2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
7
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.