Measure of the set-theoretic support of a smooth function

Question

I am trying to prove some functional inequalities. In this context, I stumbled upon the following interrogation.

Let $f \in C^\infty([0,1];\mathbb{R})$. Is is true that $$\mu \Big(\left\{ f \neq 0 \right\}\Big) = \mu \Big(\overline{\left\{ f \neq 0\right\}} \Big)$$ where $\mu$ denotes the Lebesgue measure?

I know that there are open sets $E$ such that $\mu(E) \neq \mu(\overline{E})$ but I was wondering if the added structure of the set-theoretic support for a smooth function prevented the usual counter-examples based on Cantor sets.

Answer 1

Nice question. So, every closed set is the zero set of a smooth function: see here. Therefore your counterexample with open $E$ will yield a counterexample also in your case, by taking a smooth $f$ that vanishes precisely on $\mathbb R\setminus E$.