The Mizar article:

Properties of Caratheodor's Measure

by
Jozef Bialas

Received June 25, 1992

Copyright (c) 1992 Association of Mizar Users

MML identifier: MEASURE4
[ MML identifier index ]


environ

 vocabulary SUPINF_1, MEASURE1, RLVECT_1, ARYTM_3, ARYTM_1, SETFAM_1, FUNCT_1,
      BOOLE, TARSKI, RELAT_1, SUPINF_2, PROB_1, MEASURE3, MEASURE4;
 notation TARSKI, XBOOLE_0, ENUMSET1, ZFMISC_1, SUBSET_1, NUMBERS, XREAL_0,
      RELAT_1, FUNCT_1, FUNCT_2, REAL_1, NAT_1, SETFAM_1, SUPINF_1, SUPINF_2,
      MEASURE1, MEASURE3;
 constructors ENUMSET1, NAT_1, REAL_1, SUPINF_2, PROB_2, MEASURE3, MEMBERED,
      XBOOLE_0;
 clusters SUBSET_1, FUNCT_1, SUPINF_1, MEASURE1, RELSET_1, NAT_1, MEMBERED,
      ZFMISC_1, XBOOLE_0, NUMBERS, ORDINAL2;
 requirements REAL, NUMERALS, SUBSET, BOOLE, ARITHM;
 definitions TARSKI, MEASURE3, XBOOLE_0;
 theorems TARSKI, ENUMSET1, FUNCT_1, FUNCT_2, ZFMISC_1, NAT_1, SUPINF_1,
      SUPINF_2, MEASURE1, MEASURE2, MEASURE3, RELSET_1, SETFAM_1, XBOOLE_0,
      XBOOLE_1, XCMPLX_1;
 schemes NAT_1, FUNCT_2, XBOOLE_0;

begin :: Some theorems about R_eal numbers

 reserve x,y,z for R_eal,
         A,B,X for set,
         S for sigma_Field_Subset of X;

theorem Th1:
   0.<=' x & 0.<=' y & 0.<=' z implies (x + y) + z = x + (y + z)
proof
   assume A1: 0.<=' x & 0.<=' y & 0.<=' z;
A2:(x in REAL or x in {-infty,+infty}) &
   (y in REAL or y in {-infty,+infty}) &
   (z in REAL or z in {-infty,+infty}) by SUPINF_1:def 6,XBOOLE_0:def 2;
   per cases by A1,A2,SUPINF_2:19,TARSKI:def 2;
   suppose
A3:x is Real & y is Real & z is Real;
     x + y is Real & y + z is Real
   proof not x = -infty & not x = +infty &
      not y = -infty & not y = +infty &
      not z = -infty & not z = +infty by A3,SUPINF_1:31;
      then not x + y = -infty & not x + y = +infty &
      not y + z = -infty & not y + z = +infty by SUPINF_2:8,9;
      hence thesis by MEASURE3:2;
   end;
   then consider a,b,c,d,e being Real such that
A4:x = a & y = b & z = c & x + y = d & y + z = e by A3;
     (x + y) + z = d + c by A4,SUPINF_2:1
              .= (a + b) + c by A4,SUPINF_2:1
              .= a + (b + c) by XCMPLX_1:1
              .= a + e by A4,SUPINF_2:1
              .= x + (y + z) by A4,SUPINF_2:1;
   hence thesis;
   suppose
A5:x = +infty;
A6:not (y + z) = -infty by A1,SUPINF_2:9,19;
     (x + y) + z = +infty + z by A1,A5,SUPINF_2:19,def 2
              .= +infty by A1,SUPINF_2:19,def 2
              .= +infty + (y + z) by A6,SUPINF_2:def 2;
   hence thesis by A5;
   suppose
A7:y = +infty;
   then (x + y) + z = +infty + z by A1,SUPINF_2:19,def 2
              .= +infty by A1,SUPINF_2:19,def 2
              .= x + (+infty) by A1,SUPINF_2:19,def 2
              .= x + (+infty + z) by A1,SUPINF_2:19,def 2;
   hence thesis by A7;
suppose
A8:z = +infty;
     not (x + y) = -infty by A1,SUPINF_2:9,19;
   then (x + y) + z = +infty by A8,SUPINF_2:def 2
              .= x + (+infty) by A1,SUPINF_2:19,def 2
              .= x + (y + (+infty)) by A1,SUPINF_2:19,def 2;
   hence thesis by A8;
end;

theorem Th2:
   (not x = -infty & not x = +infty) implies (y + x <=' z iff y <=' z - x)
proof
   assume
A1:not x = -infty & not x = +infty;
A2:(x in REAL or x in {-infty,+infty}) & (y in REAL or y in {-infty,+infty}) &
   (z in REAL or z in {-infty,+infty}) by SUPINF_1:def 6,XBOOLE_0:def 2;
   hereby assume
A3:   y + x <=' z;
        (y + x) - x = y
      proof
        per cases by A2,TARSKI:def 2;
        suppose y in REAL;
        then consider a,b being Real such that
     A4:a = x & b = y by A1,A2,TARSKI:def 2;
          y + x = b + a by A4,SUPINF_2:1;
        then consider c being Real such that
     A5:c = y + x;
          (y + x) - x = c - a by A4,A5,SUPINF_2:5
                   .= (b + a) - a by A4,A5,SUPINF_2:1
                   .= b + (a - a) by XCMPLX_1:29
                   .= b + 0 by XCMPLX_1:14
                   .= y by A4;
        hence thesis;
        suppose
     A6:y = -infty or y = +infty;
          now per cases by A6;
          suppose A7: y = -infty;
          hence (y + x) - x = -infty - x by A1,SUPINF_2:def 2
                   .= y by A1,A7,SUPINF_2:7;
          suppose A8: y = +infty;
          hence (y + x) - x = +infty - x by A1,SUPINF_2:def 2
                   .= y by A1,A8,SUPINF_2:6;
        end;
        hence thesis;
      end;
      hence y <=' z - x by A1,A3,SUPINF_2:15;
   end;
   assume
A9:y <=' z - x;
     (z - x) + x = z
   proof
      per cases by A2,TARSKI:def 2;
      suppose z in REAL;
      then consider a,b being Real such that
  A10:a = x & b = z by A1,A2,TARSKI:def 2;
        z - x = b - a by A10,SUPINF_2:5;
      then consider c being Real such that
  A11:c = z - x;
      thus (z - x) + x = c + a by A10,A11,SUPINF_2:1
                 .= (b - a) + a by A10,A11,SUPINF_2:5
                 .= b - (a - a) by XCMPLX_1:37
                 .= b - 0 by XCMPLX_1:14
                 .= z by A10;
      suppose
  A12:z = -infty;
      hence (z - x) + x = -infty + x by A1,SUPINF_2:7
                 .= z by A1,A12,SUPINF_2:def 2;
      suppose
  A13:z = +infty;
      hence (z - x) + x = +infty + x by A1,SUPINF_2:6
                 .= z by A1,A13,SUPINF_2:def 2;
   end;
   hence thesis by A1,A9,SUPINF_2:14;
end;

theorem
     (0. <=' x & 0. <=' y) implies x + y = y + x;

