Erdős problem 1193

Let $A\subset \mathbb{N}$ and let $g(n)$ be a non-decreasing function of $n$ which is always $>0$.

Is the lower density of $$\{ n : 1_A\ast 1_A(n)=g(n)\}$$ always $0$?

The answer is trivially no to both questions: indeed if $A=\mathbb{N}$ (assuming $0\in\mathbb{N}$) then $1_A\ast 1_A(n)=n+1$ for all $n$. Presumably Erdős had some additional restrictions on either $g$ or $A$ in mind, but these are not recorded in [Er80].