;

        assume z in ( rng Xf);

        then

        consider i be object such that

         A4: i in ( dom Xf) & z = (Xf . i) by FUNCT_1:def 3;

        reconsider i as Nat by A4;

        z = ( chi ((f . i), [:X1, X2:])) by A3, A4;

        hence z in ( Funcs ( [:X1, X2:], ExtREAL )) by FUNCT_2: 8;

      end;

      then ( rng Xf) c= ( Funcs ( [:X1, X2:], ExtREAL ));

      then

      reconsider Xf as FinSequence of ( Funcs ( [:X1, X2:], ExtREAL )) by FINSEQ_1:def 4;

      now

        let n be Nat;

        assume n in ( dom Xf);

        then (Xf . n) = ( chi ((f . n), [:X1, X2:])) by A3;

        then ( rng (Xf . n)) c= { 0 , 1} by FUNCT_3: 39;

        then not -infty in ( rng (Xf . n));

        hence (Xf . n) is without-infty by MESFUNC5:def 3;

      end;

      then Xf is without_-infty-valued;

      then

      reconsider Xf as summable FinSequence of ( Funcs ( [:X1, X2:], ExtREAL ));

      take f, A, B, Xf;

      defpred P[ Nat] means $1 in ( dom f) implies (( Partial_Sums Xf) . $1) = ( chi (( Union (f | $1)), [:X1, X2:]));

      

       A9: ( dom Xf) = ( dom f) by A3, FINSEQ_3: 29;

      

       A5: P[ 0 ] by FINSEQ_3: 24;

      

       A6: for k be Nat st P[k] holds P[(k + 1)]

      proof

        let k be Nat;

        assume

         A7: P[k];

        assume

         A8: (k + 1) in ( dom f);

        per cases ;

          suppose

           A10: k = 0 ;

          

          then

           A11: (( Partial_Sums Xf) . (k + 1)) = (Xf . (k + 1)) by DEF13

          .= ( chi ((f . (k + 1)), [:X1, X2:])) by A3, A8, A9;

          f is non empty by A8;

          then (f | (k + 1)) = <*(f . (k + 1))*> by A10, FINSEQ_5: 20;

          then ( rng (f | (k + 1))) = {(f . (k + 1))} by FINSEQ_1: 39;

          then ( union ( rng (f | (k + 1)))) = (f . (k + 1)) by ZFMISC_1: 25;

          hence (( Partial_Sums Xf) . (k + 1)) = ( chi (( Union (f | (k + 1))), [:X1, X2:])) by A11, CARD_3:def 4;

        end;

          suppose k <> 0 ;

          then

           A12: k >= 1 by NAT_1: 14;

          

           A13: 1 <= (k + 1) <= ( len Xf) by A8, A3, FINSEQ_3: 25;

          then

           A14: k < ( len Xf) by NAT_1: 13;

          then k < ( len ( Partial_Sums Xf)) by DEF13;

          then k in ( dom ( Partial_Sums Xf)) by A12, FINSEQ_3: 25;

          then

           A16: (( Partial_Sums Xf) /. k) = ( chi (( Union (f | k)), [:X1, X2:])) by A3, A14, A7, A12, FINSEQ_3: 25, PARTFUN1:def 6;

          

           A17: (Xf /. (k + 1)) = (Xf . (k + 1)) by A8, A9, PARTFUN1:def 6

          .= ( chi ((f . (k + 1)), [:X1, X2:])) by A8, A9, A3;

           A24:

          now

            assume (( Union (f | k)) /\ (f . (k + 1))) <> {} ;

            then

            consider z be object such that

             A18: z in (( Union (f | k)) /\ (f . (k + 1))) by XBOOLE_0:def 1;

            

             A19: z in ( Union (f | k)) & z in (f . (k + 1)) by A18, XBOOLE_0:def 4;

            then z in ( union ( rng (f | k))) by CARD_3:def 4;

            then

            consider Z be set such that

             A20: z in Z & Z in ( rng (f | k)) by TARSKI:def 4;

            consider j be Element of NAT such that

             A21: j in ( dom (f | k)) & Z = ((f | k) . j) by A20, PARTFUN1: 3;

            1 <= j <= ( len (f | k)) by A21, FINSEQ_3: 25;

            then

             A22: 1 <= j <= k by A14, A3, FINSEQ_1: 59;

            then

             A23: Z = (f . j) by A21, FINSEQ_3: 112;

            j <> (k + 1) by A22, NAT_1: 13;

            then (f . j) misses (f . (k + 1)) by PROB_2:def 2;

            hence contradiction by A23, A19, A20, XBOOLE_0:def 4;

          end;

          

           A25: (( Partial_Sums Xf) . (k + 1)) = (( chi (( Union (f | k)), [:X1, X2:])) + ( chi ((f . (k + 1)), [:X1, X2:]))) by A16, A17, A12, A13, DEF13, NAT_1: 13;

          1 <= (k + 1) <= ( len ( Partial_Sums Xf)) by A13, DEF13;

          then (k + 1) in ( dom ( Partial_Sums Xf)) by FINSEQ_3: 25;

          then

           A26: (( Partial_Sums Xf) /. (k + 1)) = (( Partial_Sums Xf) . (k + 1)) by PARTFUN1:def 6;

          now

            let z be Element of [:X1, X2:];

            ( dom (( chi (( Union (f | k)), [:X1, X2:])) + ( chi ((f . (k + 1)), [:X1, X2:])))) = [:X1, X2:] by A25, A26, FUNCT_2:def 1;

            then

             A28: ((( Partial_Sums Xf) . (k + 1)) . z) = ((( chi (( Union (f | k)), [:X1, X2:])) . z) + (( chi ((f . (k + 1)), [:X1, X2:])) . z)) by A25, MESFUNC1:def 3;

            per cases ;

              suppose

               A31: z in ( Union (f | (k + 1)));

              then z in ( union ( rng (f | (k + 1)))) by CARD_3:def 4;

              then

              consider Z be set such that

               A29: z in Z & Z in ( rng (f | (k + 1))) by TARSKI:def 4;

              consider j be Element of NAT such that

               A30: j in ( dom (f | (k + 1))) & Z = ((f | (k + 1)) . j) by A29, PARTFUN1: 3;

              

               A36: 1 <= j <= ( len (f | (k + 1))) by A30, FINSEQ_3: 25;

              then

               A32: 1 <= j & j <= (k + 1) by A13, A3, FINSEQ_1: 59;

              then

               A33: Z = (f . j) by A30, FINSEQ_3: 112;

              now

                per cases ;

                  suppose j = (k + 1);

                  then

                   A34: z in (f . (k + 1)) by A29, A30, FINSEQ_3: 112;

                  then

                   A35: (( chi ((f . (k + 1)), [:X1, X2:])) . z) = 1 by FUNCT_3:def 3;

                   not z in ( Union (f | k)) by A24, A34, XBOOLE_0:def 4;

                  then (( chi (( Union (f | k)), [:X1, X2:])) . z) = 0 by FUNCT_3:def 3;

                  hence ((( Partial_Sums Xf) . (k + 1)) . z) = 1 by A28, A35, XXREAL_3: 4;

                end;

                  suppose j <> (k + 1);

                  then j < (k + 1) by A32, XXREAL_0: 1;

                  then

                   A37: j <= k by NAT_1: 13;

                  then j <= ( len (f | k)) by A3, A14, FINSEQ_1: 59;

                  then j in ( dom (f | k)) & Z = ((f | k) . j) by A33, A36, A37, FINSEQ_3: 25, FINSEQ_3: 112;

                  then Z in ( rng (f | k)) by FUNCT_1: 3;

                  then z in ( union ( rng (f | k))) by A29, TARSKI:def 4;

                  then

                   A38: z in ( Union (f | k)) by CARD_3:def 4;

                  then

                   A39: (( chi (( Union (f | k)), [:X1, X2:])) . z) = 1 by FUNCT_3:def 3;

                   not z in (f . (k + 1)) by A24, A38, XBOOLE_0:def 4;

                  then (( chi ((f . (k + 1)), [:X1, X2:])) . z) = 0 by FUNCT_3:def 3;

                  hence ((( Partial_Sums Xf) . (k + 1)) . z) = 1 by A28, A39, XXREAL_3: 4;

                end;

              end;

              hence ((( Partial_Sums Xf) . (k + 1)) . z) = (( chi (( Union (f | (k + 1))), [:X1, X2:])) . z) by A31, FUNCT_3:def 3;

            end;

              suppose

               A40: not z in ( Union (f | (k + 1)));

              then

               A41: (( chi (( Union (f | (k + 1))), [:X1, X2:])) . z) = 0 by FUNCT_3:def 3;

              

               A42: not z in ( union ( rng (f | (k + 1)))) by A40, CARD_3:def 4;

              

               A43: for j be Nat st 1 <= j <= (k + 1) holds not z in (f . j)

              proof

                let j be Nat;

                assume

                 B1: 1 <= j <= (k + 1);

                then 1 <= j <= ( len (f | (k + 1))) by A3, A13, FINSEQ_1: 59;

                then j in ( dom (f | (k + 1))) by FINSEQ_3: 25;

                then ((f | (k + 1)) . j) in ( rng (f | (k + 1))) by FUNCT_1: 3;

                then (f . j) in ( rng (f | (k + 1))) by B1, FINSEQ_3: 112;

                hence not z in (f . j) by A42, TARSKI:def 4;

              end;

              now

                assume z in ( Union (f | k));

                then z in ( union ( rng (f | k))) by CARD_3:def 4;

                then

                consider Z be set such that

                 A46: z in Z & Z in ( rng (f | k)) by TARSKI:def 4;

                consider j be Element of NAT such that

                 A47: j in ( dom (f | k)) & Z = ((f | k) . j) by A46, PARTFUN1: 3;

                1 <= j <= ( len (f | k)) by A47, FINSEQ_3: 25;

                then

                 A48: 1 <= j <= k by A3, A14, FINSEQ_1: 59;

                then 1 <= j <= (k + 1) by NAT_1: 13;

                then not z in (f . j) by A43;

                hence contradiction by A46, A47, A48, FINSEQ_3: 112;

              end;

              then

               A50: (( chi (( Union (f | k)), [:X1, X2:])) . z) = 0 by FUNCT_3:def 3;

              1 <= (k + 1) by NAT_1: 11;

              then not z in (f . (k + 1)) by A43;

              then (( chi ((f . (k + 1)), [:X1, X2:])) . z) = 0 by FUNCT_3:def 3;

              hence ((( Partial_Sums Xf) . (k + 1)) . z) = (( chi (( Union (f | (k + 1))), [:X1, X2:])) . z) by A41, A28, A50;

            end;

          end;

          hence (( Partial_Sums Xf) . (k + 1)) = ( chi (( Union (f | (k + 1))), [:X1, X2:])) by A26, FUNCT_2:def 8;

        end;

      end;

      

       A51: for n be Nat holds P[n] from NAT_1:sch 2( A5, A6);

      thus E = ( Union f) by A2;

      ( union ( rng f)) <> {} by A1, A2, CARD_3:def 4;

      then ( dom f) <> {} by ZFMISC_1: 2, RELAT_1: 42;

      then ( Seg ( len f)) <> {} by FINSEQ_1:def 3;

      then

       A52: ( len f) in ( Seg ( len f)) by FINSEQ_3: 7;

      hence

       A53: ( len f) in ( dom f) by FINSEQ_1:def 3;

      thus ( len f) = ( len A) & ( len f) = ( len B) by A2;

      thus ( len f) = ( len Xf) by A3;

      thus for n be Nat st n in ( dom f) holds (f . n) = [:(A . n), (B . n):]

      proof

        let n be Nat;

        assume

         A54: n in ( dom f);

        then (f . n) in ( measurable_rectangles (S1,S2)) by PARTFUN1: 4;

        then (f . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

        then

        consider An be Element of S1, Bn be Element of S2 such that

         A55: (f . n) = [:An, Bn:];

        per cases ;

          suppose

           A57: (f . n) = {} ;

          then (A . n) = ( proj1 {} ) & (B . n) = ( proj2 {} ) by A2, A54;

          hence (f . n) = [:(A . n), (B . n):] by A57, ZFMISC_1: 90;

        end;

          suppose (f . n) <> {} ;

          then

           A59: ( proj1 (f . n)) = An & ( proj2 (f . n)) = Bn by A55, FUNCT_5: 9;

          ( proj1 (f . n)) = (A . n) & ( proj2 (f . n)) = (B . n) by A2, A54;

          hence (f . n) = [:(A . n), (B . n):] by A55, A59;

        end;

      end;

      thus for n be Nat st n in ( dom Xf) holds (Xf . n) = ( chi ((f . n), [:X1, X2:])) by A3;

      

       A60: (( Partial_Sums Xf) . ( len Xf)) = ( chi (( Union (f | ( len f))), [:X1, X2:])) by A51, A3, A53

      .= ( chi (( Union f), [:X1, X2:])) by FINSEQ_1: 58;

      hence (( Partial_Sums Xf) . ( len Xf)) = ( chi (E, [:X1, X2:])) by A2;

      thus for n be Nat, x,y be set st n in ( dom Xf) & x in X1 & y in X2 holds ((Xf . n) . (x,y)) = ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y))

      proof

        let n be Nat, x,y be set;

        assume

         Q1: n in ( dom Xf) & x in X1 & y in X2;

        then (( chi ((f . n), [:X1, X2:])) . (x,y)) = ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y)) by A2, A9;

        hence ((Xf . n) . (x,y)) = ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y)) by Q1, A3;

      end;

      thus for x be Element of X1 holds ( ProjMap1 (( chi (E, [:X1, X2:])),x)) = ( ProjMap1 ((( Partial_Sums Xf) /. ( len Xf)),x))

      proof

        let x be Element of X1;

        ( len f) = ( len ( Partial_Sums Xf)) by A3, DEF13;

        then ( len Xf) in ( dom ( Partial_Sums Xf)) by A52, A3, FINSEQ_1:def 3;

        hence thesis by A60, A2, PARTFUN1:def 6;

      end;

      let y be Element of X2;

      ( len f) = ( len ( Partial_Sums Xf)) by A3, DEF13;

      then ( len Xf) in ( dom ( Partial_Sums Xf)) by A52, A3, FINSEQ_1:def 3;

      hence thesis by A60, A2, PARTFUN1:def 6;

    end;

    theorem :: MEASUR11:75

    

     Th71: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, F be FinSequence of ( measurable_rectangles (S1,S2)) holds ( Union F) in ( sigma ( measurable_rectangles (S1,S2)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, F be FinSequence of ( measurable_rectangles (S1,S2));

      defpred P[ Nat] means $1 <= ( len F) implies ( union ( rng (F | $1))) in ( sigma ( measurable_rectangles (S1,S2)));

      

       A1: P[ 0 ] by ZFMISC_1: 2, MEASURE1: 34;

      

       A2: for k be Nat st P[k] holds P[(k + 1)]

      proof

        let k be Nat;

        assume

         A3: P[k];

        assume

         A4: (k + 1) <= ( len F);

        then (k + 1) in ( dom F) by NAT_1: 11, FINSEQ_3: 25;

        then

         A6: (F . (k + 1)) in ( measurable_rectangles (S1,S2)) by FINSEQ_2: 11;

        

         A7: ( measurable_rectangles (S1,S2)) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

        ( len (F | (k + 1))) = (k + 1) by A4, FINSEQ_1: 59;

        

        then (F | (k + 1)) = (((F | (k + 1)) | k) ^ <*((F | (k + 1)) . (k + 1))*>) by FINSEQ_3: 55

        .= ((F | k) ^ <*((F | (k + 1)) . (k + 1))*>) by NAT_1: 11, FINSEQ_1: 82

        .= ((F | k) ^ <*(F . (k + 1))*>) by FINSEQ_3: 112;

        

        then ( rng (F | (k + 1))) = (( rng (F | k)) \/ ( rng <*(F . (k + 1))*>)) by FINSEQ_1: 31

        .= (( rng (F | k)) \/ {(F . (k + 1))}) by FINSEQ_1: 39;

        

        then ( union ( rng (F | (k + 1)))) = (( union ( rng (F | k))) \/ ( union {(F . (k + 1))})) by ZFMISC_1: 78

        .= (( union ( rng (F | k))) \/ (F . (k + 1))) by ZFMISC_1: 25;

        hence ( union ( rng (F | (k + 1)))) in ( sigma ( measurable_rectangles (S1,S2))) by A4, A3, NAT_1: 13, A6, A7, MEASURE1: 34;

      end;

      for k be Nat holds P[k] from NAT_1:sch 2( A1, A2);

      then ( union ( rng (F | ( len F)))) in ( sigma ( measurable_rectangles (S1,S2)));

      then ( union ( rng F)) in ( sigma ( measurable_rectangles (S1,S2))) by FINSEQ_1: 58;

      hence ( Union F) in ( sigma ( measurable_rectangles (S1,S2))) by CARD_3:def 4;

    end;

    theorem :: MEASUR11:76

    

     Th75: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))) st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) & E <> {} holds ex F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)), A be FinSequence of S1, B be FinSequence of S2, C be summable FinSequence of ( Funcs ( [:X1, X2:], ExtREAL )), I be summable FinSequence of ( Funcs (X1, ExtREAL )), J be summable FinSequence of ( Funcs (X2, ExtREAL )) st E = ( Union F) & ( len F) in ( dom F) & ( len F) = ( len A) & ( len F) = ( len B) & ( len F) = ( len C) & ( len F) = ( len I) & ( len F) = ( len J) & (for n be Nat st n in ( dom C) holds (C . n) = ( chi ((F . n), [:X1, X2:]))) & (( Partial_Sums C) /. ( len C)) = ( chi (E, [:X1, X2:])) & (for x be Element of X1, n be Nat st n in ( dom I) holds ((I . n) . x) = ( Integral (M2,( ProjMap1 ((C /. n),x))))) & (for n be Nat, P be Element of S1 st n in ( dom I) holds (I /. n) is P -measurable) & (for x be Element of X1 holds ( Integral (M2,( ProjMap1 ((( Partial_Sums C) /. ( len C)),x)))) = ((( Partial_Sums I) /. ( len I)) . x)) & (for y be Element of X2, n be Nat st n in ( dom J) holds ((J . n) . y) = ( Integral (M1,( ProjMap2 ((C /. n),y))))) & (for n be Nat, P be Element of S2 st n in ( dom J) holds (J /. n) is P -measurable) & (for y be Element of X2 holds ( Integral (M1,( ProjMap2 ((( Partial_Sums C) /. ( len C)),y)))) = ((( Partial_Sums J) /. ( len J)) . y))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume

       A1: E in ( Field_generated_by ( measurable_rectangles (S1,S2))) & E <> {} ;

      consider F be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)), A be FinSequence of S1, B be FinSequence of S2, C be summable FinSequence of ( Funcs ( [:X1, X2:], ExtREAL )) such that

       A2: E = ( Union F) & ( len F) in ( dom F) & ( len F) = ( len A) & ( len F) = ( len B) & ( len F) = ( len C) & (for n be Nat st n in ( dom F) holds (F . n) = [:(A . n), (B . n):]) & (for n be Nat st n in ( dom C) holds (C . n) = ( chi ((F . n), [:X1, X2:]))) & (( Partial_Sums C) . ( len C)) = ( chi (E, [:X1, X2:])) & (for n be Nat, x,y be set st n in ( dom C) & x in X1 & y in X2 holds ((C . n) . (x,y)) = ((( chi ((A . n),X1)) . x) * (( chi ((B . n),X2)) . y))) & (for x be Element of X1 holds ( ProjMap1 (( chi (E, [:X1, X2:])),x)) = ( ProjMap1 ((( Partial_Sums C) /. ( len C)),x))) & (for y be Element of X2 holds ( ProjMap2 (( chi (E, [:X1, X2:])),y)) = ( ProjMap2 ((( Partial_Sums C) /. ( len C)),y))) by A1, Th70;

      

       A3: ( measurable_rectangles (S1,S2)) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

      defpred PI[ Nat, object] means ex f be Function of X1, ExtREAL st f = $2 & for x be Element of X1 holds (f . x) = ( Integral (M2,( ProjMap1 ((C /. $1),x))));

      

       I1: for n be Nat st n in ( Seg ( len F)) holds ex z be object st PI[n, z]

      proof

        let n be Nat;

        assume n in ( Seg ( len F));

        deffunc F2( Element of X1) = ( Integral (M2,( ProjMap1 ((C /. n),$1))));

        consider f be Function of X1, ExtREAL such that

         I2: for x be Element of X1 holds (f . x) = F2(x) from FUNCT_2:sch 4;

        take z = f;

        thus ex f be Function of X1, ExtREAL st f = z & for x be Element of X1 holds (f . x) = ( Integral (M2,( ProjMap1 ((C /. n),x)))) by I2;

      end;

      consider I be FinSequence such that

       I3: ( dom I) = ( Seg ( len F)) & for n be Nat st n in ( Seg ( len F)) holds PI[n, (I . n)] from FINSEQ_1:sch 1( I1);

      now

        let z be set;

        assume z in ( rng I);

        then

        consider n be object such that

         I4: n in ( dom I) & z = (I . n) by FUNCT_1:def 3;

        reconsider n as Element of NAT by I4;

        consider f be Function of X1, ExtREAL such that

         I5: f = (I . n) & for x be Element of X1 holds (f . x) = ( Integral (M2,( ProjMap1 ((C /. n),x)))) by I3, I4;

        ( dom f) = X1 & ( rng f) c= ExtREAL by FUNCT_2:def 1;

        hence z in ( Funcs (X1, ExtREAL )) by I4, I5, FUNCT_2:def 2;

      end;

      then ( rng I) c= ( Funcs (X1, ExtREAL ));

      then

      reconsider I as FinSequence of ( Funcs (X1, ExtREAL )) by FINSEQ_1:def 4;

      

       I6: for x be Element of X1, n be Nat st n in ( dom I) holds ((I . n) . x) = ( Integral (M2,( ProjMap1 ((C /. n),x))))

      proof

        let x be Element of X1, n be Nat;

        assume

         I7: n in ( dom I);

        then n in ( dom F) by I3, FINSEQ_1:def 3;

        then

        reconsider Fn = (F . n) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, PARTFUN1: 4;

        ex f be Function of X1, ExtREAL st f = (I . n) & for x be Element of X1 holds (f . x) = ( Integral (M2,( ProjMap1 ((C /. n),x)))) by I3, I7;

        hence ((I . n) . x) = ( Integral (M2,( ProjMap1 ((C /. n),x))));

      end;

       I7:

      now

        let n be Nat;

        assume

         I8: n in ( dom I);

        then

        consider f be Function of X1, ExtREAL such that

         I9: f = (I . n) & for x be Element of X1 holds (f . x) = ( Integral (M2,( ProjMap1 ((C /. n),x)))) by I3;

        

         I10: n in ( dom F) by I3, I8, FINSEQ_1:def 3;

        then

        reconsider Fn = (F . n) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, PARTFUN1: 4;

        (F . n) in ( measurable_rectangles (S1,S2)) by I10, PARTFUN1: 4;

        then (F . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

        then

        consider An be Element of S1, Bn be Element of S2 such that

         I11: (F . n) = [:An, Bn:];

        for x be Element of X1 holds 0 <= (f . x)

        proof

          let x be Element of X1;

          

           I12: n in ( dom C) by I3, I8, A2, FINSEQ_1:def 3;

          then (C /. n) = (C . n) by PARTFUN1:def 6;

          then

           I13: (C /. n) = ( chi ((F . n), [:X1, X2:])) by A2, I12;

          (f . x) = ( Integral (M2,( ProjMap1 ((C /. n),x)))) by I9;

          then

           I14: (f . x) = ((M2 . ( Measurable-X-section (Fn,x))) * (( chi (An,X1)) . x)) by I11, I13, Th65;

          (M2 . ( Measurable-X-section (Fn,x))) >= 0 & (( chi (An,X1)) . x) >= 0 by SUPINF_2: 51;

          hence 0 <= (f . x) by I14;

        end;

        then (I . n) is nonnegative Function of X1, ExtREAL by I9, SUPINF_2: 39;

        hence (I . n) is without-infty;

      end;

      then

       I15: I is without_-infty-valued;

      then

      reconsider I as summable FinSequence of ( Funcs (X1, ExtREAL ));

      defpred PJ[ Nat, object] means ex f be Function of X2, ExtREAL st f = $2 & for x be Element of X2 holds (f . x) = ( Integral (M1,( ProjMap2 ((C /. $1),x))));

      

       J1: for n be Nat st n in ( Seg ( len F)) holds ex z be object st PJ[n, z]

      proof

        let n be Nat;

        assume n in ( Seg ( len F));

        deffunc F2( Element of X2) = ( Integral (M1,( ProjMap2 ((C /. n),$1))));

        consider f be Function of X2, ExtREAL such that

         J2: for x be Element of X2 holds (f . x) = F2(x) from FUNCT_2:sch 4;

        take z = f;

        thus ex f be Function of X2, ExtREAL st f = z & for x be Element of X2 holds (f . x) = ( Integral (M1,( ProjMap2 ((C /. n),x)))) by J2;

      end;

      consider J be FinSequence such that

       J3: ( dom J) = ( Seg ( len F)) & for n be Nat st n in ( Seg ( len F)) holds PJ[n, (J . n)] from FINSEQ_1:sch 1( J1);

      now

        let z be set;

        assume z in ( rng J);

        then

        consider n be object such that

         J4: n in ( dom J) & z = (J . n) by FUNCT_1:def 3;

        reconsider n as Element of NAT by J4;

        consider f be Function of X2, ExtREAL such that

         J5: f = (J . n) & for x be Element of X2 holds (f . x) = ( Integral (M1,( ProjMap2 ((C /. n),x)))) by J3, J4;

        ( dom f) = X2 & ( rng f) c= ExtREAL by FUNCT_2:def 1;

        hence z in ( Funcs (X2, ExtREAL )) by J4, J5, FUNCT_2:def 2;

      end;

      then ( rng J) c= ( Funcs (X2, ExtREAL ));

      then

      reconsider J as FinSequence of ( Funcs (X2, ExtREAL )) by FINSEQ_1:def 4;

      

       J6: for x be Element of X2, n be Nat st n in ( dom J) holds ((J . n) . x) = ( Integral (M1,( ProjMap2 ((C /. n),x))))

      proof

        let x be Element of X2, n be Nat;

        assume

         J7: n in ( dom J);

        then n in ( dom F) by J3, FINSEQ_1:def 3;

        then

        reconsider Fn = (F . n) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, PARTFUN1: 4;

        ex f be Function of X2, ExtREAL st f = (J . n) & for x be Element of X2 holds (f . x) = ( Integral (M1,( ProjMap2 ((C /. n),x)))) by J3, J7;

        hence ((J . n) . x) = ( Integral (M1,( ProjMap2 ((C /. n),x))));

      end;

       J7:

      now

        let n be Nat;

        assume

         J8: n in ( dom J);

        then

        consider f be Function of X2, ExtREAL such that

         J9: f = (J . n) & for x be Element of X2 holds (f . x) = ( Integral (M1,( ProjMap2 ((C /. n),x)))) by J3;

        

         J10: n in ( dom F) by J3, J8, FINSEQ_1:def 3;

        then

        reconsider Fn = (F . n) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, PARTFUN1: 4;

        (F . n) in ( measurable_rectangles (S1,S2)) by J10, PARTFUN1: 4;

        then (F . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

        then

        consider An be Element of S1, Bn be Element of S2 such that

         J11: (F . n) = [:An, Bn:];

        for x be Element of X2 holds 0 <= (f . x)

        proof

          let x be Element of X2;

          

           J12: n in ( dom C) by J3, J8, A2, FINSEQ_1:def 3;

          then (C /. n) = (C . n) by PARTFUN1:def 6;

          then

           J13: (C /. n) = ( chi ((F . n), [:X1, X2:])) by A2, J12;

          (f . x) = ( Integral (M1,( ProjMap2 ((C /. n),x)))) by J9;

          then

           J14: (f . x) = ((M1 . ( Measurable-Y-section (Fn,x))) * (( chi (Bn,X2)) . x)) by J11, J13, Th65;

          (M1 . ( Measurable-Y-section (Fn,x))) >= 0 & (( chi (Bn,X2)) . x) >= 0 by SUPINF_2: 51;

          hence 0 <= (f . x) by J14;

        end;

        then (J . n) is nonnegative Function of X2, ExtREAL by J9, SUPINF_2: 39;

        hence (J . n) is without-infty;

      end;

      then

       J15: J is without_-infty-valued;

      then

      reconsider J as summable FinSequence of ( Funcs (X2, ExtREAL ));

      take F, A, B, C, I, J;

      thus E = ( Union F) & ( len F) in ( dom F) & ( len F) = ( len A) & ( len F) = ( len B) & ( len F) = ( len C) by A2;

      thus

       K1: ( len F) = ( len I) & ( len F) = ( len J) by I3, J3, FINSEQ_1:def 3;

      thus for n be Nat st n in ( dom C) holds (C . n) = ( chi ((F . n), [:X1, X2:])) by A2;

      ( len C) = ( len ( Partial_Sums C)) by DEF13;

      then ( dom F) = ( dom ( Partial_Sums C)) by A2, FINSEQ_3: 29;

      hence (( Partial_Sums C) /. ( len C)) = ( chi (E, [:X1, X2:])) by A2, PARTFUN1:def 6;

      thus for x be Element of X1, n be Nat st n in ( dom I) holds ((I . n) . x) = ( Integral (M2,( ProjMap1 ((C /. n),x)))) by I6;

      thus for n be Nat, P be Element of S1 st n in ( dom I) holds (I /. n) is P -measurable

      proof

        let n be Nat, P be Element of S1;

        assume

         I16: n in ( dom I);

        then

        consider f be Function of X1, ExtREAL such that

         I17: f = (I . n) & for x be Element of X1 holds (f . x) = ( Integral (M2,( ProjMap1 ((C /. n),x)))) by I3;

        

         I18: (I /. n) = f by I16, I17, PARTFUN1:def 6;

        

         I19: n in ( dom F) by I3, I16, FINSEQ_1:def 3;

        then (F . n) in ( measurable_rectangles (S1,S2)) by PARTFUN1: 4;

        then (F . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

        then

        consider An be Element of S1, Bn be Element of S2 such that

         I20: (F . n) = [:An, Bn:];

        reconsider Fn = (F . n) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, I19, PARTFUN1: 4;

        per cases ;

          suppose

           I21: (M2 . Bn) = +infty ;

          for x be Element of X1 holds (f . x) = (( Xchi (An,X1)) . x)

          proof

            let x be Element of X1;

            

             I22: (f . x) = ( Integral (M2,( ProjMap1 ((C /. n),x)))) by I17;

            

             I23: n in ( dom C) by I3, I16, A2, FINSEQ_1:def 3;

            then (C /. n) = (C . n) by PARTFUN1:def 6;

            then (C /. n) = ( chi ((F . n), [:X1, X2:])) by A2, I23;

            then

             I24: (f . x) = ((M2 . ( Measurable-X-section (Fn,x))) * (( chi (An,X1)) . x)) by I20, I22, Th65;

            per cases ;

              suppose

               I25: x in An;

              then (M2 . ( Measurable-X-section (Fn,x))) = +infty & (( chi (An,X1)) . x) = 1 by I20, I21, Th16, FUNCT_3:def 3;

              then (f . x) = +infty by I24, XXREAL_3: 81;

              hence (f . x) = (( Xchi (An,X1)) . x) by I25, MEASUR10:def 7;

            end;

              suppose

               I26: not x in An;

              then (( chi (An,X1)) . x) = 0 by FUNCT_3:def 3;

              then (f . x) = 0 by I24;

              hence (f . x) = (( Xchi (An,X1)) . x) by I26, MEASUR10:def 7;

            end;

          end;

          then f = ( Xchi (An,X1)) by FUNCT_2:def 8;

          hence (I /. n) is P -measurable by I18, MEASUR10: 32;

        end;

          suppose

           I27: (M2 . Bn) <> +infty ;

          (M2 . Bn) >= 0 by SUPINF_2: 51;

          then (M2 . Bn) in REAL by I27, XXREAL_0: 14;

          then

          reconsider r = (M2 . Bn) as Real;

          

           I28: ( dom ( chi (An,X1))) = X1 by FUNCT_2:def 1;

          then

           I29: ( dom f) = X1 & ( dom (r (#) ( chi (An,X1)))) = X1 by MESFUNC1:def 6, FUNCT_2:def 1;

          for x be Element of X1 st x in ( dom f) holds (f . x) = ((r (#) ( chi (An,X1))) . x)

          proof

            let x be Element of X1;

            assume x in ( dom f);

            

             I30: (f . x) = ( Integral (M2,( ProjMap1 ((C /. n),x)))) by I17;

            

             I31: n in ( dom C) by I3, I16, A2, FINSEQ_1:def 3;

            then (C /. n) = (C . n) by PARTFUN1:def 6;

            then (C /. n) = ( chi ((F . n), [:X1, X2:])) by I31, A2;

            then

             I32: (f . x) = ((M2 . ( Measurable-X-section (Fn,x))) * (( chi (An,X1)) . x)) by I20, I30, Th65;