:: Some additional theorems about measures and functions

theorem Th4:
   for S being non empty Subset-Family of X,
       F, G being Function of NAT,S,
       A being Element of S st
   for n being Element of NAT holds G.n = A /\ F.n holds
   union rng G = A /\ union rng F
proof
   let S be non empty Subset-Family of X;
   let F, G be Function of NAT,S,
       A be Element of S;
   assume
A1:for n being Element of NAT holds G.n = A /\ F.n;
   thus union rng G c= A /\ union rng F
   proof
      let r be set;
      assume r in union rng G;
      then consider E being set such that
   A2:r in E & E in rng G by TARSKI:def 4;
      consider s being set such that
   A3:s in dom G & E = G.s by A2,FUNCT_1:def 5;
      reconsider s as Element of NAT by A3,FUNCT_2:def 1;
        r in A /\ F.s by A1,A2,A3;
   then A4:r in A & r in F.s by XBOOLE_0:def 3;
        F.s in rng F by FUNCT_2:6;
      then r in union rng F by A4,TARSKI:def 4;
      hence thesis by A4,XBOOLE_0:def 3;
   end;
   let r be set;
   assume r in A /\ union rng F;
then A5:r in A & r in union rng F by XBOOLE_0:def 3;
   then consider E being set such that
A6:r in E & E in rng F by TARSKI:def 4;
   consider s being set such that
A7:s in dom F & E = F.s by A6,FUNCT_1:def 5;
   reconsider s as Element of NAT by A7,FUNCT_2:def 1;
     A /\ E = G.s by A1,A7;
then A8:r in G.s by A5,A6,XBOOLE_0:def 3;
     G.s in rng G by FUNCT_2:6;
   hence thesis by A8,TARSKI:def 4;
end;

theorem Th5:
   for S being non empty Subset-Family of X
   for F, G being Function of NAT,S st
   (G.0 = F.0 &
   for n being Element of NAT holds G.(n+1) = F.(n+1) \/ G.n) holds
   for H being Function of NAT,S st
   (H.0 = F.0 &
   for n being Element of NAT holds H.(n+1) = F.(n+1) \ G.n) holds
   union rng F = union rng H
proof
   let S be non empty Subset-Family of X;
   let F, G be Function of NAT,S;
   assume
A1:G.0 = F.0 & for n being Element of NAT holds G.(n+1) = F.(n+1) \/ G.n;
   let H be Function of NAT,S;
   assume
A2:H.0 = F.0 & for n being Element of NAT holds H.(n+1) = F.(n+1) \ G.n;
A3:for n being Element of NAT holds H.n c= F.n
   proof
      let n be Element of NAT;
   A4:n=0 implies H.n c= F.n by A2;
        (ex k being Nat st n = k + 1) implies H.n c= F.n
      proof
         given k being Nat such that
      A5:n = k + 1;
           H.n = F.n \ G.k by A2,A5;
         hence thesis by XBOOLE_1:36;
      end;
      hence thesis by A4,NAT_1:22;
   end;
A6:union rng H c= union rng F
   proof
      let r be set;
      assume r in union rng H;
      then consider E being set such that
   A7:r in E & E in rng H by TARSKI:def 4;
      consider s being set such that
   A8:s in dom H & E = H.s by A7,FUNCT_1:def 5;
      reconsider s as Element of NAT by A8,FUNCT_2:def 1;
A9:    E c= F.s by A3,A8;
        F.s in rng F by FUNCT_2:6;
      hence thesis by A7,A9,TARSKI:def 4;
   end;
    thus union rng F c= union rng H
    proof
      let r be set;
      assume r in union rng F;
      then consider E being set such that
   A10:r in E & E in rng F by TARSKI:def 4;
      consider s being set such that
   A11:s in dom F & E = F.s by A10,FUNCT_1:def 5;
      reconsider s as Element of NAT by A11,FUNCT_2:def 1;
        ex s1 being Element of NAT st r in H.s1
      proof
        defpred P[Nat] means r in F.$1;
           r in F.s by A10,A11;
     then A12:ex k being Nat st P[k];
           ex k being Nat st P[k] &
         for n being Nat st P[n] holds k <= n from Min(A12);
         then consider k being Nat such that
     A13:r in F.k & for n being Nat st r in F.n holds k <= n;
     A14:k=0 implies ex s1 being Element of NAT st r in H.s1 by A2,A13;
           (ex l being Nat st k = l + 1) implies
         (ex s1 being Element of NAT st r in H.s1)
         proof
            given l being Nat such that
        A15:k = l + 1;
        A16:not r in G.l
            proof
               assume r in G.l;
               then consider i being Nat such that
           A17:i <= l & r in F.i by A1,MEASURE2:6;
                 k <= i by A13,A17;
               hence thesis by A15,A17,NAT_1:38;
            end;
A18:        H.(l+1) = F.(l+1) \ G.l by A2;
            take k;
            thus thesis by A13,A15,A16,A18,XBOOLE_0:def 4;
         end;
         hence thesis by A14,NAT_1:22;
      end;
      then consider s1 being Element of NAT such that
  A19:r in H.s1;
        H.s1 in rng H by FUNCT_2:6;
      hence thesis by A19,TARSKI:def 4;
   end;
   thus thesis by A6;
end;

theorem Th6:
   bool X is sigma_Field_Subset of X
proof
A1:for A being set holds A in bool X implies X\A in bool X
   proof
      let A be set;
      assume A in bool X;
        X \ A c= X by XBOOLE_1:36;
      hence thesis;
   end;
     for M being N_Sub_set_fam of X st M c= bool X holds union M in bool X;
   hence thesis by A1,MEASURE1:def 3,def 9,SETFAM_1:def 7;
end;

