Since my question here seems to be too complicated (since even though I put a bounty on it nobody answered or commented) I want to break it down a little bit. Maybe it gets answerable by reducing it by a dimension and making already the choices.
Let $$f: \mathbb R\to \mathbb R\cup\{-\infty\}, \quad f(x):= \begin{cases}\log(|x|),\quad&|x|<1,\\ 0, &|x| \ge 1, \end{cases}, $$
so $f(x)=\min(0,\log(|x|))$. With $(q_n)_{n \in \mathbb N}$ we run through all rational numbers in $(0,1]$ in the following order: $$1,\tfrac{1}{2},\tfrac{1}{3},\tfrac{2}{3},\tfrac{1}{4},\tfrac{3}{4},\tfrac{1}{5},\tfrac{2}{5},\tfrac{3}{5},\dots$$
We consider the function $$g: \mathbb R \to \mathbb R\cup\{-\infty\}, \quad g(x):= \sum_{n = 1}^\infty \frac{f(x-q_n)}{2^n}.$$
My question: At which $x \in [0,1]$ do we have $g(x)=-\infty$? Only on the points in $\mathbb Q \cap (0,1]$?
Background: By Fubini–Tonelli we have $$\int_{\mathbb R} g(x) dx = \sum_{n \in \mathbb N} \frac{-2}{2^n} = -2.$$ So $|g(x)|< \infty$ almost everywhere. This is hard to imagine.