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In probability theory many results such as the existence of the regular conditional distribution only exists for standard Borel spaces (which are Borel subsets of some polish space).

Continuous stochastic processes are elements of polish spaces

A continuous stochastic processes $X = (X_t)_{t\in E}$ in $F$ can be thought of as an element of $C(E, F)$. And for $E$ a $\sigma$-compact metrizable space and $F$ a polish space we can turn $C(E,F)$ into a polish space with the help of $$ d(f,g) = \sum_{n=0}^\infty 2^{-n} (d_n(f,g)\wedge 1) \qquad \text{with} \qquad d_n(f,g) := \sup_{x\in K_n} d_F(f(x), g(x)), $$ Where $(K_n)_{n\in \mathbb N_0}$ are a sequence of increasing compact sets such that $\bigcup K_n = E$. The proof should be analogous to the proof of $C(K_n, F)$ being polish via $d_n$, which can be found as Theorem 4.19 in "Classic Descriptive Set Theory" by Kechris, Alexander with an extension to $C(E, F)$ similar to the case of $C([0,\infty)$) see e.g. Klenke (2014, Theorem 21.30).

Okay, so continuous stochastic processes are random variables in a polish space. But what about non-continuous stochastic processes?

Random generalized functions

Non-continuous stochastic processes can generally not even be evaluated point-wise so they are defined as generalized functions (except for the cadlag construction in the case of $\mathbb R$). I.e. they are defined as a stochastic process $(X_{\phi})_{\phi \in \mathcal D}$ over the test functions $\mathcal D = C_c(\mathbb R^n)$.

This results in the question: Is $ \mathcal D'$ a polish or at least standard Borel space such that $X$ is a random variable in a polish space?

While $\mathcal D=C_c(\mathbb R^n)$ is a subset of the polish space $C(\mathbb R^n)$, polish spaces are not necessarily $\sigma$-compact. And in fact they appear to be never $\sigma$ -compact if they are infinte dimensional (comments in this answer). So while $\mathcal D'$ only contains continuous linear operators and is thereby a subspace of $C(\mathcal D, \mathbb R)$ this does not accomplish what we need.

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    Since the distribution space is not metrizable, it won't be polish. – Surb Mar 24 '25 at 08:36
  • @Surb isn't there the concept of the op-norm: $|T| := \sup_{\substack{\varphi \in \mathcal D\ |\varphi|=1}} |T\varphi|$ which is finite for all elements in $\mathcal D'$ by definition? That should induce a metric – Felix Benning Mar 24 '25 at 08:42
  • Your claim that the space $\mathscr D$ of test functions is polish is wrong (there are complete separable norms on $\mathscr D$ which can be found by choosing a Hamel basis and a linear isomorphism with any separable Banach space you like -- but such norms will not yield the space of distributions as the dual). – Jochen Mar 24 '25 at 08:53
  • @Jochen fixed the statement. I only really wanted to outline an attack approach which does not work. I really only wanted to say that $\mathcal D$ is probably not $\sigma$-compact – Felix Benning Mar 24 '25 at 08:57

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