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 ${\pi_\Omega(S)}$.

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. ⬜

Notes

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.