Let X be a set, and let $f : X \rightarrow \mathbb{R}$ be a function such that the series $\Sigma_{x \in X} f(x)$ is absolutely convergent. Then the set $\{x \in X : f(x) \neq 0\}$ is at most countable.
Book's hint:
The sets $\{x \in X : f(x) > \frac{1}{n} \}$ with cardinality at most Mn, where $$M := sup \{ \Sigma_{x \in A} |f(x)| : A \subset X, A\ is \ finite\}$$
Best Answer
You’ve given no justification for your assertion that
$$\sum_{x\in A}|f(x)|=\sum_{k\in\{1,\ldots,n\}}\left|f\big(g(k)\big)\right|\ge\frac1n\cdot n=1\,,$$
which in any case doesn’t make sense. In order for $f(x)$ to make sense, $x$ must be an element of $X$, so $A$ must be a subset of $X$. You’ve not told us what $A$ is, but it seems likely that you intended it to be the $A_n$ that you defined in the previous sentence, in which case there’s no reason to think that it’s a subset of $X$: we have no reason to believe that $X$ contains any integers at all.
$\sum_{k\in\{1,\ldots,n\}}\left|f\big(g(k)\big)\right|$ makes sense for any $g:\Bbb N\to X$, but since you’ve not specified what function $g$ is, we have no reason to think that each $\left|f\big(g(k)\big)\right|\ge\frac1n$.
The point of the hint is that if there were more than $Mn$ elements $x\in X$ such that $|f(x)|\ge\frac1n$, we could take a set $A$ of more than $Mn$ of these elements and have
$$\sum_{x\in A}|f(x)|>Mn\cdot\frac1n=M\,,$$
contradicting the definition of $M$. Thus, for each $n\in\Bbb Z^+$ the set
$$S_n=\left\{x\in X:|f(x)|\ge\frac1n\right\}$$
is finite. Let $S=\bigcup_{n\in\Bbb Z^+}S_n$; this is the union of countably many countable (in fact finite) sets, so it is countable.
Now suppose that $x\in X$, and $f(x)\ne 0$; then there is an $n\in\Bbb Z^+$ such that $|f(x)|\ge\frac1n$, i.e., such that $x\in S_n\subseteq S$. Thus, there are at most countably many $x\in X$ such that $|f(x)|\ne 0$.