# 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.

Proof: The first statement is clear — the definition (1) gives that ${{\mathbb P}^*\ge{\mathbb P}}$. For any ${A\in\mathcal F}$ we can take ${B=A}$ to get ${{\mathbb P}^*(A)\le{\mathbb P}(B)={\mathbb P}(A)}$ to obtain the reverse inequality.

We show that ${{\mathbb P}^*}$ satisfies the three properties required of an ${\mathcal F}$-capacity. That it is increasing is immediate from (1). For the second condition, consider a sequence ${A_n\subseteq\Omega}$ increasing to the limit ${A=\bigcup_n A_n}$. By definition, there exists ${B_n\in\mathcal F}$ with ${A_n\subseteq B_n}$ and ${{\mathbb P}(B_n)\le{\mathbb P}^*(A_n)+1/n}$. Replacing ${B_n}$ by ${\bigcap_{m\ge n}B_m}$, if necessary, we can suppose that ${B_n}$ is an increasing sequence. Denote its limit by ${B=\bigcup_nB_n\in\mathcal F}$. Then ${A\subseteq B}$ and,

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb P}^*(A)&\displaystyle\le{\mathbb P}(B)=\lim_n{\mathbb P}(B_n)\smallskip\\ &\displaystyle\le\lim_n({\mathbb P}^*(A_n)+1/n)=\lim_n{\mathbb P}^*(A_n)\le{\mathbb P}^*(A). \end{array}$ (2)

So ${\lim_n{\mathbb P}^*(A_n)={\mathbb P}^*(A)}$. Next, consider a sequence ${A_n\in\mathcal F}$ decreasing to the limit ${A=\bigcap_n A_n}$. Then, using the first statement of the lemma and monotone convergence of the measure ${{\mathbb P}}$,

$\displaystyle \lim_n{\mathbb P}^*(A_n)=\lim_n{\mathbb P}(A_n)={\mathbb P}(A)={\mathbb P}^*(A).$

So ${{\mathbb P}^*}$ satisfies the defining properties of an ${\mathcal F}$-capacity.

Finally, for any ${A\subseteq\Omega}$, , the second statement of the lemma follows from (2) above (we can take ${A_n=A}$). ⬜

Next, recall that our definition of analytic sets made use of an auxiliary compact paved space. Before proceeding further, we will require some basic facts concerning compact pavings.

One way to construct a compact paving is to take any collection of closed subsets of a compact topological space. In fact, every compact paving arises in this way. Given a paved space ${(K,\mathcal K)}$, we can always consider the topology on K whose closed sets are generated by ${\mathcal K}$. The collection of closed sets under this topology is the closure of ${\mathcal K}$ under taking finite unions and arbitrary intersections, which we will denote by ${\mathcal{\bar K}}$. The Alexander subbase theorem states that if the original paving ${\mathcal K}$ is compact, then the topology that it generates is also compact. Alexander’s theorem is usually stated in terms of open sets rather than closed ones, in which case the collection of open sets generating the topology is known as a subbasis. The statement in terms of closed sets is equivalent, simply by taking complements.

To be precise, when I refer to finite unions’ this includes the empty union, which is just the empty set. Similarly, finite intersections’ and arbitrary intersections’ of a paving on a set X is taken to include the empty intersection, which is the whole of X.

Theorem 3 (Alexander Subbase Theorem) Let ${(K,\mathcal K)}$ be a compact paved space, and ${\mathcal{\bar K}}$ be the closure of ${\mathcal K}$ under finite unions and arbitrary intersections.

Then, ${\mathcal{\bar K}}$ is a compact paving.

Proof: Before proceeding with the proof, I note that we will make use of Zorn’s Lemma, which is equivalent to the Axiom of Choice. In fact, uncountable choice is not really required for our applications of analytic sets. For one thing, in the definition of analytic sets we could have restricted to countable compact pavings, for which the application of Zorn’s Lemma can be removed from the argument below. For another thing, we could have restricted from the start to compact paved spaces consisting of the closed subsets of a compact topological space, and the subbase theorem would not have been required. So, while this theorem is used in the approach I give here, it is not a fundamental part of the theory.

