Can convergence in distribution necessarily be realised by almost-sure convergence?

Question

Let $X$ be a Polish space. Let $(\mu_n)_{n \in \mathbb{N} \cup \{\infty\}}$ be a family of Borel probability measures $\mu_n$ on $X$ such that $\mu_n \to \mu_\infty$ weakly as $n \to \infty$. For each $k \in \mathbb{N} \cup \{\infty\}$, define the projection \begin{align*} \pi_k \colon X^{\mathbb{N} \cup \{\infty\}} \to X& \\ \pi_k\!\!\left( (x_n)_{n \in \mathbb{N} \cup \{\infty\}} \right) = x_k& \end{align*}

Does there necessarily exist a Borel probability measure $\mu$ on the product space $X^{\mathbb{N} \cup \{\infty\}}$ such that

• for each $n \in \mathbb{N} \cup \{\infty\}$, the pushforward of $\mu$ under $\pi_n$ is $\mu_n$;
• $\pi_n$ converges $\mu$-almost surely to $\pi_\infty$ as $n \to \infty$?

If so, is there a name for this result? Or a standard reference?

[I seem to recall reading somewhere that there necessarily exists $\mu$ such that the first property holds and the second holds with convergence in probability; so the point of my question is whether this convergence in probability can be strengthened to almost-sure convergence.]

Answer 1

Skorokhod's representation theorem gives representation $X_{n}\to X$ of almost sure convergence and here is a proof for it.