Proving Proposition 8.2.6 from Terence Tao's Analysis I I am currently studying Terence Tao's Analysis I and am currently stuck on trying to prove one of the propositions concerning absolutely convergent series over arbitrary sets, which he left as an exercise problem. The question is this:

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

Of course, there are more components to this proposition, but I can't solve the first one. I understand how to solve the problem in the case where $X$ is finite or countable; for reference, he defines the value of a series over an uncountable set as

We can define the value of $\sum_{x \in X} f(x)$ for any absolutely convergent series on an uncountable set $X$ by the formula $$\sum_{x \in X} f(x) = \sum_{x \in X: f(x) \ne 0} f(x),$$ since we have replaced a sum on an uncountable set $X$ by a sum on the countable set $\{x \in X: f(x) \ne 0\}$.

He defines absolute convergence as

Let $X$ be a set, and let $f: X \to \mathbb{R}$ be a function. We say that the series $$\sum_{x \in X} f(x)$$ is absolutely convergent iff $$\sup\bigg\{\sum_{x \in A} \lvert f(x) \rvert: A \subset X, A \text{ finite}\bigg\} < \infty.$$

I managed to prove the first part of the problem i.e. that the series $\sum_{x \in X} (f(x) + g(x))$ is absolutely convergent, as follows:
Let $\sup\{ \sum_{x \in A} f(x): A \subset X, A \text{ finite}\} = M$ and let $\sup\{ \sum_{x \in A} g(x): A \subset X, A \text{ finite}\} = N$. Since $\sum_{x \in X} f(x)$ and $\sum_{x \in X} g(x)$ are both absolutely convergent, we know that $M, N < \infty$. Thus for any finite subset $A \subset X$, we have $$\sum_{x \in A} \lvert f(x) + g(x) \rvert \leq \sum_{x \in A} \lvert f(x) \rvert + \sum_{x \in A} \lvert g(x) \rvert \leq M + N,$$ so $\sup\{\sum_{x \in A} \lvert f(x) + g(x) \rvert: A \subset X, A \text{ finite}\} \leq M + N$. In particular, $\sum_{x \in X} ( f(x) + g(x) )$ is absolutely convergent.
However, I'm not sure how to prove the second part of the claim; Tao indicates that it requires the axiom of choice when $X$ is uncountable, but I'm still unsure how to approach this problem. Any hints would be greatly appreciated.
 A: $A = \{x\in X: f(x)\neq 0\}, B = \{x\in X: g(x)\neq 0\}, C = \{x\in X: f(x)+g(x) \neq 0\}$ - countable sets.
$$\sum_{x\in X} (f(x)+g(x)) := \sum_{x\in C} (f(x)+g(x)) =\\\sum_{x\in C}(f(x)+g(x))+\sum_{x\in (A\cup B)\setminus C}(f(x)+g(x)) =  \sum_{x\in A\cup B}(f(x)+g(x)) $$ but also
$$\sum_{x\in X} f(x) := \sum_{x\in A} f(x) = \sum_{x\in A\cup B} f(x) $$ and similarly for $g$, so we reduced the problem to the countable case.
A: I don't think we need AOC for this one. He's probably referring to part(d).
Denote $A := \{x\in X: f(x)\neq 0\}, B := \{x\in X: g(x)\neq 0\}, C := \{x\in X: f(x)+g(x) \neq 0\}$. We know they are all at most countable. Here's my proof for the case when both $A,B$ are countably infinite.
