and we get a contradiction as before. Then f(s ,t ) is Cauchy; hence
llm f(sn,tn) exists, and we prove the limit is the same for any
n
sequence the same way as before.
This illustrates the difference between the proofs of (1)-(4) and
those of (1')-(5'): We do the same computation for the limits at
infinity, but the value of Ifl turns out to be at a point depending on
j, so we further majorize it by M. For (3') we have
j-1
Z If(si,t ) f(si+1,ti)| S If|(tl ,s ) M
i=1
J-1
SIf(si+1'ti f(s tl+1'ti+1)I Ifi(tl ,Sj ) 5 M,
i=1
so as before
j-1
(j-1)E0 < If(sl,t ) f(si+1ti+1) I 2M,
i=1
and we conclude as above. For (2'), (4'), and (5'), we follow the
same computation as in the proof of (1) and obtain
j-1
(j-1)E0 E If(sit ) f(si+1,ti+1) : f]j(sj,tj) S M
i=1
and conclude as in the proof of (1). This completes the proof of (a).
Proof of (b). Assume, now, that f is right continuous (order
sense!). We shall deal with (u) and (ui) first).