Uncountable Sum as Series

Theorem
Let $X$ be an uncountable set.

Let $f: X \to [0\,.\,.\,+\infty]$ be an extended real-valued function.

The uncountable sum:


 * $\displaystyle \sum_{x \in X} f(x) = \sup \left\{ {\sum_{x \in F}f(x) : F \subseteq X, F \text{ finite} }\right\}$

is $+\infty$, or can be expressed as a (possibly divergent) series.

Corollary
If $f: X \to [0\,.\,.\,+\infty]$ has uncountably infinite support, then:


 * $\displaystyle \sum_{x \in X} f(x) = +\infty$

Proof
Define:


 * $A_n = \left \{ { x \in X: f(x)>\dfrac 1 n, n \in \mathbb N_{\ge 1} }\right\}$

Then $\langle A_n \rangle_{n\in \mathbb N} \uparrow A$ is an exhausting sequence of sets, where:


 * $A = \left \{ { x \in X: f(x) > 0 } \right\}$

Suppose $A$ is uncountable. Then necessarily there is some $A_{n_0}$ uncountable, because otherwise $A$ would be countable union of countable sets.

Thus:


 * $\displaystyle \sum_{x \in X} f(x) = +\infty$

Otherwise, suppose then that $A$ is countably infinite.

Then there is a bijection $g: \mathbb N \leftrightarrow A$

Define $B_N = g\left[{\left\{ {1,2,\ldots,N-1,N }\right\}}\right]$

then every finite subset $F$ of $A$ is contained in some $B_N$

This implies the inequality:


 * $\displaystyle \sum_{x \in F} f(x) \le \sum_{n = 1}^N f\left({g(n)}\right) \le \sum_{x \in X} f(x)$

for each $F$ finite.

Taking the supremum of this inequality over $N$:


 * $\displaystyle \sum_{x \in F} f(x) \le \sum_{n = 1}^\infty f\left({g(n)}\right) \le \sum_{x \in X} f(x)$

Taking the supremum of this inequality over $F$:


 * $\displaystyle \sum_{x \in X} f(x) \le \sum_{n = 1}^\infty f\left({g(n)}\right) \le \sum_{x \in X} f(x)$

Thus:


 * $\displaystyle \sum_{x \in X} f(x) = \sum_{n = 1}^\infty f\left({g(n)}\right)$

for some bijection $g: \mathbb N \leftrightarrow A$