In fact, let E>0: there exists a neighborhood to the right of z
so that if r is rational and lies inside the neighborhood, then
Bz(w) Br(w)I < For any z' in this neighborhood, there exists a
similar neighborhood for it, and the intersection of these has
nonempty interior. Let r be in their intersection, r rational. We
o o
have B z(w) B(w) IB( B (w) + Br(w) B.(w)
B B(w) + IBr(w) Bz(w)I L = c. Thus B is right
continuous.
Some more properties of B are the following:
a) For wz, r>w. Then |Bq Br
o q
0 0
Bq B + B B + B B B Vq V Letting q4z, we get
IB Bw + B Brl S V V by definition of B and right continuity
of V. Letting r+w likewise, we get 13 BwI V V We
similarly have IA R (B) A R (V).
zz zz
0 2
8) For each z, B = B a.s.: In fact, for z R r rational,
z zV
0 0
r>z we have IB B I 5 V V a.s. (in fact there is a common
negligible set outside of which this holds for all r>z) letting r+z,
0 0 0
we have lB B = limlBr B I lim(V V) 0 a.s., i.e.,
rzz rez
0
B = B P-a.s.
z z
Next, we show that B has raw integrable variation |BI. Since
o 0 0
B = B P-a.s., B B = B - B a.s., and A (B) =
z z z z z z 7'
zz
AR (B ) a.s.; hence p[B z B z = B Bz, and p[AR (B)]
ZZ ZZ
zz zzS S