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Say $A$ is an uncountable set, and $\{x_\alpha\}_{\alpha \in A}$ is a set of nonnegative numbers indexed by $A$. Suppose $\sum_\alpha x_\alpha < \infty$, and let $I_n = \{\alpha \in A \mid x_\alpha > 1/n\}$. It is clear that $I_n$ is finite for every $n$. Moreover, $I = \cup_{n=1}^\infty I_n = \{\alpha \in A \mid x_\alpha > 0\}$, and thus, $A \setminus I = \{\alpha \in A \mid x_\alpha = 0\}$. Now $$ \lim_{n\to\infty} \sum_{\alpha \in A \setminus I_n} x_\alpha $$ ought to be equal to $$ \sum_{\alpha \in A\setminus I} x_\alpha = 0, $$ but I'm not sure how to show this. I thought of using the defining property of the $I_n$'s, but all I get is that for a fixed $n$, with $K \subset A \setminus I_n$ being a finite set, $$ \sum_{\alpha \in K} x_\alpha \leq \sum_{\alpha \in K} \frac{1}{n} = \frac{|K|}{n}, $$ and the right side blows up when taking the supremum over all finite subsets $K \subset I_n$ on both sides. However, if I let $n \to \infty$ before taking the supremum, I get what I want. I couldn't show that these operations commute either.

Any tips?

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If $x_\alpha$ are positive for each $\alpha \in A$, and uncountable set, then you must have $\sum_\alpha x_\alpha = \infty$.

To see this, note that at least one of the sets $A_n=\{ k | \alpha_k \ge {1 \over n} \}$ must be uncountable (since $A= \cup_n A_n$).

Addendum:

Here is a sledgehammer approach using the dominated convergence theorem: For a subset $B \subset A$ define the measure $\mu B = | \{ \alpha \in B | x_ \alpha > 0 \} | $. We are given that $\int x_\alpha d \mu(\alpha) < \infty$. Let $f_n(\alpha) = 1_{A \setminus I_n} (\alpha) x_\alpha \le x_\alpha$ and note that $f_n(\alpha) \to 0$ for all $\alpha$. Hence $\int f_n(\alpha) d \mu(\alpha) \to 0$.

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  • $\begingroup$ My mistake. I fixed it now. $\endgroup$ – Augusto Pereira Jun 5 '17 at 4:53
  • $\begingroup$ Yeah, but the issue is, @AugustoPereira, if you have uncountably many terms that are positive, you just get divergence. So you can only have at most countably many positive terms in your uncountable summation, which only differs notationally from a standard sum. $\endgroup$ – Chris Jun 5 '17 at 4:55
  • $\begingroup$ This means for every $n$, we have a set of indices $\{\alpha_i^n\}_{i=1}^\infty$ such that $x_\alpha = 0$ when $\alpha \neq \alpha_i^n$ for every $i = 1, 2, \dots$. Thus I need to show that $$ \lim_{n\to\infty} \sum_{i=1}^\infty x_{\alpha_i^n} = 0, $$ which is not at all clear to me. $\endgroup$ – Augusto Pereira Jun 5 '17 at 5:07
  • $\begingroup$ @AugustoPereira: I added a brute force answer. $\endgroup$ – copper.hat Jun 5 '17 at 15:14
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First note that $$\sum_{\alpha \in A} x_\alpha = \sum_{\alpha \in A \setminus I_n} x_\alpha + \sum_{\alpha \in I_n} x_\alpha$$

Since $I_n \subseteq I_{n+1} \to A$ and $x_\alpha \geq 0$ we can apply Lebesgue's Monotone Convergence Theorem on the last term to get $$\lim_{n\to\infty} \sum_{\alpha \in I_n} x_\alpha = \sum_{\alpha \in A} x_\alpha$$

Thus, $$\lim_{n\to\infty} \sum_{\alpha \in A \setminus I_n} x_\alpha = 0.$$

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  • $\begingroup$ What measure are you using? $\endgroup$ – Augusto Pereira Jun 5 '17 at 14:07
  • $\begingroup$ A sum is an integral with the counting measure. $\endgroup$ – md2perpe Jun 5 '17 at 14:15
  • $\begingroup$ Ok, but can you be a little more explicit, as in detailing which function you're integrating over which measure space? I'm not very familiar with measure theory arguments. $\endgroup$ – Augusto Pereira Jun 5 '17 at 14:41
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    $\begingroup$ Refering to the Wikipedia article, $X=A$, $\Sigma=\wp(A)$ (the powerset of $A$), and for $S\in\Sigma$ we have $\mu(S)=|S|$ (the number of elements in $S$). This is how a summation over a set is handled as an integral. Then I take $f(\alpha) = x_\alpha$ for $\alpha\in A$ and $f_k(\alpha) = x_\alpha$ if $\alpha \in I_k$ and $f_k(\alpha) = 0$ otherwise. $\endgroup$ – md2perpe Jun 5 '17 at 14:54
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I have found an elementary approach discussing the problem with a friend. Notice that $$ S := \sum_{\alpha \in A} x_\alpha = \sum_{\alpha \in A \setminus I} x_\alpha + \sum_{\alpha \in I} x_\alpha = \sum_{\alpha \in I} x_\alpha, $$ since $x_\alpha = 0$ for all $\alpha \in A \setminus I$. It suffices to show that $$ \lim_{n \to \infty} \sum_{\alpha \in I_n} x_\alpha = S, $$ since $$ \sum_{\alpha \in A \setminus I_n} x_\alpha = \sum_{\alpha \in A} x_\alpha - \sum_{\alpha \in I_n} x_\alpha, $$ and thus letting $n \to \infty$, noting that the limit on the left exists because the entire series is convergent, we find that $$ \lim_{n\to\infty} \sum_{\alpha \in A \setminus I_n} x_\alpha = 0. $$

So, let $\varepsilon > 0$. Then there exists a finite subset $K \subset I$ such that $$ 0 < S - \sum_{\alpha \in K} x_\alpha < \varepsilon, $$ and thus there exists $N \in \mathbb{N}$ such that $K \subset I_M$, since $K$ is finite and $I = \cup_n I_n$. Moreover, since $I_1 \subset I_2 \subset \cdots$, we in fact have that $K \subset I_n$ for all $n \geq N$. Therefore, $$ \sum_{\alpha \in K} x_\alpha \leq \sum_{\alpha \in I_n} x_\alpha \quad \forall n \geq N, $$ and finally, $$ 0 < S - \sum_{\alpha \in I_n} x_\alpha \leq S - \sum_{\alpha \in K} x_\alpha < \varepsilon \quad \forall n \geq N. $$

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