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Let $\{a_n\}$ and $\{r_n\}$ be two sequences of real numbers such that $\sum_{n=1}^\infty |a_n|< \infty.$ Prove that

$$ \displaystyle \sum_{n=1}^{\infty} \frac{a_n}{\sqrt{|x-r_n|}} $$ converges absolutely for almost every $x \in \mathbb{R}$.

Can anyone provide a useful hint to solve the problem ? I am unable to figure out how does almost every $x$ come into picture. Should I use some lebesgue integral ?

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    $\begingroup$ It seems like some property of $r_n$ should be specified. For example, what happens if $r_n \geq x$ for one or more terms $n$? $\endgroup$
    – Michael
    Commented Dec 13, 2015 at 2:07
  • $\begingroup$ @Michael: Yeah, I forgot to add the modulus. Edited it. $\endgroup$
    – pikachuchameleon
    Commented Dec 13, 2015 at 2:09
  • $\begingroup$ That does not seem to help. For example, what if $r_4=x$? $\endgroup$
    – Michael
    Commented Dec 13, 2015 at 2:09
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    $\begingroup$ @Michael that may be fine, this would automatically exclude only countably many points, the question is about almost every, I assume for all but measure zero set $\endgroup$
    – Mirko
    Commented Dec 13, 2015 at 2:11
  • $\begingroup$ @Michael: As Mirko has pointed out, you can as well remove all $\{r_n\}$ from your desired set of points for which the above property holds, since measure won't change. $\endgroup$
    – pikachuchameleon
    Commented Dec 13, 2015 at 2:13

1 Answer 1

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For any bounded measurable $A$ $$\int_A\sum_{n=1}^{\infty}\frac{|a_n|}{\sqrt{|x-r_n|}}=\sum_{n\ge 1}\int_A\frac {|a_n|}{\sqrt{|x-r_n|}}<\infty$$ Hence the sum is finite a.s. This can be extended to show a.s. absolute convergence of $$\sum_{n=1}^{\infty}a_nf_n(x)$$ for any locally uniformly integrable $f_n$.

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  • $\begingroup$ What does $E$ denote ? $\endgroup$
    – Mirko
    Commented Dec 13, 2015 at 4:23
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    $\begingroup$ I don't quite get it, can't connect the dots. I vaguely believe you need to take limit of the above inequalities as $A$ gets bigger and bigger (so $\sup A\to \mathbb R$) and then the sum may go to $\infty$. Perhaps what you have is right, but is too cryptic for me, or I overlook something related to expectations (or don't see the correct interpretation of what you are doing in terms of suitable probabilities). $\endgroup$
    – Mirko
    Commented Dec 13, 2015 at 4:30
  • $\begingroup$ @mirko I've shown that $f<\infty$ a.s. on all bounded $A_l$. Take $A_l=[l,l+1]$ to cover all of $R$. $\endgroup$
    – A.S.
    Commented Dec 13, 2015 at 4:40
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    $\begingroup$ I start to get the idea that perhaps it boils down to using that improper integral $\int_0^1\frac1{\sqrt{x}} dx$ is convergent. I assume your $f_n=\frac{|a_n|}{\sqrt{|x-r_n|}}$. I need to think about the "uniformly" or "locally uniformly" part (first I do not know the definition and need to figure it and make sense of it, may take a while, second once I know what it says I need to convince myself that it works, and that it would work even if the $r_n$ form a dense set). $\endgroup$
    – Mirko
    Commented Dec 13, 2015 at 4:52
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    $\begingroup$ The nub of the answer by A.S. is :For bounded measurable $ A$ let $f_n(x)=\sum_{j\leq n}|a_n|/\sqrt {|x-r_n|}$ for $x\in A\setminus (\{0\}\cup \{r_n\}_n). $ We have $0\leq f_n\leq f_{n+1}$ a.e. on $A$, but $\sup_n\int_Af_n(x)dx<\infty.$ $\endgroup$
    – DanielWainfleet
    Commented Dec 13, 2015 at 10:01

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