I will give a proof of the measurable section theorem, also known as measurable selection. Given a complete probability space , we denote the projection from
by
By definition, if then, for every
, there exists a
such that
. The measurable section theorem says that this choice can be made in a measurable way. That is, using
to denote the Borel sigma-algebra, if S is in the product sigma-algebra
then
and there is a measurable map
It is convenient to extend to the whole of
by setting
outside of
.

We consider measurable functions . The graph of
is
The condition that whenever
can then be expressed by stating that
. This also ensures that
is a subset of
, and
is a section of S on the whole of
if and only if
.
The proof of the measurable section theorem will make use of the properties of analytic sets and of the Choquet capacitability theorem, as described in the previous two posts. [Note: I have since posted a more direct proof which does not involve such prerequisites.] Recall that a paving on a set X denotes, simply, a collection of subsets of X. The pair
is then referred to as a paved space. Given a pair of paved spaces
and
, the product paving
denotes the collection of cartesian products
for
and
, which is a paving on
. The notation
is used for the collection of countable intersections of a paving
.
We start by showing that measurable section holds in a very simple case where, for the section of a set S, its debut will suffice. The debut is the map
We use the convention that the infimum of the empty set is . It is not clear that
is measurable, and we do not rely on this, although measurable projection can be used to show that it is measurable whenever S is in
.
Lemma 1 Let
be a measurable space,
be the collection of compact intervals in
, and
be the closure of the paving
under finite unions.
Then, the debutof any
is measurable and its graph
is contained in S.