# Transfinite series: Uncountable sums

If you sum an expression over an uncountable set $\sum_{x\in \mathbb{R}}f(x)$, then do we need $f(x)=0$ on all but a countable subset in order for the sum to have a finite value?

If not can you give an example of a function everywhere nonzero that has a transfinite sum with a finite value?

Possible keywords: Transseries, Écalle–Borel Summation, analyzable function

Transseries for beginners, GA Edgar, 2009

• I realise odds are you'll never see this, but by any chance do you happen to be poster from this thread? Be seeing you... Jul 28, 2016 at 4:33

HINT

Given $\varepsilon>0$, can you measure the set $\{x:f(x)>\varepsilon\}$?

Edit: Giving some background to the somewhat short hint.

Suppose $E\subset\mathbb{R}$ is uncountable, and let $f$ be a non-negative function defined on $E$. The usual definition of the expression $\sum_{x\in E}f(x)$ is given by $$\sum_{x\in E}f(x) =\sup_{F\subset E,\,|F|<\infty}\sum_{x\in F}f(x) \tag{1}$$ i.e. we take supremum over finite sets.

Now, let us choose $\varepsilon>0$ and consider the set $$E_\varepsilon = \{x\in E :f(x)>\varepsilon\}.$$ This set must be finite in order for the sum in (1) to be finite, because otherwise we may for each positive integer $n$ choose subsets $F_n\subset E_\varepsilon$ such that $|F_n|=n$ and then $$\sum_{x\in E}f(x)\ge \sum_{x\in F_n}f(x)>\sum_{F_n}\varepsilon =n\varepsilon.$$

Since $$\bigcup_{\varepsilon>0} E_\varepsilon =\bigcup_{n=1}^\infty E_{1/n} =\{x\in E:f(x)>0\}$$ the conclusion follows.

Note: I added the assumption $f\ge0$. In fact I do not think there is a reasonable definition for conditional convergence of this kind.

• Thanks for pointing out my stupidity! The working day must have been too long. Dec 10, 2010 at 10:09
• @Alex Bartel: I have been there too :) Dec 10, 2010 at 10:11
• For the rest of us, can somebody expand on the answer? Apr 1, 2012 at 19:47
• @AaronMcDaid I agree to your point, the hint was perhaps a bit short! I hope you like my update better... Apr 2, 2012 at 8:21
• @AaronMcDaid I am glad to hear that. Apr 2, 2012 at 13:39