As in the proof of (1), we extract an E0 > 0, and (n ,tn) such
that (s ,tn) (s,t), and for each n we have s n > sn, tn+ < tn
and If(sn+1,tn+ ) f(sn,tn)I > 0E. As in (1), for given j, conside-
the subdivisions s1 < s2 < ... < sj < s = s of [s ,s] and
t > t2 > ... t > t+ = t of [t,tl]. For i = 1,2,...,j-1, denote
R = [(siO),(s ,t )], Ri = [(0,t i ),(si+ ti)]. (See Figure 2-12.)
We have, for each I (similar to before):
lAR fI f(si+1,ti) f(s ,t ) + f(s ,0)
> If(s ,t (s ,t ) If(s ,0) f(si,) ,
hence
tl
t2
t
t I
i+1 -
R.
t (s,t)
Figure 2-12 Partition of [(0,0),(s,t)]