Let $x^{-1/2}$ for $0<x<1$ and $f(x)=0$ otherwise. Let $(r_n)$ be an enumeration of $\mathbb{Q}$ and let $g(x)=\sum_{n=1}^{\infty}2^{-n}f(x-r_n)$. Show that $g\in L_1$ and in particular, $g$ is finite a.e.
With the help of Greg, I think I have an idea. If I have that $f(x)=0$ when I take the integral term by term of the sum, it is certainly finite. On the other hand at worst case scenario I will have $x$ close to $0$ being raised to a high power so this will give me a very small number but since it is in the denominator it will give me something approaching infinity.
I know that $f$ is integrable and $f^2$ is not integrable. So maybe I could say something along the lines of $g$ is integrable because it is the sum of integrable functions?