Proof. a) Assume first that f has finite variation If|.
The proofs of limits (1)-(4) are similar; we treat (1) first. We
shall show that for any sequence (s ,tn) with s *4s, t n+t, the
sequence (f(s ,t )}neN is Cauchy in E. We shall do this by denial:
n n neN
Let (sn,t ) be a sequence as above and assume that If(s ,tn)nN is
not Cauchy--we shall reach a contradiction.
Since f(s ,t ) is not Cauchy, there exists e >0 and a subsequence
n n 0
(nk)kcN such that, for all k, we have
If(s ,t ) f(s t ) > E .
k+1 k+1 k nk
We will henceforth denote this sequence by If(sn,tn) n N, so we have
the inequality for all n.
For j given, consider the subdivisions I > s2 > ... > sj
> s = s of [s,s ] and t1 > t2 > ... > tj > tj, = t of [t,t ].
For i = 1,2,...,j-1 denote Ri = [(s +,0),(sti+1)] and
R; = [(0,ti+1),(s ,t)] (Figure 2-11). For each i, we have
IAR fI = If(s i ) f(s i 1,t + ) f(si,0) + f(si+1,0)|, so
IAR fl If(si,ti+I) f(si+1t i+1)j |f(s!,0) f(si+1,0) =>
If(si,O) f(si+1,0)I + IAf I f If(si,ti1) f(si+1,ti+1) We have
similarly
jAR fj If(siti) f(sti,+) f(O,ti) + f(O,ti+1)I
i
s If(si,ti) f(si,ti+1)I I|f(o,ti) f(0,ti+1