E arbitrary => VarR2(f) > in for all i, hence Var R(f) = + -, again a
contradiction. Hence lim Var (f) M a for any a>0, so
p++s" [(s' t),(pt')]
p++s
lir Varrst),p -](f) = 0.
By the same argument, lim Var(, ( (f) = 0 (the R
r t' s,t ,r)
case). Putting everything together, we have
lim Var C(f) = lim [Var ,-(f) + Var (f) + Var (f) + Var (f)]
u++z' Lzu u++z' z ]1 R2 3
= lim Var ,(f) + lim Var (f) + lim Var (f) + lim Var (f)
u+z' [zz u+z, R1 u4+z' 2 u+4zt 3
= Var [z,z(f) + 0 + 0 + 0) = Var[z (f),
[z,z2] [2,25%
which is what was to be proved. I
Remarks.
1) As we stated above, the converse is not true. However, if
lim Var (f) = Var ,(f), then lim Var (f) = 0, and from
u+z [z,u [,z ] uz4,Z [z ,u]
IA[z',u]f f Var [',u](f) we get that lim [z. u]fl = 0, i.e., f is
u4+z
incrementally right continuous.
2) Returning to Example 2.1.2, for f as defined there, we have
Var[z,z'](f) = 0 on any rectangle [z,z'], since RE I~ (f)R =
R EP a
a -
Z 0 = 0 for any partition P of [z,z'], so lim Var (f) =
[z,uJ
R EP U-Z- UZ Lz
a -
Var[zz'](f). However, if f is constructed from a function g that is
not right continuous, then neither is f. In fact, for all e>0,