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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 $a_i \neq 0)$


marked as duplicate by t.b., Asaf Karagila, Did, Michael Greinecker, J. M. is a poor mathematician Jul 30 '12 at 10:33

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    $\begingroup$ See also here and here. $\endgroup$ – t.b. Jul 30 '12 at 6:30
  • $\begingroup$ @t.b.: I had a hunch this was a duplicate, but I just woke up. Answering was the simpler solution than searching for a duplicate... :-P $\endgroup$ – Asaf Karagila Jul 30 '12 at 6:32
  • $\begingroup$ I remember this Transfinite series: Uncountable sums. The question appears very frequently. $\endgroup$ – AD. Jul 30 '12 at 6:38


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.

  • $\begingroup$ How do you like the hidden text? $\endgroup$ – J. M. is a poor mathematician Jul 30 '12 at 10:44
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    $\begingroup$ @J.M. First now I noted your question. Thanks for providing the hiding feature to the hint. Better late than never. $\endgroup$ – AD. Aug 22 '14 at 5:57

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 $\frac1{2^n}<a_\lambda$.

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.


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