Ad (ii). Denote L = f(s-,t+), let e > 0. There exists 6 > 0 such
that for all (s',t') with s'~~t, s-s' < 6, t'-t < 6, we have
f(s',t')-LI < j Let, then, s' < s with s-s' < 6: since f is right
continuous, there exists a point (s",t') with s' < s" < s, t' > t,
t'-t < 6, such that If(s',t) f(s",t') < But we also have
2
jf(s",t')-Lj < hence ff(s',t)-LI < gf(s',t)- f(s",t') +
If(s",t')-L < + E Thus, lim f(s',t) = L = f(s-,t+).
25 2'tts
The proof of the limit at infinity is much the same, except that
instead of s-s' < 6, there is N such that for all s' > N, the
conditions hold. We then take s' > N, s" > s, and the remainder is
the same.
Ad (iii). The proof of (iii) is the same as that of (ii), with the
roles of s and t being reversed: for E > 0 there exists 6 > 0 such
that for all s' > s, t' < t, etc. The remainder is the same.
Ad (i). This follows immediately from the definition of right
continuity: all three limits are equal to f(s,t). We should remark,
however, that it is the same to define right continuity using the open
quadrant: the limits along the horizontal and vertical paths are then
the same as the "quadrantal" limit. In fact, for E > 0, choose 6 > 0
so that for s' > s, t' > t, s'-s < 6, t'-t < 6, If(s,t)-f(s',t')j < .
Then for any s' > s with s'-s > 6, there is a similar -
2
neighborhood" for the point (s',t). Pick any point in the
intersection of these -neighborhoods," and apply the triangle
2
inequality as before. I
Our next result concerns the existence of a "Jordan
decomposition" for functions of two variables with finite variation:
~~