Uncountable Sum as Series

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Theorem

Let $X$ be an uncountable set.

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


The uncountable sum:

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

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


Corollary

Let $f: X \to \closedint 0 {+\infty}$ have uncountably infinite support.

Then:

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


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:

\(\displaystyle \displaystyle \sum_{x \mathop \in X} \map f x\) \(=\) \(\displaystyle \sup \set {\sum_{x \mathop \in F} \map f x: F \subseteq X, F \text{ finite} }\)
\(\displaystyle \) \(\ge\) \(\displaystyle \sup \set {\sum_{x \mathop \in F} \map f x: F \subseteq A_{n_0}, F \text{ finite} }\)
\(\displaystyle \) \(\ge\) \(\displaystyle \sup \set {\sum_{x \mathop \in F} \frac 1 {n_0} : F \subseteq A_{n_0}, F \text{ finite} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^{\infty} \frac 1 {n_0}\)
\(\displaystyle \) \(=\) \(\displaystyle +\infty\)

Thus:

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


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

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

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:

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

for each $F$ finite.

Taking the supremum of this inequality over $N$:

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

Taking the supremum of this inequality over $F$:

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

Thus:

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

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

$\blacksquare$


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