# How to prove that this set is countable?

Suppose that $(S, \Sigma)$ is a measurable space and $\mu$ is a finite measure on $(S, \Sigma)$. Suppose that whenever $x \in S$, the singleton $\{x\}$ belongs to $\Sigma$.

Prove that the set $\{x \in S : \mu(\{x\}) > 0 \}$ must be countable.

• if you don't give your own thoughts on the problem you're gonna have a bad time... – spaceisdarkgreen Jan 10 '17 at 23:13
• @spaceisdarkgreen I tried to prove it directly using the definition of countable, but I haven't been able to contruct a bijection. My main attempt was to write the set as a countable union or countable intersection of countable sets. – user100000000000000 Jan 10 '17 at 23:15

suppose it is not countable, Define $B_n$ as the set $\{s\in S | \mu\{s\}> \frac{1}{n}\}$. Notice that at least one of the $B_n$ must be uncountable.
Pick a numerable subset of $B_n: \{x_1,x_2,\dots\}$ and notice that $\mu (\bigcup\limits_{i=1}^\infty \{x_i\})=\sum\limits_{i=1}^\infty \mu(\{x_i\})=\infty$, contradicting the fact that $\mu$ is finite.
HINT: The set can be written $$\bigcup_{n=1}^\infty \{x\in S\mid \mu(\{x\}>1/n)\}$$