If $\sum_{x \in X} f(x) < {\infty}$ then the set $\{x | f(x) \neq 0 \}$ is countable

Question

Let $f:X \to R$ with $f(x) \ge 0$ for $x \in X$. Show that if $\sum_{x \in X} f(x) < {\infty}$ then the set $\{x / f(x) \neq 0 \}$ is countable

Could anyone help me to show this?

Answer 1

We might define $\sum_{x\in X}f(x)$ as the supremum of all sums over finite subsets of $X$. If there exists $\epsilon>0$ such that $f(x)>\epsilon$ for infinitely many $x\in X$, then this supremum is certainly $\infty$ as we can obtain a finite sum $>n\epsilon$ for arbitrary $n$. So assume that for all $\epsilon>0$ there are only finitely many $x\in X$ with $f(x)>\epsilon$. Then $$\{\,x\in X\mid f(x)\ne 0\,\}= \{\,x\in X\mid f(x)> 0\,\}=\bigcup_{n\in\mathbb N}\{\,x\in X\mid f(x)>\tfrac1n\,\}$$ is the union of countably many finite sets, hence countable.