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Let $F = (F(x) : x \in \mathbf{R}^n)$ be a family of $\mathbf{R}^k$-valued random variables indexed by $\mathbf{R}^n$ (to be clear there is a single probability space $(\Omega,\Sigma,\mathbf{P})$ such that for each $x \in \mathbf{R}^n$, we have a random variable $F(x) : (\Omega,\Sigma,\mathbf{P}) \to (\mathbf{R}^k,\text{Borel}(\mathbf{R}^k))$.

This is a 'random field in $\mathbf{R}^k$ indexed by $\mathbf{R}^n$' - i.e. we can think of it as a single random variable $F : (\Omega,\Sigma,\mathbf{P}) \to (S, B)$, where $S = (\mathbf{R}^k)^{\mathbf{R}^n}$ is the space of functions from $\mathbf{R}^n$ to $\mathbf{R}^k$ and $$ B = \Pi_{t \in \mathbf{R}^n}\text{Borel}(\mathbf{R}^k) $$ is the product of Borel sigma algebras.

Suppose I have a sequence $F_k$ of such things. What is the correct notion of 'convergence in distribution' $F_k \to F$?

Should I look at $\mu_{F_k} := \mathbf{P}\circ F_k^{-1}$ as a probability measure on $S$ and see if $\int g d\mu_{F_k} \to \int g d\mu_F$ for every bounded continuous function $g$? The reason I'm suspicious is that the topological space $(S,\text{product topology})$ is not metrizable and that throws me off what I usually think weak convergence of measures ought to be...

EDIT: It's possible that this stack exchange answer sort of resolves my confusion but does anyone know of a source? https://math.stackexchange.com/q/4698018/562248

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  • $\begingroup$ Uncountable products are quite pathological and so you may want to switch to random Schwartz distributions instead of random functions on $\mathbb{R}^n\rightarrow\mathbb{R}^k$. A space like $\mathcal{S}'(\mathbb{R}^n)$ with the strong topology is Radon so a Borel measure is automatically regular. Prokhorov's Theorem also holds as well as the Levy Continuity Theorem, i.e., the tools you might want to have in order to study weak convergence of probability measures. $\endgroup$ Commented Jul 31, 2023 at 15:11

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Maybe more a suggestion than an answer.

You could replace the space of all maps $(\mathbb{R}^k)^{\mathbb{R}^d}$ with the space of all continuous maps $C:=C(\mathbb{R}^d,\mathbb{R}^k)$. We can then endow $C$ with the topology of uniform convergence on compacta (aka compact convergence), which coincides in this case with the compact-open topology. According to mathoverflow this topology on $C$ is metrizable, and thus completely regular (see wikipedia).

On the topological space $C$ it is natural to consider $\tau$-additive probability measures and the narrow convergence of measures.

Note that, according to Fremlin Cor. 437L, narrow convergence and the vague convergence coincides for $\tau$-additive probability measures on completely regular topological spaces. So for $C$ there is no ambiguity here.

Then you could model $F$ as a measurable map $F: \Omega \to C$, where $C$ carries the Borel $\sigma$-algebra w.r.t. to the mentioned topology, such that the distribution $\mu$ of $F$ on $C$ is $\tau$-additive.

A sequence $(\mu_n)_{n \in \mathbb{N}}$ of $\tau$-additive probability measures on $C$ convergences to a $\tau$-additive probability measure $\mu$ on $C$ narrowly/weakly/vaguely if and only if for all bounded continuous functions $g: C \to \mathbb{R}$ we have for $n \to \infty$ the convergence: $$ \int g \,d\mu_n \to \int g\,d\mu.$$

Please carefully check for yourself if all arguments made are correct.

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  • $\begingroup$ Thanks for the suggestion. Unfortunately I would rather not assume they were continuous for every $\omega \in \Omega$. Another issue though is that - afaict - the only measurability that you get 'for free' is where you have the product of Borel sigma-algebras on the RHS. I am unsure if the Borel sigma algebra of the compact convergence topology is contained in this sigma-algebra?? $\endgroup$
    – SBK
    Commented Jul 28, 2023 at 11:16

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