# Proof of the Measurable Projection and Section Theorems

The aim of this post is to give a direct proof of the theorems of measurable projection and measurable section. These are generally regarded as rather difficult results, and proofs often use ideas from descriptive set theory such as analytic sets. I did previously post a proof along those lines on this blog. However, the results can be obtained in a more direct way, which is the purpose of this post. Here, I present relatively self-contained proofs which do not require knowledge of any advanced topics beyond basic probability theory.

The projection theorem states that if ${(\Omega,\mathcal F,{\mathbb P})}$ is a complete probability space, then the projection of a measurable subset of ${{\mathbb R}\times\Omega}$ onto ${\Omega}$ is measurable. To be precise, the condition is that S is in the product sigma-algebra ${\mathcal B({\mathbb R})\otimes\mathcal F}$, where ${\mathcal B({\mathbb R})}$ denotes the Borel sets in ${{\mathbb R}}$, and the projection map is denoted

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

Then, measurable projection states that ${\pi_\Omega(S)\in\mathcal{F}}$. Although it looks like a very basic property of measurable sets, maybe even obvious, measurable projection is a surprisingly difficult result to prove. In fact, the requirement that the probability space is complete is necessary and, if it is dropped, then ${\pi_\Omega(S)}$ need not be measurable. Counterexamples exist for commonly used measurable spaces such as ${\Omega= {\mathbb R}}$ and ${\mathcal F=\mathcal B({\mathbb R})}$. This suggests that there is something deeper going on here than basic manipulations of measurable sets.

By definition, if ${S\subseteq{\mathbb R}\times\Omega}$ then, for every ${\omega\in\pi_\Omega(S)}$, there exists a ${t\in{\mathbb R}}$ such that ${(t,\omega)\in S}$. The measurable section theorem — also known as measurable selection — says that this choice can be made in a measurable way. That is, if S is in ${\mathcal B({\mathbb R})\otimes\mathcal F}$ then there is a measurable section,

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

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

The graph of ${\tau}$ is

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

The condition that ${(\tau(\omega),\omega)\in S}$ whenever ${\tau < \infty}$ can alternatively 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 results described here can also be used to prove the optional and predictable section theorems which, at first appearances, also seem to be quite basic statements. The section theorems are fundamental to the powerful and interesting theory of optional and predictable projection which is, consequently, generally considered to be a hard part of stochastic calculus. In fact, the projection and section theorems are really not that hard to prove.

Let us consider how one might try and approach a proof of the projection theorem. As with many statements regarding measurable sets, we could try and prove the result first for certain simple sets, and then generalise to measurable sets by use of the monotone class theorem or similar. For example, let ${\mathcal S}$ denote the collection of all ${S\subseteq{\mathbb R}\times\Omega}$ for which ${\pi_\Omega(S)\in\mathcal F}$. It is straightforward to show that any finite union of sets of the form ${A\times B}$ for ${A\in\mathcal B({\mathbb R})}$ and ${B\in\mathcal F}$ are in ${\mathcal S}$. If it could be shown that ${\mathcal S}$ is closed under taking limits of increasing and decreasing sequences of sets, then the result would follow from the monotone class theorem. Increasing sequences are easily handled — if ${S_n}$ is a sequence of subsets of ${{\mathbb R}\times\Omega}$ then from the definition of the projection map,

 $\displaystyle \pi_\Omega\left(\bigcup\nolimits_n S_n\right)=\bigcup\nolimits_n\pi_\Omega\left(S_n\right).$

If ${S_n\in\mathcal S}$ for each n, this shows that the union ${\bigcup_nS_n}$ is again in ${\mathcal S}$. Unfortunately, decreasing sequences are much more problematic. If ${S_n\subseteq S_m}$ for all ${n\ge m}$ then we would like to use something like

 $\displaystyle \pi_\Omega\left(\bigcap\nolimits_n S_n\right)=\bigcap\nolimits_n\pi_\Omega\left(S_n\right).$ (1)

