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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.

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    $\begingroup$ if you don't give your own thoughts on the problem you're gonna have a bad time... $\endgroup$ – spaceisdarkgreen Jan 10 '17 at 23:13
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    $\begingroup$ @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. $\endgroup$ – user100000000000000 Jan 10 '17 at 23:15
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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.

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Hint: the sum of an uncountable numbers of strictly positive numbers is infinite.

Can we add an uncountable number of positive elements, and can this sum be finite?

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HINT: The set can be written $$ \bigcup_{n=1}^\infty \{x\in S\mid \mu(\{x\}>1/n)\} $$

Claim each set in the union is finite. If you can prove that, it is a countable union of finite sets and thus countable.

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