Can this set be not separable ? Can it be not compact? Let $(\mathcal X,\Sigma_X)$ be a measure space, let $\mathcal S_X=\{ q : (\mathcal X, \Sigma_X, q)\text{ is a probability space}\}$, i.e. the simplex associated to $(\mathcal X,\Sigma_X)$. We equip $\mathcal S_X$ with the weakest topology such that for all bounded measurable $f:\mathcal X\to\mathbb R$, the function $p\to p(f)\triangleq\int_{\mathcal X}f ~dp$ is continuous. This is a bounded, closed convex subset of a Hausdorff locally convex vector space. I believe that in general, this set is non separable and non compact, however I could not find an example, I think that we need $\Sigma_X$ to be big enough of an infinite set, having $\mathcal X$ infinite is not enough.
I have tried few example, including $\mathcal X=[0,1]$ and $\Sigma_X$ is the Borel $\sigma$-algebra on $[0,1]$, but I think this might actually be both compact and separable, I am not sure on how to show that either. Any input would be very useful.
 A: It's not a complete answer yet, but too big to fit a comment. As @Willam M. mentioned, usually one works with a topology generated by bounded continuous maps.
A topological space $(X,\tau)$ is called a standard Borel space if it is homeomorphic to a Borel subset of a Polish (complete separable metrizable) space. A space of all probability measures on $X$ endowed with a weak topology (as you've described) is a Borel space itself, see e.g. D. Bertsekas and S. Shreve. Stochastic optimal control: The discrete time case, Corollary 7.25.1.
Apologies for a perhaps not very canonical reference, but during my PhD time all necessary facts on the topic we were easily finding in the aforementioned book. So definitely to find an example you are interested in one should look into non-separable base spaces $X$. I'm pretty sure such examples one can find in either

*

*Classical descriptive set theory by Kechris, or


*Probability Measure on Metric Spaces by Parthasarathy
In the second book Lemma 6.1 states that $X$ is homeomorphic to $\{\delta_x,x\in X\} \subseteq \mathcal P(X)$, a topological subspace of probability measure space endowed with weak convergence. As a result, in Theorem 6.2 it is used to show that if $\mathcal P(X)$ is separable, then so is $X$. Theorem 6.4 states that $\mathcal P(X)$ is compact if and only if $X$ itself is.
For your case of bounded measurable functions generated topology I would suggest working with the fact that $p(\{x\})$ must be a continuous function for each point $\{x\}\in X$.
