Let $f$ vanishes on an open set containing 0. So there exists $l>0$ such that $f$ vanishes on $B(0,2l).$ So we can choose $g\in C_c^\infty (\mathbb{R}^n)$ (supported on $B(0,l)$) such that $f*g$ vanishes on an open set ( vanishes on $B(0,l)$).
My question: Is it remains true if we replace open set by set of positive Lebesgue measure i.e. if $f$ vanishes on an positive Lebesgue measure set around 0, then can we find a nonzero $g\in C_c^\infty (\mathbb{R}^n)$ such that $f*g$ vanishes on a set of positive Lebesgue measure?
Best Answer
As noted in Pietro Majer's comment, you can trivially take $g=0$ to get the "yes" answer.
To avoid this and make the question less trivial, one may additionally require that the function $g$ be nonnegative and nonzero. Then the answer becomes "no".
Indeed, e.g. let $$f:=1_{D},$$ where $D:=\mathbb R\setminus C$ and $C$ is a fat Cantor subset of the interval $[0,1]$, so that $C$ is a closed nowhere dense set of positive (Lebesgue) measure and hence $D$ is an everywhere dense open set.
So, $f=0$ on the set $C$ of measure $|C|>0$.
Now, take any nonnegative nonzero continuous function $g\colon\mathbb R\to\mathbb R$, so that for some real $c>0$ we have $g\ge c$ on a nonempty open interval $I$. Then, for any real $x$, $$(f*g)(x)=\int_{\mathbb R}f(x-y)g(y)\,dy \\ \ge c\int_I f(x-y)\,dy=c|D\cap(x-I)|>0,$$ because $D$ is an everywhere dense open set and hence $D\cap(x-I)$ is a nonempty open interval.
Thus, $f>0$ on $\mathbb R$.