However, this identity does not hold in general. For example, consider the decreasing sequence ${S_n=(n,\infty)\times\Omega}$. Then, ${\pi_\Omega(S_n)=\Omega}$ for all n, but ${\bigcap_nS_n}$ is empty, contradicting (1). There is some interesting history involved here. In a paper published in 1905, Henri Lebesgue claimed that the projection of a Borel subset of ${{\mathbb R}^2}$ onto ${{\mathbb R}}$ is itself measurable. This was based upon mistakenly applying (1). The error was spotted in around 1917 by Mikhail Suslin, who realised that the projection need not be Borel, and lead him to develop the theory of analytic sets.

Actually, there is at least one situation where (1) can be shown to hold. Suppose that for each ${\omega\in\Omega}$, the slices

 $\displaystyle S_n(\omega)\equiv\left\{t\in{\mathbb R}\colon(t,\omega)\in S_n\right\}$ (2)

are compact. For each ${\omega\in\bigcap_n\pi_\Omega(S_n)}$, the slices ${S_n(\omega)}$ give a decreasing sequence of nonempty compact sets, so has nonempty intersection. So, letting S be the intersection ${\bigcap_nS_n}$, the slice ${S(\omega)=\bigcap_nS_n(\omega)}$ is nonempty. Hence, ${\omega\in\pi_\Omega(S)}$, and (1) follows.

The starting point for our proof of the projection and section theorems is to consider certain special subsets of ${{\mathbb R}\times\Omega}$ where the compactness argument, as just described, can be used. The notation ${\mathcal A_\delta}$ is used to represent the collection of countable intersections, ${\bigcap_{n=1}^\infty A_n}$, of sets ${A_n}$ in ${\mathcal A}$.

Lemma 1 Let ${(\Omega,\mathcal F)}$ be a measurable space, and ${\mathcal A}$ be the collection of subsets of ${{\mathbb R}\times\Omega}$ which are finite unions ${\bigcup_kC_k\times E_k}$ over compact intervals ${C_k\subseteq{\mathbb R}}$ and ${E_k\in\mathcal F}$. Then, for any ${S\in\mathcal A_\delta}$, we have ${\pi_\Omega(S)\in\mathcal F}$, and the debut

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

is a measurable map with ${[\tau]\subseteq S}$ and ${\{\tau < \infty\}=\pi_\Omega(S)}$.

# Choquet’s Capacitability Theorem and Measurable Projection

In this post I will give a proof of the measurable projection theorem. Recall that this states that for a complete probability space ${(\Omega,\mathcal F,{\mathbb P})}$ and a set S in the product sigma-algebra ${\mathcal F\otimes\mathcal B({\mathbb R})}$, the projection, ${\pi_\Omega(S)}$, of S onto ${\Omega}$, is in ${\mathcal F}$. The previous post on analytic sets made some progress towards this result. Indeed, using the definitions and results given there, it follows quickly that ${\pi_\Omega(S)}$ is ${\mathcal F}$-analytic. To complete the proof of measurable projection, it is necessary to show that analytic sets are measurable. This is a consequence of Choquet’s capacitability theorem, which I will prove in this post. Measurable projection follows as a simple consequence.

The condition that the underlying probability space is complete is necessary and, if this condition was dropped, then the result would no longer hold. Recall that, if ${(\Omega,\mathcal F,{\mathbb P})}$ is a probability space, then the completion, ${\mathcal F_{\mathbb P}}$, of ${\mathcal F}$ with respect to ${{\mathbb P}}$ consists of the sets ${A\subseteq\Omega}$ such that there exists ${B,C\in\mathcal F}$ with ${B\subseteq A\subseteq C}$ and ${{\mathbb P}(B)={\mathbb P}(C)}$. The probability space is complete if ${\mathcal F_{\mathbb P}=\mathcal F}$. More generally, ${{\mathbb P}}$ can be uniquely extended to a measure ${\bar{\mathbb P}}$ on the sigma-algebra ${\mathcal F_{\mathbb P}}$ by setting ${\bar{\mathbb P}(A)={\mathbb P}(B)={\mathbb P}(C)}$, where B and C are as above. Then ${(\Omega,\mathcal F_{\mathbb P},\bar{\mathbb P})}$ is the completion of ${(\Omega,\mathcal F,{\mathbb P})}$.