            

             I33: ((r (#) ( chi (An,X1))) . x) = (r * (( chi (An,X1)) . x)) by I29, MESFUNC1:def 6;

            per cases ;

              suppose x in An;

              hence (f . x) = ((r (#) ( chi (An,X1))) . x) by I33, I32, I20, Th16;

            end;

              suppose not x in An;

              then

               I34: (( chi (An,X1)) . x) = 0 by FUNCT_3:def 3;

              ((r (#) ( chi (An,X1))) . x) = (r * (( chi (An,X1)) . x)) by I29, MESFUNC1:def 6;

              hence (f . x) = ((r (#) ( chi (An,X1))) . x) by I34, I32;

            end;

          end;

          then f = (r (#) ( chi (An,X1))) by I29, PARTFUN1: 5;

          hence (I /. n) is P -measurable by I18, I28, MESFUNC2: 29, MESFUNC1: 37;

        end;

      end;

      thus for x be Element of X1 holds ( Integral (M2,( ProjMap1 ((( Partial_Sums C) /. ( len C)),x)))) = ((( Partial_Sums I) /. ( len I)) . x)

      proof

        let x be Element of X1;

        defpred P2[ Nat] means $1 in ( dom I) implies ((( Partial_Sums I) . $1) . x) = ( Integral (M2,( ProjMap1 (( chi (( union ( rng (F | $1))), [:X1, X2:])),x))));

        

         I35: P2[ 0 ] by FINSEQ_3: 24;

        

         I36: for k be Nat st P2[k] holds P2[(k + 1)]

        proof

          let k be Nat;

          assume

           I37: P2[k];

          assume

           I38: (k + 1) in ( dom I);

          then

           I39: 1 <= (k + 1) <= ( len I) by FINSEQ_3: 25;

          then

           I40: k < ( len I) by NAT_1: 13;

          then

           I41: k < ( len ( Partial_Sums I)) by DEF13;

          per cases ;

            suppose

             I42: k = 0 ;

            

             I43: 1 <= ( len I) by I39, XXREAL_0: 2;

            then

            consider f be Function of X1, ExtREAL such that

             I44: f = (I . 1) & for x be Element of X1 holds (f . x) = ( Integral (M2,( ProjMap1 ((C /. 1),x)))) by I3, FINSEQ_3: 25;

            

             I45: 1 in ( Seg ( len F)) by I43, I3, FINSEQ_3: 25;

            then

             I46: 1 in ( dom C) by A2, FINSEQ_1:def 3;

            

            then

             I47: (C /. 1) = (C . 1) by PARTFUN1:def 6

            .= ( chi ((F . 1), [:X1, X2:])) by I46, A2;

            F <> {} by I38, I3;

            then (F | 1) = <*(F . 1)*> by FINSEQ_5: 20;

            then ( rng (F | 1)) = {(F . 1)} by FINSEQ_1: 39;

            then

             I49: (C /. 1) = ( chi (( union ( rng (F | 1))), [:X1, X2:])) by I47, ZFMISC_1: 25;

            (( Partial_Sums I) . (k + 1)) = f by I42, I44, DEF13;

            hence ((( Partial_Sums I) . (k + 1)) . x) = ( Integral (M2,( ProjMap1 (( chi (( union ( rng (F | (k + 1)))), [:X1, X2:])),x)))) by I44, I49, I42;

          end;

            suppose k <> 0 ;

            then

             I50: 1 <= k by NAT_1: 14;

            k <= ( len ( Partial_Sums I)) by I40, DEF13;

            then

             I51: k in ( dom ( Partial_Sums I)) by I50, FINSEQ_3: 25;

            

            then

             I52: ((( Partial_Sums I) /. k) . x) = ( Integral (M2,( ProjMap1 (( chi (( union ( rng (F | k))), [:X1, X2:])),x)))) by I37, I50, I40, FINSEQ_3: 25, PARTFUN1:def 6

            .= ( Integral (M2,( ProjMap1 (( chi (( Union (F | k)), [:X1, X2:])),x)))) by CARD_3:def 4;

            

             I53: ((( Partial_Sums I) . (k + 1)) . x) = (((( Partial_Sums I) /. k) + (I /. (k + 1))) . x) by I39, I50, NAT_1: 13, DEF13;

            ( Partial_Sums I) is without_-infty-valued by I15, Th57;

            then (( Partial_Sums I) . k) is without-infty by I50, I41, FINSEQ_3: 25;

            then

             I54: (( Partial_Sums I) /. k) is without-infty by I51, PARTFUN1:def 6;

            (I . (k + 1)) is without-infty by I7, I38;

            then (I /. (k + 1)) is without-infty by I38, PARTFUN1:def 6;

            

            then ( dom ((( Partial_Sums I) /. k) + (I /. (k + 1)))) = (( dom (( Partial_Sums I) /. k)) /\ ( dom (I /. (k + 1)))) by I54, MESFUNC5: 16

            .= (X1 /\ ( dom (I /. (k + 1)))) by FUNCT_2:def 1

            .= (X1 /\ X1) by FUNCT_2:def 1

            .= X1;

            then

             I55: ((( Partial_Sums I) . (k + 1)) . x) = (((( Partial_Sums I) /. k) . x) + ((I /. (k + 1)) . x)) by I53, MESFUNC1:def 3;

            reconsider E1 = ( Union (F | k)) as Element of ( sigma ( measurable_rectangles (S1,S2))) by Th71;

            

             I56: (k + 1) in ( dom C) & (k + 1) in ( dom F) by A2, I38, I3, FINSEQ_1:def 3;

            then

            reconsider E2 = (F . (k + 1)) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, FINSEQ_2: 11;

            

             I57: (C /. (k + 1)) = (C . (k + 1)) by I56, PARTFUN1:def 6

            .= ( chi (E2, [:X1, X2:])) by A2, I56;

            (I /. (k + 1)) = (I . (k + 1)) by I38, PARTFUN1:def 6;

            

            then

             I58: ((( Partial_Sums I) . (k + 1)) . x) = (( Integral (M2,( ProjMap1 (( chi (E1, [:X1, X2:])),x)))) + ( Integral (M2,( ProjMap1 ((C /. (k + 1)),x))))) by I6, I38, I52, I55

            .= ((M2 . ( Measurable-X-section (E1,x))) + ( Integral (M2,( ProjMap1 (( chi (E2, [:X1, X2:])),x))))) by I57, Th68

            .= ((M2 . ( Measurable-X-section (E1,x))) + (M2 . ( Measurable-X-section (E2,x)))) by Th68;

            k < (k + 1) by NAT_1: 13;

            then ( union ( rng (F | k))) misses (F . (k + 1)) by Th72;

            then

             I59: E1 misses E2 by CARD_3:def 4;

            (( union ( rng (F | k))) \/ (F . (k + 1))) = ( union ( rng (F | (k + 1)))) by Th74;

            then

             I60: (E1 \/ E2) = ( union ( rng (F | (k + 1)))) by CARD_3:def 4;

            then

            reconsider E3 = ( union ( rng (F | (k + 1)))) as Element of ( sigma ( measurable_rectangles (S1,S2)));

            (M2 . (( Measurable-X-section (E1,x)) \/ ( Measurable-X-section (E2,x)))) = (M2 . ( Measurable-X-section (E3,x))) by I60, Th20

            .= ( Integral (M2,( ProjMap1 (( chi (( union ( rng (F | (k + 1)))), [:X1, X2:])),x)))) by Th68;

            hence ((( Partial_Sums I) . (k + 1)) . x) = ( Integral (M2,( ProjMap1 (( chi (( union ( rng (F | (k + 1)))), [:X1, X2:])),x)))) by I59, I58, Th29, MEASURE1: 30;

          end;

        end;

        

         I61: for k be Nat holds P2[k] from NAT_1:sch 2( I35, I36);

        

         I62: I <> {} by A2, I3, FINSEQ_1:def 3;

        then ( len I) in ( dom I) by FINSEQ_5: 6;

        then ( len I) in ( Seg ( len I)) by FINSEQ_1:def 3;

        then ( len I) in ( Seg ( len ( Partial_Sums I))) by DEF13;

        then ( len I) in ( dom ( Partial_Sums I)) by FINSEQ_1:def 3;

        

        then

         I63: ((( Partial_Sums I) /. ( len I)) . x) = ((( Partial_Sums I) . ( len I)) . x) by PARTFUN1:def 6

        .= ( Integral (M2,( ProjMap1 (( chi (( union ( rng (F | ( len I)))), [:X1, X2:])),x)))) by I61, I62, FINSEQ_5: 6;

        E = ( union ( rng F)) by A2, CARD_3:def 4

        .= ( union ( rng (F | ( len I)))) by K1, FINSEQ_1: 58;

        hence ( Integral (M2,( ProjMap1 ((( Partial_Sums C) /. ( len C)),x)))) = ((( Partial_Sums I) /. ( len I)) . x) by A2, I63;

      end;

      thus for x be Element of X2, n be Nat st n in ( dom J) holds ((J . n) . x) = ( Integral (M1,( ProjMap2 ((C /. n),x)))) by J6;

      thus for n be Nat, P be Element of S2 st n in ( dom J) holds (J /. n) is P -measurable

      proof

        let n be Nat, P be Element of S2;

        assume

         I16: n in ( dom J);

        then

        consider f be Function of X2, ExtREAL such that

         I17: f = (J . n) & for x be Element of X2 holds (f . x) = ( Integral (M1,( ProjMap2 ((C /. n),x)))) by J3;

        

         I18: (J /. n) = f by I16, I17, PARTFUN1:def 6;

        

         I19: n in ( dom F) by J3, I16, FINSEQ_1:def 3;

        then (F . n) in ( measurable_rectangles (S1,S2)) by PARTFUN1: 4;

        then (F . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

        then

        consider An be Element of S1, Bn be Element of S2 such that

         I20: (F . n) = [:An, Bn:];

        reconsider Fn = (F . n) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, I19, PARTFUN1: 4;

        per cases ;

          suppose

           I21: (M1 . An) = +infty ;

          for x be Element of X2 holds (f . x) = (( Xchi (Bn,X2)) . x)

          proof

            let x be Element of X2;

            

             I22: (f . x) = ( Integral (M1,( ProjMap2 ((C /. n),x)))) by I17;

            

             I23: n in ( dom C) by J3, I16, A2, FINSEQ_1:def 3;

            then (C /. n) = (C . n) by PARTFUN1:def 6;

            then (C /. n) = ( chi ((F . n), [:X1, X2:])) by A2, I23;

            then

             I24: (f . x) = ((M1 . ( Measurable-Y-section (Fn,x))) * (( chi (Bn,X2)) . x)) by I20, I22, Th65;

            per cases ;

              suppose

               I25: x in Bn;

              then (M1 . ( Measurable-Y-section (Fn,x))) = +infty & (( chi (Bn,X2)) . x) = 1 by I20, I21, Th16, FUNCT_3:def 3;

              then (f . x) = +infty by I24, XXREAL_3: 81;

              hence (f . x) = (( Xchi (Bn,X2)) . x) by I25, MEASUR10:def 7;

            end;

              suppose

               I26: not x in Bn;

              then (( chi (Bn,X2)) . x) = 0 by FUNCT_3:def 3;

              then (f . x) = 0 by I24;

              hence (f . x) = (( Xchi (Bn,X2)) . x) by I26, MEASUR10:def 7;

            end;

          end;

          then f = ( Xchi (Bn,X2)) by FUNCT_2:def 8;

          hence (J /. n) is P -measurable by I18, MEASUR10: 32;

        end;

          suppose

           I27: (M1 . An) <> +infty ;

          (M1 . An) >= 0 by SUPINF_2: 51;

          then (M1 . An) in REAL by I27, XXREAL_0: 14;

          then

          reconsider r = (M1 . An) as Real;

          

           I28: ( dom ( chi (Bn,X2))) = X2 by FUNCT_2:def 1;

          then

           I29: ( dom f) = X2 & ( dom (r (#) ( chi (Bn,X2)))) = X2 by MESFUNC1:def 6, FUNCT_2:def 1;

          for x be Element of X2 st x in ( dom f) holds (f . x) = ((r (#) ( chi (Bn,X2))) . x)

          proof

            let x be Element of X2;

            assume x in ( dom f);

            

             I30: (f . x) = ( Integral (M1,( ProjMap2 ((C /. n),x)))) by I17;

            

             I31: n in ( dom C) by J3, I16, A2, FINSEQ_1:def 3;

            then (C /. n) = (C . n) by PARTFUN1:def 6;

            then (C /. n) = ( chi ((F . n), [:X1, X2:])) by I31, A2;

            then

             I32: (f . x) = ((M1 . ( Measurable-Y-section (Fn,x))) * (( chi (Bn,X2)) . x)) by I20, I30, Th65;

            

             I33: ((r (#) ( chi (Bn,X2))) . x) = (r * (( chi (Bn,X2)) . x)) by I29, MESFUNC1:def 6;

            per cases ;

              suppose x in Bn;

              hence (f . x) = ((r (#) ( chi (Bn,X2))) . x) by I33, I32, I20, Th16;

            end;

              suppose not x in Bn;

              then

               I34: (( chi (Bn,X2)) . x) = 0 by FUNCT_3:def 3;

              ((r (#) ( chi (Bn,X2))) . x) = (r * (( chi (Bn,X2)) . x)) by I29, MESFUNC1:def 6;

              hence (f . x) = ((r (#) ( chi (Bn,X2))) . x) by I34, I32;

            end;

          end;

          then f = (r (#) ( chi (Bn,X2))) by I29, PARTFUN1: 5;

          hence (J /. n) is P -measurable by I18, I28, MESFUNC2: 29, MESFUNC1: 37;

        end;

      end;

      thus for x be Element of X2 holds ( Integral (M1,( ProjMap2 ((( Partial_Sums C) /. ( len C)),x)))) = ((( Partial_Sums J) /. ( len J)) . x)

      proof

        let x be Element of X2;

        defpred P2[ Nat] means $1 in ( dom J) implies ((( Partial_Sums J) . $1) . x) = ( Integral (M1,( ProjMap2 (( chi (( union ( rng (F | $1))), [:X1, X2:])),x))));

        

         I35: P2[ 0 ] by FINSEQ_3: 24;

        

         I36: for k be Nat st P2[k] holds P2[(k + 1)]

        proof

          let k be Nat;

          assume

           I37: P2[k];

          assume

           I38: (k + 1) in ( dom J);

          then

           I39: 1 <= (k + 1) <= ( len J) by FINSEQ_3: 25;

          then

           I40: k < ( len J) by NAT_1: 13;

          then

           I41: k < ( len ( Partial_Sums J)) by DEF13;

          per cases ;

            suppose

             I42: k = 0 ;

            

             I43: 1 <= ( len J) by I39, XXREAL_0: 2;

            then

            consider f be Function of X2, ExtREAL such that

             I44: f = (J . 1) & for x be Element of X2 holds (f . x) = ( Integral (M1,( ProjMap2 ((C /. 1),x)))) by J3, FINSEQ_3: 25;

            

             I45: 1 in ( Seg ( len F)) by I43, J3, FINSEQ_3: 25;

            then

             I46: 1 in ( dom C) by A2, FINSEQ_1:def 3;

            

            then

             I47: (C /. 1) = (C . 1) by PARTFUN1:def 6

            .= ( chi ((F . 1), [:X1, X2:])) by I46, A2;

            F <> {} by I38, J3;

            then (F | 1) = <*(F . 1)*> by FINSEQ_5: 20;

            then ( rng (F | 1)) = {(F . 1)} by FINSEQ_1: 39;

            then

             I49: (C /. 1) = ( chi (( union ( rng (F | 1))), [:X1, X2:])) by I47, ZFMISC_1: 25;

            (( Partial_Sums J) . (k + 1)) = f by I42, I44, DEF13;

            hence ((( Partial_Sums J) . (k + 1)) . x) = ( Integral (M1,( ProjMap2 (( chi (( union ( rng (F | (k + 1)))), [:X1, X2:])),x)))) by I44, I49, I42;

          end;

            suppose k <> 0 ;

            then

             I50: 1 <= k by NAT_1: 14;

            k <= ( len ( Partial_Sums J)) by I40, DEF13;

            then

             I51: k in ( dom ( Partial_Sums J)) by I50, FINSEQ_3: 25;

            

            then

             I52: ((( Partial_Sums J) /. k) . x) = ( Integral (M1,( ProjMap2 (( chi (( union ( rng (F | k))), [:X1, X2:])),x)))) by I37, I50, I40, FINSEQ_3: 25, PARTFUN1:def 6

            .= ( Integral (M1,( ProjMap2 (( chi (( Union (F | k)), [:X1, X2:])),x)))) by CARD_3:def 4;

            

             I53: ((( Partial_Sums J) . (k + 1)) . x) = (((( Partial_Sums J) /. k) + (J /. (k + 1))) . x) by I39, I50, NAT_1: 13, DEF13;

            ( Partial_Sums J) is without_-infty-valued by J15, Th57;

            then (( Partial_Sums J) . k) is without-infty by I50, I41, FINSEQ_3: 25;

            then

             I54: (( Partial_Sums J) /. k) is without-infty by I51, PARTFUN1:def 6;

            (J . (k + 1)) is without-infty by J7, I38;

            then (J /. (k + 1)) is without-infty by I38, PARTFUN1:def 6;

            

            then ( dom ((( Partial_Sums J) /. k) + (J /. (k + 1)))) = (( dom (( Partial_Sums J) /. k)) /\ ( dom (J /. (k + 1)))) by I54, MESFUNC5: 16

            .= (X2 /\ ( dom (J /. (k + 1)))) by FUNCT_2:def 1

            .= (X2 /\ X2) by FUNCT_2:def 1

            .= X2;

            then

             I55: ((( Partial_Sums J) . (k + 1)) . x) = (((( Partial_Sums J) /. k) . x) + ((J /. (k + 1)) . x)) by I53, MESFUNC1:def 3;

            reconsider E1 = ( Union (F | k)) as Element of ( sigma ( measurable_rectangles (S1,S2))) by Th71;

            

             I56: (k + 1) in ( dom C) & (k + 1) in ( dom F) by A2, I38, J3, FINSEQ_1:def 3;

            then

            reconsider E2 = (F . (k + 1)) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, FINSEQ_2: 11;

            

             I57: (C /. (k + 1)) = (C . (k + 1)) by I56, PARTFUN1:def 6

            .= ( chi (E2, [:X1, X2:])) by A2, I56;

            (J /. (k + 1)) = (J . (k + 1)) by I38, PARTFUN1:def 6;

            

            then

             I58: ((( Partial_Sums J) . (k + 1)) . x) = (( Integral (M1,( ProjMap2 (( chi (E1, [:X1, X2:])),x)))) + ( Integral (M1,( ProjMap2 ((C /. (k + 1)),x))))) by J6, I38, I52, I55

            .= ((M1 . ( Measurable-Y-section (E1,x))) + ( Integral (M1,( ProjMap2 (( chi (E2, [:X1, X2:])),x))))) by I57, Th68

            .= ((M1 . ( Measurable-Y-section (E1,x))) + (M1 . ( Measurable-Y-section (E2,x)))) by Th68;

            k < (k + 1) by NAT_1: 13;

            then ( union ( rng (F | k))) misses (F . (k + 1)) by Th72;

            then

             I59: E1 misses E2 by CARD_3:def 4;

            (( union ( rng (F | k))) \/ (F . (k + 1))) = ( union ( rng (F | (k + 1)))) by Th74;

            then

             I60: (E1 \/ E2) = ( union ( rng (F | (k + 1)))) by CARD_3:def 4;

            then

            reconsider E3 = ( union ( rng (F | (k + 1)))) as Element of ( sigma ( measurable_rectangles (S1,S2)));

            (M1 . (( Measurable-Y-section (E1,x)) \/ ( Measurable-Y-section (E2,x)))) = (M1 . ( Measurable-Y-section (E3,x))) by I60, Th20

            .= ( Integral (M1,( ProjMap2 (( chi (( union ( rng (F | (k + 1)))), [:X1, X2:])),x)))) by Th68;

            hence ((( Partial_Sums J) . (k + 1)) . x) = ( Integral (M1,( ProjMap2 (( chi (( union ( rng (F | (k + 1)))), [:X1, X2:])),x)))) by I59, I58, Th29, MEASURE1: 30;

          end;

        end;

        

         I61: for k be Nat holds P2[k] from NAT_1:sch 2( I35, I36);

        

         I62: J <> {} by A2, J3, FINSEQ_1:def 3;

        then ( len J) in ( dom J) by FINSEQ_5: 6;

        then ( len J) in ( Seg ( len J)) by FINSEQ_1:def 3;

        then ( len J) in ( Seg ( len ( Partial_Sums J))) by DEF13;

        then ( len J) in ( dom ( Partial_Sums J)) by FINSEQ_1:def 3;

        

        then

         I63: ((( Partial_Sums J) /. ( len J)) . x) = ((( Partial_Sums J) . ( len J)) . x) by PARTFUN1:def 6

        .= ( Integral (M1,( ProjMap2 (( chi (( union ( rng (F | ( len J)))), [:X1, X2:])),x)))) by I61, I62, FINSEQ_5: 6;

        E = ( union ( rng F)) by A2, CARD_3:def 4

        .= ( union ( rng (F | ( len J)))) by K1, FINSEQ_1: 58;

        hence ( Integral (M1,( ProjMap2 ((( Partial_Sums C) /. ( len C)),x)))) = ((( Partial_Sums J) /. ( len J)) . x) by A2, I63;

      end;

    end;

    definition

      let X1,X2 be non empty set;

      let S1 be SigmaField of X1, S2 be SigmaField of X2;

      let F be Set_Sequence of ( sigma ( measurable_rectangles (S1,S2)));

      let n be Nat;

      :: original: .

      redefine

      func F . n -> Element of ( sigma ( measurable_rectangles (S1,S2))) ;

      coherence by MEASURE8:def 2;

    end

    definition

      let X1,X2 be non empty set;

      let S1 be SigmaField of X1, S2 be SigmaField of X2;

      let F be Function of [: NAT , ( sigma ( measurable_rectangles (S1,S2))):], ( sigma ( measurable_rectangles (S1,S2)));

      let n be Element of NAT , E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      :: original: .

      redefine

      func F . (n,E) -> Element of ( sigma ( measurable_rectangles (S1,S2))) ;

      coherence

      proof

        (F . (n,E)) in ( sigma ( measurable_rectangles (S1,S2)));

        hence thesis;

      end;

    end

    theorem :: MEASUR11:77

    

     Th76: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), V be Element of S2 st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) holds ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ V))) & (for P be Element of S1 holds F is P -measurable)

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), V be Element of S2;

      assume

       A1: E in ( Field_generated_by ( measurable_rectangles (S1,S2)));

      X1 in S1 by MEASURE1: 7;

      then [:X1, V:] in the set of all [:A, B:] where A be Element of S1, B be Element of S2;

      then

       A2: [:X1, V:] in ( measurable_rectangles (S1,S2)) by MEASUR10:def 5;

      ( measurable_rectangles (S1,S2)) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

      then

      reconsider E1 = (E /\ [:X1, V:]) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A2, FINSUB_1:def 2;

      

       A3: ( measurable_rectangles (S1,S2)) c= ( Field_generated_by ( measurable_rectangles (S1,S2))) by SRINGS_3: 21;

      per cases ;

        suppose

         A4: E1 = {} ;

        reconsider A = {} as Element of S1 by MEASURE1: 34;

         0 in REAL by XREAL_0:def 1;

        then

        reconsider F = (X1 --> 0 ) as Function of X1, ExtREAL by FUNCOP_1: 45, NUMBERS: 31;

        take F;

        thus for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ V))

        proof

          let x be Element of X1;

          

           A5: X1 = ( [#] X1) by SUBSET_1:def 3;

          (( Measurable-X-section (E,x)) /\ V) = (( X-section (E,x)) /\ ( X-section ( [:( [#] X1), V:],x))) by A5, Th16

          .= ( X-section (( {} [:X1, X2:]),x)) by A4, A5, Th21

          .= {} by Th18;

          then (M2 . (( Measurable-X-section (E,x)) /\ V)) = 0 by VALUED_0:def 19;

          hence (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ V)) by FUNCOP_1: 7;

        end;

        thus for P be Element of S1 holds F is P -measurable

        proof

          let P be Element of S1;

          for x be Element of X1 holds (F . x) = (( chi ( {} ,X1)) . x)

          proof

            let x be Element of X1;

            (( chi ( {} ,X1)) . x) = 0 by FUNCT_3:def 3;

            hence (F . x) = (( chi ( {} ,X1)) . x) by FUNCOP_1: 7;

          end;

          then F = ( chi (A,X1)) by FUNCT_2:def 8;

          hence F is P -measurable by MESFUNC2: 29;

        end;

      end;

        suppose

         A6: E1 <> {} ;

        deffunc F1( Element of X1) = (M2 . (( Measurable-X-section (E,$1)) /\ V));

        consider F be Function of X1, ExtREAL such that

         A7: for x be Element of X1 holds (F . x) = F1(x) from FUNCT_2:sch 4;

        consider f be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)), A be FinSequence of S1, B be FinSequence of S2, Xf be summable FinSequence of ( Funcs ( [:X1, X2:], ExtREAL )), If be summable FinSequence of ( Funcs (X1, ExtREAL )), Jf be summable FinSequence of ( Funcs (X2, ExtREAL )) such that

         A8: (E /\ [:X1, V:]) = ( Union f) & ( len f) in ( dom f) & ( len f) = ( len A) & ( len f) = ( len B) & ( len f) = ( len Xf) & ( len f) = ( len If) & ( len f) = ( len Jf) & (for n be Nat st n in ( dom Xf) holds (Xf . n) = ( chi ((f . n), [:X1, X2:]))) & (( Partial_Sums Xf) /. ( len Xf)) = ( chi ((E /\ [:X1, V:]), [:X1, X2:])) & (for x be Element of X1, n be Nat st n in ( dom If) holds ((If . n) . x) = ( Integral (M2,( ProjMap1 ((Xf /. n),x))))) & (for n be Nat, P be Element of S1 st n in ( dom If) holds (If /. n) is P -measurable) & (for x be Element of X1 holds ( Integral (M2,( ProjMap1 ((( Partial_Sums Xf) /. ( len Xf)),x)))) = ((( Partial_Sums If) /. ( len If)) . x)) & (for x be Element of X2, n be Nat st n in ( dom Jf) holds ((Jf . n) . x) = ( Integral (M1,( ProjMap2 ((Xf /. n),x))))) & (for n be Nat, P be Element of S2 st n in ( dom Jf) holds (Jf /. n) is P -measurable) & (for x be Element of X2 holds ( Integral (M1,( ProjMap2 ((( Partial_Sums Xf) /. ( len Xf)),x)))) = ((( Partial_Sums Jf) /. ( len Jf)) . x)) by A3, A2, A1, FINSUB_1:def 2, A6, Th75;

        take F;

        thus for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ V)) by A7;

        

         A9: ( dom If) = ( dom f) by A8, FINSEQ_3: 29;

        for x be Element of X1 holds (F . x) = ((( Partial_Sums If) /. ( len If)) . x)

        proof

          let x be Element of X1;

          ((( Partial_Sums If) /. ( len If)) . x) = ( Integral (M2,( ProjMap1 (( chi ((E /\ [:X1, V:]), [:X1, X2:])),x)))) by A8

          .= (M2 . (( Measurable-X-section (E,x)) /\ V)) by Th67;

          hence thesis by A7;

        end;

        then

         A10: F = (( Partial_Sums If) /. ( len If)) by FUNCT_2:def 8;

        let P be Element of S1;

        for n be Nat st n in ( dom If) holds (If /. n) is P -measurable by A8;

        hence F is P -measurable by A8, A9, A10, Th64;

      end;

    end;

    theorem :: MEASUR11:78

    

     Th77: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), V be Element of S1 st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) holds ex F be Function of X2, ExtREAL st (for x be Element of X2 holds (F . x) = (M1 . (( Measurable-Y-section (E,x)) /\ V))) & (for P be Element of S2 holds F is P -measurable)

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), V be Element of S1;

      assume

       A1: E in ( Field_generated_by ( measurable_rectangles (S1,S2)));

      X2 in S2 by MEASURE1: 7;

      then [:V, X2:] in the set of all [:A, B:] where A be Element of S1, B be Element of S2;

      then

       A2: [:V, X2:] in ( measurable_rectangles (S1,S2)) by MEASUR10:def 5;

      ( measurable_rectangles (S1,S2)) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

      then

      reconsider E1 = (E /\ [:V, X2:]) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A2, FINSUB_1:def 2;

      

       A3: ( measurable_rectangles (S1,S2)) c= ( Field_generated_by ( measurable_rectangles (S1,S2))) by SRINGS_3: 21;

      per cases ;

        suppose

         A4: E1 = {} ;

        reconsider A = {} as Element of S2 by MEASURE1: 34;

         0 in REAL by XREAL_0:def 1;

        then

        reconsider F = (X2 --> 0 ) as Function of X2, ExtREAL by FUNCOP_1: 45, NUMBERS: 31;

        take F;

        thus for x be Element of X2 holds (F . x) = (M1 . (( Measurable-Y-section (E,x)) /\ V))

        proof

          let x be Element of X2;

          

           A5: X2 = ( [#] X2) by SUBSET_1:def 3;

          (( Measurable-Y-section (E,x)) /\ V) = (( Y-section (E,x)) /\ ( Y-section ( [:V, ( [#] X2):],x))) by A5, Th16

          .= ( Y-section (( {} [:X1, X2:]),x)) by A4, A5, Th21

          .= {} by Th18;

          then (M1 . (( Measurable-Y-section (E,x)) /\ V)) = 0 by VALUED_0:def 19;

          hence (F . x) = (M1 . (( Measurable-Y-section (E,x)) /\ V)) by FUNCOP_1: 7;

        end;

        thus for P be Element of S2 holds F is P -measurable

        proof

          let P be Element of S2;

          for x be Element of X2 holds (F . x) = (( chi ( {} ,X2)) . x)

          proof

            let x be Element of X2;

            (( chi ( {} ,X2)) . x) = 0 by FUNCT_3:def 3;

            hence (F . x) = (( chi ( {} ,X2)) . x) by FUNCOP_1: 7;

          end;

          then F = ( chi (A,X2)) by FUNCT_2:def 8;

          hence F is P -measurable by MESFUNC2: 29;

        end;

      end;

        suppose

         A6: E1 <> {} ;

        deffunc F1( Element of X2) = (M1 . (( Measurable-Y-section (E,$1)) /\ V));

        consider F be Function of X2, ExtREAL such that

         A7: for x be Element of X2 holds (F . x) = F1(x) from FUNCT_2:sch 4;

        consider f be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)), A be FinSequence of S1, B be FinSequence of S2, Xf be summable FinSequence of ( Funcs ( [:X1, X2:], ExtREAL )), If be summable FinSequence of ( Funcs (X1, ExtREAL )), Jf be summable FinSequence of ( Funcs (X2, ExtREAL )) such that

         A8: (E /\ [:V, X2:]) = ( Union f) & ( len f) in ( dom f) & ( len f) = ( len A) & ( len f) = ( len B) & ( len f) = ( len Xf) & ( len f) = ( len If) & ( len f) = ( len Jf) & (for n be Nat st n in ( dom Xf) holds (Xf . n) = ( chi ((f . n), [:X1, X2:]))) & (( Partial_Sums Xf) /. ( len Xf)) = ( chi ((E /\ [:V, X2:]), [:X1, X2:])) & (for x be Element of X1, n be Nat st n in ( dom If) holds ((If . n) . x) = ( Integral (M2,( ProjMap1 ((Xf /. n),x))))) & (for n be Nat, P be Element of S1 st n in ( dom If) holds (If /. n) is P -measurable) & (for x be Element of X1 holds ( Integral (M2,( ProjMap1 ((( Partial_Sums Xf) /. ( len Xf)),x)))) = ((( Partial_Sums If) /. ( len If)) . x)) & (for x be Element of X2, n be Nat st n in ( dom Jf) holds ((Jf . n) . x) = ( Integral (M1,( ProjMap2 ((Xf /. n),x))))) & (for n be Nat, P be Element of S2 st n in ( dom Jf) holds (Jf /. n) is P -measurable) & (for x be Element of X2 holds ( Integral (M1,( ProjMap2 ((( Partial_Sums Xf) /. ( len Xf)),x)))) = ((( Partial_Sums Jf) /. ( len Jf)) . x)) by A3, A1, A2, FINSUB_1:def 2, A6, Th75;

        take F;

        thus for x be Element of X2 holds (F . x) = (M1 . (( Measurable-Y-section (E,x)) /\ V)) by A7;

        

         A9: ( dom Jf) = ( dom f) by A8, FINSEQ_3: 29;

        for x be Element of X2 holds (F . x) = ((( Partial_Sums Jf) /. ( len Jf)) . x)

        proof

          let x be Element of X2;

