MEASURE3

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:: Completeness of the $\sigma$-Additive Measure. Measure Theory
::  by J\'ozef Bia{\l}as
::
:: Received February 22, 1992
:: Copyright (c) 1992-2026 Association of Mizar Users
::           (Stowarzyszenie Uzytkownikow Mizara, Bialystok, Poland).
:: This code can be distributed under the GNU General Public Licence
:: version 3.0 or later, or the Creative Commons Attribution-ShareAlike
:: License version 3.0 or later, subject to the binding interpretation
:: detailed in file COPYING.interpretation.
:: See COPYING.GPL and COPYING.CC-BY-SA for the full text of these
:: licenses, or see http://www.gnu.org/licenses/gpl.html and
:: http://creativecommons.org/licenses/by-sa/3.0/.

environ

 vocabularies FUNCT_1, NUMBERS, SUBSET_1, SUPINF_2, XXREAL_0, RELAT_1,
      SUPINF_1, ORDINAL2, PROB_1, MEASURE2, MEASURE1, TARSKI, SETFAM_1, CARD_1,
      XBOOLE_0, ARYTM_3, NAT_1, ARYTM_1, REAL_1, XXREAL_2, ZFMISC_1, MEASURE3,
      FUNCT_7, REWRITE1;
 notations TARSKI, XBOOLE_0, ZFMISC_1, SUBSET_1, ORDINAL1, NUMBERS, XXREAL_0,      
      XCMPLX_0,
      XXREAL_3, XREAL_0, FUNCT_1, RELSET_1, PARTFUN1, FUNCT_2, NAT_1, SETFAM_1,
      PROB_1, XXREAL_2, SUPINF_1, SUPINF_2, MEASURE1, MEASURE2;
 constructors PARTFUN1, REAL_1, NAT_1, PROB_2, MEASURE1, MEASURE2, SUPINF_1,
      RELSET_1, XREAL_0;
 registrations SUBSET_1, ORDINAL1, FUNCT_2, NUMBERS, XREAL_0, MEMBERED,
      MEASURE1, VALUED_0, XXREAL_3, RELSET_1;
 requirements NUMERALS, SUBSET, BOOLE, ARITHM;
 definitions TARSKI, XBOOLE_0;
 equalities XBOOLE_0, XXREAL_3;
 expansions TARSKI, XBOOLE_0;
 theorems TARSKI, FUNCT_1, FUNCT_2, ZFMISC_1, NAT_1, SETFAM_1, SUPINF_2,
      MEASURE1, MEASURE2, PROB_2, XBOOLE_0, XBOOLE_1, XXREAL_0, PROB_1,
      FINSUB_1, ORDINAL1, XXREAL_2, VALUED_0, XXREAL_3, RELAT_1, XREAL_0;
 schemes NAT_1, FUNCT_2, XFAMILY;

begin

::
::             Some additional properties about R_eal numbers
::

reserve X for set;

theorem Th1:
  for F1,F2 being sequence of ExtREAL st (for n being Element
  of NAT holds Ser(F1).n <= Ser(F2).n) holds SUM(F1) <= SUM(F2)
proof
  let F1,F2 be sequence of ExtREAL;
  assume
A1: for n being Element of NAT holds Ser(F1).n <= Ser(F2).n;
A2: for x being ExtReal st x in rng Ser(F1) holds ex y being
  ExtReal st y in rng Ser(F2) & x <= y
  proof
    let x be ExtReal;
A3: dom Ser(F1) = NAT by FUNCT_2:def 1;
    assume x in rng Ser(F1);
    then consider n being object such that
A4: n in NAT and
A5: x = Ser(F1).n by A3,FUNCT_1:def 3;
    reconsider n as Element of NAT by A4;
    reconsider y = Ser(F2).n as R_eal;
    take y;
    dom Ser(F2) = NAT by FUNCT_2:def 1;
    hence thesis by A1,A5,FUNCT_1:def 3;
  end;
  SUM(F1) = sup(rng Ser F1) & SUM(F2) = sup(rng Ser F2)
    by SUPINF_2:def 13;
  hence thesis by A2,XXREAL_2:63;
end;

theorem
  for F1,F2 being sequence of ExtREAL st (for n being Element of NAT
  holds Ser(F1).n = Ser(F2).n) holds SUM(F1) = SUM(F2)
proof
  let F1,F2 be sequence of ExtREAL;
  assume
A1: for n being Element of NAT holds Ser(F1).n = Ser(F2).n;
  then for n being Element of NAT holds Ser(F2).n <= Ser(F1).n;
  then
A2: SUM(F2) <= SUM(F1) by Th1;
  for n being Element of NAT holds Ser(F1).n <= Ser(F2).n by A1;
  then SUM(F1) <= SUM(F2) by Th1;
  hence thesis by A2,XXREAL_0:1;
end;

::
::         Some additional theorems about measures and functions
::

definition
  let X be set;
  let S be SigmaField of X;
  let F be sequence of S;
  redefine func rng F -> N_Measure_fam of S;  :: redefines rng (FUNCT_1, redefinition) → rng (RELAT_1, synonym) → proj2 (XTUPLE_0, definition)
  coherence
  proof
    rng F is N_Sub_set_fam of X & rng F c= S by MEASURE1:23,RELAT_1:def 19;
    hence thesis by MEASURE2:def 1;
  end;
end;

theorem
  for S being SigmaField of X, M being sigma_Measure of S, F being
  sequence of S, A being Element of S st meet rng F c= A & (for n being
  Element of NAT holds A c= F.n) holds M.A = M.(meet rng F)
proof
  let S be SigmaField of X;
  let M be sigma_Measure of S;
  let F be sequence of S;
  let A be Element of S;
  assume that
A1: meet rng F c= A and
A2: for n being Element of NAT holds A c= F.n;
  A c= meet rng F
  proof
    let x be object;
    assume
A3: x in A;
    for Y being set st Y in rng F holds x in Y
    proof
      let Y be set;
A4:   dom F = NAT by FUNCT_2:def 1;
      assume Y in rng F;
      then ex n being object st n in NAT & Y = F.n by A4,FUNCT_1:def 3;
      then A c= Y by A2;
      hence thesis by A3;
    end;
    hence thesis by SETFAM_1:def 1;
  end;
  then
A5: M.(A) <= M.(meet rng F) by MEASURE1:31;
  M.(meet rng F) <= M.(A) by A1,MEASURE1:31;
  hence thesis by A5,XXREAL_0:1;
end;

theorem Th4:
  for S being SigmaField of X, G,F being sequence of S st (G.0
= {} & for n being Nat holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F.n )
  holds union rng G = F.0 \ meet rng F
proof
  let S be SigmaField of X;
  let G,F be sequence of S;
  assume that
A1: G.0 = {} and
A2: for n being Nat holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F. n;
A3: dom G = NAT by FUNCT_2:def 1;
  thus union rng G c= F.0 \ meet rng F
  proof
    let A be object;
    assume A in union rng G;
    then consider Z being set such that
A4: A in Z and
A5: Z in rng G by TARSKI:def 4;
    consider n being object such that
A6: n in NAT and
A7: Z = G.n by A3,A5,FUNCT_1:def 3;
    reconsider n as Element of NAT by A6;
    consider k being Nat such that
A8: n = k + 1 by A1,A4,A7,NAT_1:6;
    reconsider k as Element of NAT by ORDINAL1:def 12;
    set Y = F.k;
A9: A in F.0 \ F.k by A2,A4,A7,A8;
    then Y in rng F & not A in Y by FUNCT_2:4,XBOOLE_0:def 5;
    then
A10: not A in meet rng F by SETFAM_1:def 1;
    A in F.0 by A9,XBOOLE_0:def 5;
    hence thesis by A10,XBOOLE_0:def 5;
  end;
  let A be object;
  assume
A11: A in F.0 \ meet rng F;
  then not A in meet rng F by XBOOLE_0:def 5;
  then
A12: ex Y being set st Y in rng F & not A in Y by SETFAM_1:def 1;
  A in F.0 by A11,XBOOLE_0:def 5;
  then consider Y being set such that
A13: A in F.0 and
A14: Y in rng F and
A15: not A in Y by A12;
  dom F = NAT by FUNCT_2:def 1;
  then consider n being object such that
A16: n in NAT and
A17: Y = F.n by A14,FUNCT_1:def 3;
  reconsider n as Element of NAT by A16;
  A in F.0 \ F.n by A13,A15,A17,XBOOLE_0:def 5;
  then
A18: A in G.(n+1) by A2;
  G.(n + 1) in rng G by FUNCT_2:4;
  hence thesis by A18,TARSKI:def 4;
end;

