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Let's say you are given a function $\mu:S\rightarrow(0,1]$ and you can additionally assume $$\sum_{s\in S}\mu(s)=1$$ Does this imply that $S$ is finite?

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The range is $(0,1]$, so $\mu(s)\ne 0$ for all $s\in S$ –  user1658887 Oct 9 '12 at 9:06
    
You're right, sry. –  martini Oct 9 '12 at 9:11

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up vote 3 down vote accepted

No: let $S=\Bbb N$, and let $\mu(n)=2^{-n}$. Then $$\sum_{n\in\Bbb N}\mu(n)=\sum_{n\in\Bbb N}\frac1{2^n}=2\;.$$

You can, however, infer that $S$ is countable, as follows. For each $n\in\Bbb N$ let $$S_n=\{s\in S:\mu(s)\ge 2^{-n}\}\;;$$ then $S=\bigcup_{n\in\Bbb N}S_n$. A countable union of countable sets is countable, so if $S$ were uncountable, then some $S_n$ would be uncountable. But then $$\sum_{k\in S_n}\mu(s)\ge\sum_{s\in S_n}2^{-n}=|S_n|\cdot 2^{-n}\;,$$ which is clearly not finite. Thus, $S$ must be countable.

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