Question about the Support of a Discrete Probability Measure Suppose that $X$ is a Polish space with its Borel sigma algebra, and that $P$ is a discrete probability measure on $X$, that is to say that $P$ concentrates on a countable Borel set $C$. Suppose we are told that:
$$ \operatorname{supp}(P) = \overline{ \{x_1,x_2,...\} }, $$
where $x_i \in X$ and $\operatorname{supp}(P)$, the support of $P$, is defined as the smallest closed Borel set of full $P$ measure. Then is it necessarily true that $C = \{x_1,x_2,...\}$?
Many thanks for any help.
 A: No. Let for example $X=\mathbb R$ and consider discrete probability measure concentrated on rational numbers in $[0,1]$. Say, asign the probability $2^{-n}$ to rational number $r_n$ for some enumeration of rationals. Then 
$$\operatorname{supp}(P) = \overline{ \{r_1,r_2,...\} }=\overline{\mathbb Q\cap[0,1]}=[0,1]$$
since the rational numbers form a dense subset of the real numbers. Consider another discrete probability measure concentrated on the set $(\mathbb Q+\sqrt{2})\cap[0,1]\,$ where any rational point is shifted by $\sqrt{2}$. This set is also countable, and we can enumerate its elements $a_n$ in some order and assign the same probabilities $2^{-n}$ to $a_n$, $n\geq 1$. For this probability measure $\tilde P$
$$\operatorname{supp}(\tilde P) = \overline{ \{a_1,a_2,...\} }=\overline{(\mathbb Q+\sqrt{2})\cap[0,1]}=[0,1]$$ by the same reasons. 
And the sets $\{a_1,a_2,\ldots\}$ and $\{r_1, r_2, \ldots\}$ are disjoint since $\sqrt{2}$ is irrational number. This is counterexample to desired statement.
