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 $X\times Y$.