Proof. Case (4) has been dealt with in Theorem 4.2.1; we include it
here for completeness.
We have a measure m: M L(E,Z') with finite variation. By
Theorem 4.3.1, there exists a stochastic function B' with values also
in L(E,Z') (by 3(a)) satisfying (i) and (ii) of Theorem 4.3.1, such
that p[B'] = B' for z E R and such that for every X e L,(m) we have
z z +
= E(< 2XdB,zo>) for z0 E Z
(note: here we consider Z embedded in Z" as a norming subspace in
order to apply the theorem), and
jml( Ix ) = E( IX vld B'I ).
R-
Now, for X = 1 x with M c M, x e E, and for z0 E Z, we have
E(fl d**) = E(flMd****). In fact, E(fIMd)
E() (by 4.1.3) = E() = =
E() = E() = E(fIMd). Taking
M = [O,z]xA with A eFwe obtain
E(f1[OzxAd****) = E(1A****), and
E(f1[O,z]xAd) = E(1A); hence
E(1 ) E(1 )
for A EF so (1.1) **** = **** a.s. for x E E,
z Z. e shall prove that n cases ()-(3), B and B are
z E Z. We shall prove that In cases (1)-(3), B and B' are
**