3
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.