definition let X be set;
   let F be Function of NAT,bool X;
   redefine func rng F -> Subset-Family of X;
coherence
proof
     rng F c= bool X by RELSET_1:12;
   hence thesis by SETFAM_1:def 7;
end;
end;

definition let X be set;
   let A be Subset-Family of X;
   redefine func union A -> Element of bool X;
coherence
proof
     union A c= union bool X by ZFMISC_1:95;
   hence thesis;
end;
end;

definition let Y be set; let X,Z be non empty set;
   let F be Function of Y,X;
   let M be Function of X,Z;
  redefine func M*F -> Function of Y,Z;
coherence
proof
  thus M*F is Function of Y,Z;
end;
end;

theorem Th7:
   for a,b being R_eal holds
   ex M being Function of bool X,ExtREAL st
   for A being Element of bool X holds
   ((A = {} implies M.A = a) & (A <> {} implies M.A = b))
proof
   let a,b be R_eal;
   defpred P[set,set] means (($1 = {} implies $2 = a) &
                         ($1 <> {} implies $2 = b));
A1:for x being set st x in bool X ex y being set st y in ExtREAL & P[x,y]
   proof
      let x be set;
        x <> {} implies ex y being set st y in ExtREAL & P[x,y];
      hence thesis;
   end;
     ex F being Function of bool X,ExtREAL st
   for x being set st x in bool X holds P[x,F.x] from FuncEx1(A1);
   then consider M being Function of bool X,ExtREAL such that
A2:for x being set st x in bool X holds P[x,M.x];
   take M;
   thus thesis by A2;
end;

theorem Th8:
   ex M being Function of bool X,ExtREAL st
   for A being Element of bool X holds M.A = 0.
proof
   consider M being Function of bool X,ExtREAL such that
A1:for A being Element of bool X holds
   (A = {} implies M.A = 0.) & (A <> {} implies M.A = 0.) by Th7;
A2:for A being Element of bool X holds M.A = 0.
   proof
      let A be Element of bool X;
        A = {} implies M.A = 0.by A1;
      hence thesis by A1;
   end;
   take M;
   thus thesis by A2;
end;

canceled 2;

theorem Th11:
   ex M being Function of bool X,ExtREAL st
   M is nonnegative & M.{} = 0. &
   for A,B being Element of bool X st A c= B holds M.A <=' M.B &
   for F being Function of NAT,bool X holds
   M.(union rng F) <=' SUM(M*F)
proof
   consider M being Function of bool X,ExtREAL such that
A1:for A being Element of bool X holds M.A = 0.by Th8;
A2:for A being Element of bool X holds 0.<=' M.A by A1;
then A3:M is nonnegative by MEASURE1:def 4;
A4:{} c= X by XBOOLE_1:2;
A5:for F being Function of NAT,bool X holds M.(union rng F) <=' SUM(M*F)
   proof
      let F be Function of NAT,bool X;
   A6:M.(union rng F) = 0.by A1;
   A7:M*F is nonnegative by A3,MEASURE1:54;
   A8:for r being Nat st 0 <= r holds (M*F).r = 0.
      proof
         let r be Nat;
       dom (M*F) = NAT by FUNCT_2:def 1;
     then (M*F).r = M.(F.r) by FUNCT_1:22;
         hence thesis by A1;
      end;
   then A9:SUM(M*F) = Ser(M*F).0 by A7,SUPINF_2:67;
    Ser(M*F).0 = (M*F).0 by SUPINF_2:63;
      hence thesis by A6,A8,A9;
   end;
   take M;
     for A,B being Element of bool X holds M.A <=' M.B
   proof
      let A,B be Element of bool X;
        M.A = 0. & M.B = 0. by A1;
      hence thesis;
   end;
   hence thesis by A1,A2,A4,A5,MEASURE1:def 4;
end;

definition let X be set;
 canceled;

  mode C_Measure of X -> Function of bool X,ExtREAL means
:Def2:it is nonnegative &
       it.{} = 0. &
       for A,B being Element of bool X st A c= B holds it.A <=' it.B &
       for F being Function of NAT,bool X holds
       it.(union rng F) <=' SUM(it*F);
existence by Th11;
end;

 reserve C for C_Measure of X;

definition let X be set;
   let C be C_Measure of X;
  func sigma_Field(C) -> non empty Subset-Family of X means
:Def3:for A being Element of bool X holds
      (A in it iff for W,Z being Element of bool X holds
      (W c= A & Z c= X \ A implies C.W + C.Z <=' C.(W \/ Z)));
existence
proof
   defpred P[set] means
   for A being set holds (A = $1 implies
   for W,Z being Element of bool X holds
   (W c= A & Z c= X \ A implies C.W + C.Z <=' C.(W \/ Z)));
   consider D being set such that
A1:for y being set holds (y in D iff (y in bool X & P[y])) from Separation;
A2:for A being set holds (A in D iff A in bool X &
   for W,Z being Element of bool X holds
   (W c= A & Z c= X \ A implies C.W + C.Z <=' C.(W \/ Z)))
   proof
      let A be set;
        P[A] iff for W,Z being Element of bool X holds
      (W c= A & Z c= X \ A implies C.W + C.Z <=' C.(W \/ Z));
      hence thesis by A1;
   end;
   reconsider A = {} as Element of bool X by XBOOLE_1:2;
A3: now let W,Z be Element of bool X;
      assume W c= A & Z c= X \ A;
   then A4:W = {} by XBOOLE_1:3;
      then C.W = 0. by Def2;
      hence C.W + C.Z <=' C.(W \/ Z) by A4,SUPINF_2:18;
   end;
   then D <> {} & for A being set holds A in D implies A in bool X by A2;
   then D c= bool X by TARSKI:def 3;
   then reconsider D as non empty Subset-Family of X by A2,A3,SETFAM_1:def 7;
   take D;
   thus thesis by A2;
end;
uniqueness
proof
   let D1,D2 be non empty Subset-Family of X such that
A5:for A being Element of bool X holds
   (A in D1 iff for W,Z being Element of bool X holds
   (W c= A & Z c= X \ A implies C.W + C.Z <=' C.(W \/ Z))) and
A6:for A being Element of bool X holds
   (A in D2 iff for W,Z being Element of bool X holds
   (W c= A & Z c= X \ A implies C.W + C.Z <=' C.(W \/ Z)));
     for A being set holds A in D1 iff A in D2
   proof
      let A be set;
      hereby assume
      A7:A in D1;
         then reconsider A' = A as Element of bool X;
           for W,Z being Element of bool X holds
         (W c= A & Z c= X \ A' implies C.W + C.Z <=' C.(W \/ Z)) by A5,A7;
         hence A in D2 by A6;
      end;
      assume
   A8:A in D2;
      then reconsider A as Element of bool X;
        for W,Z being Element of bool X holds
      (W c= A & Z c= X \ A implies C.W + C.Z <=' C.(W \/ Z)) by A6,A8;
      hence thesis by A5;
   end;
   hence thesis by TARSKI:2;
end;
end;

