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Let $(a_n)$ be the sequence of rational numbers, then I would like to know why the series $$\sum_{k=1}^{\infty}2^k{\chi_{[a_k-2^{-k-1},a_k+2^{-k-1}\ \ ]}}$$ converges $\mu$-almost everywhere on $\mathbb{R}$. $\mu$ is the Lebesgue measure.

So, this must mean that the set $$A = \{x \in \mathbb{R} \:{:}\: \sum_{k=1}^{\infty}2^k{\chi_{[a_k-2^{-k-1},a_k+2^{-k-1}\ \ ]}}(x) = \infty \}$$ has a zero Lebesgue measure. So I suppose that $A$ is either a collection of rational numbers or the empty set, but I do not know how I can see this. Any help will be greatly appreciated.

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  • $\begingroup$ I've changed all $a_n$s to $a_k$s. If I made mistakes, please fix them. $\endgroup$ – Feng Shao Jul 21 at 5:50
  • $\begingroup$ @FengShao Thanks $\endgroup$ – James Jul 21 at 5:59
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Let $A_k=[a_k-2^{-k-1},a_k+2^{-k-1}]$ for all $k\in \mathbb N$, then $\mu(A_k)=\frac1{2^k}$, and thus $\sum\mu(A_k)<\infty$. By Broel-Cantelli lemma, $\mu(\limsup A_k)=0$, which means $$\mu(\{x\in \mathbb R: x \text{ belongs to infinitely many }A_k\})=0.$$ So for almost every $x\in\mathbb R$, the sum $$\sum_{k=1}^{\infty}2^k{\chi_{[a_k-2^{-k-1},a_k+2^{-k-1}\ ]}}(x)$$ is a finite sum and then converges, which concludes the proof.

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  • $\begingroup$ Thank you so much for this explanation. I can understand this so much better now. $\endgroup$ – James Jul 21 at 6:45
  • $\begingroup$ @James Glad to help:) $\endgroup$ – Feng Shao Jul 21 at 6:48
  • $\begingroup$ Just out of curiosity, would there be any $x$ (not taking the almost everywhere into account) that makes the series diverge? Thanks. $\endgroup$ – James Jul 21 at 12:39
  • $\begingroup$ @James For any $x\in \limsup A_k$, the series $\sum2^k\chi_{A_k}(x)$ diverges since there are infinitely many positive terms which are no less than 2. $\endgroup$ – Feng Shao Jul 21 at 12:55
  • $\begingroup$ Oh yes, of course. I was wondering if the set $\limsup{A_k}$ was non-empty. $\endgroup$ – James Jul 21 at 13:56

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