The problem goes as follows:
Let $r_{n}$ be an enumeration of the rationals. Using Fubini-Tonelli, show that:
$$ F(x) = \sum_{n \geq 1} \frac{1}{n^{2}} \frac{1}{|r_{n} - x |^{1/2}}$$
is finite almost everywhere.
Attempt at solution:
My idea is that I want to show the $F(x)$ is integrable. Then using Fubini-Tonelli, taking the sum outside of the integral:
$$ \int_{\mathbb{R}}F(x) dx = \sum_{n \geq 1} \frac{1}{n^{2}} \int_{\mathbb{R}}\frac{1}{|r_{n} - x |^{1/2}} dx$$ $$ = \sum_{n \geq 1} \frac{1}{n^{2}} \int_{\mathbb{R}}\frac{1}{|x |^{1/2}}dx $$by traslation invariance.
Now I know that $\frac{1}{|x |^{1/2}}$ locally integrable, but not integrable on all of $\mathbb{R}$.
I was also thinking of defining $f_{n}(x) = \frac{1}{|x |^{1/2}}$ when $ x$ in $(-n,n)$ and $0$ elsewhere. Then let $$F_{N}(x) = \sum_{n=1}^{N} \frac{1}{n^{2}} f(x-r_{n}) $$
My reasoning starts to fall apart here and I can't seem to make it work.
Any hints or solutions are greatly appreciated!
Set $f(x)=\frac{1}{|x|^{1/2}}$. Then $$F(x)=\sum_n \frac{1}{n^2} f(x-q_n).$$ You can decompose $F$ in two parts: one summing the tails of $f$, and one summing the spikes. Namely set $g(x)=f(x)\chi_{[-1,1]}$ and $h(x)=f(x)-g(x)=f(x)\chi_{[-1,1]^c}$. Then $$F(x)=\sum_n \frac{1}{n^2}g(x-q_n)+\sum_n \frac{1}{n^2}h(x-q_n)$$ and to the first sum you can apply your argument, while the second sum is uniformly bounded because $\|h\|_\infty=1$.