theorem Th12:
   for W,Z being Element of bool X holds
   C.(W \/ Z) <=' C.W + C.Z
proof
   let W,Z be Element of bool X;
   reconsider P = {} as Subset of X by XBOOLE_1:2;
   consider F being Function of NAT,bool X such that
A1:rng F = {W,Z,P} &
   F.0 = W & F.1 = Z & for n being Nat st 1 < n holds F.n = P by MEASURE1:38;
A2: union {W,Z,P} = W \/ Z
   proof
      thus union {W,Z,P} c= W \/ Z
      proof
        let x be set;
        assume x in union {W,Z,P};
        then consider Y being set such that
     A3:x in Y & Y in {W,Z,P} by TARSKI:def 4;
          x in W or x in Z or x in P by A3,ENUMSET1:13;
        hence thesis by XBOOLE_0:def 2;
      end;
      let x be set;
      assume
   A4:x in W \/ Z;
        now per cases by A4,XBOOLE_0:def 2;
         suppose A5:x in W;
         take Y = W;
         thus x in Y & Y in {W,Z,P} by A5,ENUMSET1:14;
         suppose A6:x in Z;
         take Y = Z;
         thus x in Y & Y in {W,Z,P} by A6,ENUMSET1:14;
      end;
      hence thesis by TARSKI:def 4;
   end;
     C is nonnegative by Def2;
then A7:C*F is nonnegative by MEASURE1:54;
A8:for r being Nat st 2 <= r holds (C*F).r = 0.
   proof
      let r be Nat;
      assume 2 <= r;
      then 1 + 1 <= r;
      then 1 < r by NAT_1:38;
      then F.r = {} by A1;
      then C.(F.r) = 0. by Def2;
      hence thesis by FUNCT_2:21;
   end;
     F.0 = W & F.1 = Z & F.2 = P by A1;
   then A9:(C*F).2 = C.P & (C*F).1 = C.Z & (C*F).0 = C.W by FUNCT_2:21;
   set G = C*F;
   consider y1,y2 being R_eal such that
A10:y1 = Ser(G).1 & y2 = Ser(G).0;
     Ser(G).2 = y1 + G.(1 + 1) by A10,SUPINF_2:63
             .= Ser(G).1 + 0. by A9,A10,Def2
             .= Ser(G).1 by SUPINF_2:18
             .= y2 + G.(0 + 1) by A10,SUPINF_2:63
             .= C.W + C.Z by A9,A10,SUPINF_2:63;
   then SUM(C*F) = C.W + C.Z by A7,A8,SUPINF_2:67;
   hence thesis by A1,A2,Def2;
end;

theorem
     for W,Z being Element of bool X holds C.Z + C.W = C.W + C.Z;

theorem Th14:
   for A being Element of bool X holds
   (A in sigma_Field(C) iff for W,Z being Element of bool X holds
   (W c= A & Z c= X \ A implies C.W + C.Z = C.(W \/ Z)))
proof
   let A be Element of bool X;
   hereby assume
A1:   A in sigma_Field(C);
      let W,Z be Element of bool X;
      assume W c= A & Z c= X \ A;
      then C.(W \/ Z) <=' C.W + C.Z & C.W + C.Z <=' C.(W \/ Z) by A1,Def3,Th12
;
      hence C.W + C.Z = C.(W \/ Z) by SUPINF_1:22;
   end;
   assume for W,Z being Element of bool X holds
   (W c= A & Z c= X \ A implies C.W + C.Z = C.(W \/ Z));
   then for W,Z being Element of bool X holds
   (W c= A & Z c= X \ A implies C.W + C.Z <=' C.(W \/ Z));
   hence thesis by Def3;
end;

theorem Th15:
   for W,Z being Element of bool X holds
   (W in sigma_Field(C) & Z in sigma_Field(C) & Z misses W) implies
   C.(W \/ Z) = C.W + C.Z
proof
   let W,Z be Element of bool X;
   assume
A1:W in sigma_Field(C) & Z in sigma_Field(C) & Z misses W;
     Z \ W c= X \ W by XBOOLE_1:33;
   then Z c= X \ W by A1,XBOOLE_1:83;
   hence thesis by A1,Th14;
end;

theorem Th16:
   A in sigma_Field(C) implies X \ A in sigma_Field(C)
proof
   assume
A1:A in sigma_Field(C);
A2:X \ A c= X by XBOOLE_1:109;
     for W,Z being Element of bool X holds
   W c= X \ A & Z c= X \ (X \ A) implies C.W + C.Z <=' C.(W \/ Z)
   proof
      let W,Z be Element of bool X;
      assume
   A3:W c= X \ A & Z c= X \ (X \ A);
        X \ (X \ A) = X /\ A by XBOOLE_1:48;
then Z c= A & W c= X \ A by A1,A3,XBOOLE_1:28;
      hence thesis by A1,Def3;
   end;
   hence thesis by A2,Def3;
end;