          ((( Partial_Sums Jf) /. ( len Jf)) . x) = ( Integral (M1,( ProjMap2 (( chi ((E /\ [:V, X2:]), [:X1, X2:])),x)))) by A8

          .= (M1 . (( Measurable-Y-section (E,x)) /\ V)) by Th67;

          hence thesis by A7;

        end;

        then

         A10: F = (( Partial_Sums Jf) /. ( len Jf)) by FUNCT_2:def 8;

        let P be Element of S2;

        for n be Nat st n in ( dom Jf) holds (Jf /. n) is P -measurable by A8;

        hence F is P -measurable by A8, A9, A10, Th64;

      end;

    end;

    theorem :: MEASUR11:79

    

     Th78: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))) st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) holds (for B be Element of S2 holds E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable)) })

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume

       A0: E in ( Field_generated_by ( measurable_rectangles (S1,S2)));

      let B be Element of S2;

      (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable)) by A0, Th76;

      hence thesis;

    end;

    theorem :: MEASUR11:80

    

     Th79: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E be Element of ( sigma ( measurable_rectangles (S1,S2))) st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) holds (for B be Element of S1 holds E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X2, ExtREAL st (for x be Element of X2 holds (F . x) = (M1 . (( Measurable-Y-section (E,x)) /\ B))) & (for V be Element of S2 holds F is V -measurable)) })

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume

       A0: E in ( Field_generated_by ( measurable_rectangles (S1,S2)));

      let B be Element of S1;

      (ex F be Function of X2, ExtREAL st (for x be Element of X2 holds (F . x) = (M1 . (( Measurable-Y-section (E,x)) /\ B))) & (for V be Element of S2 holds F is V -measurable)) by A0, Th77;

      hence thesis;

    end;

    theorem :: MEASUR11:81

    

     Th80: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, B be Element of S2 holds ( Field_generated_by ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable)) }

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, B be Element of S2;

      now

        let E be set;

        assume

         A1: E in ( Field_generated_by ( measurable_rectangles (S1,S2)));

        ( sigma ( measurable_rectangles (S1,S2))) = ( sigma ( DisUnion ( measurable_rectangles (S1,S2)))) by Th1

        .= ( sigma ( Field_generated_by ( measurable_rectangles (S1,S2)))) by SRINGS_3: 22;

        then ( Field_generated_by ( measurable_rectangles (S1,S2))) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

        then

        reconsider E1 = E as Element of ( sigma ( measurable_rectangles (S1,S2))) by A1;

        E1 in ( Field_generated_by ( measurable_rectangles (S1,S2))) by A1;

        hence E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable)) } by Th78;

      end;

      hence thesis;

    end;

    theorem :: MEASUR11:82

    

     Th81: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, B be Element of S1 holds ( Field_generated_by ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ B))) & (for V be Element of S2 holds F is V -measurable)) }

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, B be Element of S1;

      now

        let E be set;

        assume

         A1: E in ( Field_generated_by ( measurable_rectangles (S1,S2)));

        ( sigma ( measurable_rectangles (S1,S2))) = ( sigma ( DisUnion ( measurable_rectangles (S1,S2)))) by Th1

        .= ( sigma ( Field_generated_by ( measurable_rectangles (S1,S2)))) by SRINGS_3: 22;

        then ( Field_generated_by ( measurable_rectangles (S1,S2))) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

        then

        reconsider E1 = E as Element of ( sigma ( measurable_rectangles (S1,S2))) by A1;

        E1 in ( Field_generated_by ( measurable_rectangles (S1,S2))) by A1;

        hence E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X2, ExtREAL st (for x be Element of X2 holds (F . x) = (M1 . (( Measurable-Y-section (E,x)) /\ B))) & (for V be Element of S2 holds F is V -measurable)) } by Th79;

      end;

      hence thesis;

    end;

    begin

    definition

      let X be non empty set, S be SigmaField of X, M be sigma_Measure of S;

      :: MEASUR11:def12

      attr M is sigma_finite means ex E be Set_Sequence of S st (for n be Nat holds (M . (E . n)) < +infty ) & ( Union E) = X;

    end

    

     LM0902a: for X be non empty set, S be SigmaField of X, M be sigma_Measure of S st M is sigma_finite holds ex F be Set_Sequence of S st F is non-descending & (for n be Nat holds (M . (F . n)) < +infty ) & ( lim F) = X

    proof

      let X be non empty set, S be SigmaField of X, M be sigma_Measure of S;

      assume M is sigma_finite;

      then

      consider E be Set_Sequence of S such that

       A1: (for n be Nat holds (M . (E . n)) < +infty ) & ( Union E) = X;

      defpred P[ Nat] means (( Partial_Union E) . $1) in S;

      (( Partial_Union E) . 0 ) = (E . 0 ) by PROB_3:def 2;

      then

       A2: P[ 0 ] by MEASURE8:def 2;

      

       A3: for k be Nat st P[k] holds P[(k + 1)]

      proof

        let k be Nat;

        assume

         A4: P[k];

        

         A5: (E . (k + 1)) in S by MEASURE8:def 2;

        (( Partial_Union E) . (k + 1)) = ((E . (k + 1)) \/ (( Partial_Union E) . k)) by PROB_3:def 2;

        hence P[(k + 1)] by A4, A5, FINSUB_1:def 1;

      end;

      for n be Nat holds P[n] from NAT_1:sch 2( A2, A3);

      then

      reconsider F = ( Partial_Union E) as Set_Sequence of S by MEASURE8:def 2;

      

       A6: F is non-descending by PROB_3: 11;

      defpred Q[ Nat] means (M . (F . $1)) < +infty ;

      (F . 0 ) = (E . 0 ) by PROB_3:def 2;

      then

       A7: Q[ 0 ] by A1;

      

       A8: for k be Nat st Q[k] holds Q[(k + 1)]

      proof

        let k be Nat;

        assume

         A9: Q[k];

        (M . (E . (k + 1))) < +infty by A1;

        then

         A10: ((M . (F . k)) + (M . (E . (k + 1)))) < +infty by A9, XXREAL_3: 16, XXREAL_0: 4;

        

         A11: (M . (F . (k + 1))) = (M . ((F . k) \/ (E . (k + 1)))) by PROB_3:def 2;

        (F . k) in S & (E . (k + 1)) in S by MEASURE8:def 2;

        then (M . (F . (k + 1))) <= ((M . (F . k)) + (M . (E . (k + 1)))) by A11, MEASURE1: 33;

        hence Q[(k + 1)] by A10, XXREAL_0: 2;

      end;

      

       A12: for n be Nat holds Q[n] from NAT_1:sch 2( A7, A8);

      ( lim F) = ( Union F) by A6, SETLIM_1: 63

      .= ( Union E) by PROB_3: 15;

      hence thesis by A1, A6, A12;

    end;

    

     LM0902b: for X be non empty set, S be SigmaField of X, M be sigma_Measure of S st (ex F be Set_Sequence of S st F is non-descending & (for n be Nat holds (M . (F . n)) < +infty ) & ( lim F) = X) holds M is sigma_finite

    proof

      let X be non empty set, S be SigmaField of X, M be sigma_Measure of S;

      assume ex F be Set_Sequence of S st F is non-descending & (for n be Nat holds (M . (F . n)) < +infty ) & ( lim F) = X;

      then

      consider F be Set_Sequence of S such that

       A1: F is non-descending & (for n be Nat holds (M . (F . n)) < +infty ) & ( lim F) = X;

      

       A2: ( Partial_Union F) = F by A1, PROB_4: 15;

      defpred P[ Nat] means (( Partial_Diff_Union F) . $1) in S;

      (( Partial_Diff_Union F) . 0 ) = (F . 0 ) by PROB_3:def 3;

      then

       A3: P[ 0 ] by MEASURE8:def 2;

      

       A4: for k be Nat st P[k] holds P[(k + 1)]

      proof

        let k be Nat;

        assume P[k];

        

         A5: (( Partial_Union F) . k) in S by A2, MEASURE8:def 2;

        

         A6: (F . (k + 1)) in S by MEASURE8:def 2;

        (( Partial_Diff_Union F) . (k + 1)) = ((F . (k + 1)) \ (( Partial_Union F) . k)) by PROB_3:def 3;

        hence P[(k + 1)] by A5, A6, FINSUB_1:def 3;

      end;

      for n be Nat holds P[n] from NAT_1:sch 2( A3, A4);

      then

      reconsider E = ( Partial_Diff_Union F) as Set_Sequence of S by MEASURE8:def 2;

      defpred Q[ Nat] means (M . (E . $1)) < +infty ;

      (E . 0 ) = (F . 0 ) by PROB_3:def 3;

      then

       A7: Q[ 0 ] by A1;

      

       A8: for k be Nat st Q[k] holds Q[(k + 1)]

      proof

        let k be Nat;

        assume Q[k];

        

         A9: (E . (k + 1)) in S & (F . (k + 1)) in S by MEASURE8:def 2;

        (E . (k + 1)) = ((F . (k + 1)) \ (( Partial_Union F) . k)) by PROB_3:def 3;

        then (M . (E . (k + 1))) <= (M . (F . (k + 1))) by A9, MEASURE1: 8, XBOOLE_1: 36;

        hence Q[(k + 1)] by A1, XXREAL_0: 2;

      end;

      

       A10: for n be Nat holds Q[n] from NAT_1:sch 2( A7, A8);

      ( Union E) = ( Union F) by PROB_3: 20

      .= ( lim F) by A1, SETLIM_1: 63;

      hence M is sigma_finite by A1, A10;

    end;

    theorem :: MEASUR11:83

    for X be non empty set, S be SigmaField of X, M be sigma_Measure of S holds M is sigma_finite iff ex F be Set_Sequence of S st F is non-descending & (for n be Nat holds (M . (F . n)) < +infty ) & ( lim F) = X by LM0902a, LM0902b;

    theorem :: MEASUR11:84

    for X be set, S be semialgebra_of_sets of X, P be pre-Measure of S, M be induced_Measure of S, P holds M = (( C_Meas M) | ( Field_generated_by S))

    proof

      let X be set, S be semialgebra_of_sets of X, P be pre-Measure of S, M be induced_Measure of S, P;

      now

        let A be Element of ( Field_generated_by S);

        M is completely-additive by MEASURE9: 60;

        then (M . A) = (( C_Meas M) . A) by MEASURE8: 18;

        hence (M . A) = ((( C_Meas M) | ( Field_generated_by S)) . A) by FUNCT_1: 49;

      end;

      hence M = (( C_Meas M) | ( Field_generated_by S)) by FUNCT_2:def 8;

    end;

    begin

    theorem :: MEASUR11:85

    

     Th84: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, B be Element of S2 st (M2 . B) < +infty holds { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable)) } is MonotoneClass of [:X1, X2:]

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, B be Element of S2;

      assume

       A0: (M2 . B) < +infty ;

      set Z = { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable)) };

      now

        let A be object;

        assume A in Z;

        then ex E be Element of ( sigma ( measurable_rectangles (S1,S2))) st A = E & (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable));

        hence A in ( bool [:X1, X2:]);

      end;

      then

       A1: Z c= ( bool [:X1, X2:]);

      for A1 be SetSequence of [:X1, X2:] st A1 is monotone & ( rng A1) c= Z holds ( lim A1) in Z

      proof

        let A1 be SetSequence of [:X1, X2:];

        assume

         A2: A1 is monotone & ( rng A1) c= Z;

        

         A4: for V be set st V in ( rng A1) holds V in ( sigma ( measurable_rectangles (S1,S2)))

        proof

          let V be set;

          assume V in ( rng A1);

          then V in Z by A2;

          then ex E be Element of ( sigma ( measurable_rectangles (S1,S2))) st V = E & (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable));

          hence V in ( sigma ( measurable_rectangles (S1,S2)));

        end;

        

         A5: for n be Nat holds (A1 . n) in ( sigma ( measurable_rectangles (S1,S2)))

        proof

          let n be Nat;

          ( dom A1) = NAT by FUNCT_2:def 1;

          then n in ( dom A1) by ORDINAL1:def 12;

          hence (A1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A4, FUNCT_1: 3;

        end;

        then

        reconsider A2 = A1 as Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))) by MEASURE8:def 2;

        per cases by A2, SETLIM_1:def 1;

          suppose

           A3: A1 is non-descending;

          ( union ( rng A1)) in ( sigma ( measurable_rectangles (S1,S2))) by A4, MEASURE1: 35;

          then ( Union A1) in ( sigma ( measurable_rectangles (S1,S2))) by CARD_3:def 4;

          then

          reconsider E = ( lim A1) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, SETLIM_1: 63;

          ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable)

          proof

            defpred P[ Nat, object] means ex f1 be Function of X1, ExtREAL st $2 = f1 & (for x be Element of X1 holds (f1 . x) = (M2 . (( Measurable-X-section ((A2 . $1),x)) /\ B)) & (for V be Element of S1 holds f1 is V -measurable));

            

             A6: for n be Element of NAT holds ex f be Element of ( PFuncs (X1, ExtREAL )) st P[n, f]

            proof

              let n be Element of NAT ;

              ( dom A1) = NAT by FUNCT_2:def 1;

              then (A1 . n) in Z by A2, FUNCT_1: 3;

              then ex E1 be Element of ( sigma ( measurable_rectangles (S1,S2))) st (A1 . n) = E1 & (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E1,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable));

              then

              consider f1 be Function of X1, ExtREAL such that

               A7: (for x be Element of X1 holds (f1 . x) = (M2 . (( Measurable-X-section ((A2 . n),x)) /\ B))) & (for V be Element of S1 holds f1 is V -measurable);

              reconsider f = f1 as Element of ( PFuncs (X1, ExtREAL )) by PARTFUN1: 45;

              take f;

              thus thesis by A7;

            end;

            consider f be Function of NAT , ( PFuncs (X1, ExtREAL )) such that

             A8: for n be Element of NAT holds P[n, (f . n)] from FUNCT_2:sch 3( A6);

            

             A9: for n be Nat holds (f . n) is Function of X1, ExtREAL & (for x be Element of X1 holds ((f . n) . x) = (M2 . (( Measurable-X-section ((A2 . n),x)) /\ B)) & (for V be Element of S1 holds (f . n) is V -measurable))

            proof

              let n be Nat;

              n is Element of NAT by ORDINAL1:def 12;

              then ex f1 be Function of X1, ExtREAL st (f . n) = f1 & (for x be Element of X1 holds (f1 . x) = (M2 . (( Measurable-X-section ((A2 . n),x)) /\ B)) & (for V be Element of S1 holds f1 is V -measurable)) by A8;

              hence thesis;

            end;

            for n,m be Nat holds ( dom (f . n)) = ( dom (f . m))

            proof

              let n,m be Nat;

              (f . n) is Function of X1, ExtREAL & (f . m) is Function of X1, ExtREAL by A9;

              then ( dom (f . n)) = X1 & ( dom (f . m)) = X1 by FUNCT_2:def 1;

              hence thesis;

            end;

            then

            reconsider f as with_the_same_dom Functional_Sequence of X1, ExtREAL by MESFUNC8:def 2;

            reconsider XX1 = X1 as Element of S1 by MEASURE1: 11;

            (f . 0 ) is Function of X1, ExtREAL by A9;

            then

             A10: ( dom (f . 0 )) = XX1 by FUNCT_2:def 1;

            

             A11: for n be Nat holds (f . n) is XX1 -measurable by A9;

            

             A12: for x be Element of X1 st x in X1 holds (f # x) is convergent

            proof

              let x be Element of X1;

              assume x in X1;

              for n,m be Nat st m <= n holds ((f # x) . m) <= ((f # x) . n)

              proof

                let n,m be Nat;

                assume

                 Y1: m <= n;

                ((f # x) . m) = ((f . m) . x) & ((f # x) . n) = ((f . n) . x) by MESFUNC5:def 13;

                then

                 A13: ((f # x) . m) = (M2 . (( Measurable-X-section ((A2 . m),x)) /\ B)) & ((f # x) . n) = (M2 . (( Measurable-X-section ((A2 . n),x)) /\ B)) by A9;

                ( Measurable-X-section ((A2 . m),x)) c= ( Measurable-X-section ((A2 . n),x)) by A3, Y1, PROB_1:def 5, Th14;

                hence ((f # x) . m) <= ((f # x) . n) by A13, XBOOLE_1: 26, MEASURE1: 31;

              end;

              then (f # x) is non-decreasing by RINFSUP2: 7;

              hence (f # x) is convergent by RINFSUP2: 37;

            end;

            

             A14: ( dom ( lim f)) = X1 by A10, MESFUNC8:def 9;

            then

            reconsider F = ( lim f) as Function of X1, ExtREAL by FUNCT_2:def 1;

            take F;

            thus for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))

            proof

              let x be Element of X1;

              

               A15: (F . x) = ( lim (f # x)) by A14, MESFUNC8:def 9;

              consider G be SetSequence of X2 such that

               A16: G is non-descending & (for n be Nat holds (G . n) = ( X-section ((A1 . n),x))) by A3, Th37;

              for n be Nat holds (G . n) in S2

              proof

                let n be Nat;

                (A1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A5;

                then ( X-section ((A1 . n),x)) in S2 by Th44;

                hence (G . n) in S2 by A16;

              end;

              then

              reconsider G as Set_Sequence of S2 by MEASURE8:def 2;

              set K = (B (/\) G);

              

               A17: G is convergent & ( lim G) = ( Union G) by A16, SETLIM_1: 63;

              ( union ( rng G)) = ( X-section (( union ( rng A2)),x)) by A16, Th24;

              

              then ( Union G) = ( X-section (( union ( rng A2)),x)) by CARD_3:def 4

              .= ( X-section (( Union A2),x)) by CARD_3:def 4

              .= ( Measurable-X-section (E,x)) by A3, SETLIM_1: 63;

              then

               A18: ( lim K) = (( Measurable-X-section (E,x)) /\ B) by A17, SETLIM_2: 92;

              

               A19: ( dom K) = NAT by FUNCT_2:def 1;

              for n be object st n in NAT holds (K . n) in S2

              proof

                let n be object;

                assume n in NAT ;

                then

                reconsider n1 = n as Element of NAT ;

                (K . n1) = ((G . n1) /\ B) by SETLIM_2:def 5;

                then (K . n1) = (( Measurable-X-section ((A2 . n1),x)) /\ B) by A16;

                hence (K . n) in S2;

              end;

              then

              reconsider K2 = K as SetSequence of S2 by A19, FUNCT_2: 3;

              K2 is non-descending by A16, SETLIM_2: 22;

              then

               A20: ( lim (M2 * K2)) = (M2 . (( Measurable-X-section (E,x)) /\ B)) by A18, MEASURE8: 26;

              for n be Element of NAT holds ((f # x) . n) = ((M2 * K2) . n)

              proof

                let n be Element of NAT ;

                ((f # x) . n) = ((f . n) . x) by MESFUNC5:def 13;

                then

                 A21: ((f # x) . n) = (M2 . (( Measurable-X-section ((A2 . n),x)) /\ B)) by A9;

                (K2 . n) = ((G . n) /\ B) by SETLIM_2:def 5;

                then (K2 . n) = (( Measurable-X-section ((A2 . n),x)) /\ B) by A16;

                hence ((f # x) . n) = ((M2 * K2) . n) by A19, A21, FUNCT_1: 13;

              end;

              hence (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B)) by A15, A20, FUNCT_2: 63;

            end;

            thus for V be Element of S1 holds F is V -measurable by A10, A11, A12, MESFUNC8: 25, MESFUNC1: 30;

          end;

          hence ( lim A1) in Z;

        end;

          suppose

           A22: A1 is non-ascending;

          ( meet ( rng A1)) in ( sigma ( measurable_rectangles (S1,S2))) by A4, MEASURE1: 35;

          then ( Intersection A1) in ( sigma ( measurable_rectangles (S1,S2))) by SETLIM_1: 8;

          then

          reconsider E = ( lim A1) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A22, SETLIM_1: 64;

          defpred P[ Nat, object] means ex f1 be Function of X1, ExtREAL st $2 = f1 & (for x be Element of X1 holds (f1 . x) = (M2 . (( Measurable-X-section ((A2 . $1),x)) /\ B)) & (for V be Element of S1 holds f1 is V -measurable));

          

           A23: for n be Element of NAT holds ex f be Element of ( PFuncs (X1, ExtREAL )) st P[n, f]

          proof

            let n be Element of NAT ;

            ( dom A1) = NAT by FUNCT_2:def 1;

            then (A1 . n) in Z by A2, FUNCT_1: 3;

            then ex E1 be Element of ( sigma ( measurable_rectangles (S1,S2))) st (A1 . n) = E1 & (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E1,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable));

            then

            consider f1 be Function of X1, ExtREAL such that

             A24: (for x be Element of X1 holds (f1 . x) = (M2 . (( Measurable-X-section ((A2 . n),x)) /\ B))) & (for V be Element of S1 holds f1 is V -measurable);

            reconsider f = f1 as Element of ( PFuncs (X1, ExtREAL )) by PARTFUN1: 45;

            take f;

            thus thesis by A24;

          end;

          consider f be Function of NAT , ( PFuncs (X1, ExtREAL )) such that

           A25: for n be Element of NAT holds P[n, (f . n)] from FUNCT_2:sch 3( A23);

          

           A26: for n be Nat holds (f . n) is Function of X1, ExtREAL & (for x be Element of X1 holds ((f . n) . x) = (M2 . (( Measurable-X-section ((A2 . n),x)) /\ B)) & (for V be Element of S1 holds (f . n) is V -measurable))

          proof

            let n be Nat;

            n is Element of NAT by ORDINAL1:def 12;

            then ex f1 be Function of X1, ExtREAL st (f . n) = f1 & (for x be Element of X1 holds (f1 . x) = (M2 . (( Measurable-X-section ((A2 . n),x)) /\ B)) & (for V be Element of S1 holds f1 is V -measurable)) by A25;

            hence thesis;

          end;

          for n,m be Nat holds ( dom (f . n)) = ( dom (f . m))

          proof

            let n,m be Nat;

            (f . n) is Function of X1, ExtREAL & (f . m) is Function of X1, ExtREAL by A26;

            then ( dom (f . n)) = X1 & ( dom (f . m)) = X1 by FUNCT_2:def 1;

            hence thesis;

          end;

          then

          reconsider f as with_the_same_dom Functional_Sequence of X1, ExtREAL by MESFUNC8:def 2;

          reconsider XX1 = X1 as Element of S1 by MEASURE1: 11;

          (f . 0 ) is Function of X1, ExtREAL by A26;

          then

           A27: ( dom (f . 0 )) = XX1 by FUNCT_2:def 1;

          

           A28: for n be Nat holds (f . n) is XX1 -measurable by A26;

          

           A29: for x be Element of X1 st x in X1 holds (f # x) is convergent

          proof

            let x be Element of X1 such that x in X1;

            for n,m be Nat st m <= n holds ((f # x) . n) <= ((f # x) . m)

            proof

              let n,m be Nat;

              assume

               Y1: m <= n;

              ((f # x) . m) = ((f . m) . x) & ((f # x) . n) = ((f . n) . x) by MESFUNC5:def 13;

              then

               A30: ((f # x) . m) = (M2 . (( Measurable-X-section ((A2 . m),x)) /\ B)) & ((f # x) . n) = (M2 . (( Measurable-X-section ((A2 . n),x)) /\ B)) by A26;

              ( Measurable-X-section ((A2 . n),x)) c= ( Measurable-X-section ((A2 . m),x)) by Th14, A22, Y1, PROB_1:def 4;

              hence ((f # x) . n) <= ((f # x) . m) by A30, MEASURE1: 31, XBOOLE_1: 26;

            end;

            then (f # x) is non-increasing by RINFSUP2: 7;

            hence (f # x) is convergent by RINFSUP2: 36;

          end;

          

           A31: ( dom ( lim f)) = X1 by A27, MESFUNC8:def 9;

          then

          reconsider F = ( lim f) as Function of X1, ExtREAL by FUNCT_2:def 1;

          

           A32: for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))

          proof

            let x be Element of X1;

            ( lim (f # x)) = (M2 . (( Measurable-X-section (E,x)) /\ B))

            proof

              consider G be SetSequence of X2 such that

               A33: G is non-ascending & (for n be Nat holds (G . n) = ( X-section ((A1 . n),x))) by A22, Th39;

              for n be Nat holds (G . n) in S2

              proof

                let n be Nat;

                (A1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A5;

                then ( X-section ((A1 . n),x)) in S2 by Th44;

                hence (G . n) in S2 by A33;

              end;

              then

              reconsider G as Set_Sequence of S2 by MEASURE8:def 2;

              set K = (B (/\) G);

              

               A34: G is convergent & ( lim G) = ( Intersection G) by A33, SETLIM_1: 64;

              ( meet ( rng G)) = ( X-section (( meet ( rng A2)),x)) by A33, Th25;

              

              then ( Intersection G) = ( X-section (( meet ( rng A2)),x)) by SETLIM_1: 8

              .= ( X-section (( Intersection A2),x)) by SETLIM_1: 8

              .= ( Measurable-X-section (E,x)) by A22, SETLIM_1: 64;

              then

               A35: ( lim K) = (( Measurable-X-section (E,x)) /\ B) by A34, SETLIM_2: 92;

              (K . 0 ) = ((G . 0 ) /\ B) by SETLIM_2:def 5;

              then (K . 0 ) = (( Measurable-X-section ((A2 . 0 ),x)) /\ B) by A33;

              then (M2 . (K . 0 )) <= (M2 . B) by XBOOLE_1: 17, MEASURE1: 31;

              then

               A36: (M2 . (K . 0 )) < +infty by A0, XXREAL_0: 2;

              

               A37: ( dom K) = NAT by FUNCT_2:def 1;

              for n be object st n in NAT holds (K . n) in S2

              proof

                let n be object;

                assume n in NAT ;

                then

                reconsider n1 = n as Element of NAT ;

                (K . n1) = ((G . n1) /\ B) by SETLIM_2:def 5;

                then (K . n1) = (( Measurable-X-section ((A2 . n1),x)) /\ B) by A33;

                hence (K . n) in S2;

              end;

              then

              reconsider K2 = K as SetSequence of S2 by A37, FUNCT_2: 3;

              K2 is non-ascending by A33, SETLIM_2: 21;

              then

               A38: ( lim (M2 * K2)) = (M2 . (( Measurable-X-section (E,x)) /\ B)) by A35, A36, MEASURE8: 31;

              for n be Element of NAT holds ((f # x) . n) = ((M2 * K2) . n)

              proof

                let n be Element of NAT ;

                ((f # x) . n) = ((f . n) . x) by MESFUNC5:def 13;

                then

                 A39: ((f # x) . n) = (M2 . (( Measurable-X-section ((A2 . n),x)) /\ B)) by A26;

                (K2 . n) = ((G . n) /\ B) by SETLIM_2:def 5;

                then (K2 . n) = (( Measurable-X-section ((A2 . n),x)) /\ B) by A33;

                hence ((f # x) . n) = ((M2 * K2) . n) by A37, A39, FUNCT_1: 13;

              end;

              hence ( lim (f # x)) = (M2 . (( Measurable-X-section (E,x)) /\ B)) by A38, FUNCT_2: 63;

            end;

            hence (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B)) by A31, MESFUNC8:def 9;

          end;

          for V be Element of S1 holds F is V -measurable by A27, A28, A29, MESFUNC8: 25, MESFUNC1: 30;

          hence ( lim A1) in Z by A32;

        end;

      end;

      hence thesis by A1, PROB_3: 69;

    end;

    theorem :: MEASUR11:86

    

     Th85: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, B be Element of S1 st (M1 . B) < +infty holds { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ B))) & (for V be Element of S2 holds F is V -measurable)) } is MonotoneClass of [:X1, X2:]

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, B be Element of S1;

      assume

       A0: (M1 . B) < +infty ;

      set Z = { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ B))) & (for V be Element of S2 holds F is V -measurable)) };

      now

        let A be object;

        assume A in Z;

        then ex E be Element of ( sigma ( measurable_rectangles (S1,S2))) st A = E & (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ B))) & (for V be Element of S2 holds F is V -measurable));

        hence A in ( bool [:X1, X2:]);

      end;

      then

       A1: Z c= ( bool [:X1, X2:]);

      for A1 be SetSequence of [:X1, X2:] st A1 is monotone & ( rng A1) c= Z holds ( lim A1) in Z

      proof

        let A1 be SetSequence of [:X1, X2:];

        assume

         A2: A1 is monotone & ( rng A1) c= Z;

        

         A4: for V be set st V in ( rng A1) holds V in ( sigma ( measurable_rectangles (S1,S2)))

        proof

          let V be set;

          assume V in ( rng A1);

          then V in Z by A2;

          then ex E be Element of ( sigma ( measurable_rectangles (S1,S2))) st V = E & (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ B))) & (for V be Element of S2 holds F is V -measurable));

          hence V in ( sigma ( measurable_rectangles (S1,S2)));

        end;

        

         A5: for n be Nat holds (A1 . n) in ( sigma ( measurable_rectangles (S1,S2)))

        proof

          let n be Nat;

          ( dom A1) = NAT by FUNCT_2:def 1;

          then n in ( dom A1) by ORDINAL1:def 12;

          hence (A1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A4, FUNCT_1: 3;

        end;

        then

        reconsider A2 = A1 as Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))) by MEASURE8:def 2;

        per cases by A2, SETLIM_1:def 1;

          suppose

           A3: A1 is non-descending;

          ( union ( rng A1)) in ( sigma ( measurable_rectangles (S1,S2))) by A4, MEASURE1: 35;

          then ( Union A1) in ( sigma ( measurable_rectangles (S1,S2))) by CARD_3:def 4;

          then

          reconsider E = ( lim A1) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, SETLIM_1: 63;

          ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ B))) & (for V be Element of S2 holds F is V -measurable)

          proof

            defpred P[ Nat, object] means ex f1 be Function of X2, ExtREAL st $2 = f1 & (for y be Element of X2 holds (f1 . y) = (M1 . (( Measurable-Y-section ((A2 . $1),y)) /\ B)) & (for V be Element of S2 holds f1 is V -measurable));

            

             A6: for n be Element of NAT holds ex f be Element of ( PFuncs (X2, ExtREAL )) st P[n, f]

            proof

              let n be Element of NAT ;

              ( dom A1) = NAT by FUNCT_2:def 1;

              then (A1 . n) in Z by A2, FUNCT_1: 3;

              then ex E1 be Element of ( sigma ( measurable_rectangles (S1,S2))) st (A1 . n) = E1 & (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E1,y)) /\ B))) & (for V be Element of S2 holds F is V -measurable));

              then

              consider f1 be Function of X2, ExtREAL such that

               A7: (for y be Element of X2 holds (f1 . y) = (M1 . (( Measurable-Y-section ((A2 . n),y)) /\ B))) & (for V be Element of S2 holds f1 is V -measurable);

              reconsider f = f1 as Element of ( PFuncs (X2, ExtREAL )) by PARTFUN1: 45;

              take f;

              thus thesis by A7;

            end;

            consider f be Function of NAT , ( PFuncs (X2, ExtREAL )) such that

             A8: for n be Element of NAT holds P[n, (f . n)] from FUNCT_2:sch 3( A6);

            

             A9: for n be Nat holds (f . n) is Function of X2, ExtREAL & (for y be Element of X2 holds ((f . n) . y) = (M1 . (( Measurable-Y-section ((A2 . n),y)) /\ B)) & (for V be Element of S2 holds (f . n) is V -measurable))

            proof

              let n be Nat;

              n is Element of NAT by ORDINAL1:def 12;

              then ex f1 be Function of X2, ExtREAL st (f . n) = f1 & (for y be Element of X2 holds (f1 . y) = (M1 . (( Measurable-Y-section ((A2 . n),y)) /\ B)) & (for V be Element of S2 holds f1 is V -measurable)) by A8;

              hence thesis;

            end;

            for n,m be Nat holds ( dom (f . n)) = ( dom (f . m))

            proof

              let n,m be Nat;

              (f . n) is Function of X2, ExtREAL & (f . m) is Function of X2, ExtREAL by A9;

              then ( dom (f . n)) = X2 & ( dom (f . m)) = X2 by FUNCT_2:def 1;

              hence thesis;

            end;

            then

            reconsider f as with_the_same_dom Functional_Sequence of X2, ExtREAL by MESFUNC8:def 2;

            reconsider XX2 = X2 as Element of S2 by MEASURE1: 11;

            (f . 0 ) is Function of X2, ExtREAL by A9;

            then

             A10: ( dom (f . 0 )) = XX2 by FUNCT_2:def 1;

            

             A11: for n be Nat holds (f . n) is XX2 -measurable by A9;

            

             A12: for y be Element of X2 st y in X2 holds (f # y) is convergent

            proof

              let y be Element of X2 such that y in X2;

              for n,m be Nat st m <= n holds ((f # y) . m) <= ((f # y) . n)

              proof

                let n,m be Nat;

                assume

                 Y1: m <= n;

