Let $\{q_n\}$ be an enumeration of $\mathbb{Q}$, then put $f(x) = \sum_{n=1}^{\infty} e^{-n^2(x-q_n)^2}$, this is well defined as a function of $\mathbb{R} \to \mathbb{R} \cup \{\infty\}$. The idea is to prove that this is finite a.e and however is unbounded for every interval. I am stuck on this last part.
What I do: To prove that $f$ is finite a.e I integrate the function on $\mathbb{R}$, using monotone convergence theorem. Is easy to see that this yields a finite result and hence the integrating has to be finite a.e.
Now I am stuck on proving that this is locally unbounded, what I've tried is, given an interval, take a rational number on it, $p/q$, and then try to find a term in the series of $f(p/q)$ that is as bigger as I want.
Hint: given any interval $I_0$, find a nested sequence of intervals $I_0 \supset I_1 \supset I_2$ such that, for every $j\ge0$, there exists $n(j)$ such that $$ I_j \subset \bigg[ q_{n(j)}-\frac1{{n(j)}^2},q_{n(j)}+\frac1{{n(j)}^2} \bigg]. $$ The density of the rational numbers in $\Bbb R$ should help you construct such intervals. Do you see why we have $f(x) \ge je^{-1}$ for $x\in I_j$?