Consider a probability space and a subset S of . The projection is the set of such that there exists a with . We can ask whether there exists a map
such that . From the definition of the projection, values of satisfying this exist for each individual . By invoking the axiom of choice, then, we see that functions with the required property do exist. However, to be of use for probability theory, it is important that should be measurable. Whether or not there are measurable functions with the required properties is a much more difficult problem, and is answered affirmatively by the measurable selection theorem. For the question to have any hope of having a positive answer, we require S to be measurable, so that it lies in the product sigma-algebra , with denoting the Borel sigma-algebra on . Also, less obviously, the underlying probability space should be complete. Throughout this post, will be assumed to be a complete probability space.
It is convenient to extend to the whole of by setting for outside of . Then, is a map to the extended nonnegative reals for which precisely when is in . Next, the graph of , denoted by , is defined to be the set of with . The property that whenever is expressed succinctly by the inclusion . With this notation, the measurable selection theorem is as follows.
Theorem 1 (Measurable Selection) For any , there exists a measurable such that and
As noted above, if it wasn’t for the measurability requirement then this theorem would just be a simple application of the axiom of choice. Requiring to be measurable, on the other hand, makes the theorem much more difficult to prove. For instance, it would not hold if the underlying probability space was not required to be complete. Note also that, stated as above, measurable selection implies that the projection of S is equal to a measurable set , so the measurable projection theorem is an immediate corollary. I will leave the proof of Theorem 1 for a later post, together with the proofs of the section theorems stated below.
A closely related problem is the following. Given a measurable space and a measurable function, , does there exist a measurable right-inverse on the image of ? This is asking for a measurable function, , from to such that . In the case where is the Borel space , Theorem 1 says that it does exist. If S is the graph then will be the required right-inverse. In fact, as all uncountable Polish spaces are Borel-isomorphic to each other and, hence, to , this result applies whenever is a Polish space together with its Borel sigma-algebra. (more…)