theorem Th17:
   A in sigma_Field(C) & B in sigma_Field(C) implies A \/ B in sigma_Field(C)
proof
   assume
A1:A in sigma_Field(C) & B in sigma_Field(C);
   then reconsider A,B as Element of bool X;
   set D = A \/ B;
     for W,Z being Element of bool X holds
   W c= D & Z c= X \ D implies C.W + C.Z = C.(W \/ Z)
   proof
      let W,Z be Element of bool X;
      assume
A2:   W c= D & Z c= X \ D;
      set W1 = W /\ A;
  A3:W \ A c= X \ A by XBOOLE_1:33;
      set W2 = W \ A;
        (X \ A) /\ (X \ B) c= X \ A by XBOOLE_1:17;
      then X \ (A \/ B) c= X \ A by XBOOLE_1:53;
   then A4:W1 c= A & Z c= X \ A by A2,XBOOLE_1:1,17;
A5:   W2 \/ Z c= X \ A \/ Z by A3,XBOOLE_1:9;
      set Z1 = W2 \/ Z;
  A6:Z1 c= X \ A by A4,A5,XBOOLE_1:12;
  A7:W = W1 \/ W2 by XBOOLE_1:51;
  A8:C.(W \/ Z) = C.((W1 \/ W2) \/ Z) by XBOOLE_1:51
               .= C.(W1 \/ Z1) by XBOOLE_1:4
               .= C.W1 + C.Z1 by A1,A4,A6,Th14;
        W \ A c= B \/ A \ A by A2,XBOOLE_1:33;
      then W \ A c= B \ A & B \ A c= B by XBOOLE_1:36,40;
  then A9:W2 c= B by XBOOLE_1:1;
        Z c= (X \ A) /\ (X \ B) & (X \ A) /\ (X \ B) c= X \ B
      by A2,XBOOLE_1:17,53;
      then Z c= X \ B by XBOOLE_1:1;
  then A10:C.W2 + C.Z <=' C.Z1 by A1,A9,Def3;
        C is nonnegative by Def2;
  then A11:0.<=' C.W2 & 0.<=' C.Z by MEASURE1:def 4;
      then 0. + 0. <=' C.W2 + C.Z by MEASURE1:4;
  then A12:0. <=' C.W2 + C.Z by SUPINF_2:18;
        C is nonnegative by Def2;
  then A13:0. <=' C.W1 by MEASURE1:def 4;
      then C.W1 + (C.W2 + C.Z) <=' C.(W \/ Z) by A8,A10,A12,MEASURE1:4;
  then A14:(C.W1 + C.W2) + C.Z <=' C.(W \/ Z) by A11,A13,Th1;
        C is nonnegative by Def2;
  then A15:0.<=' C.W & 0.<=' C.Z & C.Z <=' C.Z by MEASURE1:def 4;
        C.W <=' C.W1 + C.W2 by A7,Th12;
      then C.W + C.Z <=' (C.W1 + C.W2) + C.Z by A15,MEASURE1:4;
  then A16:C.W + C.Z <=' C.(W \/ Z) by A14,SUPINF_1:29;
        C.(W \/ Z) <=' C.W + C.Z by Th12;
      hence thesis by A16,SUPINF_1:22;
   end;
   hence thesis by Th14;
end;

theorem Th18:
   A in sigma_Field(C) & B in sigma_Field(C) implies A /\ B in sigma_Field(C)
proof
   assume
A1:A in sigma_Field(C) & B in sigma_Field(C);
   then A /\ B c= X /\ X by XBOOLE_1:27;
then A2:A /\ B = X /\ (A /\ B) by XBOOLE_1:28
        .= X \ (X \(A /\ B)) by XBOOLE_1:48
        .= X \ ((X \ A) \/ (X \ B)) by XBOOLE_1:54;
     X \ A in sigma_Field(C) & X \ B in sigma_Field(C) by A1,Th16;
   then (X \ A) \/ (X \ B) in sigma_Field(C) by Th17;
   hence thesis by A2,Th16;
end;

theorem Th19:
   A in sigma_Field(C) & B in sigma_Field(C) implies A \ B in sigma_Field(C)
proof
   assume
A1:A in sigma_Field(C) & B in sigma_Field(C);
     for x being set holds x in A \ B iff x in A /\ (X \ B)
   proof
      let x be set;
      hereby assume x in A \ B;
         then x in A & not x in B by XBOOLE_0:def 4;
         then x in A & x in (X \ B) by A1,XBOOLE_0:def 4;
         hence x in A /\ (X \ B) by XBOOLE_0:def 3;
      end;
      assume x in A /\ (X \ B);
      then x in A & x in (X \ B) by XBOOLE_0:def 3;
      then x in A & (x in X & not x in B) by XBOOLE_0:def 4;
      hence thesis by XBOOLE_0:def 4;
   end;
then A2:A \ B = A /\ (X \ B) by TARSKI:2;
     X \ B in sigma_Field(C) by A1,Th16;
   hence thesis by A1,A2,Th18;
end;

theorem Th20:
   for N being Function of NAT,S holds
   for A being Element of S holds
   ex F being Function of NAT,S st
   for n being Element of NAT holds F.n = A /\ N.n
proof
   let N be Function of NAT,S;
   let A be Element of S;
   defpred P[set,set] means ($1 in NAT implies $2 = A /\ N.$1);
A1:for x being set st x in NAT ex y being set st y in S & P[x,y]
   proof
      let x be set;
      assume x in NAT;
      then reconsider x as Element of NAT;
      consider y being set such that
   A2:y = A /\ N.x;
      take y;
      thus thesis by A2;
   end;
     ex F being Function of NAT,S st
   for x being set st x in NAT holds P[x,F.x] from FuncEx1(A1);
   then consider F being Function of NAT,S such that
A3:for x being set st x in NAT holds P[x,F.x];
   take F;
   thus thesis by A3;
end;

