= AR (B) (property (5) following Definition 1.5.5), so by 1.5.7,
zz
IBz, BzI and IAR (B)I are measurable; hence the finite sums we
zz
use to compute Var [,(B (.0)), Var,t](B(0 ), and
Var[(0,),(st)(B) are also measurable.
Moreover, since B is right continuous, we can compute the
variation using partitions consisting of rational points: the first
two terms from the one-dimensional result, the third by Proposition
2.2.5. Each of these limits can then be taken along a sequence, so
Var [0s](B( ,0)), Var [ ,t](B (0 .)), Var [( ,0),(st)](B) are all
^[O~sD ,O) C[OtJ (0,*) ^[(0,0), (st)]
measurable; hence ]BI ,t = B(0,0) + Var[0,s](B(.,) +
Var [,t]B((,.) + Var[(0,O),(s,t)](B) is measurable, i.e., IBI is a raw
process, evidently increasing.
We show now that B has integrable variation. We shall show, in
fact, that IBI 5 V. We proceed with each term of IBI separately:
1) Since B(0,0)l V(0,0) we have IB((0,00) a.s.
2) Let o: 0 sO < s1 < ... < s = s be a partition of [O,s].
m-1 m-1
We have E IB( 0) B( ) | 5 E (V V ( ) =
i=O (si+1,0) (si0) i-0 (s+1) (si 0)
(sV ) V(0O ) + (V S V, ) + .. + (Vs) V )
SV(s, V(0,0), P-a.s. Now, let (o ) nN be a sequence of parti-
tions of [O,s] such that Var (B ) = lim Var [s(B *).
[0,s] (*,n) [0,s] (*,0);on
n
Each term is dominated by V(s, V(,0) a.s., and V(s,0) V
does not depend on n, so taking limits we have Var[,s] (B ( 0)
(s,0) (0,0) a.s.