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?



Problem regarding vanishing set of convolution
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](https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set) 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 $\mathbb R$.


