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.
Best Answer
$A = \{x\in X: f(x)\neq 0\}, B = \{x\in X: g(x)\neq 0\}, C = \{x\in X: f(x)+g(x) \neq 0\}$ - 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.