Probabilists often work with Polish spaces, though it is not always very clear where this assumption is needed.
Question: What can go wrong when doing probability on non-Polish spaces?
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Sign up to join this communityProbabilists often work with Polish spaces, though it is not always very clear where this assumption is needed.
Question: What can go wrong when doing probability on non-Polish spaces?
One simple thing that can go wrong is purely related to the size of the space (Polish spaces are all size $\leq 2^{\aleph_0}$). When spaces are large enough product measures become surprisingly badly behaved. Consider Nedoma's pathology:
Let $X$ be a measurable space with $|X| > 2^{\aleph_0}$. Then the diagonal in $X^2$ is not measurable.
We'll prove this by way of a theorem:
Let $U \subseteq X^2$ be measurable. $U$ can be written as a union of at most $2^{\aleph_0}$ subsets of the form $A \times B$.
Proof: First note that we can find some countable collection $(A_i)_{i\ge 0}$ of subsets of $X$, such that $U \subseteq \sigma(\{A_i \times A_j:i,j\ge 0\})$, where $\sigma(\cdot)$ denotes the $\sigma$-algebra generated by the given subsets (proof: The set of $V$ such that we can find such $A_i$ is a $\sigma$-algebra containing the basis sets).
For $x \in \{0, 1\}^\mathbb{N}$ define $B_x = \bigcap \{ A_i : x_i = 1 \} \cap \bigcap \{ A_i^c : x_i = 0 \}$.
Consider all subsets of $X^2$ which can be written as a (possibly uncountable) union of $B_x \times B_y$ for some $y$. This is a $\sigma$-algebra and obviously contains all the $A_i \times A_j$, so contains $U$.
But now we're done. There are at most $2^{\aleph_0}$ of the $B_x$, and each is certainly measurable in $X$, so $U$ can be written as a union of $2^{\aleph_0}$ subsets of the form $A \times B$.
QED
Corollary: The diagonal is not measurable.
Evidently the diagonal cannot be written as a union of at most $2^{\aleph_0}$ rectangles, as they would all have to be single points, and the diagonal has size $|X| > 2^{\aleph_0}$.
Separability is a key technical property used to avoid measure-theoretic difficulties for processes with uncountable index sets. The general problem is that measures are only countably additive and $\sigma$-algebras are closed under only countably many primitive set operations. In a variety of scenarios, uncountable collections of measure zero events can bite you; separability ensures you can use a countable sequence as a proxy for the entire process without losing probabilistic content. Here are two examples.
Weak convergence: the classical theory of weak convergence utilizes Borel-measurable maps. When dealing with some function-valued random elements, such as cadlag functions endowed with the supremum norm, Borel-measurability fails to hold. See the motivation for Weak Convergence and Empirical Processes. The $J1$ topology is basically a hack which ensures the function space is separable and thereby avoids measurability issues. The parallel theory of weak convergence described in the book embraces non-measurability.
Existence of stochastic processes with nice properties: a key property of Brownian motion is continuity of the sample paths. Continuity, however, is a property involving uncountably many indices. The existence of a continuous version of a process can be ensured with separable modifications. See this lecture and the one that follows.
Metrizability allows us to introduce concepts such as convergence in probability. Completeness (the Cauchy convergence kind, not the null subsets kind) makes it easier to conduct analysis.
There's already been some good responses, but I think it's worth adding a very simple example showing what can go wrong if you don't use Polish spaces.
Consider $\mathbb{R}$ under its usual topology, and let X be a non-Lebesgue measurable set. e.g., a Vitali set. Using the subspace topology on X, the diagonal $D\subseteq\mathbb{R}\times X$, $D=\{(x,x)\colon x\in X\}$ is Borel Measurable. However, its projection onto $\mathbb{R}$ is X, which is not Lebesgue measurable. Problems like this are avoided by keeping to Polish spaces. A measurable function between Polish spaces always take Borel sets to analytic sets which are, at least, universally measurable.
The space X in this example is a separable metrizable metric space, whereas Polish spaces are separable completely metrizable spaces. So things can go badly wrong if just the completeness requirement is dropped.
Below is a copy of an answer I gave here https://stats.stackexchange.com/questions/2932/metric-spaces-and-the-support-of-a-random-variable/20769#20769
Here are some technical conveniences of separable metric spaces
(a) If $X$ and $X'$ take values in a separable metric space $(E,d)$ then the event $\{X=X'\}$ is measurable, and this allows to define random variables in the elegant way: a random variable is the equivalence class of $X$ for the "almost surely equals" relation (note that the normed vector space $L^p$ is a set of equivalence class)
(b) The distance $d(X,X')$ between the two $E$-valued r.v. $X, X'$ is measurable; in passing this allows to define the space $L^0$ of random variables equipped with the topology of convergence in probability
(c) Simple r.v. (those taking only finitely many values) are dense in $L^0$
And some techical conveniences of complete separable (Polish) metric spaces :
(d) Existence of the conditional law of a Polish-valued r.v.
(e) Given a morphism between probability spaces, a Polish-valued r.v. on the first probability space always has a copy in the second one
(f) Doob-Dynkin functional representation: if $Y$ is a Polish-valued r.v. measurable w.r.t. the $\sigma$-field $\sigma(X)$ generated by a random element $X$ in any measurable space, then $Y = h(X)$ for some measurable function $h$.
We know by Ulam's theorem that a Borel measure on a Polish space is necessarily tight. If we just assume that the metric space is separable, we have that each Borel probability measure on $X$ is tight if and only if $X$ is universally measurable (that is, given a probability measure $\mu$ on the metric completion $\widehat X$, there are two measurable subsets $S_1$ and $S_2$ of $\widehat X$ such that $S_1\subset X\subset S_2$ and $\mu(S_1)=\mu(S_2)$. So a probability measure is not necessarily tight (take $S\subset [0,1]$ of inner Lebesgue measure $0$ and outer measure $1$), see Dudley's book Real Analysis and Probability.
An other issue related to tightness. We know by Prokhorov theorem that if $(X,d)$ is Polish and if for all sequence of Borel probability measures $\{\mu_n\}$ we can extract a subsequence which converges in law, then $\{\mu_n\}$ is necessarily uniformly tight. It may be not true if we remove the assumption of "Polishness". And it may be problematic when we want results as "$\mu_n\to \mu$ in law if and only if there is uniform tightness and convergence of finite dimensional laws."
Google "image measure catastrophe" with quotation marks.
It can also be useful to have the set of Borel probability measures on $X$ (with weak* convergence, a.k.a. convergence in law) to be metrizable, for instance to be able to treat the convergences sequentially. For this you need the space $X$ to be separable and metrizable (see Lévy-Prohorov metric).
Fun fact: you can find a non-separable Banach space and a Gaussian probability measure on it which gives measure $0$ to every ball of radius $1$. (In particular your intuition about notions like the "support" of a measure goes pretty badly wrong.)
Consider i.i.d. $\xi_n$ and take as your norm $|\xi|^2 = \sup_{k\ge 0}2^{-k}\sum_{n=1}^{2^{k}} |\xi_n|^2$. This is almost surely finite by Borel-Cantelli and almost surely at least $1$ by the law of large numbers. The fact that it gives measure $0$ to every ball of radius $1$ is left as an exercise. This norm isn't even very exotic: if you interpret the $\xi_n$'s as Fourier coefficients, then $B$ is really just the Besov space $B^{1/2}_{2,\infty}$.
As far as I remember, the projection of a measurable set may fail to be measurable, so something very natural may become not an event. Besides, constructing conditional probabilities as measures on sections becomes problematic. Perhaps, there are more reasons but these two are already good enough.