Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of real numbers, and let $(b_n)_{n \in \mathbb{N}}$ be a sequence of positive numbers. Show that if $\sum_{n \in \mathbb{N}}\sqrt{b_n}<\infty$ then $\sum_{n \in \mathbb{N}}\frac{b_n}{|x-a_n|}<\infty$ m-a.e. $x$, where m denote Lebesgue measure on $\mathbb{R}$.
My attempt: Let $A_n=\{x \in \mathbb{R}:|x-a_n|\leq\sqrt{b_n}\}$. Then if $x \in A_n \Rightarrow a_n-\sqrt{b_n}\leq x\leq a_n+\sqrt{b_n}$
$\Rightarrow A_n \subset [a_n-\sqrt{b_n},a_n+\sqrt{b_n}]=B_n \Rightarrow m(A_n)\leq m(B_n)=2\sqrt{b_n}$
$\Rightarrow \sum_{n \in \mathbb{N}}m(A_n)\leq \sum_{n \in \mathbb{N}}2\sqrt{b_n}<\infty \Rightarrow m(\limsup A_n)=0$, by Borel Cantelli lemma
$\Rightarrow |x-a_n|>\sqrt{b_n}$ m-a.e. $x$
$\Rightarrow \sum_{n \in \mathbb{N}}\frac{b_n}{|x-a_n|} \leq \sum_{n \in \mathbb{N}}\sqrt{b_n}<\infty$
Can anyone check my work? I'm not very sure about last two steps in my proof. Many thanks!
The first part of your proof looks good.
For the second-to-last step, after you conclude $m(\limsup A_n) = 0$, you can be a little more precise. For each fixed $x\in [\limsup A_n]^\complement = \liminf A_n^\complement$, there exists $N(x)$ such that for all $n> N(x)$, $|x-a_n|>\sqrt b_n$ (you can verify this by checking the definitions). Therefore, for this fixed $x$, $\sum_{n\in\mathbb N}\frac{b_n}{|x-a_n|} \le C(x) + \sum_{n> N(x)}\sqrt{b_n} < \infty$. The constant $C(x)<\infty$ depends on the first $N(x)$ terms of the series $\sum b_n/|x-a_n|$. As $x\in\liminf A_n^\complement$ was arbitrary, and $m(\limsup A_n) = 0$, this finishes the proof.