theorem Th21:
   sigma_Field(C) is sigma_Field_Subset of X
proof
A1:for A being set holds A in sigma_Field(C) implies X\A in sigma_Field(C)
   by Th16;
A2:C is nonnegative by Def2;
     for M being N_Sub_set_fam of X holds
   M c= sigma_Field(C) implies union M in sigma_Field(C)
   proof
      let M be N_Sub_set_fam of X;
      assume
   A3:M c= sigma_Field(C);
        for W,Z being Element of bool X holds
      (W c= union M & Z c= X \ union M implies C.W + C.Z <=' C.(W \/ Z))
      proof
         let W,Z be Element of bool X;
         assume
      A4:W c= union M & Z c= X \ union M;
         reconsider S = bool X as sigma_Field_Subset of X by Th6;
         consider F being Function of NAT,bool X such that
      A5:rng F = M by SUPINF_2:def 14;
      A6:for n being Element of NAT holds F.n in sigma_Field(C)
         proof
            let n be Element of NAT;
              F.n in M by A5,FUNCT_2:6;
            hence thesis by A3;
         end;
         consider G being Function of NAT,S such that
      A7:G.0 = F.0 & for n being Element of NAT holds
         G.(n+1) = F.(n+1) \/ G.n by MEASURE2:5;
     defpred P[Nat] means G.$1 in sigma_Field(C);
     A8: P[0] by A6,A7;
    A9: for k being Nat st P[k] holds P[k+1]
         proof
           let k be Nat;
           assume
       A10:G.k in sigma_Field(C);
       A11:G.(k+1) = F.(k+1) \/ G.k by A7;
             F.(k+1) in sigma_Field(C) by A6;
           hence thesis by A10,A11,Th17;
         end;
A12:      for n being Element of NAT holds P[n] from Ind(A8,A9);
         consider B being Function of NAT,S such that
     A13:B.0 = F.0 & for n being Element of NAT holds
         B.(n+1) = F.(n+1) \ G.n by MEASURE2:9;
     A14:union rng F = union rng B by A7,A13,Th5;
         consider Q being Function of NAT,S such that
     A15:for n being Element of NAT holds Q.n = W /\ B.n by Th20;
     A16:union rng Q = W /\ union rng B by A15,Th4;
         consider QQ being Function of NAT,S such that
     A17:QQ.0 = Q.0 & for n being Element of NAT holds
         QQ.(n+1) = Q.(n+1) \/ QQ.n by MEASURE2:5;
         reconsider Q,QQ,F,G as Function of NAT,bool X;
A18:       C*Q is nonnegative by A2,MEASURE1:54;
     A19:QQ.0 = W /\ F.0 by A13,A15,A17;
     A20:F.0 in sigma_Field(C) by A6;
     A21:QQ.0 c= F.0 by A19,XBOOLE_1:17;
     defpred P[Nat] means  C.(Z \/ QQ.$1) = C.Z + Ser(C*Q).$1;
     F.0 in rng F by FUNCT_2:6;
         then F.0 c= union rng F by ZFMISC_1:92;
         then X \ union rng F c= X \ F.0 by XBOOLE_1:34;
         then A22: Z c= X \ F.0 by A4,A5,XBOOLE_1:1;
           Ser(C*Q).0 = (C*Q).0 by SUPINF_2:63
                   .= C.(QQ.0) by A17,FUNCT_2:21;
    then A23: P[0] by A20,A21,A22,Th14;
    A24: for k being Nat st P[k] holds P[k+1]
         proof
            let k be Nat;
            assume
        A25:C.(Z \/ QQ.k) = C.Z + Ser(C*Q).k;
A26:        QQ.(k+1) = QQ.k \/ Q.(k+1) by A17;
              Q.(k+1) = W /\ B.(k+1) by A15
                   .= W /\ (F.(k+1) \ G.k) by A13;
        then A27:Q.(k+1) c= F.(k+1) \ G.k by XBOOLE_1:17;
              F.(k+1) in sigma_Field(C) & G.k in sigma_Field(C) by A6,A12;
        then A28:F.(k+1) \ G.k in sigma_Field(C) by Th19;
              F.(k+1) in rng F by FUNCT_2:6;
        then A29:F.(k+1) c= union rng F by ZFMISC_1:92;
              F.(k+1) \ G.k c= F.(k+1) by XBOOLE_1:36;
            then F.(k+1) \ G.k c= union rng F by A29,XBOOLE_1:1;
            then X \ union rng F c= X \ (F.(k+1) \ G.k) by XBOOLE_1:34;
        then A30:Z c= X \ (F.(k+1) \ G.k) by A4,A5,XBOOLE_1:1;
        defpred P[Nat] means QQ.$1 c= G.$1;
              QQ.0 = W /\ F.0 by A13,A15,A17;
       then A31: P[0] by A7,XBOOLE_1:17;
         for n being Nat holds QQ.n misses (F.(n+1) \ G.n)
            proof
               let n be Nat;
          A32: for n being Nat st P[n] holds P[n+1]
            proof let n be Nat;
                  assume
              A33:QQ.n c= G.n;
              A34:QQ.(n+1) = Q.(n+1) \/ QQ.n by A17
                       .= (W /\ B.(n+1)) \/ QQ.n by A15
                       .= (W /\ (F.(n+1) \ G.n)) \/ QQ.n by A13;
                    F.(n+1) \ G.n c= F.(n+1) by XBOOLE_1:36;
              then A35:W /\ (F.(n+1) \ G.n) c= W /\ F.(n+1)
                  by XBOOLE_1:26;
                    W /\ F.(n+1) c= F.(n+1) by XBOOLE_1:17;
                  then W /\ (F.(n+1) \ G.n) c= F.(n+1)
                  by A35,XBOOLE_1:1;
                  then QQ.(n+1) c= F.(n+1) \/ G.n by A33,A34,XBOOLE_1:13;
                  hence QQ.(n+1) c= G.(n+1) by A7;
               end;
                 for n being Nat holds P[n] from Ind(A31,A32);
           then A36:QQ.n c= G.n;
                 G.n misses (F.(n+1) \ G.n) by XBOOLE_1:79;
               hence thesis by A36,XBOOLE_1:63;
            end;
            then QQ.k misses (F.(k+1) \ G.k);
        then A37:QQ.k /\ (F.(k+1) \ G.k) = {} by XBOOLE_0:def 7;
              QQ.k c= X \ (F.(k+1) \ G.k)
            proof
               let z be set;
               assume
           A38:z in QQ.k;
               then not z in F.(k+1) \ G.k by A37,XBOOLE_0:def 3;
               hence thesis by A38,XBOOLE_0:def 4;
            end;
then A39:        Z \/ QQ.k c= X \ (F.(k+1) \ G.k) by A30,XBOOLE_1:8;
        A40:0. <=' C.Z & 0. <=' Ser(C*Q).k & 0. <=' (C*Q).(k+1)
            by A2,A18,SUPINF_2:58,59;
              C.(Q.(k+1)) + C.(Z \/ QQ.k) = C.((Z \/ QQ.k) \/ Q.(k+1)) by A27,
A28,A39,Th14
                        .= C.(Z \/ QQ.(k+1)) by A26,XBOOLE_1:4;
            then C.(Z \/ QQ.(k+1)) = (C.Z + Ser(C*Q).k) + (C*Q).(k+1) by A25,
FUNCT_2:21
                        .= C.Z + (Ser(C*Q).k + (C*Q).(k+1)) by A40,Th1
                        .= C.Z + Ser(C*Q).(k+1) by SUPINF_2:63;
            hence thesis;
         end;
     A41:for n being Element of NAT holds P[n] from Ind(A23,A24);
     defpred Q[Nat] means QQ.$1 c= W;
           QQ.0 = W /\ B.0 by A15,A17;
    then A42: Q[0] by XBOOLE_1:17;
    A43: for n being Nat st Q[n] holds Q[n+1]
         proof
            let n be Nat;
            assume
        A44:QQ.n c= W;
        A45:QQ.(n+1) = Q.(n+1) \/ QQ.n by A17
                    .= (W /\ B.(n+1)) \/ QQ.n by A15;
              W /\ B.(n+1) c= W by XBOOLE_1:17;
            then QQ.(n+1) c= W \/ W by A44,A45,XBOOLE_1:13;
            hence thesis;
         end;
A46:     for n being Nat holds Q[n] from Ind(A42,A43);
     A47:union rng Q = W by A4,A5,A14,A16,XBOOLE_1:28;
     A48:for n being Element of NAT holds Ser(C*Q).n + C.Z <=' C.(Z \/ W)
         proof
            let n be Element of NAT;
        A49:Ser(C*Q).n + C.Z = C.(Z \/ QQ.n) by A41;
              QQ.n c= W by A46;
            then Z \/ QQ.n c= Z \/ W by XBOOLE_1:9;
            hence thesis by A49,Def2;
         end;
     A50:C.Z = +infty implies C.W + C.Z <=' C.(W \/ Z)
         proof
            assume
        A51:C.Z = +infty;
              0. <=' C.W by A2,MEASURE1:def 4;
        then A52:C.W + C.Z = +infty by A51,SUPINF_2:19,def 2;
              Z c= W \/ Z by XBOOLE_1:7;
            hence thesis by A51,A52,Def2;
         end;
           (not C.Z = -infty & not C.Z = +infty) implies C.W + C.Z <=' C.(W \/
Z
)
         proof
            assume
        A53:not C.Z = -infty & not C.Z = +infty;
        A54:for n being Element of NAT holds Ser(C*Q).n <=' C.(Z \/ W) - C.Z
            proof
               let n be Element of NAT;
                 Ser(C*Q).n + C.Z <=' C.(Z \/ W) by A48;
               hence thesis by A53,Th2;
            end;
            defpred P[set,set] means ($1 = 0 implies
            $2 = C.(Z \/ W) - C.Z ) & ($1 <> 0 implies $2 = 0.);
        A55:for x being set st x in NAT ex y being set st y in ExtREAL & P[x,y]
            proof
               let x be set;
               assume x in NAT;
               then reconsider x as Element of NAT;
                 x <> 0 implies ex y being set st y in ExtREAL & P[x,y];
               hence thesis;
            end;
              ex R being Function of NAT,ExtREAL st
            for x being set st x in NAT holds
            P[x,R.x] from FuncEx1(A55);
            then consider R being Function of NAT,ExtREAL such that
        A56:for x being set st x in NAT holds P[x,R.x];
              for n being Nat holds 0. <=' R.n
            proof
               let n be Nat;
               per cases;
               suppose
           A57:n = 0;
                 Z c= Z \/ W by XBOOLE_1:7;
               then C.Z <=' C.(Z \/ W) by Def2;
      then A58:C.Z - C.Z <=' C.(Z \/ W) - C.Z by A53,SUPINF_2:15;
                 C.Z in REAL or C.Z in {-infty,+infty}
               by SUPINF_1:def 6,XBOOLE_0:def 2;
               then consider y being Real such that
           A59:y = C.Z by A53,TARSKI:def 2;
           A60:C.Z - C.Z = y - y by A59,SUPINF_2:5;
                 y - y = 0 by XCMPLX_1:14;
               hence thesis by A56,A57,A58,A60,SUPINF_2:def 1;
               suppose n <> 0;
               hence thesis by A56;
            end;
        then A61:R is nonnegative by SUPINF_2:58;
A62:        for n being Nat holds Ser(C*Q).n <=' Ser(R).n
            proof
              defpred P[Nat] means Ser(R).$1 = C.(Z \/ W) - C.Z;
               let n be Nat;
                 Ser(R).0 = R.0 by SUPINF_2:63;
          then A63: P[0] by A56;