theorem Th5:
  for S being SigmaField of X, G,F being sequence of S st (G.0
= {} & for n being Nat holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F.n )
  holds meet rng F = F.0 \ union rng G
proof
  let S be SigmaField of X;
  let G,F be sequence of S;
  assume that
A1: G.0 = {} and
A2: for n being Nat holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F .n;
A3: for n being Nat holds F.n c= F.0
  proof
    defpred P[Nat] means F.$1 c= F.0;
A4: for k being Nat st P[k] holds P[k+1]
    proof
      let k be Nat;
      assume
A5:   F.k c= F.0;
      F.(k+1) c= F.k by A2;
      hence thesis by A5,XBOOLE_1:1;
    end;
A6: P[0];
    thus for n being Nat holds P[n] from NAT_1:sch 2(A6,A4);
  end;
A7: meet rng F c= F.0
  proof
    set X = the Element of rng F;
    let A be object;
    dom F = NAT by FUNCT_2:def 1;
    then ex n being object st n in NAT & F.n = X by FUNCT_1:def 3;
    then
A8: X c= F.0 by A3;
    assume A in meet rng F;
    then A in X by SETFAM_1:def 1;
    hence thesis by A8;
  end;
A9: F.0 /\ meet rng F = F.0 \ (F.0 \ meet rng F) by XBOOLE_1:48;
  union rng G = F.0 \ meet rng F by A1,A2,Th4;
  hence thesis by A7,A9,XBOOLE_1:28;
end;

theorem Th6:
  for S being SigmaField of X, M being sigma_Measure of S, G,F
being sequence of S st (M.(F.0) <+infty & G.0 = {} & for n being Nat
 holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F.n ) holds M.(meet rng F) = M.(F.0)
  - M.(union rng G)
proof
  let S be SigmaField of X, M be sigma_Measure of S, G,F be sequence of S;
  assume that
A1: M.(F.0) <+infty and
A2: G.0 = {} & for n being Nat holds G.(n+1) = F.0 \ F.n & F.
  (n+1) c= F .n;
A3: union rng G = F.0 \ meet rng F by A2,Th4;
A4: M.(F.0 \ union rng G) = M.(meet rng F) by A2,Th5;
  M.(F.0 \ meet rng F) <> +infty by A1,MEASURE1:31,XBOOLE_1:36;
  then M.(union rng G) <+infty by A3,XXREAL_0:4;
  hence thesis by A3,A4,MEASURE1:32,XBOOLE_1:36;
end;

theorem Th7:
  for S being SigmaField of X, M being sigma_Measure of S, G,F
being sequence of S st (M.(F.0) <+infty & G.0 = {} & for n being Nat
 holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F.n ) holds M.(union rng G) = M.(F.0
  ) - M.(meet rng F)
proof
  let S be SigmaField of X, M be sigma_Measure of S, G,F be sequence of S;
  assume that
A1: M.(F.0) <+infty and
A2: G.0 = {} & for n being Nat holds G.(n+1) = F.0 \ F.n & F.
  (n+1) c= F .n;
A3: meet rng F = F.0 \ union rng G by A2,Th5;
A4: M.(F.0 \ meet rng F) = M.(union rng G) by A2,Th4;
  M.(F.0 \ union rng G) <> +infty by A1,MEASURE1:31,XBOOLE_1:36;
  then M.(meet rng F) <+infty by A3,XXREAL_0:4;
  hence thesis by A3,A4,MEASURE1:32,XBOOLE_1:36;
end;

theorem
  for S being SigmaField of X, M being sigma_Measure of S, G,F being
  sequence of S st (M.(F.0) <+infty & G.0 = {} & for n being Nat
  holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F.n ) holds M.(meet rng F) = M.(F.0) -
  sup(rng (M*G))
proof
  let S be SigmaField of X, M be sigma_Measure of S, G,F be sequence of S;
  assume that
A1: M.(F.0) <+infty and
A2: G.0 = {} & for n being Nat holds G.(n+1) = F.0 \ F.n & F.
  (n+1) c= F .n;
  for n being Nat holds G.n c= G.(n+1) by A2,MEASURE2:13;
  then M.(union rng G) = sup(rng (M*G)) by MEASURE2:23;
  hence thesis by A1,A2,Th6;
end;

theorem Th9:
  for S being SigmaField of X, M being sigma_Measure of S, G,F
being sequence of S st (M.(F.0) <+infty & G.0 = {} & for n being Nat
   holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F.n )
holds M.(F.0) in REAL & inf(rng (M*F)) in REAL & sup(rng (M*G)) in REAL
proof
  let S be SigmaField of X, M be sigma_Measure of S, G,F be sequence of S;
  assume that
A1: M.(F.0) <+infty and
A2: G.0 = {} and
A3: for n being Nat holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F .n;
  reconsider P = {} as Element of S by PROB_1:4;
A4: 0 in REAL by XREAL_0:def 1;
  M.P <= M.(F.0) by MEASURE1:31,XBOOLE_1:2;
  then 0. <= M.(F.0) by VALUED_0:def 19;
  hence
A5: M.(F.0) in REAL by A1,XXREAL_0:46,A4;
  for x being ExtReal st x in rng(M*G) holds x <= M.(F.0)
  proof
    let x be ExtReal;
A6: dom (M*G) = NAT by FUNCT_2:def 1;
    assume x in rng(M*G);
    then consider n being object such that
A7: n in NAT and
A8: (M*G).n = x by A6,FUNCT_1:def 3;
    reconsider n as Element of NAT by A7;
A9: x = M.(G.n) by A6,A8,FUNCT_1:12;
A10: (ex k being Nat st n = k + 1) implies x <= M.(F.0)
    proof
      given k being Nat such that
A11:  n = k + 1;
      reconsider k as Element of NAT by ORDINAL1:def 12;
      G.n = F.0 \ F.k by A3,A11;
      hence thesis by A9,MEASURE1:31,XBOOLE_1:36;
    end;
    n = 0 implies x <= M.(F.0) by A2,A9,MEASURE1:31,XBOOLE_1:2;
    hence thesis by A10,NAT_1:6;
  end;
  then M.(F.0) is UpperBound of rng(M*G) by XXREAL_2:def 1;
  then
A12: sup(rng(M*G)) <= M.(F.0) by XXREAL_2:def 3;
  for x being ExtReal st x in rng(M*F) holds 0.<= x
  proof
    let x be ExtReal;
A13: dom (M*F) = NAT by FUNCT_2:def 1;
A14: (M*F) is nonnegative by MEASURE2:1;
    assume x in rng(M*F);
    then ex n being object st n in NAT & (M*F).n = x by A13,FUNCT_1:def 3;
    hence thesis by A14,SUPINF_2:39;
  end;
  then 0. is LowerBound of rng(M*F) by XXREAL_2:def 2;
  then
A15: inf(rng(M*F)) >= In(0,REAL) by XXREAL_2:def 4;
  ex x being R_eal st x in rng(M*F) & x = M.(F.0)
  proof
    take (M*F).0;
    dom (M*F) = NAT by FUNCT_2:def 1;
    hence thesis by FUNCT_1:12,FUNCT_2:4;
  end;
  then inf(rng(M*F)) <= M.(F.0) by XXREAL_2:3;
  hence inf(rng(M*F)) in REAL by A5,A15,XXREAL_0:45;
  In(0,REAL) <= sup(rng(M*G))
  proof
    set x = (M*G).0;
    for x being R_eal st x in rng(M*G) holds 0.<= x
    proof
      let x be R_eal;
A16:  dom (M*G) = NAT by FUNCT_2:def 1;
A17:  (M*G) is nonnegative by MEASURE2:1;
      assume x in rng(M*G);
      then ex n being object st n in NAT & (M*G).n = x by A16,FUNCT_1:def 3;
      hence thesis by A17,SUPINF_2:39;
    end;
    then
A18: 0. <= x by FUNCT_2:4;
    x <= sup rng(M*G) by FUNCT_2:4,XXREAL_2:4;
    hence thesis by A18,XXREAL_0:2;
  end;
  hence thesis by A5,A12,XXREAL_0:45;
end;

