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.

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1 January 19

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.

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