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(<f 2X dB ,z0>) 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 <m(X),z0> =

 E(<f 2X dB ,z0>). (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 <m(M)x,z0> = <m(1Mx),z0>

E((<IMx)dB ,Z0>) = E(fIMd<B v,Zo>) (by 4.1.3) = E(1A<BzX,Z>)
 0
=E(A <Bzx,zo>) = <m(M)x,z0>. Then, for x c E, z0 E Z

<m(M)x,z > = <m(M)x,z0>; 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 <m(X),z0> = <m(X),z > = E(<X dB ,z0>), 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 <Bx,z > 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