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Let $X$=[0,1] denote the unit interval on the real line. Assume that $f:X \to X$ is a mapping such that $\sum_{x \in X}$ $f(x)$ $<$ $\infty$. Prove that the set {${x \in X | f(x) > 0}$} is at most countable.

I have no clue where to start with this problem.

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    You should probably tell us how you define sums whose index sets are infinite. I assume it is as follows: $$ \sum_{x\in X}f(x)=\sup\left\{ \sum_{x\in\hat{X}}f(x)\colon\hat{X}\text{ is a finite subset of }X\right\} . $$ – 2017-02-01
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    @parsiad Yes this is the definition I am using. – 2017-02-01
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    How many $x$ can have $f(x) \geqslant \varepsilon$ for $\varepsilon > 0$? – 2017-02-01
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    @DanielFischer I am essentially trying to prove that only countably many such $x$ exist. – 2017-02-01
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    You're asking about $f(x) > 0$. As a step towards that goal, consider $f(x) \geqslant \varepsilon$ for some fixed $\varepsilon > 0$. – 2017-02-01

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Let $E_{c}=\{x\in X\colon f(x)\geq c\}$.

Certainly, it is true that $E_{c}$ is finite for any $c>0$ since otherwise we can find a sequence $(x_{n})_{n}$ such that $f(x_{n})\geq c$ for each $n$ and hence $$ \sum_{x\in X}f(x)\geq\sum_{n}f(x_{n})=\infty. $$

Now, note that $$ \left\{ x\in X\colon f(x)>0\right\} =\bigcup_{n}E_{1/n}. $$ What can you conclude?

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    Since this set is a finite union of countable sets, it must also be countable. Is this correct? – 2017-02-01
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    Yes. It's better, however, to use the weaker claim is that *a countable union of finite sets is countable*. – 2017-02-01
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    Ok. Thanks for the help! – 2017-02-01