$\Sigma_{\alpha \in A}x_{\alpha}\lt \infty$, show that $x_{\alpha}=0$ for all but at most countably many $\alpha \in A$

Question

If ($x_\alpha)_{\alpha \in A}$ is a collection of numbers $x_{\alpha}\in [0, \infty)$ such that $\Sigma_{\alpha \in A}x_{\alpha}\lt \infty$, show that $x_{\alpha}= 0$ for all but at most countably many $\alpha \in A$, even if $A$ itself is uncountable.

This is exercise 0.0.1 in Terrence Tao's "An Introduction to Measure Theory", haha.

Can somebody give me some insight here? Haven't been in the analysis mindset for awhile. I think proceeding by contradiction might be a good strategy.

If there is a sequence $x_{\alpha_i}$ that converges to some number $a\neq0$, then the sum would diverge. Therefore $x_\alpha\to0$. But then for every $n$ there can only be finitely many $x_\alpha$ outside $(-1/n,1/n)$. Taking the union of those for all $n\in\mathbb{N}$ gives countably many and the total of those that are different from zero. — logarithm, May 05 '19 at 17:26

score 5 · Accepted Answer · answered May 05 '19 at 17:29

5

The number of terms that are $\ge \frac1n$ is finite, and the set of all nonzero terms is the union of countably many such sets.

answered May 05 '19 at 17:29

hmakholm left over Monica

291,611

$\Sigma_{\alpha \in A}x_{\alpha}\lt \infty$, show that $x_{\alpha}=0$ for all but at most countably many $\alpha \in A$

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