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 1Let 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.