TITLE: Polish spaces in probability
QUESTION [62 upvotes]: 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?

REPLY [7 votes]: 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 $B^{1/2}_{2,\infty}$.