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Let $F:\mathbb R^n\to\mathbb R$ is a non-constant continuous function. Is it true that $Leb[(x_1,...,x_n)\in\mathbb R^n:F(x_1,...,x_n)=0]=0?$ Here $Leb$ denotes Lebesgue measure.

I don't know if this is a well known result. I have heard something like graph of a continuous function has Lebesgue measure 0. Is this related to that? I don't even know how to prove this latter fact so if you can include a proof I would be delighted.

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  • $\begingroup$ It's true if $F$ is real analytic. Ekesh's example below shows that $C^\infty$ is not enough. $\endgroup$ – Nate Eldredge Jan 23 '19 at 17:23
  • $\begingroup$ Graph $\Gamma(f) = \{ (x, f(x)) \in \mathbb{R}^{n+1} : x \in \mathbb{R}^n \}$ of a measurable function $f : \mathbb{R}^n \to \mathbb{R}$ has measure zero in $\mathbb{R}^{n+1}$. But this has nothing to do with the measure of level-sets of $f$. $\endgroup$ – Sangchul Lee Jan 23 '19 at 18:51
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The statement is false.

A counterexample is $$f(x) = \begin{cases} 0 & \text{ if } x \leq 0 \\ e^{-1/x^{2}} & \text { if } x > 0. \end{cases} $$

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  • $\begingroup$ This example is even $C^\infty$. A simpler example could just be $f(x) = x$ when $x > 0$ (and still $0$ when $x \le 0$). $\endgroup$ – Nate Eldredge Jan 23 '19 at 17:21

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