We need to show that if ${\mathcal C}$ is a subcollection of ${\mathcal{\bar K}}$ with empty intersection then there is a finite subset of ${\mathcal C}$ with empty intersection. We use proof by contradiction, so suppose that every finite subset of ${\mathcal C}$ has nonempty intersection. The idea is to look at ${\mathcal C^\prime=\mathcal K\cap\mathcal C}$ and make use of compactness of ${\mathcal K}$. Although it is immediate that finite subsets of ${\mathcal C^\prime\subseteq\mathcal C}$ also have nonempty intersection, it is not necessarily the case that the intersection of ${\mathcal C^\prime}$ is empty.

Instead we suppose that ${\mathcal C}$ has been chosen maximal subject to the properties it has empty intersection and that all finite subsets of ${\mathcal C}$ have nonempty intersection. The existence of such a maximal ${\mathcal C}$ is a consequence of Zorn’s lemma. It then just needs to be shown that ${\mathcal C^\prime}$ also has empty intersection, and compactness of ${\mathcal K}$ will give the contradiction.

Consider a finite union ${A=\bigcup_{i=1}^nA_n\in\mathcal C}$ where each ${A_i}$ is in ${\mathcal{\bar K}}$. We can show that at least one of the ${A_i}$ is also in ${\mathcal C}$. If not then, by maximality, there exists a finite subset of ${\mathcal C\cup\{A_i\}}$ with empty intersection. As this must necessarily contain ${A_i}$, it can be written as ${\mathcal S_i\cup\{A_i\}}$. Then, ${\mathcal S_i}$ is a finite subset of ${\mathcal C}$ such that ${(\bigcap\mathcal S_i)\cap A_i=\emptyset}$. So, ${(\bigcup_i\mathcal S_i)\cup\{A\}}$ is a finite subset of ${\mathcal C}$ with empty intersection, contradicting the choice of ${\mathcal C}$. Hence, at least one of the ${A_i}$ must be in ${\mathcal C}$.

Now, consider any ${A\in\mathcal C}$. As ${A\in\mathcal{\bar K}}$, it can be written as an intersection ${A=\bigcap_{i\in I}A_i}$ where each ${A_i}$ is a finite union of elements of ${\mathcal K}$. For each i, the inclusion ${A\subseteq A_i}$ implies that the intersection of any finite subset ${\mathcal S}$ of ${\mathcal C\cup\{A_i\}}$ contains the intersection of ${(\mathcal S\setminus\{A_i\})\cup\{A\}\subseteq\mathcal C}$ and, hence, is nonempty. By maximality, this shows that ${A_i\in\mathcal C}$. As ${A_i}$ is a finite union of elements of ${\mathcal K}$, at least one of which must be in ${\mathcal C}$ (by what we showed above), there is a ${B_i\in\mathcal K\cap\mathcal C=\mathcal C^\prime}$ with ${B_i\subseteq A_i}$. So,

$\displaystyle \bigcap\mathcal C^\prime\subseteq\bigcap_{i\in I}B_i\subseteq\bigcap_{i\in I}A_i=A.$

As this holds for all ${A\in\mathcal C}$, it follows that ${\bigcap\mathcal C^\prime\subseteq\bigcap\mathcal C}$ is empty, as required. ⬜

Recall from the discussion in the previous post that the difficulty in giving an elementary proof of measurable projection is the following: if ${S_n}$ is a sequence of subsets of a product space ${X\times K}$, decreasing to a limit S, then the projections ${\pi_X(S_n)}$ need not decrease to ${\pi_X(S)}$. That is, projection is not continuous along decreasing sequences. However, we do obtain continuity so long as we restrict to sets arising from the product of a compact paving on K. This is the third statement of the lemma below, and is the point where compactness enters the theory in a fundamental way. Recall that we use ${\mathcal E_\sigma}$ and ${\mathcal E_\delta}$ to represent, respectively, the collections of countable unions and countable intersections of ${\mathcal E}$.

