3) The same argument gives Var,t](B ,) V(t) V(0,0)
a.s.
4) Let oxT be a grill on [(0,0),(s,t)]: we have E IAR(B)I
REOXT
R 6RV) = V(s,t) (0,t) V(s,0) (0,0).
To see this last equality, consider for each w the measure mV(w)
on B(R2) associated with V(w). Consider the half-open rectangles
R associated with the closed rectangles R of oxT. Then
A 6R(V(w)) = m(w) R) = m (w)((0,0),(s,t)]) = V(s,) V(0,t)
REOXT ) XT
- Vs, + V(,0). This holds a.s. Taking supremum along a sequence
(sO) (0,0)
of grills as before, we have Var[(0,O) (st)](B) 5 V(s,t) V(t)
- V(s,) + V0,0) a.s. Adding up (1)-(4), we obtain B(st)
B(0,0) + Var[0,s](B(.,0)) + Var[0,t](B(0,.)) + ar[( ,0),(s,t)](B)
V(0,0) + (V(s,0) (0,0) (V 0,t) (0,0)) + (s,t)- (s,0)
- VOt) + V(00)) t) a.s. Thus, for (s,t) E R2, IBI(s.t)
V(s,t) a.s. Since BI is increasing, and 3(n,n) V(n,n) a.s., IBI
(st) (n,n) (n,n)
is finite outside an evanescent set*; hence IBI is right continuous
outside an evanescent set. Then, since both |BI and V are right
continuous, IB|I V outside an evanescent set. In particular,
IBI| 5 V so B has integrable variation, and this completes the proof
of (1).
SMore precisely, for example is a negligible set N such that
n
w V Nn => IB (n,n)() ( V(n,n)(w); hence BI ( V(n,n)
outside Nn. Then NO = J Nn is negligible and BI < outside
this set. n