for $\nu$ a probability measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ the set ${x\in \mathbb{R} ; \nu(x) > 0}$ is at most countable Given a probability measure $\nu$ on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$, how do I show that the set (call it $S$) of all $x\in \mathbb{R}$ where $\nu(x)>0$ holds is at most countable?
I thought about utilizing countable additivity of measures and the fact that we have $\nu(A) < 1$ for all countable subsets $A\subset S$. How do I conclude rigorously?

Let $S_n:=\{x,\nu(\{x\})\geq n^{-1}\}$. Using $\sigma$-additivity, we have that $S_n$ is finite (it contains actually at most $n$ elements, as $\nu$ is a probability measure). Then $S=\bigcup_{n\geq 1}S_n$ is countable as an union of such sets. 