Lemma 4 Let ${(X,\mathcal E)}$ and ${(K,\mathcal K)}$ be paved spaces such that ${\mathcal E}$ is closed under finite unions and pairwise intersections, and ${\mathcal K}$ is compact. Let ${\mathcal F}$ be the closure of the paving ${\mathcal{E\times K}}$ under finite unions and pairwise intersections. Then,

• ${\pi_X(S)\in\mathcal E}$ for all ${S\in\mathcal F}$.
• ${\pi_X(S)\in\mathcal E_\delta}$ for all ${S\in\mathcal F_\delta}$.
• if ${S_n\in\mathcal F}$ decreases to a limit S, then ${\pi_X(S_n)}$ decreases to ${\pi_X(S)}$.
• If I is an ${\mathcal E}$-capacity then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle I\circ\pi_X\colon\mathcal P(X\times K)\rightarrow{\mathbb R}^+,\smallskip\\ &\displaystyle S\mapsto I(\pi_X(S)) \end{array}$

is an ${\mathcal F}$-capacity.

Proof: Note that in the statement of the lemma, ${\mathcal E}$ is only required to be closed under pairwise intersections’ rather than `finite intersections’. The difference is simply that we are not requiring ${\mathcal E}$ to contain the whole set X (i.e., the empty intersection). This is just a very small matter, and we state the lemma in this way as it will be a slight convenience when applying the result.

Without loss of generality, by theorem 3, we may suppose that ${\mathcal K}$ is closed under finite unions and arbitrary intersections. It follows that the intersection of a pair of sets in ${\mathcal{E\times K}}$ is again in ${\mathcal{E\times K}}$,

$\displaystyle \left(A\times L\right)\cap\left(B\times M\right)=\left(A\cap B\right)\times\left(L\cap M\right).$

So, ${\mathcal F}$ consists of finite unions of ${\mathcal{E\times K}}$.

Clearly, if ${S=A\times L}$ is in ${\mathcal{E\times K}}$, then the projection ${\pi_X(S)}$ is equal to E if L is nonempty, and is the empty set otherwise. In either case, ${\pi_X(S)\in\mathcal E}$. Now, any ${S\in\mathcal F}$ is of the form ${\bigcup_{i=1}^nS_i}$ for ${S_i\in\mathcal{E\times K}}$ so, as ${\mathcal E}$ is closed under finite unions, ${\pi_X(S)=\bigcup_i\pi_X(S_i)}$ is in ${\mathcal E}$, proving the first statement of the lemma.

Next, consider a sequence ${S_n\in\mathcal F}$ decreasing to a limit S. Then, ${\pi_X(S_n)\supseteq\pi_X(S)}$ is also decreasing. It just needs to be shown that

 $\displaystyle \bigcap_n\pi_X(S_n)\subseteq\pi_X(S).$ (3)

This is the point where compactness enters the theory in a fundamental way. For any ${x\in\bigcap_n\pi_X(S_n)}$, consider the slices

$\displaystyle S_n(x)=\left\{y\in K\colon(x,y)\in S_n\right\}.$

The condition ${x\in\pi_X(S_n)}$ is equivalent to ${S_n(x)}$ being nonempty. Furthermore, for any n, if we write ${S_n}$ as ${\bigcup_{i=1}^m A_i\times K_i}$ then we have,

$\displaystyle S_n(x)=\bigcup_{i\colon x\in A_i}K_i\in\mathcal K.$

So, ${S_n(x)}$ is a decreasing sequence of nonempty sets in ${\mathcal K}$. By compactness, the intersection ${\bigcap_nS_n(x)=S(x)}$ is nonempty, so ${x\in\pi_X(S)}$ as required. This proves (3), giving the third statement of the lemma.

By definition, any ${S\in\mathcal F_\delta}$ is the intersection of a sequence ${S_n\in\mathcal F}$. Without loss of generality, we can suppose that ${S_n}$ is decreasing (replace ${S_n}$ by ${\bigcap_{m\le n}S_m}$ if required). By the third statement of the lemma which was proved above,

$\displaystyle \pi_X(S)=\bigcap_n\pi_X(S_n)\in\mathcal E_\delta,$

proving the second statement.

