Proof of Measurable Section

I will give a proof of the measurable section theorem, also known as measurable selection. Given a complete probability space {(\Omega,\mathcal F,{\mathbb P})}, we denote the projection from {\Omega\times{\mathbb R}} by

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\pi_\Omega\colon \Omega\times{\mathbb R}\rightarrow\Omega,\smallskip\\ &\displaystyle\pi_\Omega(\omega,t)=\omega. \end{array}

By definition, if {S\subseteq\Omega\times{\mathbb R}} then, for every {\omega\in\pi_\Omega(S)}, there exists a {t\in{\mathbb R}} such that {(\omega,t)\in S}. The measurable section theorem says that this choice can be made in a measurable way. That is, using {\mathcal B({\mathbb R})} to denote the Borel sigma-algebra, if S is in the product sigma-algebra {\mathcal F\otimes\mathcal B({\mathbb R})} then {\pi_\Omega(S)\in\mathcal F} and there is a measurable map

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\tau\colon\pi_\Omega(S)\rightarrow{\mathbb R},\smallskip\\ &\displaystyle(\omega,\tau(\omega))\in S. \end{array}

It is convenient to extend {\tau} to the whole of {\Omega} by setting {\tau=\infty} outside of {\pi_\Omega(S)}.

measurable section
Figure 1: A section of a measurable set

We consider measurable functions {\tau\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\}}. The graph of {\tau} is

\displaystyle  [\tau]=\left\{(\omega,\tau(\omega))\colon\tau(\omega)\in{\mathbb R}\right\}\subseteq\Omega\times{\mathbb R}.

The condition that {(\omega,\tau(\omega))\in S} whenever {\tau < \infty} can then be expressed by stating that {[\tau]\subseteq S}. This also ensures that {\{\tau < \infty\}} is a subset of {\pi_\Omega(S)}, and {\tau} is a section of S on the whole of {\pi_\Omega(S)} if and only if {\{\tau < \infty\}=\pi_\Omega(S)}.

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 {\mathcal E} on a set X denotes, simply, a collection of subsets of X. The pair {(X,\mathcal E)} is then referred to as a paved space. Given a pair of paved spaces {(X,\mathcal E)} and {(Y,\mathcal F)}, the product paving {\mathcal E\times\mathcal F} denotes the collection of cartesian products {A\times B} for {A\in\mathcal E} and {B\in\mathcal F}, which is a paving on {X\times Y}. The notation {\mathcal E_\delta} is used for the collection of countable intersections of a paving {\mathcal E}.

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

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle D(S)\colon\Omega\rightarrow{\mathbb R}\cup\{\pm\infty\},\smallskip\\ &\displaystyle \omega\mapsto\inf\left\{t\in{\mathbb R}\colon (\omega,t)\in S\right\}. \end{array}

We use the convention that the infimum of the empty set is {\infty}. It is not clear that {D(S)} 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 {\mathcal F\otimes\mathcal B({\mathbb R})}.

Lemma 1 Let {(\Omega,\mathcal F)} be a measurable space, {\mathcal K} be the collection of compact intervals in {{\mathbb R}}, and {\mathcal E} be the closure of the paving {\mathcal{F\times K}} under finite unions.

Then, the debut {D(S)} of any {S\in\mathcal E_\delta} is measurable and its graph {[D(S)]} is contained in S.

Proof: By construction, {D(S)(\omega)} is in the closure of the slice

\displaystyle  S(\omega)\equiv\left\{t\in{\mathbb R}\colon(\omega,t)\in S\right\} (1)

whenever this is nonempty, However, {S(\omega)} is an intersection of finite unions of compact intervals so is compact, hence {D(S)(\omega)\in S(\omega)} whenever {D(S) < \infty}. Equivalently, {[D(S)]\subseteq S}.

As {\mathcal K} is a compact paving, lemma 4 of the post on Choquet’s capacitability theorem says that the projection {\pi_\Omega(S)} is in {\mathcal F_\delta=\mathcal F}. To see that the debut of S is measurable, it needs to be shown that {\{D(S) \le t\}} is in {\mathcal F} for all {t\in{\mathbb R}}. However, as the slices {S(\omega)} are closed,

\displaystyle  \{D(S) \le t\}=\pi_\Omega\left(S\cap\left(\Omega\times(-\infty,t]\right)\right).

As {S\cap\left(\Omega\times(-\infty,t]\right)} is in {\mathcal E_\delta}, this shows that {\{D(S)\le t\}} is in {\mathcal F} as required.

The proof of the section theorem for arbitrary measurable sets S will proceed by approximation from below by sets in {\mathcal E_\delta}, and applying lemma 1. Recall that {{\mathbb P}^*} refers to the outer measure on subsets of {\Omega},

\displaystyle  {\mathbb P}^*(A)=\inf\left\{{\mathbb P}(B)\colon B\in\mathcal F,A\subseteq B\right\}.

Lemma 2 Let {(\Omega,\mathcal F,{\mathbb P})} be a probability space and assume the hypotheses of lemma 1. For any {S\in\mathcal F\otimes\mathcal B({\mathbb R})} and {\epsilon > 0}, there exists {S^\prime\subseteq S} in {\mathcal E_\delta} satisfying

\displaystyle  {\mathbb P}\left(\pi_\Omega(S^\prime)\right)\ge{\mathbb P}^*\left(\pi_\Omega(S)\right)-\epsilon.

