Transfinite series: Uncountable sums

Solution 1:

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.