                ((f # y) . m) = ((f . m) . y) & ((f # y) . n) = ((f . n) . y) by MESFUNC5:def 13;

                then

                 A13: ((f # y) . m) = (M1 . (( Measurable-Y-section ((A2 . m),y)) /\ B)) & ((f # y) . n) = (M1 . (( Measurable-Y-section ((A2 . n),y)) /\ B)) by A9;

                ( Measurable-Y-section ((A2 . m),y)) c= ( Measurable-Y-section ((A2 . n),y)) by A3, Y1, PROB_1:def 5, Th15;

                hence ((f # y) . m) <= ((f # y) . n) by A13, XBOOLE_1: 26, MEASURE1: 31;

              end;

              then (f # y) is non-decreasing by RINFSUP2: 7;

              hence (f # y) is convergent by RINFSUP2: 37;

            end;

            

             A14: ( dom ( lim f)) = X2 by A10, MESFUNC8:def 9;

            then

            reconsider F = ( lim f) as Function of X2, ExtREAL by FUNCT_2:def 1;

            take F;

            thus for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ B))

            proof

              let y be Element of X2;

              

               A15: (F . y) = ( lim (f # y)) by A14, MESFUNC8:def 9;

              consider G be SetSequence of X1 such that

               A16: G is non-descending & (for n be Nat holds (G . n) = ( Y-section ((A1 . n),y))) by A3, Th38;

              for n be Nat holds (G . n) in S1

              proof

                let n be Nat;

                (A1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A5;

                then ( Y-section ((A1 . n),y)) in S1 by Th44;

                hence (G . n) in S1 by A16;

              end;

              then

              reconsider G as Set_Sequence of S1 by MEASURE8:def 2;

              set K = (B (/\) G);

              

               A17: G is convergent & ( lim G) = ( Union G) by A16, SETLIM_1: 63;

              ( union ( rng G)) = ( Y-section (( union ( rng A2)),y)) by A16, Th26;

              

              then ( Union G) = ( Y-section (( union ( rng A2)),y)) by CARD_3:def 4

              .= ( Y-section (( Union A2),y)) by CARD_3:def 4

              .= ( Measurable-Y-section (E,y)) by A3, SETLIM_1: 63;

              then

               A18: ( lim K) = (( Measurable-Y-section (E,y)) /\ B) by A17, SETLIM_2: 92;

              

               A19: ( dom K) = NAT by FUNCT_2:def 1;

              for n be object st n in NAT holds (K . n) in S1

              proof

                let n be object;

                assume n in NAT ;

                then

                reconsider n1 = n as Element of NAT ;

                (K . n1) = ((G . n1) /\ B) by SETLIM_2:def 5;

                then (K . n1) = (( Measurable-Y-section ((A2 . n1),y)) /\ B) by A16;

                hence (K . n) in S1;

              end;

              then

              reconsider K2 = K as SetSequence of S1 by A19, FUNCT_2: 3;

              K2 is non-descending by A16, SETLIM_2: 22;

              then

               A20: ( lim (M1 * K2)) = (M1 . (( Measurable-Y-section (E,y)) /\ B)) by A18, MEASURE8: 26;

              for n be Element of NAT holds ((f # y) . n) = ((M1 * K2) . n)

              proof

                let n be Element of NAT ;

                ((f # y) . n) = ((f . n) . y) by MESFUNC5:def 13;

                then

                 A21: ((f # y) . n) = (M1 . (( Measurable-Y-section ((A2 . n),y)) /\ B)) by A9;

                (K2 . n) = ((G . n) /\ B) by SETLIM_2:def 5;

                then (K2 . n) = (( Measurable-Y-section ((A2 . n),y)) /\ B) by A16;

                hence ((f # y) . n) = ((M1 * K2) . n) by A19, A21, FUNCT_1: 13;

              end;

              hence (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ B)) by A15, A20, FUNCT_2: 63;

            end;

            thus for V be Element of S2 holds F is V -measurable by A10, A11, A12, MESFUNC8: 25, MESFUNC1: 30;

          end;

          hence ( lim A1) in Z;

        end;

          suppose

           A22: A1 is non-ascending;

          ( meet ( rng A1)) in ( sigma ( measurable_rectangles (S1,S2))) by A4, MEASURE1: 35;

          then ( Intersection A1) in ( sigma ( measurable_rectangles (S1,S2))) by SETLIM_1: 8;

          then

          reconsider E = ( lim A1) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A22, SETLIM_1: 64;

          defpred P[ Nat, object] means ex f1 be Function of X2, ExtREAL st $2 = f1 & (for x be Element of X2 holds (f1 . x) = (M1 . (( Measurable-Y-section ((A2 . $1),x)) /\ B)) & (for V be Element of S2 holds f1 is V -measurable));

          

           A23: for n be Element of NAT holds ex f be Element of ( PFuncs (X2, ExtREAL )) st P[n, f]

          proof

            let n be Element of NAT ;

            ( dom A1) = NAT by FUNCT_2:def 1;

            then (A1 . n) in Z by A2, FUNCT_1: 3;

            then ex E1 be Element of ( sigma ( measurable_rectangles (S1,S2))) st (A1 . n) = E1 & (ex F be Function of X2, ExtREAL st (for x be Element of X2 holds (F . x) = (M1 . (( Measurable-Y-section (E1,x)) /\ B))) & (for V be Element of S2 holds F is V -measurable));

            then

            consider f1 be Function of X2, ExtREAL such that

             A24: (for x be Element of X2 holds (f1 . x) = (M1 . (( Measurable-Y-section ((A2 . n),x)) /\ B))) & (for V be Element of S2 holds f1 is V -measurable);

            reconsider f = f1 as Element of ( PFuncs (X2, ExtREAL )) by PARTFUN1: 45;

            take f;

            thus thesis by A24;

          end;

          consider f be Function of NAT , ( PFuncs (X2, ExtREAL )) such that

           A25: for n be Element of NAT holds P[n, (f . n)] from FUNCT_2:sch 3( A23);

          

           A26: for n be Nat holds (f . n) is Function of X2, ExtREAL & (for x be Element of X2 holds ((f . n) . x) = (M1 . (( Measurable-Y-section ((A2 . n),x)) /\ B)) & (for V be Element of S2 holds (f . n) is V -measurable))

          proof

            let n be Nat;

            n is Element of NAT by ORDINAL1:def 12;

            then ex f1 be Function of X2, ExtREAL st (f . n) = f1 & (for x be Element of X2 holds (f1 . x) = (M1 . (( Measurable-Y-section ((A2 . n),x)) /\ B)) & (for V be Element of S2 holds f1 is V -measurable)) by A25;

            hence thesis;

          end;

          for n,m be Nat holds ( dom (f . n)) = ( dom (f . m))

          proof

            let n,m be Nat;

            (f . n) is Function of X2, ExtREAL & (f . m) is Function of X2, ExtREAL by A26;

            then ( dom (f . n)) = X2 & ( dom (f . m)) = X2 by FUNCT_2:def 1;

            hence thesis;

          end;

          then

          reconsider f as with_the_same_dom Functional_Sequence of X2, ExtREAL by MESFUNC8:def 2;

          reconsider XX1 = X2 as Element of S2 by MEASURE1: 11;

          (f . 0 ) is Function of X2, ExtREAL by A26;

          then

           A27: ( dom (f . 0 )) = XX1 by FUNCT_2:def 1;

          

           A28: for n be Nat holds (f . n) is XX1 -measurable by A26;

          

           A29: for x be Element of X2 st x in X2 holds (f # x) is convergent

          proof

            let x be Element of X2 such that x in X2;

            for n,m be Nat st m <= n holds ((f # x) . n) <= ((f # x) . m)

            proof

              let n,m be Nat;

              assume

               Y1: m <= n;

              ((f # x) . m) = ((f . m) . x) & ((f # x) . n) = ((f . n) . x) by MESFUNC5:def 13;

              then

               A30: ((f # x) . m) = (M1 . (( Measurable-Y-section ((A2 . m),x)) /\ B)) & ((f # x) . n) = (M1 . (( Measurable-Y-section ((A2 . n),x)) /\ B)) by A26;

              ( Measurable-Y-section ((A2 . n),x)) c= ( Measurable-Y-section ((A2 . m),x)) by Th15, A22, Y1, PROB_1:def 4;

              hence ((f # x) . n) <= ((f # x) . m) by A30, MEASURE1: 31, XBOOLE_1: 26;

            end;

            then (f # x) is non-increasing by RINFSUP2: 7;

            hence (f # x) is convergent by RINFSUP2: 36;

          end;

          

           A31: ( dom ( lim f)) = X2 by A27, MESFUNC8:def 9;

          then

          reconsider F = ( lim f) as Function of X2, ExtREAL by FUNCT_2:def 1;

          

           A32: for x be Element of X2 holds (F . x) = (M1 . (( Measurable-Y-section (E,x)) /\ B))

          proof

            let x be Element of X2;

            ( lim (f # x)) = (M1 . (( Measurable-Y-section (E,x)) /\ B))

            proof

              consider G be SetSequence of X1 such that

               A33: G is non-ascending & (for n be Nat holds (G . n) = ( Y-section ((A1 . n),x))) by A22, Th40;

              for n be Nat holds (G . n) in S1

              proof

                let n be Nat;

                (A1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A5;

                then ( Y-section ((A1 . n),x)) in S1 by Th44;

                hence (G . n) in S1 by A33;

              end;

              then

              reconsider G as Set_Sequence of S1 by MEASURE8:def 2;

              set K = (B (/\) G);

              

               A34: G is convergent & ( lim G) = ( Intersection G) by A33, SETLIM_1: 64;

              ( meet ( rng G)) = ( Y-section (( meet ( rng A2)),x)) by A33, Th27;

              

              then ( Intersection G) = ( Y-section (( meet ( rng A2)),x)) by SETLIM_1: 8

              .= ( Y-section (( Intersection A2),x)) by SETLIM_1: 8

              .= ( Measurable-Y-section (E,x)) by A22, SETLIM_1: 64;

              then

               A35: ( lim K) = (( Measurable-Y-section (E,x)) /\ B) by A34, SETLIM_2: 92;

              (K . 0 ) = ((G . 0 ) /\ B) by SETLIM_2:def 5;

              then (K . 0 ) = (( Measurable-Y-section ((A2 . 0 ),x)) /\ B) by A33;

              then (M1 . (K . 0 )) <= (M1 . B) by XBOOLE_1: 17, MEASURE1: 31;

              then

               A36: (M1 . (K . 0 )) < +infty by A0, XXREAL_0: 2;

              

               A37: ( dom K) = NAT by FUNCT_2:def 1;

              for n be object st n in NAT holds (K . n) in S1

              proof

                let n be object;

                assume n in NAT ;

                then

                reconsider n1 = n as Element of NAT ;

                (K . n1) = ((G . n1) /\ B) by SETLIM_2:def 5;

                then (K . n1) = (( Measurable-Y-section ((A2 . n1),x)) /\ B) by A33;

                hence (K . n) in S1;

              end;

              then

              reconsider K2 = K as SetSequence of S1 by A37, FUNCT_2: 3;

              K2 is non-ascending by A33, SETLIM_2: 21;

              then

               A38: ( lim (M1 * K2)) = (M1 . (( Measurable-Y-section (E,x)) /\ B)) by A35, A36, MEASURE8: 31;

              for n be Element of NAT holds ((f # x) . n) = ((M1 * K2) . n)

              proof

                let n be Element of NAT ;

                ((f # x) . n) = ((f . n) . x) by MESFUNC5:def 13;

                then

                 A39: ((f # x) . n) = (M1 . (( Measurable-Y-section ((A2 . n),x)) /\ B)) by A26;

                (K2 . n) = ((G . n) /\ B) by SETLIM_2:def 5;

                then (K2 . n) = (( Measurable-Y-section ((A2 . n),x)) /\ B) by A33;

                hence ((f # x) . n) = ((M1 * K2) . n) by A37, A39, FUNCT_1: 13;

              end;

              hence ( lim (f # x)) = (M1 . (( Measurable-Y-section (E,x)) /\ B)) by A38, FUNCT_2: 63;

            end;

            hence (F . x) = (M1 . (( Measurable-Y-section (E,x)) /\ B)) by A31, MESFUNC8:def 9;

          end;

          for V be Element of S2 holds F is V -measurable by A27, A28, A29, MESFUNC8: 25, MESFUNC1: 30;

          hence ( lim A1) in Z by A32;

        end;

      end;

      hence thesis by A1, PROB_3: 69;

    end;

    theorem :: MEASUR11:87

    for X be non empty set, F be Field_Subset of X, L be SetSequence of X st ( rng L) is MonotoneClass of X & F c= ( rng L) holds ( sigma F) = ( monotoneclass F) & ( sigma F) c= ( rng L)

    proof

      let X be non empty set, F be Field_Subset of X, L be SetSequence of X;

      assume

       A1: ( rng L) is MonotoneClass of X & F c= ( rng L);

      thus ( sigma F) = ( monotoneclass F) by PROB_3: 73;

      hence ( sigma F) c= ( rng L) by A1, PROB_3:def 11;

    end;

    theorem :: MEASUR11:88

    

     Th87: for X be non empty set, F be Field_Subset of X, K be Subset-Family of X st K is MonotoneClass of X & F c= K holds ( sigma F) = ( monotoneclass F) & ( sigma F) c= K

    proof

      let X be non empty set, F be Field_Subset of X, K be Subset-Family of X;

      assume that

       A1: K is MonotoneClass of X and

       A2: F c= K;

      thus ( sigma F) = ( monotoneclass F) by PROB_3: 73;

      hence ( sigma F) c= K by A1, A2, PROB_3:def 11;

    end;

    theorem :: MEASUR11:89

    

     Th88: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, B be Element of S2 st (M2 . B) < +infty holds ( sigma ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable)) }

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, B be Element of S2;

      set K = { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ B))) & (for V be Element of S1 holds F is V -measurable)) };

      assume (M2 . B) < +infty ;

      then

       A1: K is MonotoneClass of [:X1, X2:] by Th84;

      

       A2: ( Field_generated_by ( measurable_rectangles (S1,S2))) c= K by Th80;

      ( sigma ( Field_generated_by ( measurable_rectangles (S1,S2)))) = ( sigma ( DisUnion ( measurable_rectangles (S1,S2)))) by SRINGS_3: 22

      .= ( sigma ( measurable_rectangles (S1,S2))) by Th1;

      hence thesis by A1, A2, Th87;

    end;

    theorem :: MEASUR11:90

    

     Th89: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, B be Element of S1 st (M1 . B) < +infty holds ( sigma ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ B))) & (for V be Element of S2 holds F is V -measurable)) }

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, B be Element of S1;

      set K = { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ B))) & (for V be Element of S2 holds F is V -measurable)) };

      assume (M1 . B) < +infty ;

      then

       A1: K is MonotoneClass of [:X1, X2:] by Th85;

      

       A2: ( Field_generated_by ( measurable_rectangles (S1,S2))) c= K by Th81;

      ( sigma ( Field_generated_by ( measurable_rectangles (S1,S2)))) = ( sigma ( DisUnion ( measurable_rectangles (S1,S2)))) by SRINGS_3: 22

      .= ( sigma ( measurable_rectangles (S1,S2))) by Th1;

      hence thesis by A1, A2, Th87;

    end;

    theorem :: MEASUR11:91

    

     Th90: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))) st M2 is sigma_finite holds (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . ( Measurable-X-section (E,x)))) & (for V be Element of S1 holds F is V -measurable))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume M2 is sigma_finite;

      then

      consider B be Set_Sequence of S2 such that

       A1: B is non-descending & (for n be Nat holds (M2 . (B . n)) < +infty ) & ( lim B) = X2 by LM0902a;

      defpred P[ Nat, object] means ex f1 be Function of X1, ExtREAL st $2 = f1 & (for x be Element of X1 holds (f1 . x) = (M2 . (( Measurable-X-section (E,x)) /\ (B . $1))) & (for V be Element of S1 holds f1 is V -measurable));

      

       A2: for n be Element of NAT holds ex f be Element of ( PFuncs (X1, ExtREAL )) st P[n, f]

      proof

        let n be Element of NAT ;

        reconsider Bn = (B . n) as Element of S2 by MEASURE8:def 2;

        (M2 . Bn) < +infty by A1;

        then ( sigma ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ Bn))) & (for V be Element of S1 holds F is V -measurable)) } by Th88;

        then E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E,x)) /\ Bn))) & (for V be Element of S1 holds F is V -measurable)) };

        then ex E1 be Element of ( sigma ( measurable_rectangles (S1,S2))) st E = E1 & (ex F be Function of X1, ExtREAL st (for x be Element of X1 holds (F . x) = (M2 . (( Measurable-X-section (E1,x)) /\ Bn))) & (for V be Element of S1 holds F is V -measurable));

        then

        consider f1 be Function of X1, ExtREAL such that

         A3: (for x be Element of X1 holds (f1 . x) = (M2 . (( Measurable-X-section (E,x)) /\ Bn))) & (for V be Element of S1 holds f1 is V -measurable);

        reconsider f = f1 as Element of ( PFuncs (X1, ExtREAL )) by PARTFUN1: 45;

        take f;

        f1 is Function of X1, ExtREAL & f = f1 & (for x be Element of X1 holds (f1 . x) = (M2 . (( Measurable-X-section (E,x)) /\ (B . n)))) & (for V be Element of S1 holds f1 is V -measurable) by A3;

        hence thesis;

      end;

      consider f be Function of NAT , ( PFuncs (X1, ExtREAL )) such that

       A4: for n be Element of NAT holds P[n, (f . n)] from FUNCT_2:sch 3( A2);

      

       A5: for n be Nat holds (f . n) is Function of X1, ExtREAL & (for x be Element of X1 holds ((f . n) . x) = (M2 . (( Measurable-X-section (E,x)) /\ (B . n))) & (for V be Element of S1 holds (f . n) is V -measurable))

      proof

        let n be Nat;

        n is Element of NAT by ORDINAL1:def 12;

        then ex f1 be Function of X1, ExtREAL st (f . n) = f1 & (for x be Element of X1 holds (f1 . x) = (M2 . (( Measurable-X-section (E,x)) /\ (B . n))) & (for V be Element of S1 holds f1 is V -measurable)) by A4;

        hence thesis;

      end;

      for n,m be Nat holds ( dom (f . n)) = ( dom (f . m))

      proof

        let n,m be Nat;

        (f . n) is Function of X1, ExtREAL & (f . m) is Function of X1, ExtREAL by A5;

        then ( dom (f . n)) = X1 & ( dom (f . m)) = X1 by FUNCT_2:def 1;

        hence thesis;

      end;

      then

      reconsider f as with_the_same_dom Functional_Sequence of X1, ExtREAL by MESFUNC8:def 2;

      reconsider XX1 = X1 as Element of S1 by MEASURE1: 11;

      (f . 0 ) is Function of X1, ExtREAL by A5;

      then

       A6: ( dom (f . 0 )) = XX1 by FUNCT_2:def 1;

      

       A7: for n be Nat holds (f . n) is XX1 -measurable by A5;

      

       A11: for x be Element of X1 st x in X1 holds (f # x) is convergent

      proof

        let x be Element of X1;

        assume x in X1;

        for n,m be Nat st m <= n holds ((f # x) . m) <= ((f # x) . n)

        proof

          let n,m be Nat;

          assume

           A8: m <= n;

          ((f # x) . m) = ((f . m) . x) & ((f # x) . n) = ((f . n) . x) by MESFUNC5:def 13;

          then

           A9: ((f # x) . m) = (M2 . (( Measurable-X-section (E,x)) /\ (B . m))) & ((f # x) . n) = (M2 . (( Measurable-X-section (E,x)) /\ (B . n))) by A5;

          

           A10: (( Measurable-X-section (E,x)) /\ (B . m)) c= (( Measurable-X-section (E,x)) /\ (B . n)) by A1, A8, PROB_1:def 5, XBOOLE_1: 26;

          (B . m) in S2 & (B . n) in S2 by MEASURE8:def 2;

          then (( Measurable-X-section (E,x)) /\ (B . m)) in S2 & (( Measurable-X-section (E,x)) /\ (B . n)) in S2 by MEASURE1: 11;

          hence ((f # x) . m) <= ((f # x) . n) by A9, A10, MEASURE1: 31;

        end;

        then (f # x) is non-decreasing by RINFSUP2: 7;

        hence (f # x) is convergent by RINFSUP2: 37;

      end;

      

       A12: ( dom ( lim f)) = X1 by A6, MESFUNC8:def 9;

      then

      reconsider F = ( lim f) as Function of X1, ExtREAL by FUNCT_2:def 1;

      take F;

      thus for x be Element of X1 holds (F . x) = (M2 . ( Measurable-X-section (E,x)))

      proof

        let x be Element of X1;

        ( lim (f # x)) = (M2 . ( Measurable-X-section (E,x)))

        proof

          deffunc F( Nat) = (( Measurable-X-section (E,x)) /\ (B . $1));

          set K1 = (( Measurable-X-section (E,x)) (/\) B);

          B is convergent by A1, SETLIM_1: 80;

          then ( lim K1) = (( Measurable-X-section (E,x)) /\ X2) by A1, SETLIM_2: 92;

          then

           A13: ( lim K1) = ( Measurable-X-section (E,x)) by XBOOLE_1: 28;

          

           A14: ( dom K1) = NAT by FUNCT_2:def 1;

          for n be object st n in NAT holds (K1 . n) in S2

          proof

            let n be object;

            assume n in NAT ;

            then

            reconsider n1 = n as Element of NAT ;

            

             A15: (K1 . n1) = (( X-section (E,x)) /\ (B . n1)) by SETLIM_2:def 5;

            reconsider Bn = (B . n1) as Element of S2 by MEASURE8:def 2;

            for x be Element of X1 holds (( X-section (E,x)) /\ Bn) in S2

            proof

              let x be Element of X1;

              ( X-section (E,x)) in S2 by Th44;

              hence (( X-section (E,x)) /\ Bn) in S2 by MEASURE1: 11;

            end;

            hence (K1 . n) in S2 by A15;

          end;

          then

          reconsider K1 as SetSequence of S2 by A14, FUNCT_2: 3;

          K1 is non-descending by A1, SETLIM_2: 22;

          then

           A16: ( lim (M2 * K1)) = (M2 . ( Measurable-X-section (E,x))) by A13, MEASURE8: 26;

          for n be Element of NAT holds ((f # x) . n) = ((M2 * K1) . n)

          proof

            let n be Element of NAT ;

            ((f # x) . n) = ((f . n) . x) by MESFUNC5:def 13;

            then

             A17: ((f # x) . n) = (M2 . (( Measurable-X-section (E,x)) /\ (B . n))) by A5;

            (K1 . n) = (( Measurable-X-section (E,x)) /\ (B . n)) by SETLIM_2:def 5;

            hence ((f # x) . n) = ((M2 * K1) . n) by A14, A17, FUNCT_1: 13;

          end;

          hence ( lim (f # x)) = (M2 . ( Measurable-X-section (E,x))) by A16, FUNCT_2: 63;

        end;

        hence (F . x) = (M2 . ( Measurable-X-section (E,x))) by A12, MESFUNC8:def 9;

      end;

      thus for V be Element of S1 holds F is V -measurable by A11, MESFUNC1: 30, A6, A7, MESFUNC8: 25;

    end;

    theorem :: MEASUR11:92

    

     Th91: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E be Element of ( sigma ( measurable_rectangles (S1,S2))) st M1 is sigma_finite holds (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . ( Measurable-Y-section (E,y)))) & (for V be Element of S2 holds F is V -measurable))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume M1 is sigma_finite;

      then

      consider B be Set_Sequence of S1 such that

       A1: B is non-descending & (for n be Nat holds (M1 . (B . n)) < +infty ) & ( lim B) = X1 by LM0902a;

      defpred P[ Nat, object] means ex f1 be Function of X2, ExtREAL st $2 = f1 & (for y be Element of X2 holds (f1 . y) = (M1 . (( Measurable-Y-section (E,y)) /\ (B . $1))) & (for V be Element of S2 holds f1 is V -measurable));

      

       A2: for n be Element of NAT holds ex f be Element of ( PFuncs (X2, ExtREAL )) st P[n, f]

      proof

        let n be Element of NAT ;

        reconsider Bn = (B . n) as Element of S1 by MEASURE8:def 2;

        (M1 . Bn) < +infty by A1;

        then ( sigma ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ Bn))) & (for V be Element of S2 holds F is V -measurable)) } by Th89;

        then E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E,y)) /\ Bn))) & (for V be Element of S2 holds F is V -measurable)) };

        then ex E1 be Element of ( sigma ( measurable_rectangles (S1,S2))) st E = E1 & (ex F be Function of X2, ExtREAL st (for y be Element of X2 holds (F . y) = (M1 . (( Measurable-Y-section (E1,y)) /\ Bn))) & (for V be Element of S2 holds F is V -measurable));

        then

        consider f1 be Function of X2, ExtREAL such that

         A3: (for y be Element of X2 holds (f1 . y) = (M1 . (( Measurable-Y-section (E,y)) /\ Bn))) & (for V be Element of S2 holds f1 is V -measurable);

        reconsider f = f1 as Element of ( PFuncs (X2, ExtREAL )) by PARTFUN1: 45;

        take f;

        f1 is Function of X2, ExtREAL & f = f1 & (for y be Element of X2 holds (f1 . y) = (M1 . (( Measurable-Y-section (E,y)) /\ (B . n)))) & (for V be Element of S2 holds f1 is V -measurable) by A3;

        hence thesis;

      end;

      consider f be Function of NAT , ( PFuncs (X2, ExtREAL )) such that

       A4: for n be Element of NAT holds P[n, (f . n)] from FUNCT_2:sch 3( A2);

      

       A5: for n be Nat holds (f . n) is Function of X2, ExtREAL & (for y be Element of X2 holds ((f . n) . y) = (M1 . (( Measurable-Y-section (E,y)) /\ (B . n))) & (for V be Element of S2 holds (f . n) is V -measurable))

      proof

        let n be Nat;

        n is Element of NAT by ORDINAL1:def 12;

        then ex f1 be Function of X2, ExtREAL st (f . n) = f1 & (for y be Element of X2 holds (f1 . y) = (M1 . (( Measurable-Y-section (E,y)) /\ (B . n))) & (for V be Element of S2 holds f1 is V -measurable)) by A4;

        hence thesis;

      end;

      for n,m be Nat holds ( dom (f . n)) = ( dom (f . m))

      proof

        let n,m be Nat;

        (f . n) is Function of X2, ExtREAL & (f . m) is Function of X2, ExtREAL by A5;

        then ( dom (f . n)) = X2 & ( dom (f . m)) = X2 by FUNCT_2:def 1;

        hence thesis;

      end;

      then

      reconsider f as with_the_same_dom Functional_Sequence of X2, ExtREAL by MESFUNC8:def 2;

      reconsider XX2 = X2 as Element of S2 by MEASURE1: 11;

      (f . 0 ) is Function of X2, ExtREAL by A5;

      then

       A6: ( dom (f . 0 )) = XX2 by FUNCT_2:def 1;

      

       A7: for n be Nat holds (f . n) is XX2 -measurable by A5;

      

       A11: for y be Element of X2 st y in X2 holds (f # y) is convergent

      proof

        let y be Element of X2;

        assume y in X2;

        for n,m be Nat st m <= n holds ((f # y) . m) <= ((f # y) . n)

        proof

          let n,m be Nat;

          assume

           A8: m <= n;

          ((f # y) . m) = ((f . m) . y) & ((f # y) . n) = ((f . n) . y) by MESFUNC5:def 13;

          then

           A9: ((f # y) . m) = (M1 . (( Measurable-Y-section (E,y)) /\ (B . m))) & ((f # y) . n) = (M1 . (( Measurable-Y-section (E,y)) /\ (B . n))) by A5;

          

           A10: (( Measurable-Y-section (E,y)) /\ (B . m)) c= (( Measurable-Y-section (E,y)) /\ (B . n)) by A1, A8, PROB_1:def 5, XBOOLE_1: 26;

          (B . m) in S1 & (B . n) in S1 by MEASURE8:def 2;

          then (( Measurable-Y-section (E,y)) /\ (B . m)) in S1 & (( Measurable-Y-section (E,y)) /\ (B . n)) in S1 by MEASURE1: 11;

          hence ((f # y) . m) <= ((f # y) . n) by A9, A10, MEASURE1: 31;

        end;

        then (f # y) is non-decreasing by RINFSUP2: 7;

        hence (f # y) is convergent by RINFSUP2: 37;

      end;

      

       A12: ( dom ( lim f)) = X2 by A6, MESFUNC8:def 9;

      then

      reconsider F = ( lim f) as Function of X2, ExtREAL by FUNCT_2:def 1;

      take F;

      thus for y be Element of X2 holds (F . y) = (M1 . ( Measurable-Y-section (E,y)))

      proof

        let y be Element of X2;

        deffunc F( Nat) = (( Measurable-Y-section (E,y)) /\ (B . $1));

        set K1 = (( Measurable-Y-section (E,y)) (/\) B);

        B is convergent by A1, SETLIM_1: 80;

        then ( lim K1) = (( Measurable-Y-section (E,y)) /\ X1) by A1, SETLIM_2: 92;

        then

         A13: ( lim K1) = ( Measurable-Y-section (E,y)) by XBOOLE_1: 28;

        

         A14: ( dom K1) = NAT by FUNCT_2:def 1;

        for n be object st n in NAT holds (K1 . n) in S1

        proof

          let n be object;

          assume n in NAT ;

          then

          reconsider n1 = n as Element of NAT ;

          

           A15: (K1 . n1) = (( Y-section (E,y)) /\ (B . n1)) by SETLIM_2:def 5;

          reconsider Bn = (B . n1) as Element of S1 by MEASURE8:def 2;

          for y be Element of X2 holds (( Y-section (E,y)) /\ Bn) in S1

          proof

            let y be Element of X2;

            ( Y-section (E,y)) in S1 by Th44;

            hence (( Y-section (E,y)) /\ Bn) in S1 by MEASURE1: 11;

          end;

          hence (K1 . n) in S1 by A15;

        end;

        then

        reconsider K1 as SetSequence of S1 by A14, FUNCT_2: 3;

        K1 is non-descending by A1, SETLIM_2: 22;

        then

         A16: ( lim (M1 * K1)) = (M1 . ( Measurable-Y-section (E,y))) by A13, MEASURE8: 26;

        for n be Element of NAT holds ((f # y) . n) = ((M1 * K1) . n)

        proof

          let n be Element of NAT ;

          ((f # y) . n) = ((f . n) . y) by MESFUNC5:def 13;

          then

           A17: ((f # y) . n) = (M1 . (( Measurable-Y-section (E,y)) /\ (B . n))) by A5;

          (K1 . n) = (( Measurable-Y-section (E,y)) /\ (B . n)) by SETLIM_2:def 5;

          hence ((f # y) . n) = ((M1 * K1) . n) by A14, A17, FUNCT_1: 13;

        end;

        then ( lim (f # y)) = (M1 . ( Measurable-Y-section (E,y))) by A16, FUNCT_2: 63;

        hence (F . y) = (M1 . ( Measurable-Y-section (E,y))) by A12, MESFUNC8:def 9;

      end;

      thus for V be Element of S2 holds F is V -measurable by A11, MESFUNC1: 30, A6, A7, MESFUNC8: 25;

    end;

    definition

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume

       A1: M2 is sigma_finite;

      :: MEASUR11:def13

      func Y-vol (E,M2) -> nonnegative Function of X1, ExtREAL means

      : DefYvol: (for x be Element of X1 holds (it . x) = (M2 . ( Measurable-X-section (E,x)))) & (for V be Element of S1 holds it is V -measurable);

      existence

      proof

        consider IT be Function of X1, ExtREAL such that

         A2: (for x be Element of X1 holds (IT . x) = (M2 . ( Measurable-X-section (E,x)))) & (for V be Element of S1 holds IT is V -measurable) by A1, Th90;

        now

          let x be Element of X1;

          (IT . x) = (M2 . ( Measurable-X-section (E,x))) by A2;

          hence 0. <= (IT . x) by SUPINF_2: 51;

        end;

        then

        reconsider IT as nonnegative Function of X1, ExtREAL by SUPINF_2: 39;

        take IT;

        thus thesis by A2;

      end;

      uniqueness

      proof

        let f1,f2 be nonnegative Function of X1, ExtREAL ;

        assume that

         A1: (for x be Element of X1 holds (f1 . x) = (M2 . ( Measurable-X-section (E,x)))) & (for V be Element of S1 holds f1 is V -measurable) and

         A2: (for x be Element of X1 holds (f2 . x) = (M2 . ( Measurable-X-section (E,x)))) & (for V be Element of S1 holds f2 is V -measurable);

        now

          let x be Element of X1;

          (f1 . x) = (M2 . ( Measurable-X-section (E,x))) by A1;

          hence (f1 . x) = (f2 . x) by A2;

        end;

        hence f1 = f2 by FUNCT_2: 63;

      end;

    end

    definition

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume

       A1: M1 is sigma_finite;

