Let $f :[0,1]\rightarrow \mathbb{R}$ satisfying $$\int_0^1|f(t)|dt<\infty .$$
We know that for given $\epsilon >0$ there exists a continuous function $g$ on $[0,1]$ such that $$\|f-g\|_{L^1([0,1])}<\epsilon .$$
Question: Can we find a continuous function $g$ satisfying above such that $Z(f)\subseteq Z(g)$?
($Z(f)$ is the zero set of $f$)
Edit: After comments I changed from $f\in L^1([0,1])$ to above.
Let $f=I_A$ where $A$ is the set of all irrational numbers on $[0,1]$. If such a continuous function $g$ exists for some $\epsilon <1$ then $g(x)=0$ for all rational $x$. But $g$ is continuous, so $g(x)=0$ for all $x$. Hence $1=\int f= \int |f-g| <\epsilon$, a contradiction.