The existence of a $\sigma$-compact set with full probability

Question

Let $X,Y$ be Polish spaces and $\mathcal P(X)$ the space of all Borel probability measures on $X$. Fix $\mu\in \mathcal P(X), \nu \in \mathcal P(Y)$. Let $\gamma \in \Pi(\mu, \nu)$, i.e., $\gamma \in \mathcal P(X \times Y)$ whose marginal on $X$ is $\mu$ and that on $Y$ is $\nu$. Let $\pi^X, \pi^Y$ be the projections from $X \times Y$ to $X$, $Y$ respectively.

In responding to a concern in this answer, I have come across this result, i.e.,

Theorem: Let $S$ be a Borel subset of $X \times Y$ and $\gamma \in \Pi(\mu, \nu)$ such that $\gamma (S) = 1$. There is a $\sigma$-compact subset $K$ of $S$ such that

• $\pi^X (K), \pi^Y (K)$ are $\sigma$-compact, and
• $\mu (\pi^X (K)) = \nu (\pi^Y (K)) =1$.

Could you have a check on my below attempt?

My attempt: We have $X \times Y$ is also Polish, so $\gamma$ is tight. Then there is an increasing sequence $(K_n)$ of compact subsets of $S$ such that $$ \lim_n \gamma (K_n) = \gamma (S) = 1. $$

Notice that $A_n :=\pi^X (K_n)$ and $B_n :=\pi^Y (K_n)$ are compact subsets of $X, Y$ respectively. Let $K := \cup_n K_n, A := \cup_n A_n$, and $B := \cup_n B_n$. Then $K,A,B$ are $\sigma$-compact subsets of $S, X, Y$ respectively such that $A = \pi^X (K)$ and $B = \pi^Y (K)$. Notice that $\pi^X_\sharp \gamma = \mu$ and $\pi^Y_\sharp \gamma = \nu$. We have $$ \mu(A) = \lim_n \mu(A_n) = \lim_n \gamma ((A_n \times Y)) \ge \lim_n \gamma (K_n) = 1. $$

Similarly, we get $\nu(B)=1$. This completes the proof.

Answer 1

I only know how to prove this if $S$ turns out to be Polish itself. For the sake of transparency, I'll assume that $S\subseteq X\times Y$ is open or closed.

Theorem: Let $S\subseteq X\times Y$ be open or closed. Further, let $\gamma\in\Pi(\mu,\nu)$ be such that $\gamma(S)=1$. There exists a $\sigma$-compact subset $K$ of $S$ such that $\pi^X(K),\pi^Y(K)$ are $\sigma$-compact and $\gamma(K)=\mu(\pi^X(K))=\nu(\pi^X(K))=1$.

To point out the required properties, I'll state two lemmas.

Lemma 1: The subspace $S\subseteq X\times Y$ is Polish and its $\sigma$-algebra is the Borel algebra.

Proof: The space $X\times Y$ is Polish because it is the product of Polish spaces. The space $S\subseteq X\times Y$ is Polish because $S\subseteq X\times Y$ is open or closed. The Borel algebra of $S$ coincides with the subspace $\sigma$-algebra simply because it is the subspace of a space with Borel algebra.

The following result I can show for Polish spaces.

Lemma 2: Let $Z$ be a Polish space and $\gamma\in\mathcal P(Z)$. Then there exists an increasing sequence $K_n$ of compact subsets of $Z$ with $\lim_{n\rightarrow}\gamma(K_n)=1$.

Proof: The set $\{\gamma\}\subseteq\mathcal P(Z)$, where $\mathcal P(Z)$ is equipped with the total variation distance, is compact and closed, so using Lemma 9.1.1 we obtain that $\gamma$ is tight with Theorem 9.1.9. So, for $n\in\mathbb Z_{>0}$ we obtain a compact set $K'_n\subseteq Z$ such that $\gamma(K'_n)\ge 1-1/n$. The set $K_n=\bigcup_{i=1}^nK'_n$ is compact since it is the finite union of compact sets, and the sequence $(K_n)_n$ is increasing with $\lim_{n\rightarrow\infty}\gamma(K_n)\ge\lim_{n\rightarrow\infty}(1-1/n)=1$.

Now, let $\gamma'\in\mathcal P(S)$ be given by $\gamma'(\mathcal E)=\gamma(\mathcal E)$, which is well-defined because $S\subseteq X\times Y$ is measurable and hence its $\sigma$-algebra is a sub-$\sigma$-algebra. Now, we can combine Lemma 1 and Lemma 2 to obtain an increasing sequence $K_n$ of compact subsets of $S$ with $\lim_{n\rightarrow\infty}\gamma'(K_n)=1$, yielding $\lim_{n\rightarrow\infty}\gamma(K_n)=1$. The sets $A_n=\pi^X(K_n)$, $B_n=\pi^Y(K_n)$ are compact because the projections are continuous. Since $X,Y$ are metric spaces, they are Hausdorff and hence $A_n,B_n$ are measurable. By definition, the sets $K=\bigcup_n K_n$, $A=\bigcup_n A_n$, $B=\bigcup_n B_n$ are $\sigma$-compact, and also measurable, being countable unions.

Notice that $\mu(A)\ge\mu(A_n)=\gamma(A_n\times Y)\ge\gamma(K_n)$ for all $n$. This monotonicity gives $\gamma(K)=\mu(A)=\nu(B)=1$. Finally, we directly verify that $A=\pi^X(K)$ and $B=\pi^Y(K)$. This completes the proof.

In a nutshell: Your proof makes sense to me, I only fail to understand how you obtain an increasing sequence of compact subsets of $S$ unless $S$ is Polish. Otherwise, I only get it for $X\times Y$.

Answer 2

I don't have enough points to comment. A Borel subset of a Polish space is a Lusin space, i.e., the continuous bijective image of a Polish space. Thus it admits a finer topology which is Polish. Use that Polish topology to find the sequence of compact sets. Since the original topology is coarser, they are still compact in the original topology.

Another approach is: Borel in Polish is Lusin, Lusin is Radon, in particular every probability measure is tight.

But the direct way is: Polish is Radon, thus the probability of S is the supremum of the probabilities of its compact subsets. This solves the gap in the original attempt.