By Tao's definition 8.2.5, for the absolutely convergent series on an uncountable set $X$: $$ \sum_{x \in X} f(x) = \sum_{x \in X: f(x) \neq 0}f(x) $$
It's easy to verify the equality also holds for countable and finite sets, and we will use this conclusion later.
We want: $$ \sum_{x \in C} (f(x) + g(x)) = \sum_{x \in A} f(x) + \sum_{x \in B} g(x) $$
Observe $C \subset A \cup B $, thus $$\sum_{x \in C} (f(x) + g(x)) = \sum_{x \in A \cup B} (f(x) + g(x)) = \sum_{n=0}^{\infty}f(s(n))+g(s(n))$$, where $s: N \to A \cup B$ is a bijection. Now we have to break it down to partial sum $S_N$.
Let $S_N$ be the partial sum and $S_N := \sum_{n=0}^{N} f(s(n)) + g(s(n)) $. Since it's finite, we can split it by Lemma 7.1.4(c).
$$ S_N = \sum_{n=0}^{N} f(s(n)) + \sum_{n=0}^{N} g(s(n))$$
Notice the two series are absolutely convergent:
$$\sum_{x \in A} f(x) = \sum_{x \in A \cup B} f(x) = \sum_{n=0}^{\infty} f(s(n)) = L_1$$
$$\sum_{x \in B} g(x) = \sum_{x \in A \cup B} g(x) = \sum_{n=0}^{\infty} g(s(n)) = L_2$$
where $L_1,L_2$ are some real numbers.
Thus, $\exists N_1 $ such that $\forall n\geq N_1, \sum_{n=0}^{N} f(s(n))$ is $\epsilon / 2$-close to L1. Similarly for the second term, $\exists N_2 $ such that $\forall n\geq N_1, \sum_{n=0}^{N} f(s(n))$ is $\epsilon / 2$-close to L2. Let $N_0 = max(N_1,N_2)$, then $\forall n \geq N_0, S_N$ is $\epsilon$-close to $L_1+L_2$. Which means $\sum_{x \in C} (f(x) + g(x))$ converges to $L_1+L_2$.
A: One must first prove Proposition $8.2.6(c)$ in the book which states that if $X = X_{1}\cup X_{2}$ for disjoint sets $X_{1}$ and $X_{2}$ then $\sum_{x\in X}f(x)$ is absolutely convergent if and only if $\sum_{x\in X_{1}}f(x)$ and $\sum_{x\in X_{2}}f(x)$ are both absolutely convergent and
$$\sum_{x\in X}f(x) = \sum_{x\in X_{1}}f(x)+\sum_{x\in X_{2}}f(x).$$
Let $$A = \{x\in X: f(x)\neq 0\},$$ $$B = \{x\in X: g(x)\neq 0\},$$ $$C = \{x\in X: f(x)+g(x)\neq 0\},$$ $$D = \{x\in X: f(x)\neq 0,\:g(x)\neq 0,\:f(x)+g(x) = 0\}.$$
We note that $A\cup B = C\cup D$, $C\cap D = \varnothing$ and all sets are at most countable. Therefore, using the result written above we have
$$\sum_{x\in A\cup B}(f(x)+g(x)) = \sum_{x\in C}(f(x)+g(x))+\underbrace{\sum_{x\in D}(f(x)+g(x))}_{ = 0}.$$
Letting $h$ be any bijection from $N$ to $A\cup B$ where $N$ is either $\mathbb{N}$ or $\{k\in \mathbb{N}: 1\leq k\leq n\}$ for some $n$ we have
$$\sum_{x\in A\cup B}(f(x)+g(x)) = \sum_{n\in N}(f(h(n))+g(h(n))) = \sum_{n\in N}f(h(n))+\sum_{n\in N}g(h(n)) = \sum_{x\in A\cup B}f(x)+\sum_{x\in A\cup B}g(x)$$
where the second equality is obtained using the property for finite or countable sums. Again using Proposition $8.2.6(c)$ (stated above) we get
$$\sum_{x\in A\cup B}f(x) = \sum_{x\in A}f(x)+\underbrace{\sum_{x\in A\cup B-A}f(x)}_{ = 0}$$
and similarly $\sum_{x\in A\cup B}g(x) = \sum_{x\in B}g(x)$. Therefore, we have proved that
$$\sum_{x\in C}(f(x)+g(x)) = \sum_{x\in A}f(x)+\sum_{x\in B}g(x)$$
which is the desired result.