theorem Th10:
  for S being SigmaField of X, M being sigma_Measure of S, G,F
being sequence of S st (M.(F.0) <+infty & G.0 = {} & for n being Nat
 holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F.n ) holds sup rng (M*G) = M.(F.0)
  - inf rng (M*F)
proof
  let S be SigmaField of X, M be sigma_Measure of S, G,F be sequence of S;
  assume that
A1: M.(F.0) <+infty and
A2: G.0 = {} and
A3: for n being Nat holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F .n;
  set l = M.(F.0) - inf rng (M*F);
  for x being ExtReal st x in rng (M*G) holds x <= l
  proof
    let x be ExtReal;
A4: dom (M*G) = NAT by FUNCT_2:def 1;
    assume x in rng (M*G);
    then consider n being object such that
A5: n in NAT and
A6: (M*G).n = x by A4,FUNCT_1:def 3;
    M*G is nonnegative by MEASURE2:1;
    then x >= In(0,REAL) by A5,A6,SUPINF_2:39;
    then
A7: x > -infty by XXREAL_0:2,12;
    reconsider n as Element of NAT by A5;
A8: n = 0 implies G.n c= F.0 by A2;
A9: dom (M*F) = NAT by FUNCT_2:def 1;
A10: n = 0 implies M.(F.0 \ G.n) in rng (M*F)
    proof
      assume
A11:  n = 0;
      M.(F.0) = (M*F).0 by A9,FUNCT_1:12;
      hence thesis by A2,A11,FUNCT_2:4;
    end;
A12: (ex k being Nat st n = k + 1) implies M.(F.0 \ G.n) in rng (M*F)
    proof
        defpred P[Nat] means F.$1 c= F.0;
A13:    for k being Nat st P[k] holds P[k+1]
        proof
          let k be Nat;
          assume
A14:      F.k c= F.0;
          F.(k+1) c= F.k by A3;
          hence thesis by A14,XBOOLE_1:1;
        end;
A15:    P[0];
A16:  for n being Nat holds P[n] from NAT_1:sch 2(A15,A13);
      given k being Nat such that
A17:  n = k + 1;
      reconsider k as Element of NAT by ORDINAL1:def 12;
A18:  M.(F.k) = (M*F).k by A9,FUNCT_1:12;
      F.0 \ G.n = F.0 \ ( F.0 \ F.k) by A3,A17
        .= F.0 /\ F.k by XBOOLE_1:48
        .= F.k by A16,XBOOLE_1:28;
      hence thesis by A18,FUNCT_2:4;
    end;
A19: (ex k being Nat st n = k + 1) implies G.n c= F.0
    proof
      given k being Nat such that
A20:  n = k + 1;
      reconsider k as Element of NAT by ORDINAL1:def 12;
      G.n = F.0 \ F.k by A3,A20;
      hence thesis by XBOOLE_1:36;
    end;
A21: x = M.(G.n) by A4,A6,FUNCT_1:12;
    then x <> +infty by A1,A8,A19,MEASURE1:31,NAT_1:6;
    then
A22: x in REAL by A7,XXREAL_0:14;
    reconsider x as R_eal by XXREAL_0:def 1;
    M.(F.0) in REAL & inf(rng(M*F)) in REAL by A1,A2,A3,Th9;
    then consider a,b,c being Real such that
A23: a = M.(F.0) and
A24: b = x and
A25: c = inf(rng (M*F)) by A22;
    M.(F.0) - x = a - b by A23,A24;
    then
A26: (M.(F.0) - x) + x = (a - b) + b by A24
      .= M.(F.0) by A23;
    inf(rng (M*F)) + x = c + b by A24,A25;
    then
A27: inf(rng (M*F)) + x - inf(rng (M*F)) = b + c - c by A25
      .= x by A24;
    M.(F.0) - x = M.(F.0 \ G.n) by A21,A8,A19,A22,MEASURE1:32,NAT_1:6
,XXREAL_0:9;
    then inf(rng (M*F)) <= M.(F.0) - x by A10,A12,NAT_1:6,XXREAL_2:3;
    then inf(rng (M*F)) + x <= M.(F.0) by A26,XXREAL_3:36;
    hence thesis by A27,XXREAL_3:37;
  end;
  then
A28: l is UpperBound of rng (M*G) by XXREAL_2:def 1;
A29: for n being Nat holds G.n c= G.(n+1) by A2,A3,MEASURE2:13;
  for y being UpperBound of rng (M*G) holds l <= y
  proof
    let y be UpperBound of rng (M*G);
    l <= y
    proof
      for x being ExtReal st x in rng (M*F) holds M.(meet rng F) <= x
      proof
        let x be ExtReal;
A30:    dom (M*F) = NAT by FUNCT_2:def 1;
        assume x in rng (M*F);
        then consider n being object such that
A31:    n in NAT and
A32:    (M*F).n = x by A30,FUNCT_1:def 3;
        reconsider n as Element of NAT by A31;
A33:    meet rng F c= F.n by FUNCT_2:4,SETFAM_1:3;
        x = M.(F.n) by A30,A32,FUNCT_1:12;
        hence thesis by A33,MEASURE1:31;
      end;
      then M.(meet rng F) is LowerBound of rng (M*F) by XXREAL_2:def 2;
      then
A34:  M.(meet rng F) <= inf(rng (M*F)) by XXREAL_2:def 4;
      set Q = union rng G;
      sup rng(M*G) = M.Q by A29,MEASURE2:23;
      then
A35:  M.Q <= y by XXREAL_2:def 3;
      M.(F.0) - M.(meet rng F) = M.(union rng G) by A1,A2,A3,Th7;
      then l <= M.Q by A34,XXREAL_3:37;
      hence thesis by A35,XXREAL_0:2;
    end;
    hence thesis;
  end;
  hence thesis by A28,XXREAL_2:def 3;
end;

theorem Th11:
  for S being SigmaField of X, M being sigma_Measure of S, G,F
being sequence of S st (M.(F.0) <+infty & G.0 = {} & for n being Nat
 holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F.n ) holds inf(rng (M*F)) = M.(F.0)
  - sup(rng (M*G))
proof
  let S be SigmaField of X, M be sigma_Measure of S, G,F be sequence of S;
  assume that
A1: M.(F.0) <+infty and
A2: G.0 = {} and
A3: for n being Nat holds G.(n+1) = F.0 \ F.n & F.(n+1) c= F .n;
  set l = M.(F.0) - sup(rng (M*G));
  for x being ExtReal st x in rng (M*F) holds l <= x
  proof
    let x be ExtReal;
    assume
A4: x in rng (M*F);
    x <> +infty implies l <=x
    proof
A5:   dom (M*F) = NAT by FUNCT_2:def 1;
      then consider n being object such that
A6:   n in NAT and
A7:   (M*F).n = x by A4,FUNCT_1:def 3;
      M*F is nonnegative by MEASURE2:1;
      then
A8:   0. <= x by A6,A7,SUPINF_2:39;
      assume
A9:   x <> +infty;
      reconsider x as R_eal by XXREAL_0:def 1;
      x <= +infty by XXREAL_0:3;
      then x < +infty by A9,XXREAL_0:1;
      then
A10:  x in REAL by A8,XXREAL_0:14,46;
      M.(F.0) in REAL & sup(rng(M*G)) in REAL by A1,A2,A3,Th9;
      then consider a,b,c being Real such that
A11:  a = M.(F.0) and
A12:  b = x and
A13:  c = sup(rng (M*G)) by A10;
      sup(rng (M*G)) + x = c + b by A12,A13;
      then
A14:  sup(rng (M*G)) + x - sup(rng (M*G)) = b + c - c by A13
        .= x by A12;
      reconsider n as Element of NAT by A6;
A15:  dom (M*G) = NAT by FUNCT_2:def 1;
A16:  M.(F.0) - x <= sup(rng (M*G))
      proof
        set k = n + 1;
A17:    for n being Nat holds F.n c= F.0
        proof
          defpred P[Nat] means F.$1 c= F.0;
A18:      for k being Nat st P[k] holds P[k+1]
          proof
            let k be Nat;
            assume
A19:        F.k c= F.0;
            F.(k+1) c= F.k by A3;
            hence thesis by A19,XBOOLE_1:1;
          end;
A20:      P[0];
          thus for k being Nat holds P[k] from NAT_1:sch 2(A20,A18);
        end;
        then M.(F.n) <= M.(F.0) by MEASURE1:31;
        then
A21:    M.(F.n) <+infty by A1,XXREAL_0:2;
        M.(F.0) - x = M.(F.0) - M.(F.n) by A5,A7,FUNCT_1:12
          .= M.(F.0 \ F.n) by A17,A21,MEASURE1:32
          .= M.(G.(n+1)) by A3;
        then M.(F.0) - x = (M*G).k by A15,FUNCT_1:12;
        hence thesis by FUNCT_2:4,XXREAL_2:4;
      end;
      M.(F.0) - x = a - b by A11,A12;
      then (M.(F.0) - x) + x = (a - b) + b by A12
        .= M.(F.0) by A11;
      then M.(F.0) <= sup(rng (M*G)) + x by A16,XXREAL_3:36;
      hence thesis by A14,XXREAL_3:37;
    end;
    hence thesis by XXREAL_0:4;
  end;
  then
A22: l is LowerBound of rng (M*F) by XXREAL_2:def 2;
  for y being LowerBound of rng (M*F) holds y <= l
  proof
A23: inf(rng (M*F)) in REAL by A1,A2,A3,Th9;
    sup(rng (M*G)) in REAL & M.(F.0) in REAL by A1,A2,A3,Th9;
    then consider a,b,c being Real such that
A24: a = sup(rng (M*G)) and
A25: b = M.(F.0) and
A26: c = inf(rng (M*F)) by A23;
    sup(rng (M*G)) + inf(rng (M*F)) = a + c by A24,A26;
    then
A27: sup(rng (M*G)) + inf(rng (M*F)) - sup(rng (M*G)) = c + a - a by A24
      .= inf(rng (M*F)) by A26;
    let y be LowerBound of rng (M*F);
    consider s,t,r being R_eal such that
    s = sup(rng (M*G)) and
    t = M.(F.0) - inf(rng (M*F)) and
A28: r = inf(rng (M*F));
A29: sup(rng (M*G)) = M.(F.0) - inf(rng (M*F)) by A1,A2,A3,Th10;
    M.(F.0) - inf(rng (M*F)) = b - c by A25,A26;
    then M.(F.0) - inf(rng (M*F)) + r = b - c + c by A26,A28
      .= M.(F.0) by A25;
    hence thesis by A29,A28,A27,XXREAL_2:def 4;
  end;
  hence thesis by A22,XXREAL_2:def 4;
end;

