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 debut of any is measurable and its graph is contained in S.