Finally, let ${I\colon\mathcal P(X)\rightarrow{\mathbb R}^+}$ be an ${\mathcal E}$-capacity and consider ${I\circ\pi_X}$. If ${S_n\subseteq X\times K}$ increases to a limit S, then, ${\pi_X(S_n)}$ increases to ${\pi_X(S)}$. So, as I is a capacity, ${I\circ\pi_X(S_n)}$ increases to ${I\circ\pi_X(S)}$. Next, if ${S_n\in\mathcal F}$ decreases to a limit S then, by the first and third statements of the lemma, ${\pi_X(S_n)}$ is a sequence in ${\mathcal E}$ decreasing to ${\pi_X(S)}$. The fact that I is an ${\mathcal E}$-capacity means that ${I\circ\pi_X(S_n)}$ decreases to ${I\circ\pi_X(S)}$. Hence, ${I\circ\pi_X}$ is an ${\mathcal F}$-capacity as required. ⬜

Given a paved space ${(X,\mathcal E)}$ and an ${\mathcal E}$-capacity, I, a set ${A\subseteq X}$ is called I-capacitable if it can be approximated from below in an appropriate sense. There are a couple of equivalent ways of stating this, given by the following lemma.

Lemma 5 Let ${(X,\mathcal E)}$ be a paved space closed under finite unions and I be an ${\mathcal E}$-capacity. For any ${A\subseteq X}$, the following are equivalent.

• There exists ${B\in\mathcal E_{\delta\sigma}}$ such that ${B\subseteq A}$ and ${I(B)=I(A)}$.
• For all ${\epsilon > 0}$, there exists ${B\in\mathcal E_\delta}$ with ${B\subseteq A}$ and,
 $\displaystyle I(B)\ge I(A)-\epsilon.$ (4)

Proof: If the first statement holds then, by definition of ${\mathcal E_{\delta\sigma}}$, we can write ${B=\bigcup_n B_n}$ for a sequence ${B_n\in\mathcal E_\delta}$. Using the fact that the capacity I is continuous along increasing sequences,

$\displaystyle I\left(\bigcup\nolimits_{m\le n}B_m\right)\rightarrow I(B)=I(A).$

So, for ${\epsilon > 0}$, we can take ${B=\bigcup_{m\le n}B_m}$ for sufficiently large n, and (4) will be satisfied. Furthermore, since ${\mathcal E}$ is closed under finite unions, B will be in ${\mathcal E_\delta}$ as required.

For the converse, suppose that, the second statement holds. So, there is a sequence ${B_n\subseteq A}$ in ${\mathcal E_\delta}$ with ${I(B_n)\ge I(A)-1/n}$. Then, ${B=\bigcup_nB_n}$ is in ${\mathcal E_{\delta\sigma}}$ and ${I(B)=I(A)}$, as claimed. ⬜

We now prove the capacitability theorem, which states that analytic sets are capacitable. This was first stated in the generality given here by Sion (On capacitability and measurability, 1963).

Theorem 6 (Choquet’s Capacitability Theorem) Let ${\mathcal E}$ be a paving closed under finite unions and pairwise intersections, and I be an ${\mathcal E}$-capacity. Then, for any ${\mathcal E}$-analytic set A, there exists ${B\in\mathcal E_{\delta\sigma}}$ such that ${B\subseteq A}$ and ${I(B)=I(A)}$.

Proof: We start by showing that any ${A\in\mathcal E_{\sigma\delta}}$ can be approximated from below by a ${B\in\mathcal E_{\delta\sigma}}$ as in the statement of the theorem, before extending to all analytic sets.

By definition, ${A=\bigcap_n A_n}$ for some sequence ${A_n\in\mathcal E_\sigma}$. Then, ${A_n=\bigcup_mA_{nm}}$ for some ${A_{nm}\in\mathcal E}$. The idea is that, by truncating the union, we can approximate ${A_n}$ from below by members of ${\mathcal E}$, so that A is approximated from below by an element of ${\mathcal E_\delta}$ satisfying (4).

