
theorem Th6:
for X,Y be non empty set, f be Function of X,Y, S be SigmaField of X,
    M be sigma_Measure of S st f is bijective holds
 ex M1 be sigma_Measure of CopyField(f,S)
    st M1 = M * ((.:f) |S)"
     & for s being Element of CopyField(f,S) holds
        ex t be Element of S st s = f.:t & M1.s = M.t
proof
    let X,Y be non empty set, f be Function of X,Y, S be SigmaField of X,
        M be sigma_Measure of S;
    assume
A1: f is bijective;
    reconsider S0 = S as Field_Subset of X;
    reconsider M0 = M as Measure of S0;
    consider F be Function of CopyField(f,S0),S0 such that
A2: F = ((.:f) |S0)" & rng F = S0 & dom F = CopyField(f,S0)
  & F is bijective by A1,Th4;
    consider G be Function of S0,CopyField(f,S0) such that
A3: G = (.:f) |S0 & dom G = S0 & rng G = CopyField(f,S0)
  & G is bijective by A1,Th4;

A4: CopyField(f,S0) = .:f.:S0 by A1,Def1; then
A5: CopyField(f,S0) = CopyField(f,S) by A1,Def2; then
    reconsider F as Function of CopyField(f,S),S;
    reconsider G as Function of S,CopyField(f,S) by A1,A4,Def2;

    consider M1 be Measure of CopyField(f,S0) such that
A6: M1 = M * ((.:f) |S0)"
  & for s being Element of CopyField(f,S0) holds
     ex t be Element of S0 st s = f.:t & M1.s = M.t by Th5,A1;
    reconsider M1 as Measure of CopyField(f,S) by A5;

    for s being Sep_Sequence of CopyField(f,S) holds
     SUM(M1*s) = M1.(union rng s)
    proof
     let s be Sep_Sequence of CopyField(f,S);
     defpred P[Nat,set] means s.$1 = f.:$2 & M1.(s.$1) = M.$2;

A7:  for x being Element of NAT ex y being Element of S st P[x,y] by A5,A6;

     consider t being Function of NAT,S such that
A8: for n being Element of NAT holds P[n,t.n] from FUNCT_2:sch 3(A7);
     reconsider t as sequence of S;

     for m, n being object st m <> n holds t.m misses t.n
     proof
      let m, n be object;
      assume
A9:   m <> n;
      per cases;
      suppose not m in dom t or not n in dom t; then
       t.m = {} or t.n = {} by FUNCT_1:def 2;
       hence t.m misses t.n;
      end;
      suppose m in dom t & n in dom t; then
       reconsider m1 = m, n1 = n as Element of NAT;

A10:   s.n1 = f.:(t.n1) & M1.(s.n1) = M.(t.n1) by A8;
A11:   s.m1 = f.:(t.m1) & M1.(s.m1) = M.(t.m1) by A8;

       s is disjoint_valued; then
A12:   s.m misses s.n by A9;

       now assume t.m1 /\ t.n1 <> {}; then
        consider x be object such that
A13:    x in t.m1 /\ t.n1 by XBOOLE_0:def 1;
A14:    x in t.m1 by XBOOLE_0:def 4,A13;
        t.m1 in S; then
        t.m1 c= X; then
        t.m1 c= dom f by FUNCT_2:def 1; then
        f.x in f.:(t.m1 /\ t.n1) by A13,A14,FUNCT_1:def 6;
        hence contradiction by A1,A10,A11,A12,FUNCT_1:62;
       end;
       hence t.m misses t.n;
      end;
     end; then
     t is disjoint_valued; then
     reconsider t as Sep_Sequence of S;

A15: dom s = NAT & dom t = NAT by FUNCT_2:def 1;

     for n be Element of NAT holds (M1*s).n = (M*t).n
     proof
      let n be Element of NAT;
A16:  s.n = f.:(t.n) & M1.(s.n) = M.(t.n) by A8;
      (M1*s).n = M1.(s.n) by A15,FUNCT_1:13;
      hence (M1*s).n = (M*t).n by A16,A15,FUNCT_1:13;
     end; then
A17: M1*s = M*t by FUNCT_2:63;

     for z be object holds z in union rng s iff z in f.:(union rng t)
     proof
      let z be object;
      hereby assume z in union rng s; then
       consider Y being set such that
A18:   z in Y & Y in rng s by TARSKI:def 4;
       consider n be Element of NAT such that
A19:   n in dom s & Y = s.n by A18,PARTFUN1:3;
A20:   z in f.:(t.n) by A8,A18,A19;

       t.n in rng t by A15,FUNCT_1:def 3; then
       f.:(t.n) c= f.:(union rng t) by RELAT_1:123,ZFMISC_1:74;
       hence z in f.:(union rng t) by A20;
      end;

      assume z in f.:(union rng t); then
      consider x be object such that
A21:  x in dom f & x in union rng t & z=f.x by FUNCT_1:def 6;
      consider Y being set such that
A22:  x in Y & Y in rng t by A21,TARSKI:def 4;

      consider n be Element of NAT such that
A23:  n in dom t & Y=t.n by A22,PARTFUN1:3;
      z in f.:(t.n) by A21,A22,A23,FUNCT_1:def 6; then
A24:  z in s.n by A8;

      s.n in rng s by A15,FUNCT_1:3;
      hence z in union rng s by A24,TARSKI:def 4;
     end; then
A25: union rng s = f.:(union rng t) by TARSKI:2;

     F*G = id (dom G) by A2,A3,FUNCT_1:39; then
A26: F*G = id (dom M) by A3,FUNCT_2:def 1;
     M1*G = M*(F*G) by A2,A6,RELAT_1:36; then
     M1*G = M by A26,RELAT_1:52; then
     M.(union rng t) = M1.(G.(union rng t)) by A3,FUNCT_1:13; then
     M.(union rng t) = M1.((.:f).(union rng t)) by A3,FUNCT_1:49; then
     M.(union rng t) = M1.(union rng s) by A1,A25,Th1;
     hence thesis by A17,MEASURE1:def 6;
    end; then
    reconsider M1 as sigma_Measure of CopyField(f,S) by MEASURE1:def 6;
    take M1;
    thus thesis by A5,A6;
end;
