Let $P_{n}(n=1,2, \ldots)$ on $([0,1], \mathcal{B}([0,1]))$, where $\mathcal{B}([0,1])$ is the Borel $\sigma$-algebra on $[0,1]$, be a sequence of measures which converges weakly to the finite measure $P$ (i.e. $P_{n} \Rightarrow P$). My question is, for every $x \in supp\;P$, does there exist an $N$ such that for all $n \geq N$ $x \in supp\;P_n$?
My approach: $x \in supp\;P$, therefore $P((x-\delta,x+\delta))>0$ for all $\delta >0$. If $x \in C(P)$, i.e. in the continuity set of $P$, there exists $K_{x,\delta}$ such that for all $n \geq K_{x,\delta}$, $P_n((x-\delta,x+\delta))>0$. But it may so happen that for all such $n$ there exists some $\delta_n \in (0,\delta)$ such that $P_n((x-\delta_n,x+\delta_n))=0$. How do we know that there exists some $N$ such that for all $\delta >0$, $P_n((x-\delta,x+\delta))>0$ for all $n \geq N$? And this is just for points of continuity of $P$. I'm not sure how to argue for other points in the support of $P$ (such as the ones with atoms, if any).
If not "convergence" (I'm using this a bit informally, of course supports are sets and so the term convergence does not directly apply), can we say something weaker about what happens to the supports as a sequence of measures converges? They must be "somehow coming closer" but I'm not able to describe exactly how.
Best Answer
The answer is negative.
Consider Dirac measures $P_n = \delta_{\frac{1}n} \to P = \delta_0 $. Then $supp P_n = \{ \frac1{n} \}$ and $supp P = \{ 0 \}$.