m(M) c L(E,F). As for complements, m(MC) = m(R2 X) m(M), and
m(R2xn)x = lim m([(0,0),(n,n)]xQ)x c F by closure of F in Z' as
n+-
before. Thus, if m(M) E L(E,F), then m(M0) E L(E,F). Then CC M C
is an algebra, so M = o(C)CM by the monotone class theorem, i.e., m
takes values in L(E,F).
c) Suppose that for x c E, v E R the function B x is F-
v
measurable and almost separably valued (in particular, if F is
separable, then B x is separably valued and weakly measurable by (li);
v
hence B x is F-measurable). Then for every A s F, x E E, the function
v
1 B x is integrable; in fact, B x is F-measurable, by assumption
AV v
and we have lB x 1 B iJ\ x| E L (P). We also have, as before,
= E(fi[O,v]xAd) = E() =
= (again, we can move the
expectation inside since 1.B x is integrable). Since this holds for
Av
all z E Z, we conclude m([0,v]xA)x = E(1 B x) E F. Then
m(EO,v]xA) E L(E,F), and we conclude by the same monotone class
argument as in (b). I
Remarks.
1) This theorem shows that if B has values in L(E,F), then mB
has values in a subspace of L(E,F"). Moreover, we do not have in
general Im B -= I.. Later we will establish some conditions
sufficient for equality.
2) The correspondence B + m is not injective. For an example
involving measures associated with functions, see Dinculeanu [6,
p. 273].