theorem Th1:
  |. R_EAL f .| = R_EAL abs f
proof
A1: now
    let x be Element of X;
    assume x in dom |. R_EAL f .|;
    then (|. R_EAL f .|).x = |. (R_EAL f).x .| by MESFUNC1:def 10;
    then (|. R_EAL f .|).x = |.f.x qua Complex.| by EXTREAL1:12;
    then (|. R_EAL f .|).x = |.f.|.x
      by VALUED_1:18;
    hence (|. R_EAL f .|).x = (R_EAL abs f).x;
  end;
  dom |. R_EAL f .| = dom R_EAL f by MESFUNC1:def 10
    .= dom abs f by VALUED_1:def 11;
  hence thesis by A1,PARTFUN1:5;
end;
