Uncountable Sum as Series

Theorem
Let $X$ be an uncountable set.

Let $f: X \to \closedint 0 {+\infty}$ be an extended real-valued function.

The uncountable sum:


 * $\ds \sum_{x \mathop \in X} \map f x = \sup \set {\sum_{x \mathop \in F} \map f x : F \subseteq X, F \text{ finite} }$

is:
 * if $f$ has uncountably infinite support, then $+\infty$
 * Otherwise, can be expressed as a (possibly divergent) series.

Proof
Define:


 * $A_n = \set {x \in X: \map f x >\dfrac 1 n, n \in \N_{\ge 1} }$

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


 * $A = \set {x \in X: \map f x > 0}$

Suppose $A$ is uncountable.

From Countable Union of Countable Sets is Countable, necessarily there is some $A_{n_0}$ which is uncountable.

Then:

Thus:


 * $\ds \sum_{x \mathop \in X} \map f x = +\infty$

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

Then there is a bijection $g: \N \to A$

Define $B_N = g \sqbrk {\set {1, 2, \ldots, N - 1, N} }$

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

This implies the inequality:


 * $\ds \sum_{x \mathop \in F} \map f x \le \sum_{n \mathop = 1}^N \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$

for each $F$ finite.

Taking the supremum of this inequality over $N$:


 * $\ds \sum_{x \mathop \in F} \map f x \le \sum_{n \mathop = 1}^\infty \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$

Taking the supremum of this inequality over $F$:


 * $\ds \sum_{x \mathop \in X} \map f x \le \sum_{n \mathop = 1}^\infty \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$

Thus:


 * $\ds \sum_{x \mathop \in X} \map f x = \sum_{n \mathop = 1}^\infty \map f {\map g n}$

for some bijection $g: \N \to A$