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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?

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2 Answers 2

up vote 1 down vote accepted

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.

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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.

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