theorem
  for S being SigmaField of X, M being sigma_Measure of S, F being
sequence of S st (for n being Nat holds F.(n+1) c= F.n) & M.(F.0
  ) <+infty holds M.(meet rng F) = inf(rng (M*F))
proof
  let S be SigmaField of X, M be sigma_Measure of S, F be sequence of S;
  assume that
A1: for n being Nat holds F.(n+1) c= F.n and
A2: M.(F.0) <+infty;
  consider G being sequence of S such that
A3: G.0 = {} & for n being Nat holds G.(n+1) = F.0 \ F.n by
MEASURE2:9;
A4: union rng G = F.0 \ meet rng F by A1,A3,Th4;
A5: M.(F.0 \ union rng G) = M.(meet rng F) by A1,A3,Th5;
A6: for A being Element of S st A = union rng G holds M.(meet rng F) = M.(F.
  0) - M.A
  proof
    let A be Element of S;
    assume
A7: A = union rng G;
    M.(F.0 \ meet rng F) <> +infty by A2,MEASURE1:31,XBOOLE_1:36;
    then M.A <+infty by A4,A7,XXREAL_0:4;
    hence thesis by A4,A5,A7,MEASURE1:32,XBOOLE_1:36;
  end;
  for n being Nat holds G.n c= G.(n+1) by A1,A3,MEASURE2:13;
  then M.(union rng G) = sup(rng (M*G)) by MEASURE2:23;
  then M.(meet rng F) = M.(F.0) - sup(rng (M*G)) by A6;
  hence thesis by A1,A2,A3,Th11;
end;

theorem Th13:
  for S being SigmaField of X, M being Measure of S, F being
  Sep_Sequence of S holds SUM(M*F) <= M.(union rng F)
proof
  let S be SigmaField of X, M be Measure of S, F be Sep_Sequence of S;
  set T = rng F;
  consider G being sequence of S such that
A1: G.0 = F.0 and
A2: for n being Nat holds G.(n+1) = F.(n+1) \/ G.n by MEASURE2:4;
  {} is Subset of X by XBOOLE_1:2;
  then consider H being sequence of bool X such that
A3: rng H = {union T,{}} and
A4: H.0 = union T and
A5: for n being Nat st 0 < n holds H.n = {} by MEASURE1:19;
  rng H c= S
  proof
    let a be object;
    assume a in rng H;
    then a = union T or a = {} by A3,TARSKI:def 2;
    hence thesis by PROB_1:4;
  end;
  then reconsider H as sequence of S by FUNCT_2:6;
  defpred P[Nat] means Ser(M*F).$1 = M.(G.$1);
A6: dom (M*F) = NAT by FUNCT_2:def 1;
A7: for n being Nat holds G.n /\ F.(n+1) = {}
  proof
    let n be Nat;
A8: for n being Nat holds for k being Element of NAT st n < k
    holds G.n /\ F.k = {}
    proof
      defpred P[Nat] means
   for k being Element of NAT st $1 < k
      holds G.$1 /\ F.k = {};
A9:   for n being Nat st P[n] holds P[n+1]
      proof
        let n be Nat;
        assume
A10:    for k being Element of NAT st n < k holds G.n /\ F.k = {};
        let k be Element of NAT;
        assume
A11:    n+1 < k;
        then
A12:    n < k by NAT_1:13;
        F.(n+1) misses F.k by A11,PROB_2:def 2;
        then
A13:    F.(n+1) /\ F.k = {};
        G.(n+1) /\ F.k = (F.(n+1) \/ G.n) /\ F.k by A2
          .= (F.(n+1) /\ F.k) \/ (G.n /\ F.k) by XBOOLE_1:23;
        hence thesis by A10,A12,A13;
      end;
A14:  P[0]
      by PROB_2:def 2,A1,XBOOLE_0:def 7;
      thus for n being Nat holds P[n] from NAT_1:sch 2(A14, A9);
    end;
    n < n + 1 by NAT_1:13;
    hence thesis by A8;
  end;
A15: for k being Nat st P[k] holds P[k+1]
  proof
    let k be Nat;
    G.k /\ F.(k+1) = {} by A7;
    then
A16: G.k misses F.(k+1);
    assume Ser(M*F).k = M.(G.k);
    then Ser(M*F).(k+1) = M.(G.k) + (M*F).(k+1) by SUPINF_2:def 11;
    then Ser(M*F).(k+1) = M.(G.k) + M.(F.(k+1)) by A6,FUNCT_1:12
      .= M.(F.(k+1) \/ G.k) by A16,MEASURE1:def 3
      .= M.(G.(k+1)) by A2;
    hence thesis;
  end;
  Ser(M*F).0 = (M*F).0 by SUPINF_2:def 11;
  then
A17: P[0] by A1,A6,FUNCT_1:12;
A18: for n being Nat holds P[n] from NAT_1:sch 2(A17,A15 );
  defpred P[Nat] means Ser(M*H).$1 = M.(union T);
A19: for n being Nat st P[n] holds P[n+1]
  proof
    let n be Nat;
    0 <= n by NAT_1:2;
    then 0 < n + 1 by NAT_1:13;
    then
A20: H.(n+1) = {} by A5;
    dom (M*H) = NAT by FUNCT_2:def 1;
    then (M*H).(n+1) = M.({}) by A20,FUNCT_1:12;
    then
A21: (M*H).(n+1) = 0. by VALUED_0:def 19;
    assume Ser(M*H).n = M.(union T);
    then Ser(M*H).(n+1) = M.(union T) + (M*H).(n+1) by SUPINF_2:def 11;
    hence thesis by A21,XXREAL_3:4;
  end;
  Ser(M*H).0 = (M*H).0 & dom (M*H) = NAT by FUNCT_2:def 1,SUPINF_2:def 11;
  then
A22: P[0] by A4,FUNCT_1:12;
A23: for n being Nat holds P[n] from NAT_1:sch 2(A22,A19 );
A24: for r being Element of NAT st 1 <= r holds (M*H).r = 0.
  proof
    let r be Element of NAT;
    assume 1 <= r;
    then 0 + 1 <= r;
    then 0 < r by NAT_1:13;
    then
A25: H.r = {} by A5;
    dom (M*H) = NAT by FUNCT_2:def 1;
    then (M*H).r = M.({}) by A25,FUNCT_1:12;
    hence thesis by VALUED_0:def 19;
  end;
A26: for n being Nat holds G.n c= union T
  proof
    defpred P[Nat] means G.$1 c= union T;
A27: for n being Nat st P[n] holds P[n+1]
    proof
      let n be Nat;
      assume
A28:  G.n c= union T;
      G.(n+1) = F.(n+1) \/ G.n & F.(n+1) c= union T by A2,FUNCT_2:4,ZFMISC_1:74
;
      hence thesis by A28,XBOOLE_1:8;
    end;
A29: P[0] by A1,FUNCT_2:4,ZFMISC_1:74;
    thus for n being Nat holds P[n] from NAT_1:sch 2(A29,A27 );
  end;
A30: for n being Element of NAT holds Ser(M*F).n <= Ser(M*H).n
  proof
    let n be Element of NAT;
    Ser(M*F).n = M.(G.n) by A18;
    then Ser(M*F).n <= M.(union T) by A26,MEASURE1:8;
    hence thesis by A23;
  end;
  M*H is nonnegative by MEASURE1:25;
  then SUM(M*H) = Ser(M*H).1 by A24,SUPINF_2:48;
  then SUM(M*H) = M.(union T) by A23;
  hence thesis by A30,Th1;
end;

