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The sum of an uncountable number of positive numbers

Consider $\sum_{\lambda \in \Lambda} a_{\lambda}$ . Here all $a_\lambda $ is non-negative. Then I want to prove that if $\sum_{\lambda \in \Lambda} a_{\lambda} < \infty $ then there exists at most countable set $ \Lambda_0 \subset \Lambda$ such that $\lambda \notin \Lambda_0 \Rightarrow a_{\lambda}=0.$

(This means that if the summation converges then there are only at most countable $a_i$'s such that $a_i \neq 0)$

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    I remember this [Transfinite series: Uncountable sums](http://math.stackexchange.com/questions/13781/transfinite-series-uncountable-sums). The question appears very frequently.2012-07-30

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HINT

Given $\varepsilon>0$, can you measure the set $\{\lambda:a_\lambda>\varepsilon\}$?


Giving some background to the somewhat short hint. Suppose $E\subset\mathbb{R}$ is uncountable, and let $\lambda\mapsto a_\lambda$ be a non-negative function defined on $E$. The usual definition of the expression $\sum_{\lambda\in E}a_\lambda$ is given by $\sum_{\lambda\in E}a_\lambda =\sup_{F\subset E,\,|F|<\infty}\sum_{\lambda\in F}a_\lambda \tag{1}$ i.e. we take supremum over finite sets. Now, let us choose $\varepsilon>0$ and consider the set $E_\varepsilon = \{\lambda\in E :a_\lambda>\varepsilon\}.$ 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 $\sum_{\lambda\in E}a_\lambda\ge \sum_{\lambda\in F_n}a_\lambda>\sum_{F_n}\varepsilon =n\varepsilon.$ Since $\bigcup_{\varepsilon>0} E_\varepsilon =\bigcup_{n=1}^\infty E_{1/n} =\{\lambda\in E:a_\lambda>0\}$ the conclusion follows. Remark: If we exclude the assumption $a_\lambda\ge0$. I do not think there is a reasonable definition for conditional convergence of this kind.

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    @J.M. First now I noted your question. Thanks for providing the hiding feature to the hint. Better late than never.2014-08-22
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Suppose the series is convergent.

There cannot be more than finitely many elements larger than $\frac12$, as that would contradict convergence; and there cannot be more than finitely many elements larger than $\frac14$, for the same reason...

In particular for every $n\in\mathbb N$ there are only finitely many $a_\lambda$ such that $\frac1{2^n}.

Now we have a countable union of finite sets of nonzero elements, so this is $\Lambda_0$, the rest has to be zero since all the elements are non-negative.