# Open sets in the topology of weak convergence

I do have various questions regarding the topic of probability measures on polish spaces in general, thus I am trying to divide them in “small” subquestions. Hence, this is my first question on this issue.

Notation:

• $\Omega$ is a Polish space;
• $\mathcal{B}(\Omega)$ is the Borel $\sigma$-algebra on $\Omega$;
• $\Delta (\Omega)$ is the set of probability measures on $\mathcal{B}(\Omega)$.

Quite often I read that $\mathcal{B}(\Omega)$ is endowed with the topology of weak convergence, where the topology of weak convergence states that for every real function $f$ bounded, continuous on $\Omega$, $$\lim_{n \to \infty} \int f d\mu_n = \int f d\mu.$$

Here there are my basic questions:

• How does the topology of weak convergence creates open sets in $\Delta (\Omega)$?
• How do those open sets look like?

As always, any feedback regarding anything will be more than welcome.

Thank you for your time.

PS: Maybe the question looks naive, but I am trying to see why, if $\Omega$ is Polish, then $\Delta (\Omega)$ endowed with the topology of weak convergence is Polish as well (of course, any feedback regarding this issue is welcome as well!). Thus I would like to see how the topology of weak convergence actuallys works.

• PPS: I am wondering about the title. I am not sure it is completely correct, but I do not think it is misleading. – Kolmin Feb 27 '15 at 15:18
• It's the minimal topology on $\Delta(\Omega)$ that makes the all functions $I(f): \Delta(\Omega) \rightarrow \mathbb{R}$ continuous, where $f$ is bounded continuous on $\Omega$, and $I(f)$ maps a measure $\mu$ to $\int f d\mu$. – Henno Brandsma Feb 27 '15 at 15:22
• math.leidenuniv.nl/~vangaans/jancol1.pdf gives an explicit metric for this topology. They look like nice notes on the subject. – Henno Brandsma Feb 27 '15 at 15:51
• Yes, it is shown in those notes that they define the same topology. The completeness is not shown, I think, except for the compact case (when $\Delta(\Omega)$ is also compact). – Henno Brandsma Feb 27 '15 at 16:14
• Why is it the weak star topology? $\Delta(\Omega)$ is not a topological vector space, so weak star has no meaning here. That is only defined for topologies on dual topological vector spaces, AFAIK. – Henno Brandsma Feb 27 '15 at 16:15