Here is an exercise in probability/measure theory. Suppose we are given a sequence of product probability measures $(\mu_{n})_{n \in \mathbb{N}}$ on $\mathbb{R}^{\mathbb{Z}}$. That they are product measures means that $\mu_{n}(\prod_{i} A_{i}) = \prod_{i} \mu_{n}^{(i)}(A_{i})$ for some sequence of probability measures $\{\mu_{n}^{(i)}\}_{i \in \mathbb{Z}}$ on $\mathbb{R}$.
A natural question is: if, for each $i \in \mathbb{Z}$, $(\mu_{n}^{(i)})_{n \in \mathbb{N}}$ converges weakly to some probability measure $\mu^{(i)}$, does $(\mu_{n})_{n \in \mathbb{N}}$ converge to the associated measure $\mu = \prod_{i} \mu^{(i)}$?
The answer is yes. The only solution that comes to mind is the following: for each $i$, we can construct a probability space $(\Omega^{(i)},\mathcal{F}^{(i)},\mathbb{P}^{(i)})$ with random variables $(X^{(i)}_{n})_{n \in \mathbb{N}}$ and $X$ such that $X^{(i)}_{n} \to X^{(i)}$ almost surely under $\mathbb{P}^{(i)}$, $\mu^{(i)}_{n}$ is the law of $X^{(i)}_{n}$ for all $n$, and $\mu^{(i)}$ is the law of $X$.
If you look at the product probability space $(\Omega,\mathcal{F},\mathbb{P})$ with $\Omega = \times_{i \in \mathbb{Z}} \Omega^{(i)}$, $\mathcal{F} = \otimes_{i \in \mathbb{Z}} \mathcal{F}^{(i)}$, and $\mathbb{P}^{(i)} = \prod_{i \in \mathbb{Z}} \mathbb{P}^{(i)}$ and random variables $X_{n} = (X^{(i)}_{n})_{i \in \mathbb{Z}}$ and $X = (X^{(i)})_{i \in \mathbb{Z}}$, then the convergence $X_{n} \to X$ in $\mathbb{R}^{\mathbb{Z}}$ holds $\mathbb{P}$-almost surely, $X_{n}$ has law $\mu_{n}$, and $X$ has law $\mu$. Therefore, it follows that $(\mu_{n})_{n \in \mathbb{N}}$ converges weakly to $\mu$ as probability measures on $\mathbb{R}^{\mathbb{Z}}$.
So the question is natural and the answer is as one expects, but I don't know of any other proof. The problem I have is the proof above is probabilistic. Is there more of an analytic proof? I feel as though there should be other, relatively easy arguments. (A wrinkle, in connection with my other question, is that tightness is not obvious here — hence Prokhorov's Theorem is not obviously applicable.)
More generally, is convergence of finite-dimensional marginals enough for weak convergence in $\mathbb{R}^{\mathbb{Z}}$? The proof above is not quite strong enough to prove it, but it's a start.
Best Answer
My answer is based on Douglas Dow's answer to my other question.
I'll show that if $(\mu_{j})_{j \in \mathbb{N}}$ is a sequence of probability measures on $\mathbb{R}^{\mathbb{Z}}$ for which the finite dimensional marginals all converge weakly, then $(\mu_{j})_{j \in \mathbb{N}}$ itself converges weakly in $\mathbb{R}^{\mathbb{Z}}$. Note that it suffices to prove $(\mu_{j})_{j \in \mathbb{N}}$ is tight. (See, e.g., Billingsley's book.)
For each $N \in \mathbb{N}$ and $j \in \mathbb{N}$, let $\mu^{(N)}_{j}$ be the marginal of $\mu_{j}$ on $\mathbb{R}^{[-N,N]}$, that is, \begin{equation*} \mu^{(N)}_{j}(A) = \mu_{j}(\{x \in \mathbb{R}^{\mathbb{Z}} \, \mid \, (x_{-N},x_{-N+1},\dots,x_{N-1},x_{N}) \in A\}) \quad \text{for} \, \, A \subseteq \mathbb{R}^{[-N,N]}. \end{equation*}
Fix $\epsilon > 0$. We need to find a compact set $K(\epsilon) \subseteq \mathbb{R}^{\mathbb{Z}}$ such that \begin{equation*} \mu_{j}(K(\epsilon)) < \epsilon. \end{equation*}
By assumption, given any $N \in \mathbb{N}$, the sequence $(\mu^{(N)}_{j})_{j \in \mathbb{N}}$ converges weakly in $\mathbb{R}^{[-N,N]}$. In particular, it is tight. It follows that there is a $M_{N} = M_{N}(\epsilon) \geq 1$ such that \begin{equation*} \mu^{(N)}_{j}([-M_{N},M_{N}]^{[-N,N]}) \geq 1 - \epsilon 2^{-N} \quad \text{for all} \, \, j \in \mathbb{N}. \end{equation*}
Let $K(\epsilon) = \bigcap_{N = 1}^{\infty} [-M_{N},M_{N}]^{[-N,N]} \subseteq \mathbb{R}^{\mathbb{Z}}$. (I abuse notation and consider $[-M_{N},M_{N}]^{[-N,N]}$ as a subset of $\mathbb{R}^{\mathbb{Z}}$ in the natural way.) Observe that, no matter the choice of $j \in \mathbb{N}$, we have \begin{equation*} \mu_{j}(\mathbb{R}^{\mathbb{Z}} \setminus K(\epsilon)) \leq \sum_{N = 1}^{\infty} \mu_{j}^{(N)}(\mathbb{R}^{\mathbb{Z}} \setminus [-M_{N},M_{N}]^{[-N,N]}) < \epsilon \sum_{N = 1}^{\infty} 2^{-N} = \epsilon. \end{equation*} Further, we know that \begin{equation*} K(\epsilon) \subseteq \prod_{N = 1}^{\infty} [-M_{N},M_{N}] \end{equation*} so $K(\epsilon)$ (being closed) is compact.