Here are some general considerations about measurable multifunctions.
Lemma 47.1.5Suppose f : K
(ω)
× Ω → X,K ⊆ X. Here X is Polish space,separable complete metric space, and
(Ω,ℱ )
is a measurable space. Also ω → K
(ω)
is a measurable multifunction as in Theorem 47.1.3. Also suppose ω → f
(x,ω)
ismeasurable and x → f
(x,ω)
is continuous. Also suppose that K
(ω)
≡ f
(K (ω ),ω)
.Then you can conclude that ω →K
(ω )
is a measurable multifunction. If K
(ω)
iscompact, then it is also strongly measurable.
Proof:Let
{xn(ω)}
be a countable dense subset of K
(ω )
, each x_{n} measurable.
Then if U is open,
∞ − 1
{ω : K (ω)∩ U ⁄= ∅} = ∪n=1f (xn(⋅),⋅) (U)
and each of the sets in the union is measurable. The latter claim follows from
the continuity of f
(⋅,ω )
. If x
(ω)
is measurable, then we can express it as the
limit of simple functions s_{n} for which ω → f
(sn(ω),ω)
is clearly measurable.
Then f
(x (ω),ω)
is the limit of f
(sn(ω),ω)
. The reason for the equality is as
follows. It is clear that the right side is contained in the left. Now if K
(ω)
∩ U≠∅,
then by definition, f
(x,ω)
∈ U for some x ∈ K
(ω )
but then by continuity,
f
(xn (ω),ω)
∈ U also for some x_{n}
(ω)
close to x. Thus the two sets are actually equal.
Thus ω →K
(ω)
is measurable. If K
(ω)
has compact values it will be strongly
measurable. ■
This lemma gives an easy example of a measurable multifunction having compact
values. In fact this is the one of most interest in what follows. However, we also have the
following general result. It gives the existence of a measurable ε net. This is
formulated in Banach space because it is convenient to add. It could also be
formulated in Polish space with a little more difficulty. One just defines things a little
differently.
Proposition 47.1.6Let ω →K
(ω)
be a measurable multifunction where K
(ω)
is a precompact set. Recall this means its closure is compact. Also, it must have an ε net foreach ε > 0. Then for each ε > 0, there exists N
(ω )
and measurable functionsy_{j},j = 1,2,
⋅⋅⋅
,N
(ω)
, y_{j}
(ω)
∈K
(ω )
, such that
∪Nj=1B (yj(ω),ε) ⊇ K (ω)
for each ω. Also ω → N
(ω )
is measurable.
Proof: Suppose that ω →K
(ω)
is a measurable multifunction having compact
values in X a Banach space. Let
{σn(ω)}
be the measurable selections such that for each
ω,