Almost Sure

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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