# Almost Sure

## 10 January 19

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

[caption_id=”sectionpic” align=”aligncenter” width=”450″] Figure 1: A section of a measurable set[/caption] 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 ${\{\tau < \infty\}=\pi_\Omega(S)}$.

## 2 January 19

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

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.

## 24 December 18

### Analytic Sets

We will shortly give a proof of measurable projection and, also, of the section theorems. Starting with the projection theorem, recall that this states that if ${(\Omega,\mathcal F,{\mathbb P})}$ is a complete probability space, then the projection of any measurable subset of ${\Omega\times{\mathbb R}}$ onto ${\Omega}$ is measurable. To be precise, the condition is that S is in the product sigma-algebra ${\mathcal{F}\otimes\mathcal B({\mathbb R})}$, where ${\mathcal B({\mathbb R})}$ denotes the Borel sets in ${{\mathbb R}}$, and ${\pi\colon\Omega\times{\mathbb R}\rightarrow\Omega}$ is the projection ${\pi(\omega,t)=\omega}$. Then, ${\pi(S)\in\mathcal{F}}$. 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 ${\pi(S)}$ 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.

The techniques which will be used to prove the projection theorem involve analytic sets, which will be introduced in this post, with the proof of measurable projection to follow in the next post. [Note: I have since posted a more direct proof of measurable projection and section, which does not make use of analytic sets.] These results can also be used to prove the optional and predictable section theorems which, at first appearances, 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, although the method given here does require stepping outside of the usual setup used in probability and involves something more like descriptive set theory. (more…)

## 29 November 16

### The Section Theorems

Consider a probability space ${(\Omega,\mathcal{F},{\mathbb P})}$ and a subset S of ${{\mathbb R}_+\times\Omega}$. The projection ${\pi_\Omega(S)}$ is the set of ${\omega\in\Omega}$ such that there exists a ${t\in{\mathbb R}_+}$ with ${(t,\omega)\in S}$. We can ask whether there exists a map

$\displaystyle \tau\colon\pi_\Omega(S)\rightarrow{\mathbb R}_+$

such that ${(\tau(\omega),\omega)\in S}$. From the definition of the projection, values of ${\tau(\omega)}$ satisfying this exist for each individual ${\omega}$. By invoking the axiom of choice, then, we see that functions ${\tau}$ with the required property do exist. However, to be of use for probability theory, it is important that ${\tau}$ 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 ${\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}}$, with ${\mathcal{B}({\mathbb R}_+)}$ denoting the Borel sigma-algebra on ${{\mathbb R}_+}$. Also, less obviously, the underlying probability space should be complete. Throughout this post, ${(\Omega,\mathcal{F},{\mathbb P})}$ will be assumed to be a complete probability space.

It is convenient to extend ${\tau}$ to the whole of ${\Omega}$ by setting ${\tau(\omega)=\infty}$ for ${\omega}$ outside of ${\pi_\Omega(S)}$. Then, ${\tau}$ is a map to the extended nonnegative reals ${\bar{\mathbb R}_+={\mathbb R}_+\cup\{\infty\}}$ for which ${\tau(\omega) < \infty}$ precisely when ${\omega}$ is in ${\pi_\Omega(S)}$. Next, the graph of ${\tau}$, denoted by ${[\tau]}$, is defined to be the set of ${(t,\omega)\in{\mathbb R}_+\times\Omega}$ with ${t=\tau(\omega)}$. The property that ${(\tau(\omega),\omega)\in S}$ whenever ${\tau(\omega) < \infty}$ is expressed succinctly by the inclusion ${[\tau]\subseteq S}$. With this notation, the measurable selection theorem is as follows.

Theorem 1 (Measurable Selection) For any ${S\in\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}}$, there exists a measurable ${\tau\colon\Omega\rightarrow\bar{\mathbb R}_+}$ such that ${[\tau]\subseteq S}$ and

 $\displaystyle \left\{\tau < \infty\right\}=\pi_\Omega(S).$ (1)

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 ${\tau}$ 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 ${\{\tau < \infty\}}$, 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 ${(X,\mathcal{E})}$ and a measurable function, ${f\colon X\rightarrow\Omega}$, does there exist a measurable right-inverse on the image of ${f}$? This is asking for a measurable function, ${g}$, from ${f(X)}$ to ${X}$ such that ${f(g(\omega))=\omega}$. In the case where ${(X,\mathcal{E})}$ is the Borel space ${({\mathbb R}_+,\mathcal{B}({\mathbb R}_+))}$, Theorem 1 says that it does exist. If S is the graph ${\{(t,f(t))\colon t\in{\mathbb R}_+\}}$ then ${\tau}$ will be the required right-inverse. In fact, as all uncountable Polish spaces are Borel-isomorphic to each other and, hence, to ${{\mathbb R}_+}$, this result applies whenever ${(X,\mathcal{E})}$ is a Polish space together with its Borel sigma-algebra. (more…)

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