# Weak convergence of product prob. measures

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.

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.