Proof: By lemmas 2 and 4 of the previous post, {{\mathbb P}^*\circ\pi_\Omega} is an {\mathcal E}-capacity. The result follows by the Choquet capacitability theorem. ⬜

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 {\pi_\Omega(S)} up to a {{\mathbb P}}-null set.

Theorem 3 (Measurable Section) Let {(\Omega,\mathcal F,{\mathbb P})} be a probability space and {S\in\mathcal F\otimes\mathcal B({\mathbb R})}. Then, there exists a measurable {\tau\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\}}, such that {[\tau]\subseteq S} and {\pi_\Omega(S)\setminus\{\tau < \infty\}} is {{\mathbb P}}-null.

Proof: Let {\mathcal E} be as in lemma 1. By the previous lemma, there exists {S_n\subseteq S} in {\mathcal E_\delta} such that

\displaystyle  {\mathbb P}(\pi_\Omega(S_n))\ge{\mathbb P}^*(\pi_\Omega(S))-1/n.

By convention, we also take {S_0=\emptyset}. Lemma 1 says that the debut of {S_n}, {\tau_n=D(S_n)}, is measurable and {[\tau_n]\subseteq S_n}. By construction, {\{\tau_n < \infty\}=\pi_\Omega(S_n)}. Define

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\tau\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\},\smallskip\\ &\displaystyle\tau(\omega)=\begin{cases} \tau_n(\omega),&{\rm\ for\ }\omega\in\pi_\Omega(S_n)\setminus\pi_\Omega(S_{n-1})\\ \infty,&{\rm\ for\ }\omega\in\Omega\setminus\bigcup_n\pi_\Omega(S_n). \end{cases} \end{array}

This is measurable with graph {[\tau]} contained in S and,

\displaystyle  {\mathbb P}(\tau < \infty)={\mathbb P}\left(\bigcup\nolimits_n\pi_\Omega(S_n)\right)\ge{\mathbb P}^*(\pi_\Omega(S)).

By lemma 2 of the previous post, there exists {B\in\mathcal F} containing {\pi_\Omega(S)} with {{\mathbb P}(B)={\mathbb P}^*(\pi_\Omega(S))}. So, {B\setminus\{\tau < \infty\}} has zero probability and contains {\pi_\Omega(S)\setminus\{\tau < \infty\}}, which is {{\mathbb P}}-null as required. ⬜

Finally, we state the theorem for complete probability spaces, in which case the section is defined on all of {\pi_\Omega(S)}, and not just up to a {{\mathbb P}}-null set.

Theorem 4 (Measurable Section) Let {(\Omega,\mathcal F,{\mathbb P})} be a complete probability space and {S\in\mathcal F\otimes\mathcal B({\mathbb R})}. Then, there exists a measurable {\tau\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\}}, such that {[\tau]\subseteq S} and {\{\tau < \infty\}=\pi_\Omega(S)}.

Proof: By theorem 3 there exists a measurable map {\tau_0\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\}} such that {[\tau_0]\subseteq S} and {\pi_\Omega(S)\setminus\{\tau_0 < \infty\}} is {{\mathbb P}}-null. Define {\tau\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\}} by

\displaystyle  \displaystyle\tau(\omega)=\begin{cases} \tau_0(\omega),&{\rm\ if\ }\tau_0(\omega) < \infty,\\ \infty,&{\rm\ if\ }\omega\not\in\pi_\Omega(S),\\ t\in S(\omega),&{\rm\ if\ }\omega\in\pi_\Omega(S)\setminus\{\tau_0 < \infty\}. \end{cases}

Here, {S(\omega)} represents the slice of S as defined by (1). We do not care about which t is chosen in the third case but, as {S(\omega)} is nonempty on {\pi_\Omega(S)}, a choice does exist. By construction, {[\tau]\subseteq S}, {\{\tau < \infty\}=\pi_\Omega(S)}, and {\tau=\tau_0} almost surely. As {\tau_0} is measurable, completeness of the probability space implies that {\tau} 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 {(\Omega,\mathcal F)}, a set {A\subseteq\Omega} is universally {\mathcal F}-measurable if it is in every completion {\mathcal F_{\mathbb P}} of {\mathcal F} with respect to probability measures {{\mathbb P}}. The universally measurable projection theorem says that {\pi_\Omega(S)} is universally measurable for all {S\in\mathcal F\otimes\mathcal B({\mathbb R})}.

It is natural to ask if the section theorem can also be generalised in such a way. Using {\mathcal{\bar F}} to denote the sigma-algebra of all universally {\mathcal F}-measurable sets, a map {\tau\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\}} is said to be universally measurable if {\tau^{-1}(A)\in\mathcal{\bar F}} 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 {\tau} is only universally measurable? In fact, this is true, and removes any reference to the probability measure {{\mathbb P}}.

Theorem 5 (Universally Measurable Section) Let {(\Omega,\mathcal F)} be a measurable space and {S\in\mathcal F\otimes\mathcal B({\mathbb R})}. Then, there exists a universally measurable {\tau\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\}}, such that {[\tau]\subseteq S} and {\{\tau < \infty\}=\pi_\Omega(S)}.

If {(\Omega,\mathcal F,{\mathbb P})} is a complete probability space, then {\mathcal F} 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 {\tau}. 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.

One thought on “Proof of Measurable Section

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s