      :: MEASUR11:def14

      func X-vol (E,M1) -> nonnegative Function of X2, ExtREAL means

      : DefXvol: (for y be Element of X2 holds (it . y) = (M1 . ( Measurable-Y-section (E,y)))) & (for V be Element of S2 holds it is V -measurable);

      existence

      proof

        consider IT be Function of X2, ExtREAL such that

         A2: (for y be Element of X2 holds (IT . y) = (M1 . ( Measurable-Y-section (E,y)))) & (for V be Element of S2 holds IT is V -measurable) by A1, Th91;

        now

          let y be Element of X2;

          (IT . y) = (M1 . ( Measurable-Y-section (E,y))) by A2;

          hence 0. <= (IT . y) by SUPINF_2: 51;

        end;

        then

        reconsider IT as nonnegative Function of X2, ExtREAL by SUPINF_2: 39;

        take IT;

        thus thesis by A2;

      end;

      uniqueness

      proof

        let f1,f2 be nonnegative Function of X2, ExtREAL ;

        assume that

         A1: (for y be Element of X2 holds (f1 . y) = (M1 . ( Measurable-Y-section (E,y)))) & (for V be Element of S2 holds f1 is V -measurable) and

         A2: (for y be Element of X2 holds (f2 . y) = (M1 . ( Measurable-Y-section (E,y)))) & (for V be Element of S2 holds f2 is V -measurable);

        now

          let y be Element of X2;

          (f1 . y) = (M1 . ( Measurable-Y-section (E,y))) by A1;

          hence (f1 . y) = (f2 . y) by A2;

        end;

        hence f1 = f2 by FUNCT_2: 63;

      end;

    end

    theorem :: MEASUR11:93

    

     Th92: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E1,E2 be Element of ( sigma ( measurable_rectangles (S1,S2))) st M2 is sigma_finite & E1 misses E2 holds ( Y-vol ((E1 \/ E2),M2)) = (( Y-vol (E1,M2)) + ( Y-vol (E2,M2)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E1,E2 be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume that

       A1: M2 is sigma_finite and

       A2: E1 misses E2;

      

       A3: ( dom ( Y-vol ((E1 \/ E2),M2))) = X1 & ( dom ( Y-vol (E1,M2))) = X1 & ( dom ( Y-vol (E2,M2))) = X1 by FUNCT_2:def 1;

      then

       A4: ( dom (( Y-vol (E1,M2)) + ( Y-vol (E2,M2)))) = (X1 /\ X1) by MESFUNC5: 22;

      for x be Element of X1 st x in ( dom ( Y-vol ((E1 \/ E2),M2))) holds (( Y-vol ((E1 \/ E2),M2)) . x) = ((( Y-vol (E1,M2)) + ( Y-vol (E2,M2))) . x)

      proof

        let x be Element of X1;

        assume x in ( dom ( Y-vol ((E1 \/ E2),M2)));

        

         A6: (( Y-vol ((E1 \/ E2),M2)) . x) = (M2 . ( Measurable-X-section ((E1 \/ E2),x))) & (( Y-vol (E1,M2)) . x) = (M2 . ( Measurable-X-section (E1,x))) & (( Y-vol (E2,M2)) . x) = (M2 . ( Measurable-X-section (E2,x))) by A1, DefYvol;

        ( Measurable-X-section ((E1 \/ E2),x)) = (( Measurable-X-section (E1,x)) \/ ( Measurable-X-section (E2,x))) by Th20;

        then (( Y-vol ((E1 \/ E2),M2)) . x) = ((( Y-vol (E1,M2)) . x) + (( Y-vol (E2,M2)) . x)) by A6, A2, Th29, MEASURE1: 30;

        hence thesis by A4, MESFUNC1:def 3;

      end;

      hence thesis by A4, A3, PARTFUN1: 5;

    end;

    theorem :: MEASUR11:94

    

     Th93: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E1,E2 be Element of ( sigma ( measurable_rectangles (S1,S2))) st M1 is sigma_finite & E1 misses E2 holds ( X-vol ((E1 \/ E2),M1)) = (( X-vol (E1,M1)) + ( X-vol (E2,M1)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E1,E2 be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume that

       A1: M1 is sigma_finite and

       A2: E1 misses E2;

      

       A3: ( dom ( X-vol ((E1 \/ E2),M1))) = X2 & ( dom ( X-vol (E1,M1))) = X2 & ( dom ( X-vol (E2,M1))) = X2 by FUNCT_2:def 1;

      then

       A4: ( dom (( X-vol (E1,M1)) + ( X-vol (E2,M1)))) = (X2 /\ X2) by MESFUNC5: 22;

      for x be Element of X2 st x in ( dom ( X-vol ((E1 \/ E2),M1))) holds (( X-vol ((E1 \/ E2),M1)) . x) = ((( X-vol (E1,M1)) + ( X-vol (E2,M1))) . x)

      proof

        let x be Element of X2;

        assume x in ( dom ( X-vol ((E1 \/ E2),M1)));

        

         A6: (( X-vol ((E1 \/ E2),M1)) . x) = (M1 . ( Measurable-Y-section ((E1 \/ E2),x))) & (( X-vol (E1,M1)) . x) = (M1 . ( Measurable-Y-section (E1,x))) & (( X-vol (E2,M1)) . x) = (M1 . ( Measurable-Y-section (E2,x))) by A1, DefXvol;

        ( Measurable-Y-section ((E1 \/ E2),x)) = (( Measurable-Y-section (E1,x)) \/ ( Measurable-Y-section (E2,x))) by Th20;

        then (( X-vol ((E1 \/ E2),M1)) . x) = ((( X-vol (E1,M1)) . x) + (( X-vol (E2,M1)) . x)) by A6, A2, Th29, MEASURE1: 30;

        hence thesis by A4, MESFUNC1:def 3;

      end;

      hence thesis by A4, A3, PARTFUN1: 5;

    end;

    theorem :: MEASUR11:95

    

     Th94: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E1,E2 be Element of ( sigma ( measurable_rectangles (S1,S2))) st M2 is sigma_finite & E1 misses E2 holds ( Integral (M1,( Y-vol ((E1 \/ E2),M2)))) = (( Integral (M1,( Y-vol (E1,M2)))) + ( Integral (M1,( Y-vol (E2,M2)))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E1,E2 be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume that

       A1: M2 is sigma_finite and

       A2: E1 misses E2;

      reconsider XX1 = X1 as Element of S1 by MEASURE1: 7;

      

       a3: ( Y-vol ((E1 \/ E2),M2)) = (( Y-vol (E1,M2)) + ( Y-vol (E2,M2))) by A1, A2, Th92;

      

       A3: ( dom ( Y-vol (E1,M2))) = XX1 & ( Y-vol (E1,M2)) is XX1 -measurable by A1, DefYvol, FUNCT_2:def 1;

      

       A4: ( dom ( Y-vol (E2,M2))) = XX1 & ( Y-vol (E2,M2)) is XX1 -measurable by A1, DefYvol, FUNCT_2:def 1;

      

       A5: ( dom ( Y-vol ((E1 \/ E2),M2))) = XX1 & ( Y-vol ((E1 \/ E2),M2)) is XX1 -measurable by A1, DefYvol, FUNCT_2:def 1;

      reconsider Y1 = ( Y-vol (E1,M2)) as PartFunc of X1, ExtREAL ;

      reconsider Y2 = ( Y-vol (E2,M2)) as PartFunc of X1, ExtREAL ;

      ex Z be Element of S1 st Z = ( dom (( Y-vol (E1,M2)) + ( Y-vol (E2,M2)))) & ( integral+ (M1,(( Y-vol (E1,M2)) + ( Y-vol (E2,M2))))) = (( integral+ (M1,(Y1 | Z))) + ( integral+ (M1,(Y2 | Z)))) by A3, A4, MESFUNC5: 78;

      

      then ( Integral (M1,( Y-vol ((E1 \/ E2),M2)))) = (( integral+ (M1,( Y-vol (E1,M2)))) + ( integral+ (M1,( Y-vol (E2,M2))))) by a3, A5, MESFUNC5: 88

      .= (( Integral (M1,( Y-vol (E1,M2)))) + ( integral+ (M1,( Y-vol (E2,M2))))) by A3, MESFUNC5: 88;

      hence thesis by A4, MESFUNC5: 88;

    end;

    theorem :: MEASUR11:96

    

     Th95: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E1,E2 be Element of ( sigma ( measurable_rectangles (S1,S2))) st M1 is sigma_finite & E1 misses E2 holds ( Integral (M2,( X-vol ((E1 \/ E2),M1)))) = (( Integral (M2,( X-vol (E1,M1)))) + ( Integral (M2,( X-vol (E2,M1)))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E1,E2 be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume that

       A1: M1 is sigma_finite and

       A2: E1 misses E2;

      reconsider XX2 = X2 as Element of S2 by MEASURE1: 7;

      

       a3: ( X-vol ((E1 \/ E2),M1)) = (( X-vol (E1,M1)) + ( X-vol (E2,M1))) by A1, A2, Th93;

      

       A3: ( dom ( X-vol (E1,M1))) = XX2 & ( X-vol (E1,M1)) is XX2 -measurable by A1, DefXvol, FUNCT_2:def 1;

      

       A4: ( dom ( X-vol (E2,M1))) = XX2 & ( X-vol (E2,M1)) is XX2 -measurable by A1, DefXvol, FUNCT_2:def 1;

      

       A5: ( dom ( X-vol ((E1 \/ E2),M1))) = XX2 & ( X-vol ((E1 \/ E2),M1)) is XX2 -measurable by A1, DefXvol, FUNCT_2:def 1;

      reconsider V1 = ( X-vol (E1,M1)) as PartFunc of X2, ExtREAL ;

      reconsider V2 = ( X-vol (E2,M1)) as PartFunc of X2, ExtREAL ;

      ex Z be Element of S2 st Z = ( dom (( X-vol (E1,M1)) + ( X-vol (E2,M1)))) & ( integral+ (M2,(( X-vol (E1,M1)) + ( X-vol (E2,M1))))) = (( integral+ (M2,(V1 | Z))) + ( integral+ (M2,(V2 | Z)))) by A3, A4, MESFUNC5: 78;

      

      then ( Integral (M2,( X-vol ((E1 \/ E2),M1)))) = (( integral+ (M2,( X-vol (E1,M1)))) + ( integral+ (M2,( X-vol (E2,M1))))) by a3, A5, MESFUNC5: 88

      .= (( Integral (M2,( X-vol (E1,M1)))) + ( integral+ (M2,( X-vol (E2,M1))))) by A3, MESFUNC5: 88;

      hence thesis by A4, MESFUNC5: 88;

    end;

    theorem :: MEASUR11:97

    

     Th96: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2 st E = [:A, B:] & M2 is sigma_finite holds ((M2 . B) = +infty implies ( Y-vol (E,M2)) = ( Xchi (A,X1))) & ((M2 . B) <> +infty implies ex r be Real st r = (M2 . B) & ( Y-vol (E,M2)) = (r (#) ( chi (A,X1))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2;

      assume that

       A1: E = [:A, B:] and

       A2: M2 is sigma_finite;

      hereby

        assume

         A3: (M2 . B) = +infty ;

        for x be Element of X1 holds (( Y-vol (E,M2)) . x) = (( Xchi (A,X1)) . x)

        proof

          let x be Element of X1;

          

           A4: (( Y-vol (E,M2)) . x) = (M2 . ( Measurable-X-section (E,x))) by A2, DefYvol

          .= ((M2 . B) * (( chi (A,X1)) . x)) by A1, Th48;

          per cases ;

            suppose

             A5: x in A;

            then (( chi (A,X1)) . x) = 1 by FUNCT_3:def 3;

            then (( Y-vol (E,M2)) . x) = +infty by A3, A4, XXREAL_3: 81;

            hence (( Y-vol (E,M2)) . x) = (( Xchi (A,X1)) . x) by A5, MEASUR10:def 7;

          end;

            suppose

             A6: not x in A;

            then (( chi (A,X1)) . x) = 0 by FUNCT_3:def 3;

            then (( Y-vol (E,M2)) . x) = 0 by A4;

            hence (( Y-vol (E,M2)) . x) = (( Xchi (A,X1)) . x) by A6, MEASUR10:def 7;

          end;

        end;

        hence ( Y-vol (E,M2)) = ( Xchi (A,X1)) by FUNCT_2:def 8;

      end;

      assume

       P1: (M2 . B) <> +infty ;

      (M2 . B) >= 0 by SUPINF_2: 51;

      then (M2 . B) in REAL by P1, XXREAL_0: 14;

      then

      reconsider r = (M2 . B) as Real;

      take r;

      ( dom (r (#) ( chi (A,X1)))) = ( dom ( chi (A,X1))) by MESFUNC1:def 6;

      then

       A8: ( dom (r (#) ( chi (A,X1)))) = X1 by FUNCT_3:def 3;

      then

       P2: ( dom ( Y-vol (E,M2))) = ( dom (r (#) ( chi (A,X1)))) by FUNCT_2:def 1;

      for x be Element of X1 st x in ( dom ( Y-vol (E,M2))) holds (( Y-vol (E,M2)) . x) = ((r (#) ( chi (A,X1))) . x)

      proof

        let x be Element of X1;

        assume x in ( dom ( Y-vol (E,M2)));

        (( Y-vol (E,M2)) . x) = (M2 . ( Measurable-X-section (E,x))) by A2, DefYvol

        .= (r * (( chi (A,X1)) . x)) by A1, Th48;

        hence (( Y-vol (E,M2)) . x) = ((r (#) ( chi (A,X1))) . x) by A8, MESFUNC1:def 6;

      end;

      hence r = (M2 . B) & ( Y-vol (E,M2)) = (r (#) ( chi (A,X1))) by P2, PARTFUN1: 5;

    end;

    theorem :: MEASUR11:98

    

     Th97: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2 st E = [:A, B:] & M1 is sigma_finite holds ((M1 . A) = +infty implies ( X-vol (E,M1)) = ( Xchi (B,X2))) & ((M1 . A) <> +infty implies ex r be Real st r = (M1 . A) & ( X-vol (E,M1)) = (r (#) ( chi (B,X2))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2;

      assume that

       A1: E = [:A, B:] and

       A2: M1 is sigma_finite;

      hereby

        assume

         A3: (M1 . A) = +infty ;

        for x be Element of X2 holds (( X-vol (E,M1)) . x) = (( Xchi (B,X2)) . x)

        proof

          let x be Element of X2;

          

           A4: (( X-vol (E,M1)) . x) = (M1 . ( Measurable-Y-section (E,x))) by A2, DefXvol

          .= ((M1 . A) * (( chi (B,X2)) . x)) by A1, Th50;

          per cases ;

            suppose

             A5: x in B;

            then (( chi (B,X2)) . x) = 1 by FUNCT_3:def 3;

            then (( X-vol (E,M1)) . x) = +infty by A3, A4, XXREAL_3: 81;

            hence (( X-vol (E,M1)) . x) = (( Xchi (B,X2)) . x) by A5, MEASUR10:def 7;

          end;

            suppose

             A6: not x in B;

            then (( chi (B,X2)) . x) = 0 by FUNCT_3:def 3;

            then (( X-vol (E,M1)) . x) = 0 by A4;

            hence (( X-vol (E,M1)) . x) = (( Xchi (B,X2)) . x) by A6, MEASUR10:def 7;

          end;

        end;

        hence ( X-vol (E,M1)) = ( Xchi (B,X2)) by FUNCT_2:def 8;

      end;

      assume

       P1: (M1 . A) <> +infty ;

      (M1 . A) >= 0 by SUPINF_2: 51;

      then (M1 . A) in REAL by P1, XXREAL_0: 14;

      then

      reconsider r = (M1 . A) as Real;

      take r;

      ( dom (r (#) ( chi (B,X2)))) = ( dom ( chi (B,X2))) by MESFUNC1:def 6;

      then

       A8: ( dom (r (#) ( chi (B,X2)))) = X2 by FUNCT_3:def 3;

      then

       P2: ( dom ( X-vol (E,M1))) = ( dom (r (#) ( chi (B,X2)))) by FUNCT_2:def 1;

      for x be Element of X2 st x in ( dom ( X-vol (E,M1))) holds (( X-vol (E,M1)) . x) = ((r (#) ( chi (B,X2))) . x)

      proof

        let x be Element of X2;

        assume x in ( dom ( X-vol (E,M1)));

        (( X-vol (E,M1)) . x) = (M1 . ( Measurable-Y-section (E,x))) by A2, DefXvol

        .= (r * (( chi (B,X2)) . x)) by A1, Th50;

        hence (( X-vol (E,M1)) . x) = ((r (#) ( chi (B,X2))) . x) by A8, MESFUNC1:def 6;

      end;

      hence r = (M1 . A) & ( X-vol (E,M1)) = (r (#) ( chi (B,X2))) by P2, PARTFUN1: 5;

    end;

    theorem :: MEASUR11:99

    

     Th98: for X be non empty set, S be SigmaField of X, M be sigma_Measure of S, A be Element of S, r be Real st r >= 0 holds ( Integral (M,(r (#) ( chi (A,X))))) = (r * (M . A))

    proof

      let X be non empty set, S be SigmaField of X, M be sigma_Measure of S, A be Element of S, r be Real;

      assume

       A1: r >= 0 ;

      reconsider XX = X as Element of S by MEASURE1: 7;

      

       A2: ( dom ( chi (A,X))) = XX & ( chi (A,X)) is XX -measurable by FUNCT_3:def 3, MESFUNC2: 29;

      then

       A3: ( dom (r (#) ( chi (A,X)))) = XX & (r (#) ( chi (A,X))) is XX -measurable by MESFUNC1:def 6, MESFUNC1: 37;

      ( Integral (M,( chi (A,X)))) = (M . A) by MESFUNC9: 14;

      then ( integral+ (M,( chi (A,X)))) = (M . A) by A2, MESFUNC5: 88;

      then ( integral+ (M,(r (#) ( chi (A,X))))) = (r * (M . A)) by A1, A2, MESFUNC5: 86;

      hence ( Integral (M,(r (#) ( chi (A,X))))) = (r * (M . A)) by A1, A3, MESFUNC5: 20, MESFUNC5: 88;

    end;

    theorem :: MEASUR11:100

    

     Th99: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F be FinSequence of ( sigma ( measurable_rectangles (S1,S2))), n be Nat st M2 is sigma_finite & F is FinSequence of ( measurable_rectangles (S1,S2)) holds (( product_sigma_Measure (M1,M2)) . (F . n)) = ( Integral (M1,( Y-vol ((F . n),M2))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F be FinSequence of ( sigma ( measurable_rectangles (S1,S2))), n be Nat;

      assume that

       A1: M2 is sigma_finite and

       A2: F is FinSequence of ( measurable_rectangles (S1,S2));

      reconsider XX1 = X1 as Element of S1 by MEASURE1: 7;

       not n in ( dom F) implies (F . n) in ( measurable_rectangles (S1,S2))

      proof

        assume not n in ( dom F);

        then (F . n) = {} by FUNCT_1:def 2;

        hence (F . n) in ( measurable_rectangles (S1,S2)) by SETFAM_1:def 8;

      end;

      then (F . n) in ( measurable_rectangles (S1,S2)) by A2, PARTFUN1: 4;

      then (F . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

      then

      consider P be Element of S1, Q be Element of S2 such that

       d4: (F . n) = [:P, Q:];

      

       d5: (( product_sigma_Measure (M1,M2)) . (F . n)) = ((M1 . P) * (M2 . Q)) by d4, Th10;

      per cases ;

        suppose

         d8: (M1 . P) = 0 & (M2 . Q) = +infty ;

        then (( product_sigma_Measure (M1,M2)) . (F . n)) = 0 & ( Y-vol ((F . n),M2)) = ( Xchi (P,X1)) by A1, d4, d5, Th96;

        hence (( product_sigma_Measure (M1,M2)) . (F . n)) = ( Integral (M1,( Y-vol ((F . n),M2)))) by d8, MEASUR10: 33;

      end;

        suppose (M1 . P) = 0 & (M2 . Q) <> +infty ;

        then ex r be Real st r = (M2 . Q) & ( Y-vol ((F . n),M2)) = (r (#) ( chi (P,X1))) by A1, d4, Th96;

        hence (( product_sigma_Measure (M1,M2)) . (F . n)) = ( Integral (M1,( Y-vol ((F . n),M2)))) by d5, Th98, SUPINF_2: 51;

      end;

        suppose

         d6: (M1 . P) <> 0 & (M2 . Q) = +infty ;

        (M1 . P) >= 0 by SUPINF_2: 51;

        then

         d7: (( product_sigma_Measure (M1,M2)) . (F . n)) = +infty by d5, d6, XXREAL_3:def 5;

        ( Y-vol ((F . n),M2)) = ( Xchi (P,X1)) by A1, d4, d6, Th96;

        hence (( product_sigma_Measure (M1,M2)) . (F . n)) = ( Integral (M1,( Y-vol ((F . n),M2)))) by d7, d6, MEASUR10: 33;

      end;

        suppose (M1 . P) <> 0 & (M2 . Q) <> +infty ;

        then ex r be Real st r = (M2 . Q) & ( Y-vol ((F . n),M2)) = (r (#) ( chi (P,X1))) by A1, d4, Th96;

        hence (( product_sigma_Measure (M1,M2)) . (F . n)) = ( Integral (M1,( Y-vol ((F . n),M2)))) by d5, Th98, SUPINF_2: 51;

      end;

    end;

    theorem :: MEASUR11:101

    

     Th100: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F be FinSequence of ( sigma ( measurable_rectangles (S1,S2))), n be Nat st M1 is sigma_finite & F is FinSequence of ( measurable_rectangles (S1,S2)) holds (( product_sigma_Measure (M1,M2)) . (F . n)) = ( Integral (M2,( X-vol ((F . n),M1))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F be FinSequence of ( sigma ( measurable_rectangles (S1,S2))), n be Nat;

      assume that

       A1: M1 is sigma_finite and

       A2: F is FinSequence of ( measurable_rectangles (S1,S2));

      reconsider XX2 = X2 as Element of S2 by MEASURE1: 7;

       not n in ( dom F) implies (F . n) in ( measurable_rectangles (S1,S2))

      proof

        assume not n in ( dom F);

        then (F . n) = {} by FUNCT_1:def 2;

        hence (F . n) in ( measurable_rectangles (S1,S2)) by SETFAM_1:def 8;

      end;

      then (F . n) in ( measurable_rectangles (S1,S2)) by A2, PARTFUN1: 4;

      then (F . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by MEASUR10:def 5;

      then

      consider P be Element of S1, Q be Element of S2 such that

       d4: (F . n) = [:P, Q:];

      

       d5: (( product_sigma_Measure (M1,M2)) . (F . n)) = ((M1 . P) * (M2 . Q)) by d4, Th10;

      per cases ;

        suppose

         d8: (M2 . Q) = 0 & (M1 . P) = +infty ;

        then (( product_sigma_Measure (M1,M2)) . (F . n)) = 0 & ( X-vol ((F . n),M1)) = ( Xchi (Q,X2)) by A1, d4, d5, Th97;

        hence (( product_sigma_Measure (M1,M2)) . (F . n)) = ( Integral (M2,( X-vol ((F . n),M1)))) by d8, MEASUR10: 33;

      end;

        suppose (M2 . Q) = 0 & (M1 . P) <> +infty ;

        then ex r be Real st r = (M1 . P) & ( X-vol ((F . n),M1)) = (r (#) ( chi (Q,X2))) by A1, d4, Th97;

        hence (( product_sigma_Measure (M1,M2)) . (F . n)) = ( Integral (M2,( X-vol ((F . n),M1)))) by d5, Th98, SUPINF_2: 51;

      end;

        suppose

         d6: (M2 . Q) <> 0 & (M1 . P) = +infty ;

        (M2 . Q) >= 0 by SUPINF_2: 51;

        then

         d7: (( product_sigma_Measure (M1,M2)) . (F . n)) = +infty by d5, d6, XXREAL_3:def 5;

        ( X-vol ((F . n),M1)) = ( Xchi (Q,X2)) by A1, d4, d6, Th97;

        hence (( product_sigma_Measure (M1,M2)) . (F . n)) = ( Integral (M2,( X-vol ((F . n),M1)))) by d7, d6, MEASUR10: 33;

      end;

        suppose (M2 . Q) <> 0 & (M1 . P) <> +infty ;

        then ex r be Real st r = (M1 . P) & ( X-vol ((F . n),M1)) = (r (#) ( chi (Q,X2))) by A1, d4, Th97;

        hence (( product_sigma_Measure (M1,M2)) . (F . n)) = ( Integral (M2,( X-vol ((F . n),M1)))) by d5, Th98, SUPINF_2: 51;

      end;

    end;

    theorem :: MEASUR11:102

    

     Th102: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F be disjoint_valued FinSequence of ( sigma ( measurable_rectangles (S1,S2))), n be Nat st M2 is sigma_finite & F is FinSequence of ( measurable_rectangles (S1,S2)) holds (( product_sigma_Measure (M1,M2)) . ( Union F)) = ( Integral (M1,( Y-vol (( Union F),M2))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F be disjoint_valued FinSequence of ( sigma ( measurable_rectangles (S1,S2))), n be Nat;

      assume that

       A1: M2 is sigma_finite and

       A2: F is FinSequence of ( measurable_rectangles (S1,S2));

      

       A3: (F | ( len F)) = F by FINSEQ_1: 58;

      defpred P[ Nat] means (( product_sigma_Measure (M1,M2)) . ( Union (F | $1))) = ( Integral (M1,( Y-vol (( Union (F | $1)),M2))));

      ( union ( rng (F | 0 ))) = {} by ZFMISC_1: 2;

      then

       A4: ( Union (F | 0 )) = {} by CARD_3:def 4;

       not 0 in ( dom F) by FINSEQ_3: 24;

      then (F . 0 ) = {} by FUNCT_1:def 2;

      then

       P1: P[ 0 ] by A4, A1, A2, Th99;

      

       P2: for k be Nat st P[k] holds P[(k + 1)]

      proof

        let k be Nat;

        assume

         P3: P[k];

        

         A6: k <= (k + 1) by NAT_1: 13;

        per cases ;

          suppose ( len F) >= (k + 1);

          then ( len (F | (k + 1))) = (k + 1) by FINSEQ_1: 59;

          

          then (F | (k + 1)) = (((F | (k + 1)) | k) ^ <*((F | (k + 1)) . (k + 1))*>) by FINSEQ_3: 55

          .= ((F | k) ^ <*((F | (k + 1)) . (k + 1))*>) by A6, FINSEQ_1: 82

          .= ((F | k) ^ <*(F . (k + 1))*>) by FINSEQ_3: 112;

          

          then ( rng (F | (k + 1))) = (( rng (F | k)) \/ ( rng <*(F . (k + 1))*>)) by FINSEQ_1: 31

          .= (( rng (F | k)) \/ {(F . (k + 1))}) by FINSEQ_1: 38;

          

          then ( union ( rng (F | (k + 1)))) = (( union ( rng (F | k))) \/ ( union {(F . (k + 1))})) by ZFMISC_1: 78

          .= (( union ( rng (F | k))) \/ (F . (k + 1))) by ZFMISC_1: 25

          .= (( Union (F | k)) \/ (F . (k + 1))) by CARD_3:def 4;

          then

           A8: ( Union (F | (k + 1))) = (( Union (F | k)) \/ (F . (k + 1))) by CARD_3:def 4;

          

          then (( product_sigma_Measure (M1,M2)) . ( Union (F | (k + 1)))) = ((( product_sigma_Measure (M1,M2)) . ( Union (F | k))) + (( product_sigma_Measure (M1,M2)) . (F . (k + 1)))) by Th101, Th12

          .= (( Integral (M1,( Y-vol (( Union (F | k)),M2)))) + ( Integral (M1,( Y-vol ((F . (k + 1)),M2))))) by A1, A2, P3, Th99;

          hence P[(k + 1)] by A1, A8, Th101, Th94;

        end;

          suppose ( len F) < (k + 1);

          then (F | (k + 1)) = F & (F | k) = F by FINSEQ_3: 49, NAT_1: 13;

          hence P[(k + 1)] by P3;

        end;

      end;

      for k be Nat holds P[k] from NAT_1:sch 2( P1, P2);

      hence thesis by A3;

    end;

    theorem :: MEASUR11:103

    

     Th103: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F be disjoint_valued FinSequence of ( sigma ( measurable_rectangles (S1,S2))), n be Nat st M1 is sigma_finite & F is FinSequence of ( measurable_rectangles (S1,S2)) holds (( product_sigma_Measure (M1,M2)) . ( Union F)) = ( Integral (M2,( X-vol (( Union F),M1))))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, F be disjoint_valued FinSequence of ( sigma ( measurable_rectangles (S1,S2))), n be Nat;

      assume that

       A1: M1 is sigma_finite and

       A2: F is FinSequence of ( measurable_rectangles (S1,S2));

      

       A3: (F | ( len F)) = F by FINSEQ_1: 58;

      defpred P[ Nat] means (( product_sigma_Measure (M1,M2)) . ( Union (F | $1))) = ( Integral (M2,( X-vol (( Union (F | $1)),M1))));

      ( union ( rng (F | 0 ))) = {} by ZFMISC_1: 2;

      then

       A4: ( Union (F | 0 )) = {} by CARD_3:def 4;

       not 0 in ( dom F) by FINSEQ_3: 24;

      then (F . 0 ) = {} by FUNCT_1:def 2;

      then

       P1: P[ 0 ] by A4, A1, A2, Th100;

      

       P2: for k be Nat st P[k] holds P[(k + 1)]

      proof

        let k be Nat;

        assume

         P3: P[k];

        

         A6: k <= (k + 1) by NAT_1: 13;

        per cases ;

          suppose ( len F) >= (k + 1);

          then ( len (F | (k + 1))) = (k + 1) by FINSEQ_1: 59;

          

          then (F | (k + 1)) = (((F | (k + 1)) | k) ^ <*((F | (k + 1)) . (k + 1))*>) by FINSEQ_3: 55

          .= ((F | k) ^ <*((F | (k + 1)) . (k + 1))*>) by A6, FINSEQ_1: 82

          .= ((F | k) ^ <*(F . (k + 1))*>) by FINSEQ_3: 112;

          

          then ( rng (F | (k + 1))) = (( rng (F | k)) \/ ( rng <*(F . (k + 1))*>)) by FINSEQ_1: 31

          .= (( rng (F | k)) \/ {(F . (k + 1))}) by FINSEQ_1: 38;

          

          then ( union ( rng (F | (k + 1)))) = (( union ( rng (F | k))) \/ ( union {(F . (k + 1))})) by ZFMISC_1: 78

          .= (( union ( rng (F | k))) \/ (F . (k + 1))) by ZFMISC_1: 25

          .= (( Union (F | k)) \/ (F . (k + 1))) by CARD_3:def 4;

          then

           A8: ( Union (F | (k + 1))) = (( Union (F | k)) \/ (F . (k + 1))) by CARD_3:def 4;

          

          then (( product_sigma_Measure (M1,M2)) . ( Union (F | (k + 1)))) = ((( product_sigma_Measure (M1,M2)) . ( Union (F | k))) + (( product_sigma_Measure (M1,M2)) . (F . (k + 1)))) by Th101, Th12

          .= (( Integral (M2,( X-vol (( Union (F | k)),M1)))) + ( Integral (M2,( X-vol ((F . (k + 1)),M1))))) by A1, A2, P3, Th100;

          hence P[(k + 1)] by A1, A8, Th101, Th95;

        end;

          suppose ( len F) < (k + 1);

          then (F | (k + 1)) = F & (F | k) = F by FINSEQ_3: 49, NAT_1: 13;

          hence P[(k + 1)] by P3;

        end;

      end;

      for k be Nat holds P[k] from NAT_1:sch 2( P1, P2);

      hence thesis by A3;

    end;

    theorem :: MEASUR11:104

    

     Th104: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))) st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) & M2 is sigma_finite holds for V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2 st V = [:A, B:] holds E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M1,( Y-vol ((E /\ V),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) }

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume that

       A1: E in ( Field_generated_by ( measurable_rectangles (S1,S2))) and

       A2: M2 is sigma_finite;

      let V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2;

      assume

       A3: V = [:A, B:];

      V in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by A3;

      then

       A5: V in ( measurable_rectangles (S1,S2)) by MEASUR10:def 5;

      ( measurable_rectangles (S1,S2)) c= ( Field_generated_by ( measurable_rectangles (S1,S2))) by SRINGS_3: 21;

      then

       A6: (E /\ V) in ( Field_generated_by ( measurable_rectangles (S1,S2))) by A1, A5, FINSUB_1:def 2;

      reconsider XX1 = X1 as Element of S1 by MEASURE1: 7;

