It is well known that if $X$ is, say, compact and metric, then the set of probability measures on the Borel subsets of $X$ endowed with the usual topology of weak convergence of measures has as a dense subset the set of finitely supported probability measures on $X$. Is it known whether the result is true when the relevant topology is the topology of setwise convergence of probability measures?

$\begingroup$ If you ask a question here it is not only a matter of politeness to show some reaction if your question is answered. $\endgroup$ – Jochen Wengenroth May 30 '15 at 12:11

$\begingroup$ @Wengenroth. I apologize for being slow. Many thanks for your answer. I haven't had a chance to look at your answer carefully and there were a couple of steps that were not entirely clear to me. Also, I am not familiar with the website (for instance, how do I ask a question that contains mathematical symbols in response to an answer). $\endgroup$ – mo15 Jun 1 '15 at 20:17
I think that the answer is yes but not very useful. Every neighbourhood of a probability measure $P$ contains a set of the form $$ \lbrace \mu: \mu(A_k) P(A_k)< \varepsilon, k=1,\ldots,n\rbrace $$ for finitely many Borel sets $A_1,\ldots,A_n$ and $\varepsilon>0$. We have to show that such a set contains a probability measure of finite support. For this we "disjointify" the sets $A_1,\ldots,A_n$ in the usual way: Set $B_k^1=A_k$ and $B_k^0=X\setminus A_k$ and, for $e \in \lbrace 0,1\rbrace^n$, define $A^e =B_1^{e_1} \cap \cdots\cap B_n^{e_n}$. Then choose from each nonempty $A^e$ a point $x_e$ and define $\mu=\sum_e P(A^e) \delta_{x_e}$ where $\delta_x$ denotes the Dirac measure. Since each $A_k$ is the disjoint union of all $A^e$ which are contained in $A_k$ you even get $P(A_k)=\mu(A_k)$ for all $k=1,\ldots,n$.
However, I don't think that the topology is very useful. It is very nonmetrizable and although discrete probability measures are dense they are not sequentially dense. Moreover, if $x_n \to x$ in $X$ (but all $x_n$ are distinct from $x$) the sequence $\delta_{x_n}$ does not converge to $\delta_x$ (it does not converge at all). This is probably not what you want.

$\begingroup$ Isn't the topology of setwise convergence of probability measures metrized by the following metric? $\rho(\mu,\nu)=\sup_{B\text{ measurable}}\mu(B)\nu(B)$. $\endgroup$ – mo15 Mar 15 '18 at 21:38

$\begingroup$ I see that my question is answered here: mathoverflow.net/questions/240282/… $\endgroup$ – mo15 Mar 15 '18 at 21:48