TITLE: The sum of an uncountable number of positive numbers
QUESTION [174 upvotes]: Claim: If $(x_\alpha)_{\alpha\in A}$ is a collection of real numbers $x_\alpha\in [0,\infty]$
such that $\sum_{\alpha\in A}x_\alpha<\infty$, then $x_\alpha=0$ for all but at most countably many $\alpha\in A$ ($A$ need not be countable).
Proof: Let $\sum_{\alpha\in A}x_\alpha=M<\infty$. Consider $S_n=\{\alpha\in A \mid x_\alpha>1/n\}$.
Then $M\geq\sum_{\alpha\in S_n}x_\alpha>\sum_{\alpha\in S_n}1/n=\frac{N}{n}$, where $N\in\mathbb{N}\cup\{\infty\}$ is the number of elements in $S_n$.
Thus $S_n$ has at most $Mn$ elements. 
Hence $\{\alpha\in A \mid x_\alpha>0\}=\bigcup_{n\in\mathbb{N}}S_n$ is countable as the countable union of finite sets. $\square$
First, is my proof correct? Second, are there more concise/elegant proofs?

REPLY [9 votes]: A more explicit formulation of the statement in the title is this. Let $A$ be some uncountable family of non-negative numbers. If $S$ is to be the "sum of $A$", then by any reasonable definition of "sum", surely $S$ must be greater than or equal to the sum of any finite sub-family of $A$. In that sense we can prove that any for any reasonable definition of "sum", the sum of $A$ must be infinite (unless all but countably many elements of $A$ are zero) by showing:

For any $M$, there exists a finite sub-family $B$ of $A$ such that the sum of $B$ is at least $M$.

Proof: Assume that $A^+$, the positive members of $A$, is uncountable (otherwise the theorem is obviously false). $A^+=\bigcup_n A_n$, where $A_n=\{a\in A | a \geq \frac 1 n\}$. Since the union of countably many finite sets is countable, one of the $A_n$ must be infinite. Grab as many elements as you need from that set to get a sum greater than $M$.