all z rational. Then B is determined uniquely outside this evanescent
set; we already showed that p[B ] = B for every v E R2
Proof of (3):
0
a), b): If one of these is satisfied, then B takes values in
L(E,F); hence B takes values in L(E,F) since L(E,F) Is closed in
L(E,Z') for the metric topology (recall that B = lim B ).
z r
r+z
r rational
2
c): Assume E is separable, F e RNP. Let x E E, z E R The
measure p: F* F defined by R(A) = m (A)x for A E F is o-additive
(since ImZ(A)xj < ImZ(A)I-Ixl so An + > AImZA ) + 0 =>
|p(An) + 0; p is evidently additive) and has finite variation
|p| S IImZl'Ix. i S lzxj. Then p << P since Imlz << P, as we showed
already. Since F e RNP, there is a Bochner-integrable function
B'z E LF such that p(A) = E(1 B' ) for A E F. We can choose
x,z F A x,z
B' separably valued; therefore, we can consider F separable. More
precisely, let S be a countable dense set in E. For x E S, we have
lim B' = B' a.s. In fact, for A EF E(1 B' ) = p(A) =
x,u x,z A x,u
u+z
u rational
mu(A)x = m([0,u]xA)x + m([0,z]xA)x mz(A)x = E(1 AB' ); hence
B' + B' a.s. Then IB' (w): x E S, z rational} is separable;
x,u x,z x,z
hence (B' (w): x E E, z rational is separable. For any z,
x,z
B' lim B' a.s., and we modify B' on the exceptional set;
xz x,u x,z
u rational
we can do this and still get a R-N derivative of u. Hence we can take
F separable. We can then choose Z separable in F', norming for F.