
theorem Th95:
  for X be non empty set, S be SigmaField of X, M be
sigma_Measure of S, f be PartFunc of X,ExtREAL, E,A be Element of S st E = dom
  f & f is E-measurable & M.A =0 holds Integral(M,f|(E\A)) = Integral(M,f)
proof
  let X be non empty set, S be SigmaField of X, M be sigma_Measure of S, f be
  PartFunc of X,ExtREAL, E,A be Element of S such that
A1: E = dom f and
A2: f is E-measurable and
A3: M.A =0;
  set B = E\A;
A4: dom f=dom(max+f) by MESFUNC2:def 2;
A5: max-f is nonnegative by Lm1;
A6: max+f is nonnegative by Lm1;
A7: dom f=dom(max-f) by MESFUNC2:def 3;
  Integral(M,f|B) = integral+(M,(max+f)|B) -integral+(M,max-(f|B)) by Th28
    .= integral+(M,(max+f)|B) -integral+(M,(max-f)|B) by Th28
    .=integral+(M,max+f) -integral+(M,(max-f)|B) by A1,A2,A3,A4,A6,Th84,
MESFUNC2:25;
  hence thesis by A1,A2,A3,A7,A5,Th84,MESFUNC2:26;
end;
