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?

Given $n\in\mathbb N$, consider the set
$$A_n=\{x\in\mathbb R:\nu(\{x\})\geq\tfrac{1}{n}\}$$
It must be finite; otherwise, the probability of $A_n$ would be infinite since $\nu$ is additive. Thus, $A=\cup_{n\in\mathbb N}A_n$ is countable as a countable  union of finite sets, but it is clear that
$$A=\{x\in\mathbb R:\nu(\{x\})>0\}$$
so you are done.
