
theorem Th102:
for X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2,
  M1 be sigma_Measure of S1, M2 be sigma_Measure of S2,
  F be disjoint_valued FinSequence of sigma measurable_rectangles(S1,S2),
  n be Nat
st M2 is sigma_finite
 & F is FinSequence of measurable_rectangles(S1,S2)
holds
   product_sigma_Measure(M1,M2).(Union F) = Integral(M1,Y-vol(Union F,M2))
proof
   let X1,X2 be non empty set, S1 be SigmaField of X1, S2 be SigmaField of X2,
   M1 be sigma_Measure of S1, M2 be sigma_Measure of S2,
   F be disjoint_valued FinSequence of sigma measurable_rectangles(S1,S2),
   n be Nat;
   assume that
A1: M2 is sigma_finite and
A2: F is FinSequence of measurable_rectangles(S1,S2);
A3:F|(len F) = F by FINSEQ_1:58;
   defpred P[Nat] means
    product_sigma_Measure(M1,M2).(Union(F|$1))
     = Integral(M1,Y-vol(Union(F|$1),M2));
   union rng(F|0) = {} by ZFMISC_1:2; then
A4:Union(F|0) = {} by CARD_3:def 4;
   not 0 in dom F by FINSEQ_3:24; then
   F.0 = {} by FUNCT_1:def 2; then
P1:P[0] by A4,A1,A2,Th99;
P2:for k be Nat st P[k] holds P[k+1]
   proof
    let k be Nat;
    assume P3: P[k];
A6: k <= k+1 by NAT_1:13;
    per cases;
    suppose len F >= k+1; then
     len(F|(k+1)) = k+1 by FINSEQ_1:59; then
     F|(k+1) = ((F|(k+1))|k) ^ <* (F|(k+1)).(k+1) *> by FINSEQ_3:55
      .= F|k ^ <* (F|(k+1)).(k+1) *> by A6,FINSEQ_1:82
      .= F|k ^ <* F.(k+1) *> by FINSEQ_3:112; then
     rng(F|(k+1)) = rng(F|k) \/ rng <* F.(k+1) *> by FINSEQ_1:31
      .= rng(F|k) \/ {F.(k+1)} by FINSEQ_1:38; then
     union rng(F|(k+1)) = union rng(F|k) \/ union {F.(k+1)} by ZFMISC_1:78
      .= union rng(F|k) \/ F.(k+1) by ZFMISC_1:25
      .= Union(F|k) \/ F.(k+1) by CARD_3:def 4; then
A8:  Union(F|(k+1)) = Union(F|k) \/ F.(k+1) by CARD_3:def 4; then
     product_sigma_Measure(M1,M2).(Union(F|(k+1)))
      = product_sigma_Measure(M1,M2).(Union(F|k))
         + product_sigma_Measure(M1,M2).(F.(k+1)) by Th101,Th12
     .= Integral(M1,Y-vol(Union(F|k),M2))
         + Integral(M1,Y-vol(F.(k+1),M2)) by A1,A2,P3,Th99;
     hence P[k+1] by A1,A8,Th101,Th94;
    end;
    suppose len F < k+1; then
     (F|(k+1)) = F & F|k = F by FINSEQ_3:49,NAT_1:13;
     hence P[k+1] by P3;
    end;
   end;
   for k be Nat holds P[k] from NAT_1:sch 2(P1,P2);
   hence thesis by A3;
end;