For any finite sequence of positive integers ${p_1,p_2,\ldots,p_r}$, write

$\displaystyle B_{p_1p_2\cdots p_r} = \bigcap_{n=1}^r\bigcup_{m=1}^{p_r}A_{nm}.$

These are in ${\mathcal E}$, as it is closed under finite unions and pairwise intersections. Fixing any ${\epsilon > 0}$, we can inductively define an infinite sequence ${p_1,p_2,\ldots}$ such that

 $\displaystyle I(A\cap B_{p_1\cdots p_r}) > I(A) - \epsilon$ (5)

for all r. First, for r equal to 1, note that ${A\cap B_{p_1}}$ increases to A as ${p_1\rightarrow\infty}$. Hence, as the capacity I is continuous along increasing sequences, (5) will be satisfied for large enough ${p_1}$. Next, suppose that ${p_1,\ldots,p_{k-1}}$ have been chosen satisfying (5) for ${r=k-1}$. Letting ${p_k}$ increase to infinity and using continuity of I along increasing sequences,

$\displaystyle I(A\cap B_{p_1\cdots p_{k-1}p_k})\rightarrow I(A\cap B_{p_1\cdots p_{k-1}}) > I(A) - \epsilon.$

So, (5) holds for ${r=k}$ so long as ${p_k}$ is chosen large enough. Hence, the sequence ${p_1,p_2,\ldots}$ can be inductively constructed to satisfy (5).

We now set,

$\displaystyle B=\lim_{r\rightarrow\infty}B_{p_1\cdots p_r}=\bigcap_{r=1}^\infty B_{p_1\cdots p_r}\in\mathcal E_\delta.$

By construction, ${B\subseteq A}$ and, using the continuity of I along decreasing sequences in ${\mathcal E}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle I(B)&\displaystyle=\lim_{r\rightarrow\infty}I(B_{p_1\cdots p_r})\smallskip\\ &\displaystyle\ge\lim_{r\rightarrow\infty}I(A\cap B_{p_1\cdots p_r})\smallskip\\ &\displaystyle\ge I(A)-\epsilon. \end{array}$

So, (4) is satisfied, as required.

Now, consider any ${\mathcal E}$-analytic set A. By definition, there is a compact paved space ${(K,\mathcal K)}$ and an ${S\in(\mathcal{E\times K})_{\sigma\delta}}$ such that A is the projection ${\pi_X(S)}$. Letting ${\mathcal F}$ be the closure of ${\mathcal{E\times K}}$ under finite unions and pairwise intersections, Lemma 4 states that ${I\circ\pi_X}$ is an ${\mathcal F}$-capacity. So, by what we have shown above, for any ${\epsilon > 0}$, there exists an ${S^\prime\in\mathcal F_\delta}$ such that ${S^\prime\subseteq S}$ and,

$\displaystyle I(\pi_X(S^\prime))\ge I(\pi_X(S))-\epsilon = I(A)-\epsilon.$

So, (4) is satisfied for ${B=\pi_X(S^\prime)}$ which, by Lemma 4, is in ${\mathcal E_\delta}$. ⬜

An immediate consequence of the capacitability theorem is the measurability of analytic sets.

Corollary 7 Let ${(\Omega,\mathcal F,{\mathbb P})}$ be a complete probability space. Then, every ${\mathcal F}$-analytic set is in ${\mathcal F}$. That is,

$\displaystyle \mathcal{A(F)}=\mathcal F.$

Proof: Let A be ${\mathcal F}$-analytic. By Lemma 2, there exists ${B\in\mathcal F}$ such that ${A\subseteq B}$ and ${{\mathbb P}(B)={\mathbb P}^*(A)}$. By Theorem 6, there is a ${C\in\mathcal F_{\delta\sigma}}$ such that ${C\subseteq A}$ and ${{\mathbb P}^*(C)={\mathbb P}^*(A)}$.

So, since ${\mathcal F}$ is a sigma-algebra, both B and C are in ${\mathcal F}$. As ${C\subseteq A\subseteq B}$ and ${{\mathbb P}(C)={\mathbb P}(B)}$, this shows that A is in the completion of ${\mathcal F}$. By the completeness assumption, ${A\in\mathcal F}$. ⬜

Measurable projection follows by combining the capacitability theorem with the properties of analytic sets established in the previous post.

