theorem Th22:
  f|A is bounded implies
    for D,D1 ex D2 st D<=D2 & D1<=D2 & rng D2=rng D1 \/ rng D & 0<=
  upper_sum(f,D)-upper_sum(f,D2) & 0<=upper_sum(f,D1)-upper_sum(f,D2)
  proof
    assume
A1: f|A is bounded;
    let D,D1;
    consider D2 such that
A3: D<=D2 and
A4: D1<=D2 and
A5: rng D2=rng D1 \/ rng D by Th4;
A6: upper_sum(f,D1)-upper_sum(f,D2)>=0 by A1,A4,INTEGRA1:45,XREAL_1:48;
    upper_sum(f,D)-upper_sum(f,D2)>=0 by A1,A3,INTEGRA1:45,XREAL_1:48;
    hence thesis by A3,A4,A5,A6;
  end;
