theorem Th26:
  f a.e.cpfunc= g,M implies a(#)f a.e.cpfunc= a(#)g,M
proof
  assume f a.e.cpfunc= g,M;
  then consider E being Element of S such that
A1: M.E = 0 and
A2: f|E`=g|E`;
  (a(#)f)|E` = a(#)(g|E`) by A2,RFUNCT_1:49
    .= (a(#)g)|E` by RFUNCT_1:49;
  hence thesis by A1;
end;
