Consider a sequence $(P_n)_{n\in\mathbb N}$ of probability measures on $(\mathbb R^{\mathbb N}, \mathcal B)$, where $\mathbb R^{\mathbb N}$ is the countable Cartesian product of $\mathbb R$, and $\mathcal B$ is the $\sigma$-algebra generated by the topology of pointwise convergence, i.e. the topology generated by the metric $$d(x,y) = \sum_{n=1}^\infty \frac{\min(1,\vert x_n - y_n\vert)}{2^n}.$$ The corresponding metric space is complete, and the generated topology is separable. Therefore, every probability measure on $(\mathbb R^{\mathbb N},\mathcal B)$ is tight.
Let $P$ be a (tight) probability measure on $(\mathbb R^{\mathbb N},\mathcal B)$. We have the following theorem:
Theorem 1:$(P_n)_{n\in\mathbb N}$ converges to $P$ iff $(\pi_k\#P_n)_{n\in\mathbb N}$ converges to $\pi_k\#P$ for all $k\in\mathbb N$.
Proof. (See Example 2.6 in Billingsley "Convergence of probability measures" (2nd edition).)
Here $\pi_k$ denotes the coordinate projection of $\mathbb R^{\mathbb N}$ onto $\mathbb R^k$ and $\pi_k\#P$ the pushforward probability measure of $P$ under $\pi_k$.
Now suppose that $(\pi_k\#P_n)_{n\in\mathbb N}$ converges to $\pi_k\#P$ for all $k\in\mathbb N$. By Theorem 1, $(P_n)_{n\in\mathbb N}$ converges to $P$. Since $P_n$ is tight for every $n\in\mathbb N$, and $P$ is also tight, a Theorem due to Le Cam yields that $(P_n)_{n\in\mathbb N}$ is uniformly tight. (Is this correct?)
I would like to prove the "$\Leftarrow$"-direction of Theorem 1 just by using tightness + weak convergence of finte dimensional marginals. To this end, I would have to show uniform tightness of $(P_n)_{n\in\mathbb N}$. However, I don't know how to approach this. What we have:
- $P_n$ is tight for each $n\in\mathbb N$
- $\pi_k\#P_n$ converges to $\pi_k\#P$ for all $k\in\mathbb N$
To show: for all $\epsilon>0$ there is $K\subset\mathbb R^{\mathbb N}$ compact such that $$\sup_{n\in\mathbb N}P_n(\mathbb R^{\mathbb N}\setminus K) \leq \epsilon.$$
My idea: for any $k\in\mathbb N$, we have that $\pi_k\#P_n$ converges to $\pi_k\#P$ by assumption. Hence, for any $\epsilon>0$ there is $A\subset\mathbb R^k$ compact such that $$\sup_{n\in\mathbb N}\,(\pi_k\#P_n)(\mathbb R^k\setminus A) \leq \epsilon.$$ I thought that the considered topology over $\mathbb R^{\mathbb N}$ allows to somehow "embed" a compact set $K$ in $\pi_k^{-1}(A)$ in the sense that $$P_n(\mathbb R^{\mathbb N}\setminus K)\leq P_n(\pi_k^{-1}(\mathbb R^k\setminus A)) = (\pi_k\#P_n)(\mathbb R^k\setminus A) < \epsilon.$$ I don't know how to make it rigorouos though.
Edit 2: In an earlier version of this post, my proof idea had two fundamental flaws. First, I used the wrong characterization of compact sets in $\mathbb R^{\mathbb N}$ and second, I claimed that $\pi_k^{-1}(A)$ can be a subset of a compact set, which is not true as $pi_k^{-1}(A)$ is the pre-image of a closed set under a continuous function, and hence also closed. If $\pi_k^{-1}(A)$ was a closed subset of a compact set, it would be compact itself, but $\pi_k^{-1}(A)$ is clearly not compact. Therefore, I have to give up on the idea mentioned above. I will add a bounty to this question to draw some attention to it as I need some more clues on how to proceed.