We already showed that Iml = pl.B, which establishes the second
equality in (1). As for the first, we note that by Theorem 4.2.3,
E() is defined for z0 e Z. To prove the equality, we
R+
note first, that by Theorem 4.2.2 there exists a stochastic measure
m: M + L(E,Z') corresponding to B satisfying =
E(). (Note that since Z' is the dual of a Banach space,
m has values in the same space of operators as B; cf. statement (3a)
of this theorem.) We shall show that m m; as both are o-additive it
will suffice, as above, to prove for sets of the form M [O,z]xA,
A E F.
Let M = [O,z]xA, A E F; let x E E, z E Z (which is a norming
subspace of (Z')'). We have =
E(() = E(fIMd**) (by 4.1.3) = E(1A)
0
=E(A ) = . Then, for x c E, z0 E Z
= ; hence m(M) = m(M) for M = [0,z]xA. As
2
before, we conclude that m = m on all of B(R )xF. Then for
X E LE(m) we have = = E(), which
completes the proof of (1).
Proof of (2): If Bx is separably valued for x E E, then Bx is
measurable for x E E since it is weakly measurable by (ii). If B is
separably valued, then B is measurable since is measurable for
all x E E, z0 E Z, by (11) [6, Proposition 24, p. 106].
Proof of (4): If p is a lifting of P, we can choose B
uniquely a.s. for all z; n particular, outside an evanescent set for
uniquely a.s. for all z; in particular, outside an evanescent set for
**