In measurable projection, then, it needs to be shown that if ${A\subseteq\Omega}$ is the projection of a set in ${\mathcal F\otimes\mathcal B({\mathbb R})}$, then A is in the completion of ${\mathcal F}$. That is, we need to find sets ${B,C\in\mathcal F}$ with ${B\subseteq A\subseteq C}$ with ${{\mathbb P}(B)={\mathbb P}(C)}$. In fact, it is always possible to find a ${C\supseteq A}$ in ${\mathcal F}$ which minimises ${{\mathbb P}(C)}$, and its measure is referred to as the outer measure of A. For any probability measure ${{\mathbb P}}$, we can define an outer measure on the subsets of ${\Omega}$, ${{\mathbb P}^*\colon\mathcal P(\Omega)\rightarrow{\mathbb R}^+}$ by approximating ${A\subseteq\Omega}$ from above,

 $\displaystyle {\mathbb P}^*(A)\equiv\inf\left\{{\mathbb P}(B)\colon B\in\mathcal F, A\subseteq B\right\}.$ (1)

Similarly, we can define an inner measure by approximating A from below,

$\displaystyle {\mathbb P}_*(A)\equiv\sup\left\{{\mathbb P}(B)\colon B\in\mathcal F, B\subseteq A\right\}.$

It can be shown that A is ${\mathcal F}$-measurable if and only if ${{\mathbb P}_*(A)={\mathbb P}^*(A)}$. We will be concerned primarily with the outer measure ${{\mathbb P}^*}$, and will show that that if A is the projection of some ${S\in\mathcal F\otimes\mathcal B({\mathbb R})}$, then A can be approximated from below in the following sense: there exists ${B\subseteq A}$ in ${\mathcal F}$ for which ${{\mathbb P}^*(B)={\mathbb P}^*(A)}$. From this, it will follow that A is in the completion of ${\mathcal F}$.

It is convenient to prove the capacitability theorem in slightly greater generality than just for the outer measure ${{\mathbb P}^*}$. The only properties of ${{\mathbb P}^*}$ that are required is that it is a capacity, which we now define. Recall that a paving ${\mathcal E}$ on a set X is simply any collection of subsets of X, and we refer to the pair ${(X,\mathcal E)}$ as a paved space.

Definition 1 Let ${(X,\mathcal E)}$ be a paved space. Then, an ${\mathcal E}$-capacity is a map ${I\colon\mathcal P(X)\rightarrow{\mathbb R}}$ which is increasing, continuous along increasing sequences, and continuous along decreasing sequences in ${\mathcal E}$. That is,

• if ${A\subseteq B}$ then ${I(A)\le I(B)}$.
• if ${A_n\subseteq X}$ is increasing in n then ${I(A_n)\rightarrow I(\bigcup_nA_n)}$ as ${n\rightarrow\infty}$.
• if ${A_n\in\mathcal E}$ is decreasing in n then ${I(A_n)\rightarrow I(\bigcap_nA_n)}$ as ${n\rightarrow\infty}$.

As was claimed above, the outer measure ${{\mathbb P}^*}$ defined by (1) is indeed a capacity.

Lemma 2 Let ${(\Omega,\mathcal F,{\mathbb P})}$ be a probability space. Then,

• ${{\mathbb P}^*(A)={\mathbb P}(A)}$ for all ${A\in\mathcal F}$.
• For all ${A\subseteq\Omega}$, there exists a ${B\in\mathcal F}$ with ${A\subseteq B}$ and ${{\mathbb P}^*(A)={\mathbb P}(B)}$.
• ${{\mathbb P}^*}$ is an ${\mathcal F}$-capacity.