Theorem 8 (Measurable Projection) Let ${(\Omega,\mathcal F,{\mathbb P})}$ be a complete probability space. Then, ${\pi_\Omega(S)\in\mathcal F}$ for all ${S\in\mathcal F\otimes\mathcal B({\mathbb R})}$.

Proof: Let ${\mathcal K}$ be the collection of compact intervals in ${{\mathbb R}}$, which is a compact paving generating the sigma-algebra ${\mathcal B({\mathbb R})}$. As the complement of any compact interval is a countable union of compact intervals, it follows from corollary 4 of the post on analytic sets that the sigma-algebra generated by ${\mathcal{F\times K}}$ consists of ${\mathcal{F\times K}}$-analytic sets,

$\displaystyle \mathcal F\otimes\mathcal B({\mathbb R})=\sigma\left(\mathcal{F\times K}\right)\subseteq\mathcal{A(F\times K)}.$

In particular, S is ${\mathcal{F\times K}}$-analytic. Then, theorem 2 of the post on analytic sets states that ${\pi_\Omega(S)}$ is ${\mathcal F}$-analytic and corollary 7 above says that ${\pi_\Omega(S)\in\mathcal F}$. ⬜

#### Universal Measurability

Finally, we note that the reference to a probability measure can be removed from the statement of the projection theorem. Rather than referring to a complete probability space, a universally complete sigma-algebra can be used. On a measurable space ${(\Omega,\mathcal F)}$, a set ${A\subseteq\Omega}$ is said to be universally measurable if it is in the completion of ${\mathcal F}$ with respect to every probability measure on ${(\Omega,\mathcal F)}$. To remove ambiguity, we will also say that A is universally ${\mathcal F}$-measurable. The universal completion of ${\mathcal F}$ is the collection of universally ${\mathcal F}$-measurable sets,

$\displaystyle \mathcal{\bar F}=\bigcap_\mu\mathcal F_\mu.$

The intersection here is taken over all probability measures ${\mu}$ on ${(\Omega,\mathcal F)}$. Then, ${\mathcal F}$ and ${(\Omega,\mathcal F)}$ are referred to as universally complete if every universally measurable set is in ${\mathcal F}$. That is, if ${\mathcal{\bar F}=\mathcal F}$. In particular, if ${(\Omega,\mathcal F,{\mathbb P})}$ is a complete probability space then it is immediate that every universally ${\mathcal F}$-measurable set is in ${\mathcal F}$. Corollary 7 above can be generalised using universal measurability.

Corollary 9 Let ${(\Omega,\mathcal F)}$ be a measurable space. Then, every ${\mathcal F}$-analytic set is universally ${\mathcal F}$-measurable. That is,

$\displaystyle \mathcal{A(F)}\subseteq\mathcal{\bar F}.$

Proof: By corollary 7, every ${\mathcal F}$-analytic set is in ${\mathcal F_\mu}$ for any probability measure ${\mu}$ on ${(\Omega,\mathcal F)}$ and, hence, is universally ${\mathcal F}$-measurable. ⬜

Similarly, measurable projection can be generalized.

Theorem 10 (Universally Measurable Projection) Let ${(\Omega,\mathcal F)}$ be a measurable space. Then, ${\pi_\Omega(S)}$ is universally ${\mathcal F}$-measurable, for all ${S\in\mathcal F\otimes\mathcal B({\mathbb R})}$.

Proof: If ${\mu}$ is a probability measure on ${(\Omega,\mathcal F)}$ then, as ${\mathcal F\subseteq\mathcal F_\mu}$, the set S is ${\mathcal F_\mu\otimes\mathcal B({\mathbb R})}$ measurable. By theorem 8, ${\pi_\Omega(S)\in\mathcal F_\mu}$. This holds for all such probability measures ${\mu}$, so ${\pi_\Omega(S)}$ is in ${\mathcal{\bar F}=\mathcal F}$. ⬜