      (E /\ V) in ( DisUnion ( measurable_rectangles (S1,S2))) by A6, SRINGS_3: 22;

      then (E /\ V) in { W where W be Subset of [:X1, X2:] : ex G be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st W = ( Union G) } by SRINGS_3:def 3;

      then

      consider W be Subset of [:X1, X2:] such that

       A11: (E /\ V) = W & ex G be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st W = ( Union G);

      consider G be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) such that

       A12: (E /\ V) = ( Union G) by A11;

      

       A13: G in (( measurable_rectangles (S1,S2)) * ) by FINSEQ_1:def 11;

      ( measurable_rectangles (S1,S2)) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

      then (( measurable_rectangles (S1,S2)) * ) c= (( sigma ( measurable_rectangles (S1,S2))) * ) by FINSEQ_1: 62;

      then

      reconsider G as disjoint_valued FinSequence of ( sigma ( measurable_rectangles (S1,S2))) by A13, FINSEQ_1:def 11;

      ( Integral (M1,( Y-vol (( Union G),M2)))) = (( product_sigma_Measure (M1,M2)) . ( Union G)) by A2, Th102;

      hence E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M1,( Y-vol ((E /\ V),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) } by A12;

    end;

    theorem :: MEASUR11:105

    

     Th105: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))) st E in ( Field_generated_by ( measurable_rectangles (S1,S2))) & M1 is sigma_finite holds for V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2 st V = [:A, B:] holds E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M2,( X-vol ((E /\ V),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) }

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume that

       A1: E in ( Field_generated_by ( measurable_rectangles (S1,S2))) and

       A2: M1 is sigma_finite;

      let V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2;

      assume V = [:A, B:];

      then V in the set of all [:A, B:] where A be Element of S1, B be Element of S2;

      then

       A5: V in ( measurable_rectangles (S1,S2)) by MEASUR10:def 5;

      ( measurable_rectangles (S1,S2)) c= ( Field_generated_by ( measurable_rectangles (S1,S2))) by SRINGS_3: 21;

      then

       A6: (E /\ V) in ( Field_generated_by ( measurable_rectangles (S1,S2))) by A1, A5, FINSUB_1:def 2;

      reconsider XX2 = X2 as Element of S2 by MEASURE1: 7;

      (E /\ V) in ( DisUnion ( measurable_rectangles (S1,S2))) by A6, SRINGS_3: 22;

      then (E /\ V) in { W where W be Subset of [:X1, X2:] : ex G be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st W = ( Union G) } by SRINGS_3:def 3;

      then

      consider W be Subset of [:X1, X2:] such that

       A11: (E /\ V) = W & ex G be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) st W = ( Union G);

      consider G be disjoint_valued FinSequence of ( measurable_rectangles (S1,S2)) such that

       A12: (E /\ V) = ( Union G) by A11;

      

       A13: G in (( measurable_rectangles (S1,S2)) * ) by FINSEQ_1:def 11;

      ( measurable_rectangles (S1,S2)) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

      then (( measurable_rectangles (S1,S2)) * ) c= (( sigma ( measurable_rectangles (S1,S2))) * ) by FINSEQ_1: 62;

      then

      reconsider G as disjoint_valued FinSequence of ( sigma ( measurable_rectangles (S1,S2))) by A13, FINSEQ_1:def 11;

      ( Integral (M2,( X-vol (( Union G),M1)))) = (( product_sigma_Measure (M1,M2)) . ( Union G)) by A2, Th103;

      hence E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M2,( X-vol ((E /\ V),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) } by A12;

    end;

    theorem :: MEASUR11:106

    

     Th106: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2 st M2 is sigma_finite & V = [:A, B:] holds ( Field_generated_by ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M1,( Y-vol ((E /\ V),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) }

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2;

      assume

       A1: M2 is sigma_finite & V = [:A, B:];

      let E be object;

      assume

       A2: E in ( Field_generated_by ( measurable_rectangles (S1,S2)));

      ( sigma ( measurable_rectangles (S1,S2))) = ( sigma ( DisUnion ( measurable_rectangles (S1,S2)))) by Th1

      .= ( sigma ( Field_generated_by ( measurable_rectangles (S1,S2)))) by SRINGS_3: 22;

      then ( Field_generated_by ( measurable_rectangles (S1,S2))) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

      then

      reconsider E1 = E as Element of ( sigma ( measurable_rectangles (S1,S2))) by A2;

      E1 in ( Field_generated_by ( measurable_rectangles (S1,S2))) by A2;

      hence thesis by A1, Th104;

    end;

    theorem :: MEASUR11:107

    

     Th107: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2 st M1 is sigma_finite & V = [:A, B:] holds ( Field_generated_by ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M2,( X-vol ((E /\ V),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) }

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2;

      assume

       A1: M1 is sigma_finite & V = [:A, B:];

      let E be object;

      assume

       A2: E in ( Field_generated_by ( measurable_rectangles (S1,S2)));

      ( sigma ( measurable_rectangles (S1,S2))) = ( sigma ( DisUnion ( measurable_rectangles (S1,S2)))) by Th1

      .= ( sigma ( Field_generated_by ( measurable_rectangles (S1,S2)))) by SRINGS_3: 22;

      then ( Field_generated_by ( measurable_rectangles (S1,S2))) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

      then

      reconsider E1 = E as Element of ( sigma ( measurable_rectangles (S1,S2))) by A2;

      E1 in ( Field_generated_by ( measurable_rectangles (S1,S2))) by A2;

      hence thesis by A1, Th105;

    end;

    theorem :: MEASUR11:108

    

     Th108: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E,V be Element of ( sigma ( measurable_rectangles (S1,S2))), P be Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))), x be Element of X1 st P is non-descending & ( lim P) = E holds ex K be SetSequence of S2 st K is non-descending & (for n be Nat holds (K . n) = (( Measurable-X-section ((P . n),x)) /\ ( Measurable-X-section (V,x)))) & ( lim K) = (( Measurable-X-section (E,x)) /\ ( Measurable-X-section (V,x)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E,V be Element of ( sigma ( measurable_rectangles (S1,S2))), P be Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))), x be Element of X1;

      assume that

       A1: P is non-descending and

       A2: ( lim P) = E;

      

       A4: for n be Nat holds (P . n) in ( sigma ( measurable_rectangles (S1,S2)));

      reconsider P1 = P as SetSequence of [:X1, X2:];

      consider G be SetSequence of X2 such that

       A5: G is non-descending & (for n be Nat holds (G . n) = ( X-section ((P1 . n),x))) by A1, Th37;

      for n be Nat holds (G . n) in S2

      proof

        let n be Nat;

        (P1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A4;

        then ( X-section ((P1 . n),x)) in S2 by Th44;

        hence (G . n) in S2 by A5;

      end;

      then

      reconsider G as Set_Sequence of S2 by MEASURE8:def 2;

      set K = (( Measurable-X-section (V,x)) (/\) G);

      

       A6: G is convergent & ( lim G) = ( Union G) by A5, SETLIM_1: 63;

      ( union ( rng G)) = ( X-section (( union ( rng P)),x)) by A5, Th24;

      

      then

       A7: ( Union G) = ( X-section (( union ( rng P)),x)) by CARD_3:def 4

      .= ( X-section (( Union P),x)) by CARD_3:def 4

      .= ( Measurable-X-section (E,x)) by A1, A2, SETLIM_1: 63;

      

       A8: ( dom K) = NAT by FUNCT_2:def 1;

      for n be object st n in NAT holds (K . n) in S2

      proof

        let n be object;

        assume n in NAT ;

        then

        reconsider n1 = n as Element of NAT ;

        (K . n1) = ((G . n1) /\ ( Measurable-X-section (V,x))) by SETLIM_2:def 5;

        then (K . n1) = (( Measurable-X-section ((P . n1),x)) /\ ( Measurable-X-section (V,x))) by A5;

        hence (K . n) in S2;

      end;

      then

      reconsider K as SetSequence of S2 by A8, FUNCT_2: 3;

      

       A9: for n be Nat holds (K . n) = (( Measurable-X-section ((P . n),x)) /\ ( Measurable-X-section (V,x)))

      proof

        let n be Nat;

        (K . n) = ((G . n) /\ ( Measurable-X-section (V,x))) by SETLIM_2:def 5;

        hence (K . n) = (( Measurable-X-section ((P . n),x)) /\ ( Measurable-X-section (V,x))) by A5;

      end;

      take K;

      thus thesis by A9, A7, A6, A5, SETLIM_2: 22, SETLIM_2: 92;

    end;

    theorem :: MEASUR11:109

    

     Th109: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E,V be Element of ( sigma ( measurable_rectangles (S1,S2))), P be Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))), y be Element of X2 st P is non-descending & ( lim P) = E holds ex K be SetSequence of S1 st K is non-descending & (for n be Nat holds (K . n) = (( Measurable-Y-section ((P . n),y)) /\ ( Measurable-Y-section (V,y)))) & ( lim K) = (( Measurable-Y-section (E,y)) /\ ( Measurable-Y-section (V,y)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E,V be Element of ( sigma ( measurable_rectangles (S1,S2))), P be Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))), x be Element of X2;

      assume that

       A1: P is non-descending and

       A2: ( lim P) = E;

      

       A4: for n be Nat holds (P . n) in ( sigma ( measurable_rectangles (S1,S2)));

      reconsider P1 = P as SetSequence of [:X1, X2:];

      consider G be SetSequence of X1 such that

       A5: G is non-descending & (for n be Nat holds (G . n) = ( Y-section ((P1 . n),x))) by A1, Th38;

      for n be Nat holds (G . n) in S1

      proof

        let n be Nat;

        (P1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A4;

        then ( Y-section ((P1 . n),x)) in S1 by Th44;

        hence (G . n) in S1 by A5;

      end;

      then

      reconsider G as Set_Sequence of S1 by MEASURE8:def 2;

      set K = (( Measurable-Y-section (V,x)) (/\) G);

      

       A6: G is convergent & ( lim G) = ( Union G) by A5, SETLIM_1: 63;

      ( union ( rng G)) = ( Y-section (( union ( rng P)),x)) by A5, Th26;

      

      then

       A7: ( Union G) = ( Y-section (( union ( rng P)),x)) by CARD_3:def 4

      .= ( Y-section (( Union P),x)) by CARD_3:def 4

      .= ( Measurable-Y-section (E,x)) by A1, A2, SETLIM_1: 63;

      

       A8: ( dom K) = NAT by FUNCT_2:def 1;

      for n be object st n in NAT holds (K . n) in S1

      proof

        let n be object;

        assume n in NAT ;

        then

        reconsider n1 = n as Element of NAT ;

        (K . n1) = ((G . n1) /\ ( Measurable-Y-section (V,x))) by SETLIM_2:def 5;

        then (K . n1) = (( Measurable-Y-section ((P . n1),x)) /\ ( Measurable-Y-section (V,x))) by A5;

        hence (K . n) in S1;

      end;

      then

      reconsider K as SetSequence of S1 by A8, FUNCT_2: 3;

      

       A9: for n be Nat holds (K . n) = (( Measurable-Y-section ((P . n),x)) /\ ( Measurable-Y-section (V,x)))

      proof

        let n be Nat;

        (K . n) = ((G . n) /\ ( Measurable-Y-section (V,x))) by SETLIM_2:def 5;

        hence (K . n) = (( Measurable-Y-section ((P . n),x)) /\ ( Measurable-Y-section (V,x))) by A5;

      end;

      take K;

      thus thesis by A9, A7, A5, A6, SETLIM_2: 22, SETLIM_2: 92;

    end;

    theorem :: MEASUR11:110

    

     Th110: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E,V be Element of ( sigma ( measurable_rectangles (S1,S2))), P be Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))), x be Element of X1 st P is non-ascending & ( lim P) = E holds ex K be SetSequence of S2 st K is non-ascending & (for n be Nat holds (K . n) = (( Measurable-X-section ((P . n),x)) /\ ( Measurable-X-section (V,x)))) & ( lim K) = (( Measurable-X-section (E,x)) /\ ( Measurable-X-section (V,x)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M2 be sigma_Measure of S2, E,V be Element of ( sigma ( measurable_rectangles (S1,S2))), P be Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))), x be Element of X1;

      assume that

       A1: P is non-ascending and

       A2: ( lim P) = E;

      

       A4: for n be Nat holds (P . n) in ( sigma ( measurable_rectangles (S1,S2)));

      reconsider P1 = P as SetSequence of [:X1, X2:];

      consider G be SetSequence of X2 such that

       A5: G is non-ascending & (for n be Nat holds (G . n) = ( X-section ((P1 . n),x))) by A1, Th39;

      for n be Nat holds (G . n) in S2

      proof

        let n be Nat;

        (P1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A4;

        then ( X-section ((P1 . n),x)) in S2 by Th44;

        hence (G . n) in S2 by A5;

      end;

      then

      reconsider G as Set_Sequence of S2 by MEASURE8:def 2;

      set K = (( Measurable-X-section (V,x)) (/\) G);

      

       A6: G is convergent & ( lim G) = ( Intersection G) by A5, SETLIM_1: 64;

      ( meet ( rng G)) = ( X-section (( meet ( rng P)),x)) by A5, Th25;

      

      then

       A7: ( Intersection G) = ( X-section (( meet ( rng P)),x)) by SETLIM_1: 8

      .= ( X-section (( Intersection P),x)) by SETLIM_1: 8

      .= ( Measurable-X-section (E,x)) by A1, A2, SETLIM_1: 64;

      

       A8: ( dom K) = NAT by FUNCT_2:def 1;

      for n be object st n in NAT holds (K . n) in S2

      proof

        let n be object;

        assume n in NAT ;

        then

        reconsider n1 = n as Element of NAT ;

        (K . n1) = ((G . n1) /\ ( Measurable-X-section (V,x))) by SETLIM_2:def 5;

        then (K . n1) = (( Measurable-X-section ((P . n1),x)) /\ ( Measurable-X-section (V,x))) by A5;

        hence (K . n) in S2;

      end;

      then

      reconsider K as SetSequence of S2 by A8, FUNCT_2: 3;

      

       A9: for n be Nat holds (K . n) = (( Measurable-X-section ((P . n),x)) /\ ( Measurable-X-section (V,x)))

      proof

        let n be Nat;

        (K . n) = ((G . n) /\ ( Measurable-X-section (V,x))) by SETLIM_2:def 5;

        hence (K . n) = (( Measurable-X-section ((P . n),x)) /\ ( Measurable-X-section (V,x))) by A5;

      end;

      take K;

      thus thesis by A9, A7, A6, A5, SETLIM_2: 21, SETLIM_2: 92;

    end;

    theorem :: MEASUR11:111

    

     Th111: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E,V be Element of ( sigma ( measurable_rectangles (S1,S2))), P be Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))), y be Element of X2 st P is non-ascending & ( lim P) = E holds ex K be SetSequence of S1 st K is non-ascending & (for n be Nat holds (K . n) = (( Measurable-Y-section ((P . n),y)) /\ ( Measurable-Y-section (V,y)))) & ( lim K) = (( Measurable-Y-section (E,y)) /\ ( Measurable-Y-section (V,y)))

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, E,V be Element of ( sigma ( measurable_rectangles (S1,S2))), P be Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))), x be Element of X2;

      assume that

       A1: P is non-ascending and

       A2: ( lim P) = E;

      

       A4: for n be Nat holds (P . n) in ( sigma ( measurable_rectangles (S1,S2)));

      reconsider P1 = P as SetSequence of [:X1, X2:];

      consider G be SetSequence of X1 such that

       A5: G is non-ascending & (for n be Nat holds (G . n) = ( Y-section ((P1 . n),x))) by A1, Th40;

      for n be Nat holds (G . n) in S1

      proof

        let n be Nat;

        (P1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A4;

        then ( Y-section ((P1 . n),x)) in S1 by Th44;

        hence (G . n) in S1 by A5;

      end;

      then

      reconsider G as Set_Sequence of S1 by MEASURE8:def 2;

      set K = (( Measurable-Y-section (V,x)) (/\) G);

      

       A6: G is convergent & ( lim G) = ( Intersection G) by A5, SETLIM_1: 64;

      ( meet ( rng G)) = ( Y-section (( meet ( rng P)),x)) by A5, Th27;

      

      then

       A7: ( Intersection G) = ( Y-section (( meet ( rng P)),x)) by SETLIM_1: 8

      .= ( Y-section (( Intersection P),x)) by SETLIM_1: 8

      .= ( Measurable-Y-section (E,x)) by A1, A2, SETLIM_1: 64;

      

       A8: ( dom K) = NAT by FUNCT_2:def 1;

      for n be object st n in NAT holds (K . n) in S1

      proof

        let n be object;

        assume n in NAT ;

        then

        reconsider n1 = n as Element of NAT ;

        (K . n1) = ((G . n1) /\ ( Measurable-Y-section (V,x))) by SETLIM_2:def 5;

        then (K . n1) = (( Measurable-Y-section ((P . n1),x)) /\ ( Measurable-Y-section (V,x))) by A5;

        hence (K . n) in S1;

      end;

      then

      reconsider K as SetSequence of S1 by A8, FUNCT_2: 3;

      

       A9: for n be Nat holds (K . n) = (( Measurable-Y-section ((P . n),x)) /\ ( Measurable-Y-section (V,x)))

      proof

        let n be Nat;

        (K . n) = ((G . n) /\ ( Measurable-Y-section (V,x))) by SETLIM_2:def 5;

        hence (K . n) = (( Measurable-Y-section ((P . n),x)) /\ ( Measurable-Y-section (V,x))) by A5;

      end;

      take K;

      thus thesis by A9, A7, A6, A5, SETLIM_2: 21, SETLIM_2: 92;

    end;

    theorem :: MEASUR11:112

    

     Th112: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2 st M2 is sigma_finite & V = [:A, B:] & (( product_sigma_Measure (M1,M2)) . V) < +infty & (M2 . B) < +infty holds { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M1,( Y-vol ((E /\ V),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) } is MonotoneClass of [:X1, X2:]

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2;

      assume that

       A01: M2 is sigma_finite and

       A02: V = [:A, B:] and

       A0: (( product_sigma_Measure (M1,M2)) . V) < +infty and

       PS2: (M2 . B) < +infty ;

      set Z = { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M1,( Y-vol ((E /\ V),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) };

      reconsider XX1 = X1 as Element of S1 by MEASURE1: 7;

      now

        let A be object;

        assume A in Z;

        then ex E be Element of ( sigma ( measurable_rectangles (S1,S2))) st A = E & ( Integral (M1,( Y-vol ((E /\ V),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V));

        hence A in ( bool [:X1, X2:]);

      end;

      then

       A1: Z c= ( bool [:X1, X2:]);

      for A1 be SetSequence of [:X1, X2:] st A1 is monotone & ( rng A1) c= Z holds ( lim A1) in Z

      proof

        let A1 be SetSequence of [:X1, X2:];

        assume

         A2: A1 is monotone & ( rng A1) c= Z;

        

         A4: for V be set st V in ( rng A1) holds V in ( sigma ( measurable_rectangles (S1,S2)))

        proof

          let W be set;

          assume W in ( rng A1);

          then W in Z by A2;

          then ex E be Element of ( sigma ( measurable_rectangles (S1,S2))) st W = E & ( Integral (M1,( Y-vol ((E /\ V),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V));

          hence W in ( sigma ( measurable_rectangles (S1,S2)));

        end;

        for n be Nat holds (A1 . n) in ( sigma ( measurable_rectangles (S1,S2)))

        proof

          let n be Nat;

          ( dom A1) = NAT by FUNCT_2:def 1;

          then n in ( dom A1) by ORDINAL1:def 12;

          hence (A1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A4, FUNCT_1: 3;

        end;

        then

        reconsider A2 = A1 as Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))) by MEASURE8:def 2;

        

         PP: for n be Nat holds ( Integral (M1,( Y-vol (((A2 . n) /\ V),M2)))) = (( product_sigma_Measure (M1,M2)) . ((A2 . n) /\ V))

        proof

          let n be Nat;

          ( dom A2) = NAT by FUNCT_2:def 1;

          then n in ( dom A2) by ORDINAL1:def 12;

          then (A2 . n) in Z by A2, FUNCT_1: 3;

          then ex E be Element of ( sigma ( measurable_rectangles (S1,S2))) st (A2 . n) = E & ( Integral (M1,( Y-vol ((E /\ V),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V));

          hence thesis;

        end;

        per cases by A2, SETLIM_1:def 1;

          suppose

           A3: A1 is non-descending;

          ( union ( rng A1)) in ( sigma ( measurable_rectangles (S1,S2))) by A4, MEASURE1: 35;

          then ( Union A1) in ( sigma ( measurable_rectangles (S1,S2))) by CARD_3:def 4;

          then

          reconsider E = ( lim A1) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, SETLIM_1: 63;

          defpred P[ Element of NAT , object] means $2 = ( Y-vol (((A2 . $1) /\ V),M2));

          

           T1: for n be Element of NAT holds ex f be Element of ( PFuncs (X1, ExtREAL )) st P[n, f]

          proof

            let n be Element of NAT ;

            reconsider f1 = ( Y-vol (((A2 . n) /\ V),M2)) as Element of ( PFuncs (X1, ExtREAL )) by PARTFUN1: 45;

            take f1;

            thus f1 = ( Y-vol (((A2 . n) /\ V),M2));

          end;

          consider F be Function of NAT , ( PFuncs (X1, ExtREAL )) such that

           T2: for n be Element of NAT holds P[n, (F . n)] from FUNCT_2:sch 3( T1);

          reconsider F as Functional_Sequence of X1, ExtREAL ;

          

           T2a: for n be Nat holds (F . n) = ( Y-vol (((A2 . n) /\ V),M2))

          proof

            let n be Nat;

            n is Element of NAT by ORDINAL1:def 12;

            hence thesis by T2;

          end;

          (F . 0 ) = ( Y-vol (((A2 . 0 ) /\ V),M2)) by T2;

          then

           T3: ( dom (F . 0 )) = XX1 & (F . 0 ) is nonnegative by FUNCT_2:def 1;

          

           T4: for n be Nat, x be Element of X1 holds ((F # x) . n) = (( Y-vol (((A2 . n) /\ V),M2)) . x)

          proof

            let n be Nat, x be Element of X1;

            ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13;

            hence ((F # x) . n) = (( Y-vol (((A2 . n) /\ V),M2)) . x) by T2a;

          end;

          

           T5: for n,m be Nat holds ( dom (F . n)) = ( dom (F . m))

          proof

            let n,m be Nat;

            (F . n) = ( Y-vol (((A2 . n) /\ V),M2)) & (F . m) = ( Y-vol (((A2 . m) /\ V),M2)) by T2a;

            then ( dom (F . n)) = XX1 & ( dom (F . m)) = XX1 by FUNCT_2:def 1;

            hence ( dom (F . n)) = ( dom (F . m));

          end;

          

           T6: for n be Nat holds (F . n) is XX1 -measurable

          proof

            let n be Nat;

            (F . n) = ( Y-vol (((A2 . n) /\ V),M2)) by T2a;

            hence (F . n) is XX1 -measurable by A01, DefYvol;

          end;

          

           T7: for n,m be Nat st n <= m holds for x be Element of X1 st x in XX1 holds ((F . n) . x) <= ((F . m) . x)

          proof

            let n,m be Nat;

            assume

             T71: n <= m;

            hereby

              let x be Element of X1;

              assume x in XX1;

              

               T72: ((A2 . n) /\ V) c= ((A2 . m) /\ V) by A3, T71, PROB_1:def 5, XBOOLE_1: 26;

              

               T73: (M2 . ( Measurable-X-section (((A2 . n) /\ V),x))) = (( Y-vol (((A2 . n) /\ V),M2)) . x) by A01, DefYvol

              .= ((F # x) . n) by T4

              .= ((F . n) . x) by MESFUNC5:def 13;

              (M2 . ( Measurable-X-section (((A2 . m) /\ V),x))) = (( Y-vol (((A2 . m) /\ V),M2)) . x) by A01, DefYvol

              .= ((F # x) . m) by T4

              .= ((F . m) . x) by MESFUNC5:def 13;

              hence ((F . n) . x) <= ((F . m) . x) by T72, T73, Th14, MEASURE1: 31;

            end;

          end;

          

           T8: for x be Element of X1 st x in XX1 holds (F # x) is convergent

          proof

            let x be Element of X1;

            assume x in XX1;

            now

              let n,m be Nat;

              assume m <= n;

              then ((F . m) . x) <= ((F . n) . x) by T7;

              then ((F # x) . m) <= ((F . n) . x) by MESFUNC5:def 13;

              hence ((F # x) . m) <= ((F # x) . n) by MESFUNC5:def 13;

            end;

            then (F # x) is non-decreasing by RINFSUP2: 7;

            hence (F # x) is convergent by RINFSUP2: 37;

          end;

          consider I be ExtREAL_sequence such that

           V2: (for n be Nat holds (I . n) = ( Integral (M1,(F . n)))) & I is convergent & ( Integral (M1,( lim F))) = ( lim I) by T3, T5, T6, T7, T8, MESFUNC8:def 2, MESFUNC9: 52;

          ( dom ( lim F)) = ( dom (F . 0 )) by MESFUNC8:def 9;

          then

           V4: ( dom ( lim F)) = ( dom ( Y-vol ((E /\ V),M2))) by T3, FUNCT_2:def 1;

          for x be Element of X1 st x in ( dom ( lim F)) holds (( lim F) . x) = (( Y-vol ((E /\ V),M2)) . x)

          proof

            let x be Element of X1;

            assume x in ( dom ( lim F));

            then

             L2: (( lim F) . x) = ( lim (F # x)) by MESFUNC8:def 9;

            consider G be SetSequence of S2 such that

             L3: G is non-descending & (for n be Nat holds (G . n) = (( Measurable-X-section ((A2 . n),x)) /\ ( Measurable-X-section (V,x)))) & ( lim G) = (( Measurable-X-section (E,x)) /\ ( Measurable-X-section (V,x))) by A3, Th108;

            for n be Element of NAT holds ((F # x) . n) = ((M2 * G) . n)

            proof

              let n be Element of NAT ;

              

               L5: ( dom G) = NAT by FUNCT_2:def 1;

              

               L4: ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13

              .= (( Y-vol (((A2 . n) /\ V),M2)) . x) by T2

              .= (M2 . ( Measurable-X-section (((A2 . n) /\ V),x))) by A01, DefYvol;

              ( Measurable-X-section (((A2 . n) /\ V),x)) = (( Measurable-X-section ((A2 . n),x)) /\ ( Measurable-X-section (V,x))) by Th21;

              then ( Measurable-X-section (((A2 . n) /\ V),x)) = (G . n) by L3;

              hence ((F # x) . n) = ((M2 * G) . n) by L4, L5, FUNCT_1: 13;

            end;

            then (F # x) = (M2 * G) by FUNCT_2: 63;

            then (( lim F) . x) = (M2 . (( Measurable-X-section (E,x)) /\ ( Measurable-X-section (V,x)))) by L2, L3, MEASURE8: 26;

            then (( lim F) . x) = (M2 . ( Measurable-X-section ((E /\ V),x))) by Th21;

            hence (( lim F) . x) = (( Y-vol ((E /\ V),M2)) . x) by A01, DefYvol;

          end;

          then

           V3: ( lim F) = ( Y-vol ((E /\ V),M2)) by V4, PARTFUN1: 5;

          set J = (V (/\) A2);

          

           E1: ( dom J) = NAT by FUNCT_2:def 1;

          for n be object st n in NAT holds (J . n) in ( sigma ( measurable_rectangles (S1,S2)))

          proof

            let n be object;

            assume n in NAT ;

            then

            reconsider n1 = n as Element of NAT ;

            (J . n) = ((A2 . n1) /\ V) by SETLIM_2:def 5;

            hence (J . n) in ( sigma ( measurable_rectangles (S1,S2)));

          end;

          then

          reconsider J as SetSequence of ( sigma ( measurable_rectangles (S1,S2))) by E1, FUNCT_2: 3;

          

           R11: J is non-descending by A3, SETLIM_2: 22;

          A2 is convergent by A3, SETLIM_1: 63;

          then

           R13: ( lim J) = (E /\ V) by SETLIM_2: 92;

          

           R3: ( product_sigma_Measure (M1,M2)) is sigma_Measure of ( sigma ( measurable_rectangles (S1,S2))) by Th2;

          then

           R4: ( dom ( product_sigma_Measure (M1,M2))) = ( sigma ( measurable_rectangles (S1,S2))) by FUNCT_2:def 1;

          ( rng J) c= ( sigma ( measurable_rectangles (S1,S2))) by RELAT_1:def 19;

          then

           R2: (( product_sigma_Measure (M1,M2)) /* J) = (( product_sigma_Measure (M1,M2)) * J) by R4, FUNCT_2:def 11;

          for n be Element of NAT holds (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n)

          proof

            let n be Element of NAT ;

            

             R21: ( dom J) = NAT by FUNCT_2:def 1;

            (I . n) = ( Integral (M1,(F . n))) by V2

            .= ( Integral (M1,( Y-vol (((A2 . n) /\ V),M2)))) by T2

            .= (( product_sigma_Measure (M1,M2)) . ((A2 . n) /\ V)) by PP

            .= (( product_sigma_Measure (M1,M2)) . (J . n)) by SETLIM_2:def 5;

            hence (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n) by R2, R21, FUNCT_1: 13;

          end;

          then ( lim I) = ( lim (( product_sigma_Measure (M1,M2)) /* J)) by FUNCT_2: 63;

          then ( lim I) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) by R13, R11, R2, R3, MEASURE8: 26;

          hence ( lim A1) in Z by V2, V3;

        end;

          suppose

           A3: A1 is non-ascending;

          ( meet ( rng A1)) in ( sigma ( measurable_rectangles (S1,S2))) by A4, MEASURE1: 35;

          then ( Intersection A1) in ( sigma ( measurable_rectangles (S1,S2))) by SETLIM_1: 8;

          then

          reconsider E = ( lim A1) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, SETLIM_1: 64;

          defpred P[ Element of NAT , object] means $2 = ( Y-vol (((A2 . $1) /\ V),M2));

          

           T1: for n be Element of NAT holds ex f be Element of ( PFuncs (X1, ExtREAL )) st P[n, f]

          proof

            let n be Element of NAT ;

            reconsider f1 = ( Y-vol (((A2 . n) /\ V),M2)) as Element of ( PFuncs (X1, ExtREAL )) by PARTFUN1: 45;

            take f1;

            thus f1 = ( Y-vol (((A2 . n) /\ V),M2));

          end;

          consider F be Function of NAT , ( PFuncs (X1, ExtREAL )) such that

           T2: for n be Element of NAT holds P[n, (F . n)] from FUNCT_2:sch 3( T1);

          reconsider F as Functional_Sequence of X1, ExtREAL ;

          

           T2a: for n be Nat holds (F . n) = ( Y-vol (((A2 . n) /\ V),M2))

          proof

            let n be Nat;

            n is Element of NAT by ORDINAL1:def 12;

            hence thesis by T2;

          end;

          (F . 0 ) = ( Y-vol (((A2 . 0 ) /\ V),M2)) by T2;

          then

           T3: ( dom (F . 0 )) = XX1 by FUNCT_2:def 1;

          

           T4: for n be Nat, x be Element of X1 holds ((F # x) . n) = (( Y-vol (((A2 . n) /\ V),M2)) . x)

          proof

            let n be Nat, x be Element of X1;

            ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13;

            hence ((F # x) . n) = (( Y-vol (((A2 . n) /\ V),M2)) . x) by T2a;

          end;

          for n,m be Nat holds ( dom (F . n)) = ( dom (F . m))

          proof

            let n,m be Nat;

            (F . n) = ( Y-vol (((A2 . n) /\ V),M2)) & (F . m) = ( Y-vol (((A2 . m) /\ V),M2)) by T2a;

            then ( dom (F . n)) = XX1 & ( dom (F . m)) = XX1 by FUNCT_2:def 1;

            hence ( dom (F . n)) = ( dom (F . m));

          end;

          then

          reconsider F as with_the_same_dom Functional_Sequence of X1, ExtREAL by MESFUNC8:def 2;

          

           T6: for n be Nat holds (F . n) is nonnegative & (F . n) is XX1 -measurable

          proof

            let n be Nat;

            (F . n) = ( Y-vol (((A2 . n) /\ V),M2)) by T2a;

            hence (F . n) is nonnegative & (F . n) is XX1 -measurable by A01, DefYvol;

          end;

          

           T7: for x be Element of X1, n,m be Nat st x in XX1 & n <= m holds ((F . n) . x) >= ((F . m) . x)

          proof

            let x be Element of X1, n,m be Nat;

            assume x in XX1 & n <= m;

            then

             T72: ((A2 . m) /\ V) c= ((A2 . n) /\ V) by A3, PROB_1:def 4, XBOOLE_1: 26;

            

             T73: (M2 . ( Measurable-X-section (((A2 . n) /\ V),x))) = (( Y-vol (((A2 . n) /\ V),M2)) . x) by A01, DefYvol

            .= ((F # x) . n) by T4

            .= ((F . n) . x) by MESFUNC5:def 13;

