Prove that the series $\sum_{x∈X}(f(x) + g(x))$ is absolutely convergent, and $ \sum_{x∈X}(f(x) + g(x)) = \sum_{x∈X}f(x) + \sum_{x∈X}g(x)$ 
Let $X$ be an arbitrary set (possibly uncountable), and let $f:X → R$ and $g: X → R$ be functions such that the series $\sum_{x∈X} f(x)$ and $\sum_{x∈X} g(x)$  are both absolutely convergent.
  Prove: The series $\sum_{x∈X}(f(x) + g(x))$ is absolutely convergent, and
  $$ \sum_{x∈X}(f(x) + g(x)) = \sum_{x∈X}f(x) + \sum_{x∈X}g(x)$$.

If the set $X$ is finite then I have the result. If it is countable, then $h: N \to X$ is a bijection and $\sum_{n=0}^{\infty} (f+g)(h(n))$ is absolutely convergent by definition. In other words, $\sum_{n=0}^{\infty} |f(h(n)+g(h(n))| = L$.
Since it is known that $\sum_{x∈X} f(x)$ and $\sum_{x∈X} g(x)$  are both absolutely convergent then $\sum_{x∈X} |f(x)| = M$ and $\sum_{x∈X} |g(x)|=K$.
I don't know how to proceed. It seems to me I found a solution for the uncountable case here Proving Proposition 8.2.6 from Terence Tao's Analysis I but still I got the problem with the countable one.(Frankly speaking I am not entirly getting the uncountable either).
 A: For any finite subset $F$ of $X$, we have
\begin{align*}
\sum_{x\in F}|f(x)+g(x)|\leq\sum_{x\in F}|f(x)|+\sum_{x\in F}|g(x)|\leq\sum_{x\in X}|f(x)|+\sum_{x\in X}|g(x)|<\infty,
\end{align*}
so 
\begin{align*}
\sup_{F\subseteq X, F~\text{finite}}\sum_{x\in F}|f(x)+g(x)|<\infty.
\end{align*}
This proves the absolute convergence of $\displaystyle\sum_{x\in X}(f(x)+g(x))$.
For any finite subset $F$ of $X$, we have
\begin{align*}
\sum_{x\in F}(f(x)+g(x))&=\sum_{x\in F}f(x)+\sum_{x\in F}g(x).
\end{align*}
Since 
\begin{align*}
\sum_{x\in F}(f(x)+g(x))&\rightarrow\sum_{x\in X}(f(x)+g(x))\\
\sum_{x\in F}f(x)&\rightarrow\sum_{x\in X}f(x)\\
\sum_{x\in F}g(x)&\rightarrow\sum_{x\in X}g(x)
\end{align*}
as nets, and addition in real numbers is continuous, we get the equality.
A: Absolute convergence 0f $\sum f(x)$ implies that $f(x)=0$ except on  a countable set $\{x_1,x_2,...\}$. There is a countable set $\{y_1,y_2,...\}$ such that $g(x)=0$ for  $ x \notin  \{y_1,y_2,...\}$. Let $\{a_1,a_2,...\}=\{x_1,x_2,...\} \cup \{y_1,y_2,...\}$. Then $\sum (f(x)+g(x)) =\sum_n [f(a_n)+g(a_n)]=\sum_n f(a_n) +\sum_n g(a_n)=\sum f(x)+\sum g(x).$ 
