For $E=[0,1]\setminus \mathbb{Q}, ~l>0$ and consider the function $$\chi_{[-l,l]}*\chi_E(x)=\int_{-l}^{l}\chi_E(x-t)dt.$$ What is the zero set of the above function (in terms of $l$ may be)?
Notations: $\mathbb{Q}$ is the set of rationals. $\chi_E$ is the characteristic function of $E.$
Edit after 1st answer: Let $f$ be any positive integrable function on $\mathbb{R}$ such that $\mathcal{Z}_f$ is the zero set of $f$. What is the zero set of $$\chi_{[-l,l]}*f(x)?$$
As Daniel Fischer points out in the comments, the answer is the same if you replace $E$ by the interval $[0,1]$. Now we can rewrite $\chi_{[0,1]}(x-t) = \chi_{[x-1,x]}(t)$, so the integral is $$ \int_{-\ell}^{\ell} \chi_{[x-1,x]}(t) \,dt = m([-\ell,\ell] \cap [x-1,x]) $$ where $m$ is Lebesgue measure. Therefore this function is equal to zero iff $x \leq -\ell$ or $x \geq \ell+1$.