            (M2 . ( Measurable-X-section (((A2 . m) /\ V),x))) = (( Y-vol (((A2 . m) /\ V),M2)) . x) by A01, DefYvol

            .= ((F # x) . m) by T4

            .= ((F . m) . x) by MESFUNC5:def 13;

            hence ((F . m) . x) <= ((F . n) . x) by T72, T73, Th14, MEASURE1: 31;

          end;

          

           M3: ( product_sigma_Measure (M1,M2)) is sigma_Measure of ( sigma ( measurable_rectangles (S1,S2))) by Th2;

          ( Integral (M1,((F . 0 ) | XX1))) = ( Integral (M1,( Y-vol (((A2 . 0 ) /\ V),M2)))) by T2a;

          then

           M1: ( Integral (M1,((F . 0 ) | XX1))) = (( product_sigma_Measure (M1,M2)) . ((A2 . 0 ) /\ V)) by PP;

          (( product_sigma_Measure (M1,M2)) . ((A2 . 0 ) /\ V)) <= (( product_sigma_Measure (M1,M2)) . V) by M3, MEASURE1: 31, XBOOLE_1: 17;

          then ( Integral (M1,((F . 0 ) | XX1))) < +infty by A0, M1, XXREAL_0: 2;

          then

          consider I be ExtREAL_sequence such that

           V2: (for n be Nat holds (I . n) = ( Integral (M1,(F . n)))) & I is convergent & ( lim I) = ( Integral (M1,( lim F))) by T3, T6, T7, MESFUN10: 18;

          ( dom ( lim F)) = ( dom (F . 0 )) by MESFUNC8:def 9;

          then

           V4: ( dom ( lim F)) = ( dom ( Y-vol ((E /\ V),M2))) by T3, FUNCT_2:def 1;

          for x be Element of X1 st x in ( dom ( lim F)) holds (( lim F) . x) = (( Y-vol ((E /\ V),M2)) . x)

          proof

            let x be Element of X1;

            assume x in ( dom ( lim F));

            then

             L2: (( lim F) . x) = ( lim (F # x)) by MESFUNC8:def 9;

            consider G be SetSequence of S2 such that

             L3: G is non-ascending & (for n be Nat holds (G . n) = (( Measurable-X-section ((A2 . n),x)) /\ ( Measurable-X-section (V,x)))) & ( lim G) = (( Measurable-X-section (E,x)) /\ ( Measurable-X-section (V,x))) by A3, Th110;

            (G . 0 ) = (( Measurable-X-section ((A2 . 0 ),x)) /\ ( Measurable-X-section (V,x))) by L3;

            then

             L31: (M2 . (G . 0 )) <= (M2 . ( Measurable-X-section (V,x))) by MEASURE1: 31, XBOOLE_1: 17;

            ( Measurable-X-section (V,x)) c= B by A02, Th16;

            then (M2 . ( Measurable-X-section (V,x))) <= (M2 . B) by MEASURE1: 31;

            then (M2 . (G . 0 )) <= (M2 . B) by L31, XXREAL_0: 2;

            then

             LL: (M2 . (G . 0 )) < +infty by PS2, XXREAL_0: 2;

            for n be Element of NAT holds ((F # x) . n) = ((M2 * G) . n)

            proof

              let n be Element of NAT ;

              

               L5: ( dom G) = NAT by FUNCT_2:def 1;

              

               L4: ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13

              .= (( Y-vol (((A2 . n) /\ V),M2)) . x) by T2

              .= (M2 . ( Measurable-X-section (((A2 . n) /\ V),x))) by A01, DefYvol;

              ( Measurable-X-section (((A2 . n) /\ V),x)) = (( Measurable-X-section ((A2 . n),x)) /\ ( Measurable-X-section (V,x))) by Th21;

              then ( Measurable-X-section (((A2 . n) /\ V),x)) = (G . n) by L3;

              hence ((F # x) . n) = ((M2 * G) . n) by L4, L5, FUNCT_1: 13;

            end;

            then (F # x) = (M2 * G) by FUNCT_2: 63;

            then (( lim F) . x) = (M2 . (( Measurable-X-section (E,x)) /\ ( Measurable-X-section (V,x)))) by L2, L3, LL, MEASURE8: 31;

            then (( lim F) . x) = (M2 . ( Measurable-X-section ((E /\ V),x))) by Th21;

            hence (( lim F) . x) = (( Y-vol ((E /\ V),M2)) . x) by A01, DefYvol;

          end;

          then

           V3: ( lim F) = ( Y-vol ((E /\ V),M2)) by V4, PARTFUN1: 5;

          set J = (V (/\) A2);

          

           E1: ( dom J) = NAT by FUNCT_2:def 1;

          for n be object st n in NAT holds (J . n) in ( sigma ( measurable_rectangles (S1,S2)))

          proof

            let n be object;

            assume n in NAT ;

            then

            reconsider n1 = n as Element of NAT ;

            (J . n) = ((A2 . n1) /\ V) by SETLIM_2:def 5;

            hence (J . n) in ( sigma ( measurable_rectangles (S1,S2)));

          end;

          then

          reconsider J as SetSequence of ( sigma ( measurable_rectangles (S1,S2))) by E1, FUNCT_2: 3;

          

           R11: J is non-ascending by A3, SETLIM_2: 21;

          A2 is convergent by A3, SETLIM_1: 64;

          then

           R13: ( lim J) = (E /\ V) by SETLIM_2: 92;

          

           R3: ( product_sigma_Measure (M1,M2)) is sigma_Measure of ( sigma ( measurable_rectangles (S1,S2))) by Th2;

          then

           R4: ( dom ( product_sigma_Measure (M1,M2))) = ( sigma ( measurable_rectangles (S1,S2))) by FUNCT_2:def 1;

          ( rng J) c= ( sigma ( measurable_rectangles (S1,S2))) by RELAT_1:def 19;

          then

           R2: (( product_sigma_Measure (M1,M2)) /* J) = (( product_sigma_Measure (M1,M2)) * J) by R4, FUNCT_2:def 11;

          ((A2 . 0 ) /\ V) c= V by XBOOLE_1: 17;

          then (J . 0 ) c= V by SETLIM_2:def 5;

          then (( product_sigma_Measure (M1,M2)) . (J . 0 )) <= (( product_sigma_Measure (M1,M2)) . V) by R3, MEASURE1: 31;

          then

           K1: (( product_sigma_Measure (M1,M2)) . (J . 0 )) < +infty by A0, XXREAL_0: 2;

          for n be Element of NAT holds (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n)

          proof

            let n be Element of NAT ;

            

             R21: ( dom J) = NAT by FUNCT_2:def 1;

            (I . n) = ( Integral (M1,(F . n))) by V2

            .= ( Integral (M1,( Y-vol (((A2 . n) /\ V),M2)))) by T2

            .= (( product_sigma_Measure (M1,M2)) . ((A2 . n) /\ V)) by PP

            .= (( product_sigma_Measure (M1,M2)) . (J . n)) by SETLIM_2:def 5;

            hence (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n) by R2, R21, FUNCT_1: 13;

          end;

          then ( lim I) = ( lim (( product_sigma_Measure (M1,M2)) /* J)) by FUNCT_2: 63;

          then ( lim I) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) by R13, R11, R2, R3, K1, MEASURE8: 31;

          hence ( lim A1) in Z by V2, V3;

        end;

      end;

      hence thesis by A1, PROB_3: 69;

    end;

    theorem :: MEASUR11:113

    

     Th113: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2 st M1 is sigma_finite & V = [:A, B:] & (( product_sigma_Measure (M1,M2)) . V) < +infty & (M1 . A) < +infty holds { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M2,( X-vol ((E /\ V),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) } is MonotoneClass of [:X1, X2:]

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2;

      assume that

       A01: M1 is sigma_finite and

       A02: V = [:A, B:] and

       A0: (( product_sigma_Measure (M1,M2)) . V) < +infty and

       PS2: (M1 . A) < +infty ;

      set Z = { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M2,( X-vol ((E /\ V),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) };

      reconsider XX2 = X2 as Element of S2 by MEASURE1: 7;

      now

        let A be object;

        assume A in Z;

        then ex E be Element of ( sigma ( measurable_rectangles (S1,S2))) st A = E & ( Integral (M2,( X-vol ((E /\ V),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V));

        hence A in ( bool [:X1, X2:]);

      end;

      then

       A1: Z c= ( bool [:X1, X2:]);

      for A1 be SetSequence of [:X1, X2:] st A1 is monotone & ( rng A1) c= Z holds ( lim A1) in Z

      proof

        let A1 be SetSequence of [:X1, X2:];

        assume

         A2: A1 is monotone & ( rng A1) c= Z;

        

         A4: for V be set st V in ( rng A1) holds V in ( sigma ( measurable_rectangles (S1,S2)))

        proof

          let W be set;

          assume W in ( rng A1);

          then W in Z by A2;

          then ex E be Element of ( sigma ( measurable_rectangles (S1,S2))) st W = E & ( Integral (M2,( X-vol ((E /\ V),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V));

          hence W in ( sigma ( measurable_rectangles (S1,S2)));

        end;

        for n be Nat holds (A1 . n) in ( sigma ( measurable_rectangles (S1,S2)))

        proof

          let n be Nat;

          ( dom A1) = NAT by FUNCT_2:def 1;

          then n in ( dom A1) by ORDINAL1:def 12;

          hence (A1 . n) in ( sigma ( measurable_rectangles (S1,S2))) by A4, FUNCT_1: 3;

        end;

        then

        reconsider A2 = A1 as Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))) by MEASURE8:def 2;

        

         PP: for n be Nat holds ( Integral (M2,( X-vol (((A2 . n) /\ V),M1)))) = (( product_sigma_Measure (M1,M2)) . ((A2 . n) /\ V))

        proof

          let n be Nat;

          ( dom A2) = NAT by FUNCT_2:def 1;

          then n in ( dom A2) by ORDINAL1:def 12;

          then (A2 . n) in Z by A2, FUNCT_1: 3;

          then ex E be Element of ( sigma ( measurable_rectangles (S1,S2))) st (A2 . n) = E & ( Integral (M2,( X-vol ((E /\ V),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V));

          hence thesis;

        end;

        per cases by A2, SETLIM_1:def 1;

          suppose

           A3: A1 is non-descending;

          ( union ( rng A1)) in ( sigma ( measurable_rectangles (S1,S2))) by A4, MEASURE1: 35;

          then ( Union A1) in ( sigma ( measurable_rectangles (S1,S2))) by CARD_3:def 4;

          then

          reconsider E = ( lim A1) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, SETLIM_1: 63;

          defpred P[ Element of NAT , object] means $2 = ( X-vol (((A2 . $1) /\ V),M1));

          

           T1: for n be Element of NAT holds ex f be Element of ( PFuncs (X2, ExtREAL )) st P[n, f]

          proof

            let n be Element of NAT ;

            reconsider f1 = ( X-vol (((A2 . n) /\ V),M1)) as Element of ( PFuncs (X2, ExtREAL )) by PARTFUN1: 45;

            take f1;

            thus f1 = ( X-vol (((A2 . n) /\ V),M1));

          end;

          consider F be Function of NAT , ( PFuncs (X2, ExtREAL )) such that

           T2: for n be Element of NAT holds P[n, (F . n)] from FUNCT_2:sch 3( T1);

          reconsider F as Functional_Sequence of X2, ExtREAL ;

          

           T2a: for n be Nat holds (F . n) = ( X-vol (((A2 . n) /\ V),M1))

          proof

            let n be Nat;

            n is Element of NAT by ORDINAL1:def 12;

            hence thesis by T2;

          end;

          (F . 0 ) = ( X-vol (((A2 . 0 ) /\ V),M1)) by T2;

          then

           T3: ( dom (F . 0 )) = XX2 & (F . 0 ) is nonnegative by FUNCT_2:def 1;

          

           T4: for n be Nat, x be Element of X2 holds ((F # x) . n) = (( X-vol (((A2 . n) /\ V),M1)) . x)

          proof

            let n be Nat, x be Element of X2;

            ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13;

            hence ((F # x) . n) = (( X-vol (((A2 . n) /\ V),M1)) . x) by T2a;

          end;

          

           T5: for n,m be Nat holds ( dom (F . n)) = ( dom (F . m))

          proof

            let n,m be Nat;

            (F . n) = ( X-vol (((A2 . n) /\ V),M1)) & (F . m) = ( X-vol (((A2 . m) /\ V),M1)) by T2a;

            then ( dom (F . n)) = XX2 & ( dom (F . m)) = XX2 by FUNCT_2:def 1;

            hence ( dom (F . n)) = ( dom (F . m));

          end;

          

           T6: for n be Nat holds (F . n) is XX2 -measurable

          proof

            let n be Nat;

            (F . n) = ( X-vol (((A2 . n) /\ V),M1)) by T2a;

            hence (F . n) is XX2 -measurable by A01, DefXvol;

          end;

          

           T7: for n,m be Nat st n <= m holds for x be Element of X2 st x in XX2 holds ((F . n) . x) <= ((F . m) . x)

          proof

            let n,m be Nat;

            assume

             T71: n <= m;

            hereby

              let x be Element of X2;

              assume x in XX2;

              

               T72: ((A2 . n) /\ V) c= ((A2 . m) /\ V) by A3, T71, PROB_1:def 5, XBOOLE_1: 26;

              

               T73: (M1 . ( Measurable-Y-section (((A2 . n) /\ V),x))) = (( X-vol (((A2 . n) /\ V),M1)) . x) by A01, DefXvol

              .= ((F # x) . n) by T4

              .= ((F . n) . x) by MESFUNC5:def 13;

              (M1 . ( Measurable-Y-section (((A2 . m) /\ V),x))) = (( X-vol (((A2 . m) /\ V),M1)) . x) by A01, DefXvol

              .= ((F # x) . m) by T4

              .= ((F . m) . x) by MESFUNC5:def 13;

              hence ((F . n) . x) <= ((F . m) . x) by T72, T73, Th15, MEASURE1: 31;

            end;

          end;

          

           T8: for x be Element of X2 st x in XX2 holds (F # x) is convergent

          proof

            let x be Element of X2;

            assume x in XX2;

            now

              let n,m be Nat;

              assume m <= n;

              then ((F . m) . x) <= ((F . n) . x) by T7;

              then ((F # x) . m) <= ((F . n) . x) by MESFUNC5:def 13;

              hence ((F # x) . m) <= ((F # x) . n) by MESFUNC5:def 13;

            end;

            then (F # x) is non-decreasing by RINFSUP2: 7;

            hence (F # x) is convergent by RINFSUP2: 37;

          end;

          consider I be ExtREAL_sequence such that

           V2: (for n be Nat holds (I . n) = ( Integral (M2,(F . n)))) & I is convergent & ( Integral (M2,( lim F))) = ( lim I) by T3, T5, T6, T7, T8, MESFUNC9: 52, MESFUNC8:def 2;

          ( dom ( lim F)) = ( dom (F . 0 )) by MESFUNC8:def 9;

          then

           V4: ( dom ( lim F)) = ( dom ( X-vol ((E /\ V),M1))) by T3, FUNCT_2:def 1;

          for x be Element of X2 st x in ( dom ( lim F)) holds (( lim F) . x) = (( X-vol ((E /\ V),M1)) . x)

          proof

            let x be Element of X2;

            assume x in ( dom ( lim F));

            then

             L2: (( lim F) . x) = ( lim (F # x)) by MESFUNC8:def 9;

            consider G be SetSequence of S1 such that

             L3: G is non-descending & (for n be Nat holds (G . n) = (( Measurable-Y-section ((A2 . n),x)) /\ ( Measurable-Y-section (V,x)))) & ( lim G) = (( Measurable-Y-section (E,x)) /\ ( Measurable-Y-section (V,x))) by A3, Th109;

            for n be Element of NAT holds ((F # x) . n) = ((M1 * G) . n)

            proof

              let n be Element of NAT ;

              

               L5: ( dom G) = NAT by FUNCT_2:def 1;

              

               L4: ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13

              .= (( X-vol (((A2 . n) /\ V),M1)) . x) by T2

              .= (M1 . ( Measurable-Y-section (((A2 . n) /\ V),x))) by A01, DefXvol;

              ( Measurable-Y-section (((A2 . n) /\ V),x)) = (( Measurable-Y-section ((A2 . n),x)) /\ ( Measurable-Y-section (V,x))) by Th21;

              then ( Measurable-Y-section (((A2 . n) /\ V),x)) = (G . n) by L3;

              hence ((F # x) . n) = ((M1 * G) . n) by L4, L5, FUNCT_1: 13;

            end;

            then (F # x) = (M1 * G) by FUNCT_2: 63;

            then (( lim F) . x) = (M1 . (( Measurable-Y-section (E,x)) /\ ( Measurable-Y-section (V,x)))) by L2, L3, MEASURE8: 26;

            then (( lim F) . x) = (M1 . ( Measurable-Y-section ((E /\ V),x))) by Th21;

            hence (( lim F) . x) = (( X-vol ((E /\ V),M1)) . x) by A01, DefXvol;

          end;

          then

           V3: ( lim F) = ( X-vol ((E /\ V),M1)) by V4, PARTFUN1: 5;

          set J = (V (/\) A2);

          

           E1: ( dom J) = NAT by FUNCT_2:def 1;

          for n be object st n in NAT holds (J . n) in ( sigma ( measurable_rectangles (S1,S2)))

          proof

            let n be object;

            assume n in NAT ;

            then

            reconsider n1 = n as Element of NAT ;

            (J . n) = ((A2 . n1) /\ V) by SETLIM_2:def 5;

            hence (J . n) in ( sigma ( measurable_rectangles (S1,S2)));

          end;

          then

          reconsider J as SetSequence of ( sigma ( measurable_rectangles (S1,S2))) by E1, FUNCT_2: 3;

          

           R11: J is non-descending by A3, SETLIM_2: 22;

          A2 is convergent by A3, SETLIM_1: 63;

          then

           R13: ( lim J) = (E /\ V) by SETLIM_2: 92;

          

           R3: ( product_sigma_Measure (M1,M2)) is sigma_Measure of ( sigma ( measurable_rectangles (S1,S2))) by Th2;

          then

           R4: ( dom ( product_sigma_Measure (M1,M2))) = ( sigma ( measurable_rectangles (S1,S2))) by FUNCT_2:def 1;

          ( rng J) c= ( sigma ( measurable_rectangles (S1,S2))) by RELAT_1:def 19;

          then

           R2: (( product_sigma_Measure (M1,M2)) /* J) = (( product_sigma_Measure (M1,M2)) * J) by R4, FUNCT_2:def 11;

          for n be Element of NAT holds (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n)

          proof

            let n be Element of NAT ;

            

             R21: ( dom J) = NAT by FUNCT_2:def 1;

            (I . n) = ( Integral (M2,(F . n))) by V2

            .= ( Integral (M2,( X-vol (((A2 . n) /\ V),M1)))) by T2

            .= (( product_sigma_Measure (M1,M2)) . ((A2 . n) /\ V)) by PP

            .= (( product_sigma_Measure (M1,M2)) . (J . n)) by SETLIM_2:def 5;

            hence (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n) by R2, R21, FUNCT_1: 13;

          end;

          then ( lim I) = ( lim (( product_sigma_Measure (M1,M2)) /* J)) by FUNCT_2: 63;

          then ( lim I) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) by R13, R11, R2, R3, MEASURE8: 26;

          hence ( lim A1) in Z by V2, V3;

        end;

          suppose

           A3: A1 is non-ascending;

          ( meet ( rng A1)) in ( sigma ( measurable_rectangles (S1,S2))) by A4, MEASURE1: 35;

          then ( Intersection A1) in ( sigma ( measurable_rectangles (S1,S2))) by SETLIM_1: 8;

          then

          reconsider E = ( lim A1) as Element of ( sigma ( measurable_rectangles (S1,S2))) by A3, SETLIM_1: 64;

          defpred P[ Element of NAT , object] means $2 = ( X-vol (((A2 . $1) /\ V),M1));

          

           T1: for n be Element of NAT holds ex f be Element of ( PFuncs (X2, ExtREAL )) st P[n, f]

          proof

            let n be Element of NAT ;

            reconsider f1 = ( X-vol (((A2 . n) /\ V),M1)) as Element of ( PFuncs (X2, ExtREAL )) by PARTFUN1: 45;

            take f1;

            thus f1 = ( X-vol (((A2 . n) /\ V),M1));

          end;

          consider F be Function of NAT , ( PFuncs (X2, ExtREAL )) such that

           T2: for n be Element of NAT holds P[n, (F . n)] from FUNCT_2:sch 3( T1);

          reconsider F as Functional_Sequence of X2, ExtREAL ;

          

           T2a: for n be Nat holds (F . n) = ( X-vol (((A2 . n) /\ V),M1))

          proof

            let n be Nat;

            n is Element of NAT by ORDINAL1:def 12;

            hence thesis by T2;

          end;

          (F . 0 ) = ( X-vol (((A2 . 0 ) /\ V),M1)) by T2;

          then

           T3: ( dom (F . 0 )) = XX2 by FUNCT_2:def 1;

          

           T4: for n be Nat, x be Element of X2 holds ((F # x) . n) = (( X-vol (((A2 . n) /\ V),M1)) . x)

          proof

            let n be Nat, x be Element of X2;

            ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13;

            hence ((F # x) . n) = (( X-vol (((A2 . n) /\ V),M1)) . x) by T2a;

          end;

          for n,m be Nat holds ( dom (F . n)) = ( dom (F . m))

          proof

            let n,m be Nat;

            (F . n) = ( X-vol (((A2 . n) /\ V),M1)) & (F . m) = ( X-vol (((A2 . m) /\ V),M1)) by T2a;

            then ( dom (F . n)) = XX2 & ( dom (F . m)) = XX2 by FUNCT_2:def 1;

            hence ( dom (F . n)) = ( dom (F . m));

          end;

          then

          reconsider F as with_the_same_dom Functional_Sequence of X2, ExtREAL by MESFUNC8:def 2;

          

           T6: for n be Nat holds (F . n) is nonnegative & (F . n) is XX2 -measurable

          proof

            let n be Nat;

            (F . n) = ( X-vol (((A2 . n) /\ V),M1)) by T2a;

            hence (F . n) is nonnegative & (F . n) is XX2 -measurable by A01, DefXvol;

          end;

          

           T7: for x be Element of X2, n,m be Nat st x in XX2 & n <= m holds ((F . n) . x) >= ((F . m) . x)

          proof

            let x be Element of X2, n,m be Nat;

            assume x in XX2 & n <= m;

            then

             T72: ((A2 . m) /\ V) c= ((A2 . n) /\ V) by A3, PROB_1:def 4, XBOOLE_1: 26;

            

             T73: (M1 . ( Measurable-Y-section (((A2 . n) /\ V),x))) = (( X-vol (((A2 . n) /\ V),M1)) . x) by A01, DefXvol

            .= ((F # x) . n) by T4

            .= ((F . n) . x) by MESFUNC5:def 13;

            (M1 . ( Measurable-Y-section (((A2 . m) /\ V),x))) = (( X-vol (((A2 . m) /\ V),M1)) . x) by A01, DefXvol

            .= ((F # x) . m) by T4

            .= ((F . m) . x) by MESFUNC5:def 13;

            hence ((F . m) . x) <= ((F . n) . x) by T72, T73, Th15, MEASURE1: 31;

          end;

          

           M3: ( product_sigma_Measure (M1,M2)) is sigma_Measure of ( sigma ( measurable_rectangles (S1,S2))) by Th2;

          ( Integral (M2,((F . 0 ) | XX2))) = ( Integral (M2,( X-vol (((A2 . 0 ) /\ V),M1)))) by T2a;

          then

           M1: ( Integral (M2,((F . 0 ) | XX2))) = (( product_sigma_Measure (M1,M2)) . ((A2 . 0 ) /\ V)) by PP;

          (( product_sigma_Measure (M1,M2)) . ((A2 . 0 ) /\ V)) <= (( product_sigma_Measure (M1,M2)) . V) by M3, MEASURE1: 31, XBOOLE_1: 17;

          then ( Integral (M2,((F . 0 ) | XX2))) < +infty by A0, M1, XXREAL_0: 2;

          then

          consider I be ExtREAL_sequence such that

           V2: (for n be Nat holds (I . n) = ( Integral (M2,(F . n)))) & I is convergent & ( lim I) = ( Integral (M2,( lim F))) by T3, T6, T7, MESFUN10: 18;

          ( dom ( lim F)) = ( dom (F . 0 )) by MESFUNC8:def 9;

          then

           V4: ( dom ( lim F)) = ( dom ( X-vol ((E /\ V),M1))) by T3, FUNCT_2:def 1;

          for x be Element of X2 st x in ( dom ( lim F)) holds (( lim F) . x) = (( X-vol ((E /\ V),M1)) . x)

          proof

            let x be Element of X2;

            assume x in ( dom ( lim F));

            then

             L2: (( lim F) . x) = ( lim (F # x)) by MESFUNC8:def 9;

            consider G be SetSequence of S1 such that

             L3: G is non-ascending & (for n be Nat holds (G . n) = (( Measurable-Y-section ((A2 . n),x)) /\ ( Measurable-Y-section (V,x)))) & ( lim G) = (( Measurable-Y-section (E,x)) /\ ( Measurable-Y-section (V,x))) by A3, Th111;

            (G . 0 ) = (( Measurable-Y-section ((A2 . 0 ),x)) /\ ( Measurable-Y-section (V,x))) by L3;

            then

             L31: (M1 . (G . 0 )) <= (M1 . ( Measurable-Y-section (V,x))) by MEASURE1: 31, XBOOLE_1: 17;

            ( Measurable-Y-section (V,x)) c= A by A02, Th16;

            then (M1 . ( Measurable-Y-section (V,x))) <= (M1 . A) by MEASURE1: 31;

            then (M1 . (G . 0 )) <= (M1 . A) by L31, XXREAL_0: 2;

            then

             LL: (M1 . (G . 0 )) < +infty by PS2, XXREAL_0: 2;

            for n be Element of NAT holds ((F # x) . n) = ((M1 * G) . n)

            proof

              let n be Element of NAT ;

              

               L5: ( dom G) = NAT by FUNCT_2:def 1;

              

               L4: ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13

              .= (( X-vol (((A2 . n) /\ V),M1)) . x) by T2

              .= (M1 . ( Measurable-Y-section (((A2 . n) /\ V),x))) by A01, DefXvol;

              ( Measurable-Y-section (((A2 . n) /\ V),x)) = (( Measurable-Y-section ((A2 . n),x)) /\ ( Measurable-Y-section (V,x))) by Th21;

              then ( Measurable-Y-section (((A2 . n) /\ V),x)) = (G . n) by L3;

              hence ((F # x) . n) = ((M1 * G) . n) by L4, L5, FUNCT_1: 13;

            end;

            then (F # x) = (M1 * G) by FUNCT_2: 63;

            then (( lim F) . x) = (M1 . (( Measurable-Y-section (E,x)) /\ ( Measurable-Y-section (V,x)))) by L2, L3, LL, MEASURE8: 31;

            then (( lim F) . x) = (M1 . ( Measurable-Y-section ((E /\ V),x))) by Th21;

            hence (( lim F) . x) = (( X-vol ((E /\ V),M1)) . x) by A01, DefXvol;

          end;

          then

           V3: ( lim F) = ( X-vol ((E /\ V),M1)) by V4, PARTFUN1: 5;

          set J = (V (/\) A2);

          

           E1: ( dom J) = NAT by FUNCT_2:def 1;

          for n be object st n in NAT holds (J . n) in ( sigma ( measurable_rectangles (S1,S2)))

          proof

            let n be object;

            assume n in NAT ;

            then

            reconsider n1 = n as Element of NAT ;

            (J . n) = ((A2 . n1) /\ V) by SETLIM_2:def 5;

            hence (J . n) in ( sigma ( measurable_rectangles (S1,S2)));

          end;

          then

          reconsider J as SetSequence of ( sigma ( measurable_rectangles (S1,S2))) by E1, FUNCT_2: 3;

          

           R11: J is non-ascending by A3, SETLIM_2: 21;

          A2 is convergent by A3, SETLIM_1: 64;

          then

           R13: ( lim J) = (E /\ V) by SETLIM_2: 92;

          

           R3: ( product_sigma_Measure (M1,M2)) is sigma_Measure of ( sigma ( measurable_rectangles (S1,S2))) by Th2;

          then

           R4: ( dom ( product_sigma_Measure (M1,M2))) = ( sigma ( measurable_rectangles (S1,S2))) by FUNCT_2:def 1;

          ( rng J) c= ( sigma ( measurable_rectangles (S1,S2))) by RELAT_1:def 19;

          then

           R2: (( product_sigma_Measure (M1,M2)) /* J) = (( product_sigma_Measure (M1,M2)) * J) by R4, FUNCT_2:def 11;

          ((A2 . 0 ) /\ V) c= V by XBOOLE_1: 17;

          then (J . 0 ) c= V by SETLIM_2:def 5;

          then (( product_sigma_Measure (M1,M2)) . (J . 0 )) <= (( product_sigma_Measure (M1,M2)) . V) by R3, MEASURE1: 31;

          then

           K1: (( product_sigma_Measure (M1,M2)) . (J . 0 )) < +infty by A0, XXREAL_0: 2;

          for n be Element of NAT holds (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n)

          proof

            let n be Element of NAT ;

            

             R21: ( dom J) = NAT by FUNCT_2:def 1;

            (I . n) = ( Integral (M2,(F . n))) by V2

            .= ( Integral (M2,( X-vol (((A2 . n) /\ V),M1)))) by T2

            .= (( product_sigma_Measure (M1,M2)) . ((A2 . n) /\ V)) by PP

            .= (( product_sigma_Measure (M1,M2)) . (J . n)) by SETLIM_2:def 5;

            hence (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n) by R2, R21, FUNCT_1: 13;

          end;

          then ( lim I) = ( lim (( product_sigma_Measure (M1,M2)) /* J)) by FUNCT_2: 63;

          then ( lim I) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) by R13, R11, R2, R3, K1, MEASURE8: 31;

          hence ( lim A1) in Z by V2, V3;

        end;

      end;

      hence thesis by A1, PROB_3: 69;

    end;

    theorem :: MEASUR11:114

    

     Th114: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2 st M2 is sigma_finite & V = [:A, B:] & (( product_sigma_Measure (M1,M2)) . V) < +infty & (M2 . B) < +infty holds ( sigma ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M1,( Y-vol ((E /\ V),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) }

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2;

      set K = { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M1,( Y-vol ((E /\ V),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) };

      assume that

       A1: M2 is sigma_finite and

       A2: V = [:A, B:] and

       A3: (( product_sigma_Measure (M1,M2)) . V) < +infty and

       A4: (M2 . B) < +infty ;

      

       A5: K is MonotoneClass of [:X1, X2:] by A1, A2, A3, A4, Th112;

      

       A6: ( Field_generated_by ( measurable_rectangles (S1,S2))) c= K by A1, A2, Th106;

      ( sigma ( Field_generated_by ( measurable_rectangles (S1,S2)))) = ( sigma ( DisUnion ( measurable_rectangles (S1,S2)))) by SRINGS_3: 22

      .= ( sigma ( measurable_rectangles (S1,S2))) by Th1;

      hence thesis by A5, A6, Th87;

    end;

    theorem :: MEASUR11:115

    

     Th115: for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2 st M1 is sigma_finite & V = [:A, B:] & (( product_sigma_Measure (M1,M2)) . V) < +infty & (M1 . A) < +infty holds ( sigma ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M2,( X-vol ((E /\ V),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) }

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, V be Element of ( sigma ( measurable_rectangles (S1,S2))), A be Element of S1, B be Element of S2;

      set K = { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M2,( X-vol ((E /\ V),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ V)) };

      assume that

       A1: M1 is sigma_finite and

       A2: V = [:A, B:] and

       A3: (( product_sigma_Measure (M1,M2)) . V) < +infty and

       A4: (M1 . A) < +infty ;

      

       A5: K is MonotoneClass of [:X1, X2:] by A1, A2, A3, A4, Th113;

      

       A6: ( Field_generated_by ( measurable_rectangles (S1,S2))) c= K by A1, A2, Th107;

      ( sigma ( Field_generated_by ( measurable_rectangles (S1,S2)))) = ( sigma ( DisUnion ( measurable_rectangles (S1,S2)))) by SRINGS_3: 22

      .= ( sigma ( measurable_rectangles (S1,S2))) by Th1;

      hence thesis by A5, A6, Th87;

    end;

    theorem :: MEASUR11:116

    

