Unbounded on every interval except null set but finite a.e

This question originates from Folland’s problem 2.25. In this problem, first given $f(x)=x^{-1/2}$ when $0<x<1$, and $0$ otherwise.Then consider $g(x)=sum_{n}2^{-n}f(x-r_{n})$, where sequence $r_{n}$ is the all the rational number. Then it requires us to prove $g$ is integrable, discontinuous everywhere and unbounded on every interval. We need to show that discontinuity and unboundedness hold even removing a Lebesgue null set.

While I can show $g$ is integrable(thus finite a.e.),unboundedness on every interval. I have doubts about why $g$ is finite a.e. Consider any irrational number $x$, there always exist a sequence of $r_{n}$ such that $x-r_{n}<3^{-2n}$, so that $2^{-n}f(x-r_{n})>(3/2)^{n}$. This shows the unboundedness of $g$, however since this holds for any irrational number,this function should not be finite a.e. as $sum_{n}(3/2)^{n}=infty$ for every irrational number $x$, thus not finite a.e.

I know there must be something wrong in my argument, but I can’t spot it. Can anyone tell what is the problem?