
theorem Th25:
for X,Y be non empty set, S be SigmaField of X, T be Function of X,Y,
 M be sigma_Measure of S, f be PartFunc of X,ExtREAL,
 g be PartFunc of Y,ExtREAL
  st T is bijective & g = f*T" & f is_simple_func_in S & f is nonnegative
  holds integral(CopyMeasure(T,M),g) = integral(M,f)
proof
    let X,Y be non empty set, S be SigmaField of X, T be Function of X,Y,
    M be sigma_Measure of S, f be PartFunc of X,ExtREAL,
    g be PartFunc of Y,ExtREAL;
    assume that
A1: T is bijective and
A2: g = f*T" and
A3: f is_simple_func_in S and
A4: f is nonnegative;

A5: g is_simple_func_in CopyField(T,S)
  & g is nonnegative by A1,A2,A3,A4,Th23,Th22;

    consider F be Finite_Sep_Sequence of S, a,x be FinSequence of ExtREAL
     such that
A6: F,a are_Re-presentation_of f
  & a.1 = 0.
  & ( for n being Nat st 2 <= n & n in dom a holds 0. < a.n & a.n < +infty )
  & dom x = dom F
  & ( for n being Nat st n in dom x holds x.n = (a.n)*((M*F).n) )
  & integral(M,f) = Sum x by A3,A4,MESFUNC3:def 2;

    consider G be Finite_Sep_Sequence of CopyField(T,S) such that
A7: G = ((.:T) |S)*F & G,a are_Re-presentation_of g by A1,A2,A6,Th24;

    set L = CopyMeasure(T,M);
    set H = (.:T) |S;
    for n being Nat st n in dom x holds x.n = (a.n)*((L*G).n)
    proof
     let n be Nat;
     assume
A8:  n in dom x; then
A9:  x.n = (a.n)*((M*F).n) by A6;

A10: (M*F).n = M.(F.n) by A8,A6,FUNCT_1:13;
A11: (L*G).n = L.(G.n) by A8,A6,A7,FUNCT_1:13;

     reconsider Gn = G.n as Element of CopyField(T,S);
     consider Kn be Element of S such that
A12: Gn = T.:Kn & L.Gn =M.Kn by A1,Def4;
A13: Gn = (.:T).Kn by A12,A1,Th1;
A14: .:T is one-to-one by A1;
A15: dom (.:T) = bool X by FUNCT_2:def 1;

     G.n = ((.:T) |S).(F.n) by A7,A8,A6,FUNCT_1:13; then
     G.n = (.:T).(F.n) by FUNCT_1:49;
     hence thesis by A9,A10,A11,A12,A13,A14,A15;
    end;
    hence thesis by A5,A6,A7,MESFUNC3:def 2;
end;