theorem
  for S being SigmaField of X, M being Measure of S st (for F being
  Sep_Sequence of S holds M.(union rng F) <= SUM(M*F)) holds M is sigma_Measure
  of S
proof
  let S be SigmaField of X, M be Measure of S;
  assume
A1: for F being Sep_Sequence of S holds M.(union rng F) <= SUM(M*F);
  for F being Sep_Sequence of S holds SUM(M*F) = M.(union rng F)
  proof
    let F be Sep_Sequence of S;
    M.(union rng F) <= SUM(M*F) & SUM(M*F) <= M.(union rng F) by A1,Th13;
    hence thesis by XXREAL_0:1;
  end;
  hence thesis by MEASURE1:def 6;
end;

::
::                Completeness of sigma_additive Measure
::

definition
  let X be set;
  let Sigma be SigmaField of X;
  let M be sigma_Measure of Sigma;
  attr M is complete means
  for A being Subset of X, B being set st
  B in Sigma & A c= B & M.B = 0. holds A in Sigma;
end;

definition
  let X be set;
  let S be SigmaField of X;
  let M be sigma_Measure of S;
  mode thin of M -> Subset of X means
    :Def2:
    ex B being set st B in S & it c= B & M.B = 0.;
  existence
  proof
    reconsider A = {} as Subset of X by XBOOLE_1:2;
    take A;
    take B={};
    thus B in S by PROB_1:4;
    thus A c= B;
    thus thesis by VALUED_0:def 19;
  end;
end;

definition
  let X be set;
  let S be SigmaField of X;
  let M be sigma_Measure of S;
  func COM(S,M) -> non empty Subset-Family of X means
  :Def3:
  for A being set
holds (A in it iff ex B being set st B in S & ex C being thin of M st A = B \/
  C );
  existence
  proof
A1: ex B being set st B in S & {} c= B & M.B = 0.
    proof
      consider B being set such that
A2:   B = {} & B in S by PROB_1:4;
      take B;
      thus thesis by A2,VALUED_0:def 19;
    end;
A3: {} is Subset of X by XBOOLE_1:2;
A4: for A being set st A = {} holds ex B being set st B in S & ex C being
    thin of M st A = B \/ C
    proof
      reconsider C = {} as thin of M by A3,A1,Def2;
      let A be set;
      consider B being set such that
A5:   B = {} and
A6:   B in S by PROB_1:4;
      assume A = {};
      then A = B \/ C by A5;
      hence thesis by A6;
    end;
    defpred P[set] means for A being set st A = $1 holds ex B being set st B
    in S & ex C being thin of M st A = B \/ C;
    consider D being set such that
A7: for y being set holds y in D iff y in bool X & P[y] from XFAMILY:
    sch 1;
A8: for A being set holds (A in D iff ex B being set st (B in S & ex C
    being thin of M st A = B \/ C) )
    proof
      let A be set;
A9:   A in D iff (A in bool X & for y being set st y = A holds ex B being
      set st (B in S & ex C being thin of M st y = B \/ C) ) by A7;
      (ex B being set st (B in S & ex C being thin of M st A = B \/ C))
      implies A in D
      proof
        assume
A10:    ex B being set st (B in S & ex C being thin of M st A = B \/ C );
        then A c= X by XBOOLE_1:8;
        hence thesis by A9,A10;
      end;
      hence thesis by A7;
    end;
A11: D c= bool X
    by A7;
    {} c= X;
    then reconsider D as non empty Subset-Family of X by A7,A11,A4;
    take D;
    thus thesis by A8;
  end;
  uniqueness
  proof
    let D1,D2 be non empty Subset-Family of X such that
A12: for A being set holds (A in D1 iff ex B being set st (B in S & ex
    C being thin of M st A = B \/ C) ) and
A13: for A being set holds (A in D2 iff ex B being set st (B in S & ex
    C being thin of M st A = B \/ C) );
    for A being object holds A in D1 iff A in D2
    proof
      let A be object;
      thus A in D1 implies A in D2
      proof
        assume A in D1;
        then ex B being set st (B in S & ex C being thin of M st A = B \/ C)
        by A12;
        hence thesis by A13;
      end;
      assume A in D2;
      then ex B being set st (B in S & ex C being thin of M st A = B \/ C) by
A13;
      hence thesis by A12;
    end;
    hence thesis by TARSKI:2;
  end;
end;

definition
  let X be set;
  let S be SigmaField of X;
  let M be sigma_Measure of S;
  let A be Element of COM(S,M);
  func MeasPart(A) -> non empty Subset-Family of X means
  :Def4:
  for B being set holds (B in it iff B in S & B c= A & A \ B is thin of M );
  existence
  proof
    defpred P[set] means for t being set st t = $1 holds t in S & t c= A & A \
    t is thin of M;
    consider D being set such that
A1: for t being set holds t in D iff t in bool X & P[t] from XFAMILY:sch 1;
A2: for B being set holds B in D iff B in S & B c= A & A \ B is thin of M
    proof
      let B be set;
      B in S & B c= A & A \ B is thin of M implies B in D
      proof
        assume that
A3:     B in S and
A4:     B c= A & A \ B is thin of M;
        for t being set st t = B holds t in S & t c= A & A \ t is thin of
        M by A3,A4;
        hence thesis by A1,A3;
      end;
      hence thesis by A1;
    end;
A5: D c= bool X
    proof
      let B be object;
      assume B in D;
      then B in S by A2;
      hence thesis;
    end;
    D <> {}
    proof
      consider B being set such that
A6:   B in S and
A7:   ex C being thin of M st A = B \/ C by Def3;
      consider C being thin of M such that
A8:   A = B \/ C by A7;
      consider E being set such that
A9:   E in S and
A10:  C c= E and
A11:  M.E = 0. by Def2;
      A \ B = C \ B by A8,XBOOLE_1:40;
      then A \ B c= C by XBOOLE_1:36;
      then A \ B c= E by A10;
      then
A12:  A \ B is thin of M by A9,A11,Def2;
      B c= A by A8,XBOOLE_1:7;
      hence thesis by A2,A6,A12;
    end;
    then reconsider D as non empty Subset-Family of X by A5;
    take D;
    thus thesis by A2;
  end;
  uniqueness
  proof
    let D1,D2 be non empty Subset-Family of X such that
A13: for B being set holds B in D1 iff B in S & B c= A & A \ B is thin of M and
A14: for B being set holds B in D2 iff B in S & B c= A & A \ B is thin of M;
    for B being object holds B in D1 iff B in D2
    proof
      let B be object;
      reconsider BB = B as set by TARSKI:1;
      thus B in D1 implies B in D2
      proof
        assume
A15:    B in D1;
        then
A16:    A \ BB is thin of M by A13;
        B in S & BB c= A by A13,A15;
        hence thesis by A14,A16;
      end;
      assume
A17:  B in D2;
      then
A18:  A \ BB is thin of M by A14;
      B in S & BB c= A by A14,A17;
      hence thesis by A13,A18;
    end;
    hence thesis by TARSKI:2;
  end;
end;

