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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.

afshi7n
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1 Answers1

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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.

Eric Wofsey
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  • I see. This is very helpful. Given this, the weakest assumption that I know which makes the answer positive (measure=0) is that $f$ is analytic. Are there other reasonable assumptions that you may know? Thanks a lot! – afshi7n Jul 11 '18 at 21:32
  • Well, I guess you could assume that $f$ is a submersion (its derivative is nonzero everywhere), so its zero set will be an $(n-1)$-manifold and thus have measure $0$. (I don't know off the top of my head exactly how much smoothness you need for this to be valid; $C^\infty$ is definitely enough, $C^2$ is probably enough, and $C^1$ may be enough.) – Eric Wofsey Jul 11 '18 at 21:41
  • Great. Thank you for your help! – afshi7n Jul 12 '18 at 16:49