Me again, I have troubles to understand the concept of absolute convergence in the general setting when the set can be uncountable. In the book what I read says:
Definition: Let $X$ be a set (possible uncountable) and let $f: X\rightarrow \mathbb{R}$ be a function. We say that $\sum _{x\in X}f(x)$ converges absolutely iff ${\; \text{ sup}}\big\{\sum _{x\in F}|f(x)|:F\subset X; \#F<\infty \big\}<\infty$.
$\sum _{x\in X}f(x):= \sum _{\{x\in X:f(x)\not=0\}}f(x)$
In one exercise of the book we showed that ${\{x\in X:f(x)\not=0\}}$ is at most countable. I hope my proof of this assertion was correct.
Lemma: Let $X$ be a set (possible uncountable) and let $f: X\rightarrow \mathbb{R}$ and $g: X\rightarrow \mathbb{R}$ be functions such that $\sum _{x\in X}f(x)$ and $\sum _{x\in X}g(x)$ converges absolutely. Then $\sum _{x\in X}f(x)+g(x)$ converges absolutely. Moreover $\sum _{x\in X}f(x)+g(x)=\sum _{x\in X}f(x)+\sum _{x\in X}g(x)$.
Proof: Let $F\subset X$ and $\#F<\infty$. Let $j: \{i\in\mathbb{N}: 1\le i \le N \}\rightarrow F$ be a bijective map (notice that this is possible since $F$ is finite and we may assumed that $\#F= N$ for $N\in \mathbb{N}$). Then, we have
\begin{align} \sum _{x\in F}|f(x)+g(x)|=\sum _{i=1}^N |f(j(i))+g(j(i))| &\le \sum _{i=1}^N |f(j(i))|+\sum _{i=1}^N |g(j(i))|\\ &=\sum _{x\in F}|f(x)|+\sum _{x\in F}|g(x)|\\ &\le \sup_{F\subset X,\,\#F<\infty}\sum _{x\in A}|f(x)|+\sup_{F\subset X,\,\#F<\infty}\sum _{x\in A}|g(x)| \end{align}
In other words, we find an upper bound for an arbitrary finite subset of $X$. Thus
$$ \sup_{F\subset X,\, F \text{ finite}} \sum _{x\in F}|f(x)+g(x)|\le \sup_{A\subset X; A \text{ finite}}\sum _{x\in A}|f(x)|+\sup_{A\subset X; A \text{ finite}}\sum _{x\in A}|g(x)|< \infty$$
$$\,\, \sup_{F\subset X; F \text{ finite}} \sum _{x\in F}|f(x)+g(x)|<\infty$$
Then the series converges absolutely.
Let $A:=\{y\in X: f(y)+g(y)\not=0 \}$ we may assume that the set is infinite, and let $h: \mathbb{N}\rightarrow A$ be a bijective map [is well-define since $A$ is a denumerable set]. Then
$$\sum _{x\in X}f(x)+g(x)=\sum _{x\in A}f(x)+g(x)=\sum _{i=0}^{\infty}f(h(i))+g(h(i))$$
On the other hand since $\sum _{x\in X}f(x)$ and $\sum _{x\in X}g(x)$ converges absolutely then both converges conditionally. Let $B:=\{y\in X: f(y)\not=0 \}$ and $C:=\{y\in X: g(y)\not=0 \}$ each one contained in $A$, i.e., $B,C\subset A$. By definition $\sum _{x\in X}f(x)=\sum _{x\in B}f(x)$ and $\sum _{x\in X}g(x)=\sum _{x\in C}g(x)$.
If we can show $\sum _{x\in B}f(x)=\sum _{x\in A}f(x)$ and $\sum _{x\in C}g(x)=\sum _{x\in A}g(x)$ we almost done. Just we will show that $\sum _{x\in B}f(x)=\sum _{x\in A}f(x)$, the other equality can be prove in a similar way.
Claim $\sum _{x\in B}f(x)=\sum _{x\in A}f(x)$. Clearly any element in $B$ lies in $A$ and each element (if there some) in $A$ but not in $B$ is zero under $f$ for the way in which $B$ was constructed. Thus we are not adding any new element.
[Here I'd like to state the result formally is kinda obvious but I really felt uncomfortable with this argument I need something more formal, but I'm not sure of how to begin.]
