Ad (ii). Denote L = f(s-,t+), let e > 0. There exists 6 > 0 such

that for all (s',t') with s'<s, t'>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: