
theorem Th57:
  for X be non empty set, S be SigmaField of X, M be sigma_Measure of S,
  f be PartFunc of X,ExtREAL st dom f in S &
   (for x be Element of X st x in dom f holds 0= f.x)
  holds (for E be Element of S st E c= dom f holds f is E-measurable)
   & Integral(M,f) = 0
proof
    let X be non empty set, S be SigmaField of X, M be sigma_Measure of S,
    f be PartFunc of X,ExtREAL;
    assume that
a1:  dom f in S and
a2:  for x be Element of X st x in dom f holds f.x = 0;
    reconsider E = dom f as Element of S by a1;
    dom(chi(0,E,X)|E) = dom(chi(0,E,X)) /\ E by RELAT_1:61; then
    dom(chi(0,E,X)|E) = X /\ E by FUNCT_2:def 1; then
a3: dom(chi(0,E,X)|E) = E by XBOOLE_1:28;

    now let x be Element of X;
     assume a4: x in dom f; then
     (chi(0,E,X)|E).x = chi(0,E,X).x by FUNCT_1:49; then
     (chi(0,E,X)|E).x = 0 by a4,Def1;
     hence f.x = (chi(0,E,X)|E).x by a2,a4;
    end; then
a4: f = chi(0,E,X)|E by a3,PARTFUN1:5;
    hence for A be Element of S st A c= dom f holds f is A-measurable
     by Th15;
    Integral(M,f) = 0 * M.E by a4,Th50;
    hence Integral(M,f) = 0;
end;
