Let B: R xR L(E,Z') be the stochastic function associated with m
for this choice of Z. We have, then, for z0 in a countable dense
subset ZOC Z, and for A EE E(1 ** = E(<1A B ,z,>)
C 7 A x,z' 0 A XZ
0
S = E( ****) =
E(1 **** = Bx,z > a.s. for each z E ZO.
Now, since ZO Is countable, there is a common negligible set;
since Z is dense in Z we have B' = B x outside this negligible
0 x,z z
set. There is then a common negligible set such that B' = B x
X,z z
for all z rational, x in a countable dense set of E. Then, by right
continuity of B and closure of L(E,F) in L(E,Z'), we have
B = B' E L(E,F) outside this negligible set, i.e., up to
evanescence. By modifying B on this evanescent set, we obtain B with
values In L(E,F). Moreover, since B x = B' B x is Integrable (in
z xz z
particular Bx is measurable and separably valued by right continuity)
and so, by 4.2.3(2), for X E L (m), we have m(X) = E( 2X dB )
+
(m Is the measure associated with B via 4.2.3), and in
particular, m(M)x = m(lMx) = E( 1MxdB ), which completes (c).
d) Assume the range of m is contained in G CL(E,F) with
G E RNP. We write G L(R,G) and apply (c): R is separable,
G E RNP. Then B has values in L(R,G) = G C L(E,F), and Ba is
measurable for a E R; hence B is measurable. Also, for E LI (m)
we have by (c) m() = E(f vdB ), which is (d), and completes the proof
of this theorem.
Remark. In the last part, once we have B measurable, we can also get
the equality by applying Theorem 4.2.1.
**