A question about an uncountable summation. 
Possible Duplicate:
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 $a_i \neq 0)$ 
 A: 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.
A: 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.

