Is a Radon measure with discrete support a sum of Dirac measures? Let $X$ be a topological space ($X$ can be assumed to be locally compact, Hausdorff and second countable if needed, but I'd like to remain general if possible) and let $\mu$ be a Radon probability measure on $X$ such that $\mathrm{supp}(\mu)$ is a discrete set. 
Must $\mu$ be a sum of weighted Dirac measures centered at points of the support?
Note that this is false for arbitrary measures, for example the Dieudonné measure on $\omega_1+1$ (which is Hausdorff and compact) is supported on the singleton $\{\omega_1\}$ but it isn't a weighted Dirac, however it isn't inner regular so in particular it isn't Radon either.
 A: We shall call a space Lindelöf, if each its open cover has a countable subcover.
Proposition. Each $\sigma$-additive measure $\mu$ with discrete support $S$ defined on a $\sigma$-algebra (containing all Borel subsets of $X$) of subsets of a Lindelöf space $X$  is a sum of weighted Dirac measures centered at points of the support.
Proof. For each point $x\in X\setminus S$ pick a neighborhood $U_x$ with zero measure. An open cover $\mathcal U=\{U_x: x\in X\setminus S\}$ has a countable subcover $\mathcal V$. Then 
$$\mu(X\setminus S)=\mu\left(\bigcup\mathcal U\right)=\mu\left(\bigcup\mathcal V\right)=\sum_{V\in\mathcal V}\mu(V)=0.$$
For each point $x\in S$ pick a neighborhood $V_x$ such that $V_x\cap S=\{x\}$. Then 
$$\mu(V_x)=\mu(V_x\cap S)+\mu(V_x\cap (X\setminus S))=\mu(\{x\}).$$
Since $\{V_x\cap S: x\in S\}$ is a disjoint open cover of a Lindelöf space $S$, it is countable.
Now let $A$ be any $\mu$-measurable subset of $X$. Then 
$$\mu(A)=\mu(A\cap S)+\mu(A\cap (X\setminus S))=\sum_{x\in A\cap S}\mu(\{x\}).$$ $\square$
