topology of setwise convergence of measures 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?
 A: 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 non-empty $A^e$ 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 non-metrizable 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.