A64:      for k being Nat st P[k] holds P[k+1]
           proof let k be Nat;
                  assume
              A65:Ser(R).k = C.(Z \/ W) - C.Z;
                  set y = Ser(R).k;
                  thus Ser(R).(k+1) = y + R.(k+1) by SUPINF_2:63
                              .= y + 0. by A56
                              .= C.(Z \/ W) - C.Z by A65,SUPINF_2:18;
               end;
         for k being Nat holds P[k] from Ind(A63,A64);
               then Ser(R).n = C.(Z \/ W) - C.Z;
               hence thesis by A54;
            end;
              now let k be Nat;
               assume 1 <= k;
               then k <> 0;
               hence R.k = 0. by A56;
            end;
        then A66:SUM(R) = Ser(R).1 by A61,SUPINF_2:67;
            set y = Ser(R).0;
              y = R.0 by SUPINF_2:63;
        then A67:y = C.(Z \/ W) - C.Z by A56;
              Ser(R).1 = y + R.(0+1) by SUPINF_2:63
                    .= y + 0. by A56
                    .= C.(Z \/ W) - C.Z by A67,SUPINF_2:18;
        then A68:SUM(C*Q) <=' C.(Z \/ W) - C.Z by A62,A66,MEASURE3:3;
              C.W <=' SUM(C*Q) by A47,Def2;
            then C.W <=' C.(Z \/ W) - C.Z by A68,SUPINF_1:29;
            hence thesis by A53,Th2;
         end;
         hence thesis by A2,A50,MEASURE1:def 4,SUPINF_2:19;
      end;
      hence thesis by Def3;
   end;
   hence thesis by A1,MEASURE1:def 3,def 9;
end;

definition let X be set;
   let C be C_Measure of X;
   cluster sigma_Field(C) -> sigma_Field_Subset-like compl-closed non empty;
  coherence by Th21;
end;

definition let X be set;
   let S be sigma_Field_Subset of X;
   let A be N_Sub_fam of S;
   redefine func union A -> Element of S;
coherence
proof
     A c= S by MEASURE2:def 1;
   hence thesis by MEASURE1:def 9;
end;
end;

