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.
Proof: By construction, is in the closure of the slice
whenever this is nonempty, However, is an intersection of finite unions of compact intervals so is compact, hence whenever . Equivalently, .
As is a compact paving, lemma 4 of the post on Choquet’s capacitability theorem says that the projection is in . To see that the debut of S is measurable, it needs to be shown that is in for all . However, as the slices are closed,
As is in , this shows that is in as required.
Lemma 2 Let be a probability space and assume the hypotheses of lemma 1. For any and , there exists in satisfying
The section theorem is a straightforward consequence of the previous two lemmas. We first state it in a way which does not require completeness of the probability space. This constructs the section on up to a -null set.
Theorem 3 (Measurable Section) Let be a probability space and . Then, there exists a measurable , such that and is -null.
Proof: Let be as in lemma 1. By the previous lemma, there exists in such that
By convention, we also take . Lemma 1 says that the debut of , , is measurable and . By construction, . Define
This is measurable with graph contained in S and,
By lemma 2 of the previous post, there exists containing with . So, has zero probability and contains , which is -null as required. ⬜
Finally, we state the theorem for complete probability spaces, in which case the section is defined on all of , and not just up to a -null set.
Theorem 4 (Measurable Section) Let be a complete probability space and . Then, there exists a measurable , such that and .
Proof: By theorem 3 there exists a measurable map such that and is -null. Define by
Here, represents the slice of S as defined by (1). We do not care about which t is chosen in the third case but, as is nonempty on , a choice does exist. By construction, , , and almost surely. As is measurable, completeness of the probability space implies that is also measurable. ⬜
As with the measurable projection theorem, measurable section can be generalised using the notional of universal measurability. Recall that, for a measurable space , a set is universally -measurable if it is in every completion of with respect to probability measures . The universally measurable projection theorem says that is universally measurable for all .
It is natural to ask if the section theorem can also be generalised in such a way. Using to denote the sigma-algebra of all universally -measurable sets, a map is said to be universally measurable if for all Borel sets A. The question is, can we remove the requirement that the probability space is complete in theorem 4, and instead require that the section is only universally measurable? In fact, this is true, and removes any reference to the probability measure .
Theorem 5 (Universally Measurable Section) Let be a measurable space and . Then, there exists a universally measurable , such that and .
If is a complete probability space, then is universally complete and theorem 5 reduces to the statement of measurable section given in theorem 4 above. So, theorem 4 is an immediate consequence of universally measurable section. In the other direction, theorem 5 is a significant generalisation of theorem 4, and does not follow directly. Proving universally measurable section requires going back to an explicit representation of the analytic set S in order to construct the map . We cannot prove it by simply applying the Choquet capacitability theorem as was done for measurable section above. I will not prove theorem 5 here and, in any case, it is not required for the applications to stochastic calculus in these notes.