theorem Th15:
  for S being SigmaField of X, M being sigma_Measure of S, F being
  sequence of COM(S,M) holds ex G being sequence of S st for n being
  Element of NAT holds G.n in MeasPart(F.n)
proof
  let S be SigmaField of X, M be sigma_Measure of S, F be sequence of COM(
  S,M);
  defpred P[Element of NAT, set] means for n being Element of NAT, y being set
  st n = $1 & y = $2 holds y in MeasPart(F.n);
A1: for t being Element of NAT ex A being Element of S st P[t,A]
  proof
    let t be Element of NAT;
    set A = the Element of MeasPart(F.t);
    reconsider A as Element of S by Def4;
    take A;
    thus thesis;
  end;
  ex G being sequence of S st for t being Element of NAT holds P[t,G.t
  ] from FUNCT_2:sch 3(A1);
  then consider G being sequence of S such that
A2: for t being Element of NAT, n being Element of NAT, y being set st n
  = t & y = G.t holds y in MeasPart(F.n);
  take G;
  thus thesis by A2;
end;

theorem Th16:
  for S being SigmaField of X, M being sigma_Measure of S, F being
sequence of COM(S,M), G being sequence of S ex H being sequence of
  bool X st for n being Element of NAT holds H.n = F.n \ G.n
proof
  let S be SigmaField of X, M be sigma_Measure of S, F be sequence of COM(
  S,M), G be sequence of S;
  defpred P[Element of NAT, set] means for n being Element of NAT, y being set
  st n = $1 & y = $2 holds y = F.n \ G.n;
A1: for t being Element of NAT ex A being Subset of X st P[t,A]
  proof
    let t be Element of NAT;
    F.t is Element of COM(S,M);
    then reconsider A = F.t \ G.t as Subset of X by XBOOLE_1:1;
    take A;
    thus thesis;
  end;
  ex H being sequence of bool X st for t being Element of NAT holds P[
  t,H.t] from FUNCT_2:sch 3(A1);
  then consider H being sequence of bool X such that
A2: for t being Element of NAT holds for n being Element of NAT for y
  being set holds (n = t & y = H.t implies y = F.n \ G.n);
  take H;
  thus thesis by A2;
end;

theorem Th17:
  for S being SigmaField of X, M being sigma_Measure of S, F being
  sequence of bool X st (for n being Element of NAT holds F.n is thin of M)
holds ex G being sequence of S st for n being Element of NAT holds F.n c= G
  .n & M.(G.n) = 0.
proof
  let S be SigmaField of X, M be sigma_Measure of S, F be sequence of bool
  X;
  defpred P[Element of NAT, set] means for n being Element of NAT, y being set
  st n = $1 & y = $2 holds y in S & F.n c= y & M.y = 0.;
  assume
A1: for n being Element of NAT holds F.n is thin of M;
A2: for t being Element of NAT ex A being Element of S st P[t,A]
  proof
    let t be Element of NAT;
    F.t is thin of M by A1;
    then consider A being set such that
A3: A in S and
A4: F.t c= A & M.A = 0. by Def2;
    reconsider A as Element of S by A3;
    take A;
    thus thesis by A4;
  end;
  ex G being sequence of S st for t being Element of NAT holds P[t,G.t
  ] from FUNCT_2:sch 3(A2);
  then consider G being sequence of S such that
A5: for t being Element of NAT, n being Element of NAT, y being set st n
  = t & y = G.t holds y in S & F.n c= y & M.y = 0.;
  take G;
  thus thesis by A5;
end;

theorem Th18:
  for S being SigmaField of X, M being sigma_Measure of S, D being
  non empty Subset-Family of X st (for A being set holds (A in D iff ex B being
set st B in S & ex C being thin of M st A = B \/ C)) holds D is SigmaField of X
proof
  let S be SigmaField of X, M be sigma_Measure of S, D be non empty
  Subset-Family of X;
  assume
A1: for A being set holds A in D iff ex B being set st B in S & ex C
  being thin of M st A = B \/ C;
A2: for K being N_Sub_set_fam of X st K c= D holds union K in D
  proof
    let K be N_Sub_set_fam of X;
    consider F being sequence of bool X such that
A3: K = rng F by SUPINF_2:def 8;
    assume
A4: K c= D;
A5: for n being Element of NAT holds F.n in D
    proof
      let n be Element of NAT;
      F.n in K by A3,FUNCT_2:4;
      hence thesis by A4;
    end;
A6: for n being Element of NAT holds ex B being set st B in S & ex C
    being thin of M st F.n = B \/ C
    by A5,A1;
    for n being Element of NAT holds F.n in COM(S,M)
    proof
      let n be Element of NAT;
      ex B being set st B in S & ex C being thin of M st F.n = B \/ C by A6;
      hence thesis by Def3;
    end;
    then
A7: for n being object st n in NAT holds F.n in COM(S,M);
A8: dom F = NAT by FUNCT_2:def 1;
    then reconsider F as sequence of COM(S,M) by A7,FUNCT_2:3;
    consider G being sequence of S such that
A9: for n being Element of NAT holds G.n in MeasPart(F.n) by Th15;
    consider H be sequence of bool X such that
A10: for n being Element of NAT holds H.n = F.n \ G.n by Th16;
A11: for n being Element of NAT holds G.n in S & G.n c= F.n & F.n \ G.n is
    thin of M
    proof
      let n be Element of NAT;
      G.n in MeasPart(F.n) by A9;
      hence thesis by Def4;
    end;
    for n being Element of NAT holds H.n is thin of M
    proof
      let n be Element of NAT;
      F.n \ G.n is thin of M by A11;
      hence thesis by A10;
    end;
    then consider L being sequence of S such that
A12: for n being Element of NAT holds H.n c= L.n & M.(L.n) = 0. by Th17;
    ex B being set st B in S & ex C being thin of M st union K = B \/ C
    proof
      set B = union rng G;
      take B;
A13:  union rng G c= union rng F
      proof
        let x be object;
        assume x in union rng G;
        then consider Z being set such that
A14:    x in Z and
A15:    Z in rng G by TARSKI:def 4;
        dom G = NAT by FUNCT_2:def 1;
        then consider n being object such that
A16:    n in NAT and
A17:    Z = G.n by A15,FUNCT_1:def 3;
        reconsider n as Element of NAT by A16;
        set P = F.n;
A18:    G.n c= P by A11;
        ex P being set st P in rng F & x in P
        proof
          take P;
          thus thesis by A8,A14,A17,A18,FUNCT_1:def 3;
        end;
        hence thesis by TARSKI:def 4;
      end;
      ex C being thin of M st union K = B \/ C
      proof
        for A being set st A in rng L holds A is measure_zero of M
        proof
          let A be set;
          assume
A19:      A in rng L;
          dom L = NAT by FUNCT_2:def 1;
          then
A20:      ex n being object st n in NAT & A = L.n by A19,FUNCT_1:def 3;
          rng L c= S by MEASURE2:def 1;
          then reconsider A as Element of S by A19;
          M.A = 0. by A12,A20;
          hence thesis by MEASURE1:def 7;
        end;
        then union rng L is measure_zero of M by MEASURE2:14;
        then
A21:    M.(union rng L) = 0. by MEASURE1:def 7;
        set C = union K \ B;
A22:    union K = C \/ union rng F /\ union rng G by A3,XBOOLE_1:51
          .= B \/ C by A13,XBOOLE_1:28;
        reconsider C as Subset of X;
A23:    C c= union rng H
        proof
          let x be object;
          assume
A24:      x in C;
          then x in union rng F by A3,XBOOLE_0:def 5;
          then consider Z being set such that
A25:      x in Z and
A26:      Z in rng F by TARSKI:def 4;
          consider n being object such that
A27:      n in NAT and
A28:      Z = F.n by A8,A26,FUNCT_1:def 3;
          reconsider n as Element of NAT by A27;
A29:      not x in union rng G by A24,XBOOLE_0:def 5;
          not x in G.n
          proof
            dom G = NAT by FUNCT_2:def 1;
            then
A30:        G.n in rng G by FUNCT_1:def 3;
            assume x in G.n;
            hence thesis by A29,A30,TARSKI:def 4;
          end;
          then
A31:      x in F.n \ G.n by A25,A28,XBOOLE_0:def 5;
          ex Z being set st x in Z & Z in rng H
          proof
            take H.n;
            dom H = NAT by FUNCT_2:def 1;
            hence thesis by A10,A31,FUNCT_1:def 3;
          end;
          hence thesis by TARSKI:def 4;
        end;
        union rng H c= union rng L
        proof
          let x be object;
          assume x in union rng H;
          then consider Z being set such that
A32:      x in Z and
A33:      Z in rng H by TARSKI:def 4;
          dom H = NAT by FUNCT_2:def 1;
          then consider n being object such that
A34:      n in NAT and
A35:      Z = H.n by A33,FUNCT_1:def 3;
          reconsider n as Element of NAT by A34;
          n in dom L by A34,FUNCT_2:def 1;
          then
A36:      L.n in rng L by FUNCT_1:def 3;
          H.n c= L.n by A12;
          hence thesis by A32,A35,A36,TARSKI:def 4;
        end;
        then C c= union rng L by A23;
        then C is thin of M by A21,Def2;
        then consider C being thin of M such that
A37:    union K = B \/ C by A22;
        take C;
        thus thesis by A37;
      end;
      hence thesis;
    end;
    hence thesis by A1;
  end;
  for A being set holds A in D implies X\A in D
  proof
    let A be set;
    assume
A38: A in D;
    ex Q being set st Q in S & ex W being thin of M st X \ A = Q \/ W
    proof
      consider B being set such that
A39:  B in S and
A40:  ex C being thin of M st A = B \/ C by A1,A38;
      set P = X \ B;
      consider C being thin of M such that
A41:  A = B \/ C by A40;
      consider G being set such that
A42:  G in S and
A43:  C c= G and
A44:  M.G = 0. by Def2;
      set Q = P \ G;
A45:  X \ A = P \ C by A41,XBOOLE_1:41;
A46:  ex W being thin of M st X \ A = Q \/ W
      proof
        set W = P /\ (G \ C);
        W c= P by XBOOLE_1:17;
        then reconsider W as Subset of X by XBOOLE_1:1;
        reconsider W as thin of M by A42,A44,Def2;
        take W;
        thus thesis by A43,A45,XBOOLE_1:117;
      end;
      take Q;
      X \ B in S by A39,MEASURE1:def 1;
      hence thesis by A42,A46,MEASURE1:6;
    end;
    hence thesis by A1;
  end;
  then reconsider
  D9 = D as compl-closed sigma-additive non empty Subset-Family of
  X by A2,MEASURE1:def 1,def 5;
  D9 is SigmaField of X;
  hence thesis;
end;