definition let X be set;
   let C be C_Measure of X;
  func sigma_Meas(C) -> Function of sigma_Field(C),ExtREAL means
:Def4:for A being Element of bool X st A in sigma_Field(C) holds it.A = C.A;
existence
proof
     ex D being Function of sigma_Field(C),ExtREAL st
   (for A being Element of bool X holds
   A in sigma_Field(C) implies D.A = C.A)
   proof
     deffunc F(set) = C.$1;
   A1:for S being set st S in sigma_Field(C) holds (F(S) in ExtREAL)
        by FUNCT_2:7;
      consider D being Function of sigma_Field(C),ExtREAL such that
   A2:for S being set st S in sigma_Field(C) holds D.S = F(S) from Lambda1(A1);
      take D;
      thus thesis by A2;
   end;
   then consider D being Function of sigma_Field(C),ExtREAL such that
A3:for A being Element of bool X st A in sigma_Field(C) holds D.A = C.A;
   take D;
   thus thesis by A3;
end;
uniqueness
proof
   let D1,D2 be Function of sigma_Field(C),ExtREAL such that
A4:for A being Element of bool X holds
   A in sigma_Field(C) implies D1.A = C.A and
A5:for A being Element of bool X holds
   A in sigma_Field(C) implies D2.A = C.A;
   A6:(sigma_Field(C) = dom D1) & (sigma_Field(C) = dom D2) by FUNCT_2:def 1;
     for A being set st A in sigma_Field(C) holds D1.A = D2.A
   proof
      let A be set;
      assume
   A7:A in sigma_Field(C);
      then reconsider A as Element of bool X;
        D1.A = C.A by A4,A7
          .= D2.A by A5,A7;
      hence thesis;
   end;
   hence thesis by A6,FUNCT_1:9;
end;
end;

theorem Th22:
   sigma_Meas(C) is Measure of sigma_Field(C)
proof
A1:now let A be Element of sigma_Field(C);
      reconsider A' = A as Element of bool X;
A2:   (sigma_Meas(C)).A' = C.A' by Def4;
        C is nonnegative by Def2;
      hence 0.<=' (sigma_Meas(C)).A by A2,MEASURE1:def 4;
   end;
 {} in sigma_Field(C) by MEASURE1:45;
   then (sigma_Meas(C)).{} = C.{} by Def4;
then A3:sigma_Meas(C) is nonnegative & (sigma_Meas(C)).{} = 0.
   by A1,Def2,MEASURE1:def 4;
     now let A,B be Element of sigma_Field(C);
      assume
A4:   A misses B;
      reconsider A' = A,B' = B as Element of bool X;
   A5:(sigma_Meas(C)).A' = C.A' & (sigma_Meas(C)).B' = C.B' by Def4;
        C.(A' \/ B') = C.A' + C.B' by A4,Th15;
      hence (sigma_Meas(C)).(A \/ B) = (sigma_Meas(C)).A + (sigma_Meas(C)).B
        by A5,Def4;
   end;
   hence thesis by A3,MEASURE1:def 5;
end;

definition let X be set;
   let C be C_Measure of X;
   let A be Element of sigma_Field(C);
   redefine func C.A -> R_eal;
coherence by FUNCT_2:7;
end;

theorem Th23:
   sigma_Meas(C) is sigma_Measure of sigma_Field(C)
proof
   reconsider M = sigma_Meas(C) as Measure of sigma_Field(C) by Th22;
     for F being Sep_Sequence of sigma_Field(C) holds
   M.(union rng F) <=' SUM(M*F)
   proof
      let F be Sep_Sequence of sigma_Field(C);
      consider A being Element of bool X such that
   A1:A = union rng F;
        M.A in ExtREAL by A1,FUNCT_2:7;
      then consider a,b being R_eal such that
   A2:a = M.A & b = C.A;
      reconsider F' = F as Function of NAT,bool X by FUNCT_2:9;
A3:   C*F' is Function of NAT,ExtREAL;
        for k being set st k in NAT holds (M*F).k = (C*F).k
      proof
         let k be set;
         assume
      A4:k in NAT;
      then A5:(M*F).k = M.(F.k) by FUNCT_2:21;
      A6:F.k in sigma_Field(C) by A4,FUNCT_2:7;
         reconsider F as Function of NAT,bool X by FUNCT_2:9;
         (C*F).k = C.(F.k) by A4,FUNCT_2:21;
       hence thesis by A5,A6,Def4;
      end;
then A7:   M*F = C*F by A3,FUNCT_2:18;
      reconsider F as Function of NAT,bool X by FUNCT_2:9;
        b <=' SUM(C*F) by A1,A2,Def2;
      hence thesis by A1,A2,A7,Def4;
   end;
   hence thesis by MEASURE3:17;
end;

definition let X be set;
   let C be C_Measure of X;
   redefine func sigma_Meas(C) -> sigma_Measure of sigma_Field(C);
coherence by Th23;
end;

theorem Th24:
   for A being Element of bool X holds
   C.A = 0. implies A in sigma_Field(C)
proof
   let A be Element of bool X;
   assume
A1:C.A = 0.;
     now let W,Z be Element of bool X;
      assume
   A2:W c= A & Z c= X \ A;
        C is nonnegative by Def2;
      then 0.<=' C.W & C.W <=' 0. by A1,A2,Def2,MEASURE1:def 4;
   then A3:C.W = 0. by SUPINF_1:22;
        Z c= W \/ Z by XBOOLE_1:7;
      then C.Z <=' C.(W \/ Z) by Def2;
      hence C.W + C.Z <=' C.(W \/ Z) by A3,SUPINF_2:18;
   end;
   hence thesis by Def3;
end;

theorem
     sigma_Meas(C) is_complete sigma_Field(C)
proof
   let A be Subset of X;
   let B;
   assume that
A1:B in sigma_Field(C) and
A2:A c= B & (sigma_Meas(C)).B = 0.;
   reconsider B as Element of bool X by A1;
A3:C.B = 0. by A1,A2,Def4;
     C is nonnegative by Def2;
   then 0.<=' C.A & C.A <=' 0. by A2,A3,Def2,MEASURE1:def 4;
   then C.A = 0. by SUPINF_1:22;
   hence thesis by Th24;
end;


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