Problem; Let $\{a_n\}$ and $\{r_n\}$ be two sequences of real numbers such that $\displaystyle\sum_{n\geq 1} |a_n|<\infty$. Prove that $$\sum_{n\geq 1} \frac{a_n}{\sqrt{|x-r_n|}}$$ converges absolutely for almost every $x\in \mathbb{R}$.
(From 'Real analysis for graduate students'(by Richard F.Bass) - Chapter 11 Exercise 14)
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This problem was in the chapter about product measures, so I tried to approach this problem with counting measure; Let $f, g:\mathbb{N}\to\mathbb{R}$ functions such that $f(n)=|a_n|$, $g(n)=|r_n|$. Let $\mu$, $m$ counting measure and Lebesgue measure respectively. And let $$S_x =\sum_{n\geq 1} \frac{|a_n|}{\sqrt{|x-r_n|}}$$ Then $$\begin{aligned} \int_\mathbb{R}S_x m(dx)&=\int_\mathbb{R} \int_\mathbb{N} \frac{f(n)}{\sqrt{|x-g(n)|}}m(dx)\mu(dn) \end{aligned}$$ My idea was; if we prove this integral is convergent, then it will imply $S_x<\infty$ for almost every $x\in \mathbb{R}$. But I'm stuck with proving the finiteness of this integral. I tried to apply Fubini's theorem... but I soon realized that we need the convergence of this integral BEFORE applying Fubini.
Could you let me know whether I am going to the right way, or give some hints for this problem, please? Thanks.
Where did you see that one needs to check the convergence of an integral before applying Fubini? One can always apply Fubini, only the conclusion is stronger when any of the iterated integrals converges.
Here, one considers a bounded interval $K$ of length $2\ell$, then, for every real number $r$, $$\int_K\frac{1}{\sqrt{|x-r|}}m(dx)\leqslant2\int_0^\ell\frac{1}{\sqrt{x}}m(dx)=4\sqrt{\ell}.$$ Using this for $r=r_n$ and summing over $n$, Tonelli yields $$ \int_KS_xm(dx)\leqslant4\sqrt{\ell}\sum_n|a_n|.$$ The RHS is finite hence $S_x$ is finite $m(dx)$-almost everywhere on $K$. Considering an increasing sequence $(K_i)$ with union $\mathbb R$ allows to conclude.