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Suppose on a metric space $X$ there exists points $x_n$, Borel measures $\mu_n$ such that $x_n$ converges to $x$ and $\mu_n$ converges weakly to $\mu$. Moreover, $x_n \in supp \mu_n$ for each $n$. Does it follow that $x\in supp \mu$? Thanks.

Thanks Etienne for the counter example. I realized that in my original problem the measures are actually probability measures.

The original problem on my hand is: Given compact Polish space $X$ and $\mu \in \Delta(X\times \Delta(X))$, where $\Delta(X)$ is the set of Borel probability measures on $X$ endowed with the weak topology.
Let $B= supp marg_{\Delta(X)}\mu$.
Let $A=\{x\in X: x\in supp v \mbox{ for some } v\in B\}$.
Question: $A$ is Borel measurable ??

The way I try is to show $A$ is closed. Say $x_n\in A$, $x_n\rightarrow x$, then there exists $\mu_n \in B$ such that $x_n \in supp \mu_n$. Now since $\Delta(\Delta(X))$ is compact(by well known results) and $B$ is closed, we can pass to subsequences and assume $\mu_n\rightarrow \mu\in B$ weakly. If we can show $x\in supp \mu$ then we can show $A$ is closed hence measurable.

Closedness is much stronger than what I actually need. If this method does not go through then is there other way to show $A$ is measurable or not?

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    $\begingroup$ No.Take any point $a\in X$ and consider $\mu_n=\frac1n\delta_a$. $\endgroup$ Commented Jan 11, 2014 at 8:04
  • $\begingroup$ I see. What if we have the additional assumption that $\mu_n$ and $\mu$ are both probability measures? $\endgroup$ Commented Jan 11, 2014 at 9:35
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    $\begingroup$ Take two points $a\neq b$ in $X$. Consider $\mu_n=\frac1n\delta_a+(1-\frac1n)\delta_b$ and $x_n\equiv a$. $\endgroup$ Commented Jan 11, 2014 at 20:05
  • $\begingroup$ Could you explain what is $suppmarg_{\Delta(X)}\mu$? $\endgroup$ Commented Jan 11, 2014 at 20:09
  • $\begingroup$ It is the support of the marginal measure, marginal measure of an event is evaluated by the original measure by taking the inverse of projections. Say $\mu\in\Delta(X\times Y)$, let $v=marg_X \mu$, $v$ is defined as $v(E)=\mu(E\times Y)$. By the way, thanks for your example. So my method doesn't work... $\endgroup$ Commented Jan 11, 2014 at 20:15

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It does not seem quite clear that $A$ is Borel.

On the other hand, $A$ is an analytic set, so it is at least universally measurable, which may be enough for what you want to do with it.

To show that $A$ is analytic, it is enough to check that the set $$ E=\{ (x,\nu)\in X\times\Delta(X);\; \nu\in B\;{\rm and}\;x\in {\rm supp}(\nu)\}$$ is Borel in the Polish space $X\times\Delta (X)$, since $A$ is the projection of $E$ into $X$.

If you take a countable basis of open set $(V_n)_{n\in\mathbb N}$ for $X$, then $$(x,\nu)\in E\Leftrightarrow \nu\in B\;{\rm and}\;\forall n\; :\;\left(x\not\in V_n\;{\rm or}\; \nu(V_n)>0 \right)$$ The relation under brackets is $G_\delta$ in $X\times\Delta (X)$ (it is the intersction of a closed set and an open set); so $E$ is $G_\delta$ in $X\times\Delta(X)$, being the intersection of a closed set ($B$ is closed) and a $G_\delta$ set.

If $E$ were $F_\sigma$, i.e. $K_\sigma$ since $X$ is compact, then $A$ would be $K_\sigma$ as well and hence Borel; but I see no immediate reason for this to be true. However, the only thing that was really used above concerning $B$ is that it is Borel set; but $B$ is a rather special set, so maybe one could do better by looking at it more carefully.

I hope this helps.

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  • $\begingroup$ Thanks a lot. That really helps me brush up the math. $\endgroup$ Commented Jan 12, 2014 at 6:49

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