
theorem Th5:
for X,Y be non empty set, f be Function of X,Y, S be Field_Subset of X,
    M be Measure of S st f is bijective holds
 ex M1 be 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 Field_Subset of X,
        M be Measure of S;
    assume
A1: f is bijective; then
A2: CopyField(f,S) = .:f.:S by Def1;
    consider F be Function of CopyField(f,S),S such that
A3: F = ((.:f) |S)" & rng F = S & dom F = CopyField(f,S)
  & F is bijective by A1,Th4;

    consider G be Function of S,CopyField(f,S) such that
A4: G = (.:f) |S & dom G = S & rng G = CopyField(f,S)
  & G is bijective by A1,Th4;

    reconsider M1 = M * F as Function of CopyField(f,S),ExtREAL;

A5: {} c= X;
A6: {} in dom G by A4,PROB_1:4;
    ((.:f) |S).{} = (.:f).{} by FUNCT_1:49,PROB_1:4; then
    ((.:f) |S).{} = f.:{} by A5,A1,Th1; then
A7: {} = F.{} by FUNCT_1:34,A3,A4,A6; then

    M1.{} = M.{} by A3,FUNCT_1:13,PROB_1:4; then
A8: M1 is zeroed by VALUED_0:def 19;

A9: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 s be Element of CopyField(f,S);
     consider t be object such that
A10: t in dom (.:f) & t in S & s= (.:f).t by A2,FUNCT_1:def 6;
     reconsider t as Element of S by A10;
A11: s = G.t by A4,A10,FUNCT_1:49;
     take t;
     thus s = f.:t by A1,A10,Th1;
     M1.s = M.(F.s) by A3,FUNCT_1:13;
     hence M1.s = M.t by A3,A4,A11,FUNCT_1:34;
    end;

    for A,B being Element of CopyField(f,S) st
     A misses B & A \/ B in CopyField(f,S) holds M1.(A \/ B) = M1.A + M1.B
    proof
     let A,B be Element of CopyField(f,S);
     assume
A12: A misses B & A \/ B in CopyField(f,S);
     consider a be Element of S such that
A13: A = f.:a & M1.A = M.a by A9;
     consider b be Element of S such that
A14: B = f.:b & M1.B = M.b by A9;
     A /\ B = f.:(a /\ b) by FUNCT_1:62,A1,A13,A14; then
     A /\ B = (.:f).(a /\ b) by A1,Th1; then
     {} = G.(a /\ b) by A12,A4,FUNCT_1:49; then
A15: a misses b by A7,A3,A4,FUNCT_1:34;

     A \/ B = f.:(a \/ b) by RELAT_1:120,A13,A14; then
     A \/ B =(.:f).(a \/ b) by A1,Th1; then
     A \/ B = G.(a \/ b) by A4,FUNCT_1:49; then
     M1.(A \/ B) = M.(F.(G.(a \/ b))) by FUNCT_1:13,A3; then
     M1.(A \/ B) = M.(a \/ b) by A3,A4,FUNCT_1:34;
     hence M1.(A \/ B) = M1.A + M1.B by A13,A14,A15,MEASURE1:def 3;
    end; then
    M1 is additive;
    hence thesis by A3,A8,A9;
end;
