Suppose $f:(0,1)^n\to \mathbb{R}$ is continuous. Does the boundary of the set of its roots have Lebesgue measure 0? I guess the answer is negative, in that case, are there any reasonable conditions on $f$, e.g. Lipschitz continuity or (continuous) differentiability, that make the answer positive?
Thanks a lot, I'd appreciate any input.
Every single closed subset of $(0,1)^n$ (including, say, a fat Cantor set which has empty interior and positive measure) is the zero set of some smooth function $f:(0,1)^n\to\mathbb{R}$ (see Every closed subset $E\subseteq \mathbb{R}^n$ is the zero point set of a smooth function). So even assuming $f$ is $C^\infty$ is not enough. If you just require $f$ to be continuous, the proof is much simpler: just define $f(x)$ to be the distance between $x$ and the closed set $A$ which you want to be the zero set.