registration
  let X be set;
  let S be SigmaField of X;
  let M be sigma_Measure of S;
  cluster COM(S,M) -> sigma-additive compl-closed non empty;
  coherence
  proof
    for A being set holds A in COM(S,M) iff ex B being set st B in S & ex
    C being thin of M st A = B \/ C by Def3;
    hence thesis by Th18;
  end;
end;

theorem Th19:
  for S being SigmaField of X, M being sigma_Measure of S, B1,B2
being set st B1 in S & B2 in S holds for C1,C2 being thin of M holds B1 \/ C1 =
  B2 \/ C2 implies M.B1 = M.B2
proof
  let S be SigmaField of X, M be sigma_Measure of S, B1,B2 be set;
  assume
A1: B1 in S & B2 in S;
  let C1,C2 be thin of M;
  assume
A2: B1 \/ C1 = B2 \/ C2;
  then
A3: B1 c= B2 \/ C2 by XBOOLE_1:7;
A4: B2 c= B1 \/ C1 by A2,XBOOLE_1:7;
  consider D1 being set such that
A5: D1 in S and
A6: C1 c= D1 and
A7: M.D1 = 0. by Def2;
A8: B1 \/ C1 c= B1 \/ D1 by A6,XBOOLE_1:9;
  consider D2 being set such that
A9: D2 in S and
A10: C2 c= D2 and
A11: M.D2 = 0. by Def2;
A12: B2 \/ C2 c= B2 \/ D2 by A10,XBOOLE_1:9;
  reconsider B1,B2,D1,D2 as Element of S by A1,A5,A9;
A13: M.(B1 \/ D1) <= M.B1 + M.D1 & M.B1 + M.D1 = M.B1 by A7,MEASURE1:33
,XXREAL_3:4;
  M.B2 <= M.(B1 \/ D1) by A4,A8,MEASURE1:31,XBOOLE_1:1;
  then
A14: M.B2 <= M.B1 by A13,XXREAL_0:2;
A15: M.(B2 \/ D2) <= M.B2 + M.D2 & M.B2 + M.D2 = M.B2 by A11,MEASURE1:33
,XXREAL_3:4;
  M.B1 <= M.(B2 \/ D2) by A3,A12,MEASURE1:31,XBOOLE_1:1;
  then M.B1 <= M.B2 by A15,XXREAL_0:2;
  hence thesis by A14,XXREAL_0:1;
end;

