$\mu$ pure point measure if and only if $\mu(A)=\sum_{x\in A} \mu(\left\{x\right\})$

Question

$\mu$ is a pure point measure if and only if for every $A\in \mathcal{B}(X)$, $\mathcal{B}(X)$ sigma-algebra Baire. $X$ compact hausdorff $\mu(A)=\sum_{x\in A} \mu(\left\{x\right\})$

I have this

If $\mu(A)=0$ is hold.

If $\mu(A)>0$, then exists $x\in A: \mu(\left\{x\right\})>0$ Then $\left\{x:\mu(A)>0\right\}=\bigcup_{x\in A} \left\{x:\mu(\left\{x\right\})>0\right\}$ and I do not know how to continue ...

Answer 1

I think $\mu$ is supposed to be a finite measure. ''If" part: $\{x:\mu\{x\}>\frac 1 n\}$ is finite for each $n$ and hence $\{x:\mu\{x\}>0\}$ is an at most countable set , say $\{x_1,x_2,...\}$. Verify now that $\mu =\sum_n \mu\{x_n\}\delta_{x_n}$. (You need the fact that $\mu\{x\}=0$ if $x \notin \{x_1,x_2,...\}$). "Only if" part is obvious.