
theorem Th40:
for X be non empty set, S be SigmaField of X, M be sigma_Measure of S,
 E be Element of S, f,g be PartFunc of X,REAL st E = dom(f-g)
  & (f-g) a.e.= (X-->0)|E,M holds f|E a.e.= g|E,M
proof
    let X be non empty set, S be SigmaField of X, M be sigma_Measure of S,
    E be Element of S, f,g be PartFunc of X,REAL;
    assume that
A1:  E = dom(f-g) and
A2:  (f-g) a.e.= (X-->0)|E,M;
    consider A be Element of S such that
A3:  M.A = 0 & (f-g)|A` = ((X-->0)|E)|A` by A2,LPSPACE1:def 10;

    dom f /\ dom g = E by A1,VALUED_1:12; then
A4: dom(f|E) = E & dom(g|E) = E by XBOOLE_1:17,RELAT_1:62; then
A5: dom((f|E)|A`) = dom(g|E) /\ A` by RELAT_1:61
     .= dom((g|E)|A`) by RELAT_1:61;

    for x be Element of X st x in dom((f|E)|A`) holds
     ((f|E)|A`).x = ((g|E)|A`).x
    proof
     let x be Element of X;
     assume x in dom((f|E)|A`); then
A6:  x in dom(f|E) & x in A` by RELAT_1:57;

A7:  ((f-g)|A`).x = (f-g).x by A6,FUNCT_1:49
      .= f.x - g.x by A1,A6,A4,VALUED_1:13;

A8:  (((X-->0)|E)|A`).x = ((X-->0)|E).x by A6,FUNCT_1:49
      .= (X-->0).x by A6,A4,FUNCT_1:49
      .= 0;

     ((f|E)|A`).x = (f|E).x by A6,FUNCT_1:49
      .= g.x by A6,A8,A3,A7,A4,FUNCT_1:49
      .= (g|E).x by A6,A4,FUNCT_1:49;
     hence ((f|E)|A`).x = ((g|E)|A`).x by A6,FUNCT_1:49;
    end;
    hence f|E a.e.= g|E,M by A3,A5,PARTFUN1:5,LPSPACE1:def 10;
end;
