Almost Sure

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)}.

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.

1 Comment »

  1. […] capacities is required any more, and will not be directly used here. Instead, we will make use of measurable section theorem (which did use analytic sets and capacities in its proof) but, other than that, this post is […]

    Pingback by Proof of Optional and Predictable Section | Almost Sure — 7 January 19 @ 12:03 PM | Reply

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