A small part to a homework problem.
Let $\{r_n\}_1^\infty$ be an enumeration of the rationals, and let $f(x) = x^{-1/2}$ if $0<x<1$ and $f(x) = 0$ otherwise. Define $g(x) = \sum_{n=1}^\infty 2^{-n} f(x-r_n)$. Show that $g^2$ is not integrable on any interval.
I have shown that $g \in L^1$, which follows by using the comparison test on the series and noting that $f(x-r_n) \le 2$, noting that each term in the series is in $L^1$, and applying a few standard theorems about swapping sums and integral signs.
I have shown that $g < \infty$ a.e., which follows more or less directly from the definition.
Using $g^2 = g\cdot g$, I have shown that $g^2 < \infty$ a.e. So this rules out using that $g^2 = \infty$ on sets of non-zero Lebesgue measure. It would suffice to show that $g^2 \notin L^1$, but I have a limited arsenal of tools that state that a function is not $L^1$. I am loath to write the product of the series as a Cauchy sum. Is there some other way?
Following the comments in the question:
It is clear that every interval contains some $r_n$, and that the Cauchy sum contains a term like $4^{-1}\frac{1}{|x-r_n|}$. Therefore, on any interval $g^2 \ge \frac{4^{-n}}{|x-r_n|}$, and the right-hand side is not integrable.