theorem Th45:
  f in L1_Functions M & g in L1_Functions M & f a.e.= g,M implies
  abs f a.e.= (abs g),M & Integral(M,abs f) = Integral(M,abs g)
proof
  assume that
A1: f in L1_Functions M and
A2: g in L1_Functions M and
A3: f a.e.= g,M;
A4: ex f1 be PartFunc of X,REAL st f=f1 & ex ND be Element of S st M.ND=0 &
  dom f1 = ND` & f1 is_integrable_on M by A1;
  then consider NDf be Element of S such that
A5: M.NDf=0 and
A6: dom f = NDf` and
  f is_integrable_on M;
A7: abs f is_integrable_on M by A4,Th44;
  consider EQ being Element of S such that
A8: M.EQ = 0 and
A9: f|EQ` = g|EQ` by A3;
  (abs f)|EQ` = abs(g|EQ`) by A9,RFUNCT_1:46
    .= (abs g)|EQ` by RFUNCT_1:46;
  then
A10: abs f a.e.= (abs g),M by A8;
A11: ex g1 be PartFunc of X,REAL st g=g1 & ex ND be Element of S st M.ND=0 &
  dom g1 = ND` & g1 is_integrable_on M by A2;
  then consider NDg be Element of S such that
A12: M.NDg=0 and
A13: dom g = NDg` and
  g is_integrable_on M;
A14: abs g is_integrable_on M by A11,Th44;
  dom abs g = NDg` by A13,VALUED_1:def 11;
  then
A15: abs g in L1_Functions M by A12,A14;
  dom abs f = NDf` by A6,VALUED_1:def 11;
  then abs f in L1_Functions M by A5,A7;
  hence thesis by A15,A10,Th43;
end;
