# The sum of Infinite series

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.

• I've changed all $a_n$s to $a_k$s. If I made mistakes, please fix them. – Feng Shao Jul 21 at 5:50
• @FengShao Thanks – James Jul 21 at 5:59

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.
• Just out of curiosity, would there be any $x$ (not taking the almost everywhere into account) that makes the series diverge? Thanks. – James Jul 21 at 12:39
• @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. – Feng Shao Jul 21 at 12:55
• Oh yes, of course. I was wondering if the set $\limsup{A_k}$ was non-empty. – James Jul 21 at 13:56