To conclude we can use $h$, and we have $\sum _{i=0}^{\infty}f(h(i))+\sum _{i=0}^{\infty}g(h(i))=\sum _{i=0}^{\infty}f(h(i))+g(h(i))$ using the limit laws and we're done since the case when $A$ is finite is trivial [follows by induction in the size of $A$].
Any help? Am I correct?
Hopefully the following is the formal argument that I've found it.
*[Let $h,B,C$ and $A$ as we defined above. We may assume that $A$ is denumerable.
We claim that $A\subset B \cup C$. Suppose that $x\notin B \cup C$, i.e., $f(x)=0$ and $g(x)=0$. This means $f(x)+g(x)=0$, in other words $x\notin A$. Hence, we have shown that if $x$ fails to be in $B \cup C$ also $x\in A$ fails. And then the claim follows by contraposition.
Furthermore, $B\cup C$ is not a finite set and then we have that at least one of the set $A,B$ is infinite (otherwise we get a contradiction). Without loss of generality we may assume that $B$ is infinite. Let $E:=\{i\in \mathbb{N}: g(i)\in B \}$, which is nothing more than the inverse image of $B$ under $h$. It is not difficult to show that the set $E$ is infinite. Let $j: \mathbb{N}\rightarrow E$ be any bijection from the natural numbers to $E$ and $h\restriction _E$ be the restriction of $h$ in $E$.
We need to show that $\sum _{x\in A}f(x)$ converges to $\sum _{x\in B}f(x)$. Let $\varepsilon>0$ be given. Since $\sum _{x\in A}|f(x)|$ is absolutely convergent by hypothesis there is some $N(\varepsilon)\in \mathbb{N}$ such that for any $p,q\ge N(\varepsilon)$ we have: $$\sum _{i=p}^{q}|f(h\restriction _(j(i)))|\le \varepsilon$$
Let $N\ge N(\varepsilon)$ such that $\sum _{i=0}^{N}f(h(j(i)))$ is $\varepsilon$-close to $\sum _{i=0}^{\infty}f(h(j(i)))$. Now the sequence $(j(i))_{i=0}^N$ is clearly finite and hence bounded. Let $N'$ be an upper bound for the sequence. In particular for any $M\ge N'$ the set $\{i\in \mathbb{N}: i\le M\}$ contains $\{j(i): i\le N\}$. Then for what we said above:
\begin{align}\sum _{i=0}^{M}f(h(i))=\sum _{k\in \{i\in \mathbb{N}: i\le M\}}f(h(k)) &=\sum _{k\in \{j(i): i\le N\}}f(h(k))+\sum _{x\in X'}f(h(x))\\ &=\sum _{i=0}^N f(h(j(i)))+\sum _{x\in X'}f(h(x)) \end{align}
Where $X'=\{ i\in \mathbb{N}: i\le M\}-\{j(i): i\le N\}$.
Then we have $X'\subset \{\,j(i): N+1\le i \le M \}\cup \{i\in \mathbb{N}: i\not= j(k) \text{ and } i\le M \}$
And hence
\begin{align}\bigg|\sum _{x\in X'}f(h(x)) \bigg|\le \sum _{x\in X'}|f(h(x))| &\le \sum _{k\in \{\,j(i): N+1\le i \le M \}\cup \{i\not= j(k) \text{ and } i\le M \} }|f(h(k))|\\ &= \sum _{k\in \{\,j(i): N+1\le i \le M \}}|f(h(k))|+ \sum _{k\in \{i\not= j(k) \text{ and } i\le M \} }|f(h(k))|\\ &= \sum _{k=N+1}^{M}|f(h(j(k)))|+ \sum _{k\in \{i\not= j(k) \text{ and } i\le M \} }|f(h(k))|\le \varepsilon\\ \end{align}
This is because whenever $i\in \{i\not= j(k) \text{ and } i\le M \}$ it is no difficult to show that $h(i)\notin B$ and then is zero under $f$. And then this means $\sum _{i=0}^{M}f(h(i))$ is $\varepsilon$ close to $\sum _{i=0}^N f(h(j(i)))$ which is also $\varepsilon$ close to $\sum _{i=0}^{\infty}f(h(j(i)))$ and hence we can conclude that $\sum _{i=0}^{M}f(h(i))$ is $2\varepsilon$ close to $\sum _{i=0}^{\infty}f(h(j(i)))$. Thus $\sum _{x\in A}f(x)$ converges to $\sum _{x\in B}f(x)$]
Hopefully the former is correct.