Assume $f : X \to [0,\infty]$
I want to prove $\sum_{x \in X} f(x)<\infty \Longrightarrow \{x \in X | f(x) >0\} \text { is a countable set}$
Is it connected with finite property? Give me some help to prove it.
Assume $f : X \to [0,\infty]$
I want to prove $\sum_{x \in X} f(x)<\infty \Longrightarrow \{x \in X | f(x) >0\} \text { is a countable set}$
Is it connected with finite property? Give me some help to prove it.
Hint: Call $S=\sum\limits_{x\in X}f(x)$ and, for every positive integer $n$, $E_n=\{x\in X\mid f(x)\geqslant1/n\}$. Bound the size of $E_n$ in terms of $S$ and $n$. Then consider $E=\bigcup\limits_{n\geqslant1}E_n$.
The answer given above by Didier is 100% correct. I wanted to share another perspective, in terms of integration (as this is the approach I used when I solved this problem a couple years ago).
You could examine this using Markov's (Chebyshev's) inequality where the integral is with respect to the counting measure. In other words, let $\mu$ be the counting measure on $X$. Then $\sum_{x\in X} f(x) = \int_X f(x) \, d\mu(x) = S < \infty$ Consider $A_n = \left\{x\in X \, | \, f(x) \geq \frac{1}{n}\right\}$. If the integral (sum) is finite, then Markov's (Chebyshev's) inequality tells you immediately that each $A_n$ is finite.
In fact, Markov's (Chebyshev's) inequality tells us that $\mu(A_n) \leq n S$ To show this, replace $f(x)$ by $\frac{f(x)}{S}$ and look at what Markov's inequality says about this function.
Concluding, if each $A_n$ is finite, then $\{x\in X \, | \, f(x) > 0\} = \bigcup_{n\in\mathbb{Z}_+} A_n$ is countable.
Here is another approach:
Let $E_{\epsilon} = \{x \in X | f(x) >\epsilon \}$. The intent is to show that $E_0$ is countable. Note that $E_0 = \cup_{n=0}^{\infty} E_{\frac{1}{n}}$.
Suppose $E_0$ is uncountable. Then one of the sets $E_{\frac{1}{n'}}$ must be infinite, otherwise $E_0$ would be countable. Then we would have, for any $m$, $ \sum_{x \in X} f(x) \geq \sum_{x \in E_{\frac{1}{n'}}} f(x) \geq m \frac{1}{n'}.$ This contradicts the assumption that $ \sum_{x \in X} f(x) < \infty$, hence $E_0$ must be countable.