definition
  let X be set;
  let S be SigmaField of X;
  let M be sigma_Measure of S;
  func COM(M) -> sigma_Measure of COM(S,M) means
  :Def5:
  for B being set st B in S for C being thin of M holds it.(B \/ C) = M.B;
  existence
  proof
    set B = {};
    defpred P[object,object] means
for x,y being set st x in COM(S,M) holds (x = $1
& y = $2 implies (for B being set st B in S for C being thin of M st x = B \/ C
    holds y = M.B));
A1: ex B1 being set st B1 in S & {} c= B1 & M.B1 = 0.
    proof
      take {};
      thus thesis by PROB_1:4,VALUED_0:def 19;
    end;
    {} is Subset of X by XBOOLE_1:2;
    then reconsider C = {} as thin of M by A1,Def2;
A2: for x being object st x in COM(S,M)
    ex y being object st y in ExtREAL & P[x, y]
    proof
      let x be object;
      assume x in COM(S,M);
      then consider B being set such that
A3:   B in S & ex C being thin of M st x = B \/ C by Def3;
      take M.B;
      thus thesis by A3,Th19;
    end;
    consider comM being Function of COM(S,M),ExtREAL such that
A4: for x being object st x in COM(S,M) holds P[x,comM.x] from FUNCT_2:
    sch 1( A2);
A5: for B being set st B in S for C being thin of M holds comM.(B \/ C) = M.B
    proof
      let B be set;
      assume
A6:   B in S;
      let C be thin of M;
      B \/ C in COM(S,M) by A6,Def3;
      hence thesis by A4,A6;
    end;
A7: for F being Sep_Sequence of COM(S,M) holds SUM(comM*F) = comM.(union rng F)
    proof
      let F be Sep_Sequence of COM(S,M);
      consider G being sequence of S such that
A8:   for n being Element of NAT holds G.n in MeasPart(F.n) by Th15;
      consider H be sequence of bool X such that
A9:   for n being Element of NAT holds H.n = F.n \ G.n by Th16;
A10:  for n being Element of NAT holds G.n in S & G.n c= F.n & F.n \ G.n
      is thin of M
      proof
        let n be Element of NAT;
        G.n in MeasPart(F.n) by A8;
        hence thesis by Def4;
      end;
      for n being Element of NAT holds H.n is thin of M
      proof
        let n be Element of NAT;
        F.n \ G.n is thin of M by A10;
        hence thesis by A9;
      end;
      then consider L being sequence of S such that
A11:  for n being Element of NAT holds H.n c= L.n & M.(L.n) = 0. by Th17;
A12:  for n,m being object st n <> m holds G.n misses G.m
      proof
        let n,m be object;
A13:    dom F = NAT by FUNCT_2:def 1
          .= dom G by FUNCT_2:def 1;
        for n being object holds G.n c= F.n
        proof
          let n be object;
          per cases;
          suppose
            n in dom F;
            hence thesis by A10;
          end;
          suppose
A14:        not n in dom F;
            then F.n = {} by FUNCT_1:def 2
              .= G.n by A13,A14,FUNCT_1:def 2;
            hence thesis;
          end;
        end;
        then
A15:    G.n c= F.n & G.m c= F.m;
        assume n <> m;
        then F.n misses F.m by PROB_2:def 2;
        then F.n /\ F.m = {};
        then G.n /\ G.m = {} by A15,XBOOLE_1:3,27;
        hence thesis;
      end;
      consider B being set such that
A16:  B = union rng G;
A17:  dom F = NAT by FUNCT_2:def 1;
A18:  B c= union rng F
      proof
        let x be object;
        assume x in B;
        then consider Z being set such that
A19:    x in Z and
A20:    Z in rng G by A16,TARSKI:def 4;
        dom G = NAT by FUNCT_2:def 1;
        then consider n being object such that
A21:    n in NAT and
A22:    Z = G.n by A20,FUNCT_1:def 3;
        reconsider n as Element of NAT by A21;
        set P = F.n;
A23:    G.n c= P by A10;
        ex P being set st P in rng F & x in P
        proof
          take P;
          thus thesis by A17,A19,A22,A23,FUNCT_1:def 3;
        end;
        hence thesis by TARSKI:def 4;
      end;
A24:  ex C being thin of M st union rng F = B \/ C
      proof
        for A being set st A in rng L holds A is measure_zero of M
        proof
          let A be set;
          assume
A25:      A in rng L;
          dom L = NAT by FUNCT_2:def 1;
          then
A26:      ex n being object st n in NAT & A = L.n by A25,FUNCT_1:def 3;
          rng L c= S by MEASURE2:def 1;
          then reconsider A as Element of S by A25;
          M.A = 0. by A11,A26;
          hence thesis by MEASURE1:def 7;
        end;
        then union rng L is measure_zero of M by MEASURE2:14;
        then
A27:    M.(union rng L) = 0. by MEASURE1:def 7;
        set C = union rng F \ B;
A28:    union rng F = C \/ union rng F /\ B by XBOOLE_1:51
          .= B \/ C by A18,XBOOLE_1:28;
        reconsider C as Subset of X;
A29:    C c= union rng H
        proof
          let x be object;
          assume
A30:      x in C;
          then x in union rng F by XBOOLE_0:def 5;
          then consider Z being set such that
A31:      x in Z and
A32:      Z in rng F by TARSKI:def 4;
          consider n being object such that
A33:      n in NAT and
A34:      Z = F.n by A17,A32,FUNCT_1:def 3;
          reconsider n as Element of NAT by A33;
A35:      not x in union rng G by A16,A30,XBOOLE_0:def 5;
          not x in G.n
          proof
            dom G = NAT by FUNCT_2:def 1;
            then
A36:        G.n in rng G by FUNCT_1:def 3;
            assume x in G.n;
            hence thesis by A35,A36,TARSKI:def 4;
          end;
          then
A37:      x in F.n \ G.n by A31,A34,XBOOLE_0:def 5;
          ex Z being set st x in Z & Z in rng H
          proof
            take H.n;
            dom H = NAT by FUNCT_2:def 1;
            hence thesis by A9,A37,FUNCT_1:def 3;
          end;
          hence thesis by TARSKI:def 4;
        end;
        union rng H c= union rng L
        proof
          let x be object;
          assume x in union rng H;
          then consider Z being set such that
A38:      x in Z and
A39:      Z in rng H by TARSKI:def 4;
          dom H = NAT by FUNCT_2:def 1;
          then consider n being object such that
A40:      n in NAT and
A41:      Z = H.n by A39,FUNCT_1:def 3;
          reconsider n as Element of NAT by A40;
          n in dom L by A40,FUNCT_2:def 1;
          then
A42:      L.n in rng L by FUNCT_1:def 3;
          H.n c= L.n by A11;
          hence thesis by A38,A41,A42,TARSKI:def 4;
        end;
        then C c= union rng L by A29;
        then C is thin of M by A27,Def2;
        then consider C being thin of M such that
A43:    union rng F = B \/ C by A28;
        take C;
        thus thesis by A43;
      end;
      reconsider G as Sep_Sequence of S by A12,PROB_2:def 2;
A44:  for n being Element of NAT holds comM.(F.n) = M.(G.n)
      proof
        let n be Element of NAT;
        F.n \ G.n is thin of M by A10;
        then consider C being thin of M such that
A45:    C = F.n \ G.n;
        F.n = (F.n /\ G.n) \/ (F.n \ G.n) by XBOOLE_1:51
          .= G.n \/ C by A10,A45,XBOOLE_1:28;
        hence thesis by A5;
      end;
A46:  for n being Element of NAT holds (comM*F).n = (M*G).n
      proof
        let n be Element of NAT;
        (comM*F).n = comM.(F.n) by FUNCT_2:15
          .= M.(G.n) by A44
          .= (M*G).n by FUNCT_2:15;
        hence thesis;
      end;
      then for n being Element of NAT holds (M*G).n <= (comM*F).n;
      then
A47:  SUM(M*G) <= SUM(comM*F) by SUPINF_2:43;
      for n being Element of NAT holds (comM*F).n <= (M*G).n by A46;
      then SUM(comM*F) <= SUM(M*G) by SUPINF_2:43;
      then SUM(M*G) = M.(union rng G) & SUM(comM*F) = SUM(M*G) by A47,
MEASURE1:def 6,XXREAL_0:1;
      hence thesis by A5,A16,A24;
    end;
A48: for x being Element of COM(S,M) holds 0. <= comM.x
    proof
      let x be Element of COM(S,M);
      consider B being set such that
A49:  B in S and
A50:  ex C being thin of M st x = B \/ C by Def3;
      reconsider B as Element of S by A49;
      comM.x = M.B by A4,A50;
      hence thesis by MEASURE1:def 2;
    end;
    {} = B \/ C;
    then comM.{} = M.{} by A5,PROB_1:4
      .= 0. by VALUED_0:def 19;
    then reconsider comM as sigma_Measure of COM(S,M) by A48,A7,MEASURE1:def 2
,def 6,VALUED_0:def 19;
    take comM;
    thus thesis by A5;
  end;
  uniqueness
  proof
    let M1,M2 be sigma_Measure of COM(S,M) such that
A51: for B being set st B in S for C being thin of M holds M1.(B \/ C)
    = M.B and
A52: for B being set st B in S for C being thin of M holds M2.(B \/ C) = M.B;
    for x being object st x in COM(S,M) holds M1.x = M2.x
    proof
      let x be object;
      assume x in COM(S,M);
      then consider B being set such that
A53:  B in S & ex C being thin of M st x = B \/ C by Def3;
      M1.x = M.B by A51,A53
        .= M2.x by A52,A53;
      hence thesis;
    end;
    hence thesis by FUNCT_2:12;
  end;
end;

theorem
  for S being SigmaField of X, M being sigma_Measure of S holds COM(M)
  is complete
proof
  let S be SigmaField of X, M be sigma_Measure of S;
  for A being Subset of X, B being set st B in COM(S,M) holds (A c= B & (
  COM(M)).B = 0. implies A in COM(S,M))
  proof
    let A be Subset of X;
    let B be set;
    assume
A1: B in COM(S,M);
    assume that
A2: A c= B and
A3: (COM(M)).B = 0.;
    ex B1 being set st (B1 in S & ex C1 being thin of M st A = B1 \/ C1)
    proof
      take {};
      consider B2 being set such that
A4:   B2 in S and
A5:   ex C2 being thin of M st B = B2 \/ C2 by A1,Def3;
A6:   M.B2 = 0. by A3,A4,A5,Def5;
      consider C2 being thin of M such that
A7:   B = B2 \/ C2 by A5;
      set C1 = (A /\ B2) \/ (A /\ C2);
      consider D2 being set such that
A8:   D2 in S and
A9:   C2 c= D2 and
A10:  M.D2 = 0. by Def2;
      set O = B2 \/ D2;
      A /\ C2 c= C2 by XBOOLE_1:17;
      then
A11:  A /\ B2 c= B2 & A /\ C2 c= D2 by A9,XBOOLE_1:17;
      ex O being set st O in S & C1 c= O & M.O = 0.
      proof
        reconsider B2,D2 as Element of S by A4,A8;
        reconsider O1 = O as Element of S by A4,A8,FINSUB_1:def 1;
        take O;
        M.(B2 \/ D2) <= 0. + 0. & 0. <= M.O1 by A6,A10,MEASURE1:33,def 2;
        hence thesis by A11,XBOOLE_1:13,XXREAL_0:1;
      end;
      then
A12:  C1 is thin of M by Def2;
      A = A /\ (B2 \/ C2) by A2,A7,XBOOLE_1:28
        .= {} \/ C1 by XBOOLE_1:23;
      hence thesis by A12,PROB_1:4;
    end;
    hence thesis by Def3;
  end;
  hence thesis;
end;