     Th116: for X,Y be set, A be SetSequence of X, B be SetSequence of Y, C be SetSequence of [:X, Y:] st A is non-descending & B is non-descending & (for n be Nat holds (C . n) = [:(A . n), (B . n):]) holds C is non-descending & C is convergent & ( Union C) = [:( Union A), ( Union B):]

    proof

      let X,Y be set, A be SetSequence of X, B be SetSequence of Y, C be SetSequence of [:X, Y:];

      assume that

       A1: A is non-descending and

       A2: B is non-descending and

       A3: (for n be Nat holds (C . n) = [:(A . n), (B . n):]);

      for n,m be Nat st n <= m holds (C . n) c= (C . m)

      proof

        let n,m be Nat;

        assume n <= m;

        then (A . n) c= (A . m) & (B . n) c= (B . m) by A1, A2, PROB_1:def 5;

        then [:(A . n), (B . n):] c= [:(A . m), (B . m):] by ZFMISC_1: 96;

        then (C . n) c= [:(A . m), (B . m):] by A3;

        hence (C . n) c= (C . m) by A3;

      end;

      hence C is non-descending by PROB_1:def 5;

      hence C is convergent by SETLIM_1: 63;

      now

        let z be set;

        assume z in [:( Union A), ( Union B):];

        then

        consider x,y be object such that

         A6: x in ( Union A) & y in ( Union B) & z = [x, y] by ZFMISC_1:def 2;

        

         A7: x in ( union ( rng A)) & y in ( union ( rng B)) by A6, CARD_3:def 4;

        then

        consider A1 be set such that

         A8: x in A1 & A1 in ( rng A) by TARSKI:def 4;

        consider n be object such that

         A9: n in ( dom A) & A1 = (A . n) by A8, FUNCT_1:def 3;

        reconsider n as Nat by A9;

        consider B1 be set such that

         A10: y in B1 & B1 in ( rng B) by A7, TARSKI:def 4;

        consider m be object such that

         A11: m in ( dom B) & B1 = (B . m) by A10, FUNCT_1:def 3;

        reconsider m as Nat by A11;

        reconsider N = ( max (n,m)) as Element of NAT by ORDINAL1:def 12;

        (A . n) c= (A . N) & (B . m) c= (B . N) by A1, A2, XXREAL_0: 25, PROB_1:def 5;

        then z in [:(A . N), (B . N):] by A6, A8, A9, A10, A11, ZFMISC_1:def 2;

        then z in (C . N) & (C . N) in ( rng C) by A3, FUNCT_2: 112;

        then z in ( union ( rng C)) by TARSKI:def 4;

        hence z in ( Union C) by CARD_3:def 4;

      end;

      then

       A12: [:( Union A), ( Union B):] c= ( Union C);

      now

        let z be set;

        assume z in ( Union C);

        then z in ( union ( rng C)) by CARD_3:def 4;

        then

        consider C1 be set such that

         A13: z in C1 & C1 in ( rng C) by TARSKI:def 4;

        consider n be object such that

         A14: n in ( dom C) & C1 = (C . n) by A13, FUNCT_1:def 3;

        reconsider n as Element of NAT by A14;

        z in [:(A . n), (B . n):] by A3, A13, A14;

        then

        consider x,y be object such that

         A15: x in (A . n) & y in (B . n) & z = [x, y] by ZFMISC_1:def 2;

        (A . n) in ( rng A) & (B . n) in ( rng B) by FUNCT_2: 112;

        then x in ( union ( rng A)) & y in ( union ( rng B)) by A15, TARSKI:def 4;

        then x in ( Union A) & y in ( Union B) by CARD_3:def 4;

        hence z in [:( Union A), ( Union B):] by A15, ZFMISC_1:def 2;

      end;

      then ( Union C) c= [:( Union A), ( Union B):];

      hence ( Union C) = [:( Union A), ( Union B):] by A12;

    end;

    ::$Notion-Name

    theorem :: MEASUR11:117

    for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))) st M1 is sigma_finite & M2 is sigma_finite holds ( Integral (M1,( Y-vol (E,M2)))) = (( product_sigma_Measure (M1,M2)) . E)

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume that

       A1: M1 is sigma_finite and

       A2: M2 is sigma_finite;

      consider A be Set_Sequence of S1 such that

       A3: A is non-descending & (for n be Nat holds (M1 . (A . n)) < +infty ) & ( lim A) = X1 by A1, LM0902a;

      consider B be Set_Sequence of S2 such that

       A4: B is non-descending & (for n be Nat holds (M2 . (B . n)) < +infty ) & ( lim B) = X2 by A2, LM0902a;

      deffunc C( Element of NAT ) = [:(A . $1), (B . $1):];

      consider C be Function of NAT , ( bool [:X1, X2:]) such that

       A5: for n be Element of NAT holds (C . n) = C(n) from FUNCT_2:sch 4;

      

       A6: for n be Nat holds (C . n) = [:(A . n), (B . n):]

      proof

        let n be Nat;

        n is Element of NAT by ORDINAL1:def 12;

        hence (C . n) = [:(A . n), (B . n):] by A5;

      end;

      for n be Nat holds (C . n) in ( sigma ( measurable_rectangles (S1,S2)))

      proof

        let n be Nat;

        

         A7: (C . n) = [:(A . n), (B . n):] by A6;

        (A . n) in S1 & (B . n) in S2 by MEASURE8:def 2;

        then (C . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by A7;

        then

         A8: (C . n) in ( measurable_rectangles (S1,S2)) by MEASUR10:def 5;

        ( measurable_rectangles (S1,S2)) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

        hence (C . n) in ( sigma ( measurable_rectangles (S1,S2))) by A8;

      end;

      then

      reconsider C as Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))) by MEASURE8:def 2;

      

       a9: for n,m be Nat st n <= m holds (C . n) c= (C . m)

      proof

        let n,m be Nat;

        assume n <= m;

        then (A . n) c= (A . m) & (B . n) c= (B . m) by A3, A4, PROB_1:def 5;

        then [:(A . n), (B . n):] c= [:(A . m), (B . m):] by ZFMISC_1: 96;

        then (C . n) c= [:(A . m), (B . m):] by A6;

        hence (C . n) c= (C . m) by A6;

      end;

      then

       A9: C is non-descending by PROB_1:def 5;

      then

       a10: ( lim C) = ( Union C) by SETLIM_1: 63;

      

       a11: ( lim A) = ( Union A) & ( lim B) = ( Union B) by A3, A4, SETLIM_1: 63;

      

       A15: for n be Nat holds (( product_sigma_Measure (M1,M2)) . (C . n)) < +infty

      proof

        let n be Nat;

        

         A12: (A . n) in S1 & (B . n) in S2 & (C . n) in ( sigma ( measurable_rectangles (S1,S2))) by MEASURE8:def 2;

        (C . n) = [:(A . n), (B . n):] by A6;

        then

         A13: (( product_sigma_Measure (M1,M2)) . (C . n)) = ((M1 . (A . n)) * (M2 . (B . n))) by A12, Th10;

        (M1 . (A . n)) <> +infty & (M1 . (A . n)) <> -infty & (M2 . (B . n)) <> +infty & (M2 . (B . n)) <> -infty by A3, A4, SUPINF_2: 51;

        hence (( product_sigma_Measure (M1,M2)) . (C . n)) < +infty by A13, XXREAL_3: 69, XXREAL_0: 4;

      end;

      set C1 = (E (/\) C);

      

       A16: ( dom C1) = NAT by FUNCT_2:def 1;

      for n be object st n in NAT holds (C1 . n) in ( sigma ( measurable_rectangles (S1,S2)))

      proof

        let n be object;

        assume n in NAT ;

        then

        reconsider n1 = n as Element of NAT ;

        (C1 . n) = ((C . n1) /\ E) by SETLIM_2:def 5;

        hence (C1 . n) in ( sigma ( measurable_rectangles (S1,S2)));

      end;

      then

      reconsider C1 as SetSequence of ( sigma ( measurable_rectangles (S1,S2))) by A16, FUNCT_2: 3;

      

       A17: for n be Nat holds ( Integral (M1,( Y-vol ((E /\ (C . n)),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ (C . n)))

      proof

        let n be Nat;

        

         A18: (A . n) in S1 & (B . n) in S2 & (C . n) in ( sigma ( measurable_rectangles (S1,S2))) by MEASURE8:def 2;

        

         A19: (C . n) = [:(A . n), (B . n):] by A6;

        

         A20: (( product_sigma_Measure (M1,M2)) . (C . n)) < +infty by A15;

        (M2 . (B . n)) < +infty by A4;

        then ( sigma ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M1,( Y-vol ((E /\ (C . n)),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ (C . n))) } by A2, A18, A19, A20, Th114;

        then E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M1,( Y-vol ((E /\ (C . n)),M2)))) = (( product_sigma_Measure (M1,M2)) . (E /\ (C . n))) };

        then ex E1 be Element of ( sigma ( measurable_rectangles (S1,S2))) st E = E1 & ( Integral (M1,( Y-vol ((E1 /\ (C . n)),M2)))) = (( product_sigma_Measure (M1,M2)) . (E1 /\ (C . n)));

        hence thesis;

      end;

      defpred P[ Element of NAT , object] means $2 = ( Y-vol ((E /\ (C . $1)),M2));

      

       A21: for n be Element of NAT holds ex f be Element of ( PFuncs (X1, ExtREAL )) st P[n, f]

      proof

        let n be Element of NAT ;

        reconsider f1 = ( Y-vol ((E /\ (C . n)),M2)) as Element of ( PFuncs (X1, ExtREAL )) by PARTFUN1: 45;

        take f1;

        thus f1 = ( Y-vol ((E /\ (C . n)),M2));

      end;

      consider F be Function of NAT , ( PFuncs (X1, ExtREAL )) such that

       A22: for n be Element of NAT holds P[n, (F . n)] from FUNCT_2:sch 3( A21);

      reconsider F as Functional_Sequence of X1, ExtREAL ;

      

       A23: for n be Nat holds (F . n) = ( Y-vol ((E /\ (C . n)),M2))

      proof

        let n be Nat;

        n is Element of NAT by ORDINAL1:def 12;

        hence thesis by A22;

      end;

      reconsider XX1 = X1 as Element of S1 by MEASURE1: 7;

      reconsider X12 = [:X1, X2:] as Element of ( sigma ( measurable_rectangles (S1,S2))) by MEASURE1: 7;

      (F . 0 ) = ( Y-vol ((E /\ (C . 0 )),M2)) by A22;

      then

       A24: ( dom (F . 0 )) = XX1 & (F . 0 ) is nonnegative by FUNCT_2:def 1;

      

       A25: for n be Nat, x be Element of X1 holds ((F # x) . n) = (( Y-vol ((E /\ (C . n)),M2)) . x)

      proof

        let n be Nat, x be Element of X1;

        ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13;

        hence ((F # x) . n) = (( Y-vol ((E /\ (C . n)),M2)) . x) by A23;

      end;

      

       a26: for n,m be Nat holds ( dom (F . n)) = ( dom (F . m))

      proof

        let n,m be Nat;

        (F . n) = ( Y-vol ((E /\ (C . n)),M2)) & (F . m) = ( Y-vol ((E /\ (C . m)),M2)) by A23;

        then ( dom (F . n)) = XX1 & ( dom (F . m)) = XX1 by FUNCT_2:def 1;

        hence ( dom (F . n)) = ( dom (F . m));

      end;

      

       A27: for n be Nat holds (F . n) is XX1 -measurable

      proof

        let n be Nat;

        (F . n) = ( Y-vol ((E /\ (C . n)),M2)) by A23;

        hence (F . n) is XX1 -measurable by A2, DefYvol;

      end;

      

       A28: for n,m be Nat st n <= m holds for x be Element of X1 st x in XX1 holds ((F . n) . x) <= ((F . m) . x)

      proof

        let n,m be Nat;

        assume

         A29: n <= m;

        let x be Element of X1;

        assume x in XX1;

        

         A30: (E /\ (C . n)) c= (E /\ (C . m)) by a9, A29, XBOOLE_1: 26;

        

         A31: (M2 . ( Measurable-X-section ((E /\ (C . n)),x))) = (( Y-vol ((E /\ (C . n)),M2)) . x) by A2, DefYvol

        .= ((F # x) . n) by A25

        .= ((F . n) . x) by MESFUNC5:def 13;

        (M2 . ( Measurable-X-section ((E /\ (C . m)),x))) = (( Y-vol ((E /\ (C . m)),M2)) . x) by A2, DefYvol

        .= ((F # x) . m) by A25

        .= ((F . m) . x) by MESFUNC5:def 13;

        hence ((F . n) . x) <= ((F . m) . x) by A30, A31, Th14, MEASURE1: 31;

      end;

      

       A32: for x be Element of X1 st x in XX1 holds (F # x) is convergent

      proof

        let x be Element of X1;

        assume x in XX1;

        now

          let n,m be Nat;

          assume m <= n;

          then ((F . m) . x) <= ((F . n) . x) by A28;

          then ((F # x) . m) <= ((F . n) . x) by MESFUNC5:def 13;

          hence ((F # x) . m) <= ((F # x) . n) by MESFUNC5:def 13;

        end;

        then (F # x) is non-decreasing by RINFSUP2: 7;

        hence (F # x) is convergent by RINFSUP2: 37;

      end;

      consider I be ExtREAL_sequence such that

       A33: (for n be Nat holds (I . n) = ( Integral (M1,(F . n)))) & I is convergent & ( Integral (M1,( lim F))) = ( lim I) by A24, a26, A27, A28, A32, MESFUNC8:def 2, MESFUNC9: 52;

      ( dom ( lim F)) = ( dom (F . 0 )) by MESFUNC8:def 9;

      then

       A34: ( dom ( lim F)) = ( dom ( Y-vol (E,M2))) by A24, FUNCT_2:def 1;

      for x be Element of X1 st x in ( dom ( lim F)) holds (( lim F) . x) = (( Y-vol (E,M2)) . x)

      proof

        let x be Element of X1;

        assume x in ( dom ( lim F));

        then

         L2: (( lim F) . x) = ( lim (F # x)) by MESFUNC8:def 9;

        consider G be SetSequence of S2 such that

         L3: G is non-descending & (for n be Nat holds (G . n) = (( Measurable-X-section ((C . n),x)) /\ ( Measurable-X-section (E,x)))) & ( lim G) = (( Measurable-X-section (X12,x)) /\ ( Measurable-X-section (E,x))) by A9, a11, A3, A4, a10, A6, Th116, Th108;

        for n be Element of NAT holds ((F # x) . n) = ((M2 * G) . n)

        proof

          let n be Element of NAT ;

          

           L5: ( dom G) = NAT by FUNCT_2:def 1;

          

           L4: ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13

          .= (( Y-vol (((C . n) /\ E),M2)) . x) by A22

          .= (M2 . ( Measurable-X-section (((C . n) /\ E),x))) by A2, DefYvol;

          ( Measurable-X-section (((C . n) /\ E),x)) = (( Measurable-X-section ((C . n),x)) /\ ( Measurable-X-section (E,x))) by Th21;

          then ( Measurable-X-section (((C . n) /\ E),x)) = (G . n) by L3;

          hence ((F # x) . n) = ((M2 * G) . n) by L4, L5, FUNCT_1: 13;

        end;

        then (F # x) = (M2 * G) by FUNCT_2: 63;

        then (( lim F) . x) = (M2 . (( Measurable-X-section (E,x)) /\ ( Measurable-X-section (X12,x)))) by L2, L3, MEASURE8: 26;

        

        then (( lim F) . x) = (M2 . ( Measurable-X-section ((E /\ X12),x))) by Th21

        .= (M2 . ( Measurable-X-section (E,x))) by XBOOLE_1: 28;

        hence (( lim F) . x) = (( Y-vol (E,M2)) . x) by A2, DefYvol;

      end;

      then

       V3: ( lim F) = ( Y-vol (E,M2)) by A34, PARTFUN1: 5;

      set J = (E (/\) C);

      

       E1: ( dom J) = NAT by FUNCT_2:def 1;

      for n be object st n in NAT holds (J . n) in ( sigma ( measurable_rectangles (S1,S2)))

      proof

        let n be object;

        assume n in NAT ;

        then

        reconsider n1 = n as Element of NAT ;

        (J . n) = ((C . n1) /\ E) by SETLIM_2:def 5;

        hence (J . n) in ( sigma ( measurable_rectangles (S1,S2)));

      end;

      then

      reconsider J as SetSequence of ( sigma ( measurable_rectangles (S1,S2))) by E1, FUNCT_2: 3;

      

       R11: J is non-descending by A9, SETLIM_2: 22;

      C is convergent by A9, SETLIM_1: 63;

      

      then

       R13: ( lim J) = (E /\ ( lim C)) by SETLIM_2: 92

      .= (E /\ [:X1, X2:]) by a11, A3, A4, a10, A6, Th116

      .= E by XBOOLE_1: 28;

      

       R3: ( product_sigma_Measure (M1,M2)) is sigma_Measure of ( sigma ( measurable_rectangles (S1,S2))) by Th2;

      then

       R4: ( dom ( product_sigma_Measure (M1,M2))) = ( sigma ( measurable_rectangles (S1,S2))) by FUNCT_2:def 1;

      ( rng J) c= ( sigma ( measurable_rectangles (S1,S2))) by RELAT_1:def 19;

      then

       R2: (( product_sigma_Measure (M1,M2)) /* J) = (( product_sigma_Measure (M1,M2)) * J) by R4, FUNCT_2:def 11;

      for n be Element of NAT holds (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n)

      proof

        let n be Element of NAT ;

        

         R21: ( dom J) = NAT by FUNCT_2:def 1;

        (I . n) = ( Integral (M1,(F . n))) by A33

        .= ( Integral (M1,( Y-vol (((C . n) /\ E),M2)))) by A22

        .= (( product_sigma_Measure (M1,M2)) . ((C . n) /\ E)) by A17

        .= (( product_sigma_Measure (M1,M2)) . (J . n)) by SETLIM_2:def 5;

        hence (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n) by R2, R21, FUNCT_1: 13;

      end;

      then I = (( product_sigma_Measure (M1,M2)) /* J) by FUNCT_2: 63;

      hence ( Integral (M1,( Y-vol (E,M2)))) = (( product_sigma_Measure (M1,M2)) . E) by A33, V3, R13, R11, R2, R3, MEASURE8: 26;

    end;

    ::$Notion-Name

    theorem :: MEASUR11:118

    for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2))) st M1 is sigma_finite & M2 is sigma_finite holds ( Integral (M2,( X-vol (E,M1)))) = (( product_sigma_Measure (M1,M2)) . E)

    proof

      let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2, M1 be sigma_Measure of S1, M2 be sigma_Measure of S2, E be Element of ( sigma ( measurable_rectangles (S1,S2)));

      assume that

       A1: M1 is sigma_finite and

       A2: M2 is sigma_finite;

      consider A be Set_Sequence of S1 such that

       A3: A is non-descending & (for n be Nat holds (M1 . (A . n)) < +infty ) & ( lim A) = X1 by A1, LM0902a;

      consider B be Set_Sequence of S2 such that

       A4: B is non-descending & (for n be Nat holds (M2 . (B . n)) < +infty ) & ( lim B) = X2 by A2, LM0902a;

      deffunc C( Element of NAT ) = [:(A . $1), (B . $1):];

      consider C be Function of NAT , ( bool [:X1, X2:]) such that

       A5: for n be Element of NAT holds (C . n) = C(n) from FUNCT_2:sch 4;

      

       A6: for n be Nat holds (C . n) = [:(A . n), (B . n):]

      proof

        let n be Nat;

        n is Element of NAT by ORDINAL1:def 12;

        hence (C . n) = [:(A . n), (B . n):] by A5;

      end;

      for n be Nat holds (C . n) in ( sigma ( measurable_rectangles (S1,S2)))

      proof

        let n be Nat;

        

         A7: (C . n) = [:(A . n), (B . n):] by A6;

        (A . n) in S1 & (B . n) in S2 by MEASURE8:def 2;

        then (C . n) in the set of all [:A, B:] where A be Element of S1, B be Element of S2 by A7;

        then

         A8: (C . n) in ( measurable_rectangles (S1,S2)) by MEASUR10:def 5;

        ( measurable_rectangles (S1,S2)) c= ( sigma ( measurable_rectangles (S1,S2))) by PROB_1:def 9;

        hence (C . n) in ( sigma ( measurable_rectangles (S1,S2))) by A8;

      end;

      then

      reconsider C as Set_Sequence of ( sigma ( measurable_rectangles (S1,S2))) by MEASURE8:def 2;

      

       a9: for n,m be Nat st n <= m holds (C . n) c= (C . m)

      proof

        let n,m be Nat;

        assume n <= m;

        then (A . n) c= (A . m) & (B . n) c= (B . m) by A3, A4, PROB_1:def 5;

        then [:(A . n), (B . n):] c= [:(A . m), (B . m):] by ZFMISC_1: 96;

        then (C . n) c= [:(A . m), (B . m):] by A6;

        hence (C . n) c= (C . m) by A6;

      end;

      then

       A9: C is non-descending by PROB_1:def 5;

      then

       a10: ( lim C) = ( Union C) by SETLIM_1: 63;

      

       a11: ( lim A) = ( Union A) & ( lim B) = ( Union B) by A3, A4, SETLIM_1: 63;

      

       A15: for n be Nat holds (( product_sigma_Measure (M1,M2)) . (C . n)) < +infty

      proof

        let n be Nat;

        

         A12: (A . n) in S1 & (B . n) in S2 & (C . n) in ( sigma ( measurable_rectangles (S1,S2))) by MEASURE8:def 2;

        (C . n) = [:(A . n), (B . n):] by A6;

        then

         A13: (( product_sigma_Measure (M1,M2)) . (C . n)) = ((M1 . (A . n)) * (M2 . (B . n))) by A12, Th10;

        (M1 . (A . n)) <> +infty & (M1 . (A . n)) <> -infty & (M2 . (B . n)) <> +infty & (M2 . (B . n)) <> -infty by A3, A4, SUPINF_2: 51;

        hence (( product_sigma_Measure (M1,M2)) . (C . n)) < +infty by A13, XXREAL_3: 69, XXREAL_0: 4;

      end;

      set C1 = (E (/\) C);

      

       A16: ( dom C1) = NAT by FUNCT_2:def 1;

      for n be object st n in NAT holds (C1 . n) in ( sigma ( measurable_rectangles (S1,S2)))

      proof

        let n be object;

        assume n in NAT ;

        then

        reconsider n1 = n as Element of NAT ;

        (C1 . n) = ((C . n1) /\ E) by SETLIM_2:def 5;

        hence (C1 . n) in ( sigma ( measurable_rectangles (S1,S2)));

      end;

      then

      reconsider C1 as SetSequence of ( sigma ( measurable_rectangles (S1,S2))) by A16, FUNCT_2: 3;

      

       A17: for n be Nat holds ( Integral (M2,( X-vol ((E /\ (C . n)),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ (C . n)))

      proof

        let n be Nat;

        

         A18: (A . n) in S1 & (B . n) in S2 & (C . n) in ( sigma ( measurable_rectangles (S1,S2))) by MEASURE8:def 2;

        

         A19: (C . n) = [:(A . n), (B . n):] by A6;

        

         A20: (( product_sigma_Measure (M1,M2)) . (C . n)) < +infty by A15;

        (M1 . (A . n)) < +infty by A3;

        then ( sigma ( measurable_rectangles (S1,S2))) c= { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M2,( X-vol ((E /\ (C . n)),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ (C . n))) } by A1, A18, A19, A20, Th115;

        then E in { E where E be Element of ( sigma ( measurable_rectangles (S1,S2))) : ( Integral (M2,( X-vol ((E /\ (C . n)),M1)))) = (( product_sigma_Measure (M1,M2)) . (E /\ (C . n))) };

        then ex E1 be Element of ( sigma ( measurable_rectangles (S1,S2))) st E = E1 & ( Integral (M2,( X-vol ((E1 /\ (C . n)),M1)))) = (( product_sigma_Measure (M1,M2)) . (E1 /\ (C . n)));

        hence thesis;

      end;

      defpred P[ Element of NAT , object] means $2 = ( X-vol ((E /\ (C . $1)),M1));

      

       A21: for n be Element of NAT holds ex f be Element of ( PFuncs (X2, ExtREAL )) st P[n, f]

      proof

        let n be Element of NAT ;

        reconsider f1 = ( X-vol ((E /\ (C . n)),M1)) as Element of ( PFuncs (X2, ExtREAL )) by PARTFUN1: 45;

        take f1;

        thus f1 = ( X-vol ((E /\ (C . n)),M1));

      end;

      consider F be Function of NAT , ( PFuncs (X2, ExtREAL )) such that

       A22: for n be Element of NAT holds P[n, (F . n)] from FUNCT_2:sch 3( A21);

      reconsider F as Functional_Sequence of X2, ExtREAL ;

      

       A23: for n be Nat holds (F . n) = ( X-vol ((E /\ (C . n)),M1))

      proof

        let n be Nat;

        n is Element of NAT by ORDINAL1:def 12;

        hence thesis by A22;

      end;

      reconsider XX2 = X2 as Element of S2 by MEASURE1: 7;

      reconsider X12 = [:X1, X2:] as Element of ( sigma ( measurable_rectangles (S1,S2))) by MEASURE1: 7;

      (F . 0 ) = ( X-vol ((E /\ (C . 0 )),M1)) by A22;

      then

       A24: ( dom (F . 0 )) = XX2 & (F . 0 ) is nonnegative by FUNCT_2:def 1;

      

       A25: for n be Nat, x be Element of X2 holds ((F # x) . n) = (( X-vol ((E /\ (C . n)),M1)) . x)

      proof

        let n be Nat, x be Element of X2;

        ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13;

        hence ((F # x) . n) = (( X-vol ((E /\ (C . n)),M1)) . x) by A23;

      end;

      

       a26: for n,m be Nat holds ( dom (F . n)) = ( dom (F . m))

      proof

        let n,m be Nat;

        (F . n) = ( X-vol ((E /\ (C . n)),M1)) & (F . m) = ( X-vol ((E /\ (C . m)),M1)) by A23;

        then ( dom (F . n)) = XX2 & ( dom (F . m)) = XX2 by FUNCT_2:def 1;

        hence ( dom (F . n)) = ( dom (F . m));

      end;

      

       A27: for n be Nat holds (F . n) is XX2 -measurable

      proof

        let n be Nat;

        (F . n) = ( X-vol ((E /\ (C . n)),M1)) by A23;

        hence (F . n) is XX2 -measurable by A1, DefXvol;

      end;

      

       A28: for n,m be Nat st n <= m holds for x be Element of X2 st x in XX2 holds ((F . n) . x) <= ((F . m) . x)

      proof

        let n,m be Nat;

        assume

         A29: n <= m;

        let x be Element of X2;

        assume x in XX2;

        

         A30: (E /\ (C . n)) c= (E /\ (C . m)) by a9, A29, XBOOLE_1: 26;

        

         A31: (M1 . ( Measurable-Y-section ((E /\ (C . n)),x))) = (( X-vol ((E /\ (C . n)),M1)) . x) by A1, DefXvol

        .= ((F # x) . n) by A25

        .= ((F . n) . x) by MESFUNC5:def 13;

        (M1 . ( Measurable-Y-section ((E /\ (C . m)),x))) = (( X-vol ((E /\ (C . m)),M1)) . x) by A1, DefXvol

        .= ((F # x) . m) by A25

        .= ((F . m) . x) by MESFUNC5:def 13;

        hence ((F . n) . x) <= ((F . m) . x) by A30, A31, Th15, MEASURE1: 31;

      end;

      

       A32: for x be Element of X2 st x in XX2 holds (F # x) is convergent

      proof

        let x be Element of X2;

        assume x in XX2;

        now

          let n,m be Nat;

          assume m <= n;

          then ((F . m) . x) <= ((F . n) . x) by A28;

          then ((F # x) . m) <= ((F . n) . x) by MESFUNC5:def 13;

          hence ((F # x) . m) <= ((F # x) . n) by MESFUNC5:def 13;

        end;

        then (F # x) is non-decreasing by RINFSUP2: 7;

        hence (F # x) is convergent by RINFSUP2: 37;

      end;

      consider I be ExtREAL_sequence such that

       A33: (for n be Nat holds (I . n) = ( Integral (M2,(F . n)))) & I is convergent & ( Integral (M2,( lim F))) = ( lim I) by A24, a26, A27, A28, A32, MESFUNC8:def 2, MESFUNC9: 52;

      ( dom ( lim F)) = ( dom (F . 0 )) by MESFUNC8:def 9;

      then

       A34: ( dom ( lim F)) = ( dom ( X-vol (E,M1))) by A24, FUNCT_2:def 1;

      for x be Element of X2 st x in ( dom ( lim F)) holds (( lim F) . x) = (( X-vol (E,M1)) . x)

      proof

        let x be Element of X2;

        assume x in ( dom ( lim F));

        then

         L2: (( lim F) . x) = ( lim (F # x)) by MESFUNC8:def 9;

        consider G be SetSequence of S1 such that

         L3: G is non-descending & (for n be Nat holds (G . n) = (( Measurable-Y-section ((C . n),x)) /\ ( Measurable-Y-section (E,x)))) & ( lim G) = (( Measurable-Y-section (X12,x)) /\ ( Measurable-Y-section (E,x))) by A9, a11, A3, A4, a10, A6, Th116, Th109;

        for n be Element of NAT holds ((F # x) . n) = ((M1 * G) . n)

        proof

          let n be Element of NAT ;

          

           L5: ( dom G) = NAT by FUNCT_2:def 1;

          

           L4: ((F # x) . n) = ((F . n) . x) by MESFUNC5:def 13

          .= (( X-vol (((C . n) /\ E),M1)) . x) by A22

          .= (M1 . ( Measurable-Y-section (((C . n) /\ E),x))) by A1, DefXvol;

          ( Measurable-Y-section (((C . n) /\ E),x)) = (( Measurable-Y-section ((C . n),x)) /\ ( Measurable-Y-section (E,x))) by Th21;

          then ( Measurable-Y-section (((C . n) /\ E),x)) = (G . n) by L3;

          hence ((F # x) . n) = ((M1 * G) . n) by L4, L5, FUNCT_1: 13;

        end;

        then (F # x) = (M1 * G) by FUNCT_2: 63;

        then (( lim F) . x) = (M1 . (( Measurable-Y-section (E,x)) /\ ( Measurable-Y-section (X12,x)))) by L2, L3, MEASURE8: 26;

        

        then (( lim F) . x) = (M1 . ( Measurable-Y-section ((E /\ X12),x))) by Th21

        .= (M1 . ( Measurable-Y-section (E,x))) by XBOOLE_1: 28;

        hence (( lim F) . x) = (( X-vol (E,M1)) . x) by A1, DefXvol;

      end;

      then

       V3: ( lim F) = ( X-vol (E,M1)) by A34, PARTFUN1: 5;

      set J = (E (/\) C);

      

       E1: ( dom J) = NAT by FUNCT_2:def 1;

      for n be object st n in NAT holds (J . n) in ( sigma ( measurable_rectangles (S1,S2)))

      proof

        let n be object;

        assume n in NAT ;

        then

        reconsider n1 = n as Element of NAT ;

        (J . n) = ((C . n1) /\ E) by SETLIM_2:def 5;

        hence (J . n) in ( sigma ( measurable_rectangles (S1,S2)));

      end;

      then

      reconsider J as SetSequence of ( sigma ( measurable_rectangles (S1,S2))) by E1, FUNCT_2: 3;

      

       R11: J is non-descending by A9, SETLIM_2: 22;

      C is convergent by A9, SETLIM_1: 63;

      

      then

       R13: ( lim J) = (E /\ ( lim C)) by SETLIM_2: 92

      .= (E /\ [:X1, X2:]) by a11, A3, A4, a10, A6, Th116

      .= E by XBOOLE_1: 28;

      

       R3: ( product_sigma_Measure (M1,M2)) is sigma_Measure of ( sigma ( measurable_rectangles (S1,S2))) by Th2;

      then

       R4: ( dom ( product_sigma_Measure (M1,M2))) = ( sigma ( measurable_rectangles (S1,S2))) by FUNCT_2:def 1;

      ( rng J) c= ( sigma ( measurable_rectangles (S1,S2))) by RELAT_1:def 19;

      then

       R2: (( product_sigma_Measure (M1,M2)) /* J) = (( product_sigma_Measure (M1,M2)) * J) by R4, FUNCT_2:def 11;

      for n be Element of NAT holds (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n)

      proof

        let n be Element of NAT ;

        

         R21: ( dom J) = NAT by FUNCT_2:def 1;

        (I . n) = ( Integral (M2,(F . n))) by A33

        .= ( Integral (M2,( X-vol (((C . n) /\ E),M1)))) by A22

        .= (( product_sigma_Measure (M1,M2)) . ((C . n) /\ E)) by A17

        .= (( product_sigma_Measure (M1,M2)) . (J . n)) by SETLIM_2:def 5;

        hence (I . n) = ((( product_sigma_Measure (M1,M2)) /* J) . n) by R2, R21, FUNCT_1: 13;

      end;

      then I = (( product_sigma_Measure (M1,M2)) /* J) by FUNCT_2: 63;

      hence ( Integral (M2,( X-vol (E,M1)))) = (( product_sigma_Measure (M1,M2)) . E) by A33, V3, R13, R11, R2, R3, MEASURE8: 26;

    end;