
theorem
  for X being non empty set,
      S being SigmaField of X, M being sigma_Measure of S,
      f being PartFunc of X,ExtREAL, A being Element of S
   st f is_integrable_on M &
      (for x being object st x in dom f \ A holds f.x = 0)
   holds Integral(M,f) = Integral(M,f|A)
proof
   let X be non empty set, S be SigmaField of X, M be sigma_Measure of S,
   f be PartFunc of X,ExtREAL, A be Element of S;
   assume that
A1: f is_integrable_on M and
A2: for x being object st x in dom f \ A holds f.x = 0;
   consider E be Element of S such that
A3: E = dom f & f is E-measurable by A1,MESFUNC5:def 17;
   reconsider B = E\A as Element of S;
A4:Integral(M,f|(A\/B)) = Integral(M,f|A) + Integral(M,f|B)
      by A1,MESFUNC5:98,XBOOLE_1:79;
   A \/ B = dom f \/ A by A3,XBOOLE_1:39; then
A5:f|(A\/B) = f by RELAT_1:68,XBOOLE_1:7;
A6:B c= dom f by A3,XBOOLE_1:36;
   dom(f|B) = dom f /\ B by RELAT_1:61; then
A7:dom(f|B) = B by A3,XBOOLE_1:28,36;
   now let x be object;
    assume A9: x in dom(f|B); then
    (f|B).x = (f|B)/.x by PARTFUN1:def 6 .= f/.x by A9,PARTFUN1:80
     .= f.x by A9,A7,A6,PARTFUN1:def 6;
    hence (f|B).x = 0 by A2,A9,A7,A3;
   end; then
   Integral(M,f|B) = 0 * M.(dom(f|B)) by A7,Th27 .= 0;
   hence Integral(M,f) = Integral(M,f|A) by A4,A5,XXREAL_3:4;
end;
