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Suppose $f:\mathbb R^n\to\mathbb R$ is a smooth function and define $E_f := \{x\in\mathbb R^n\;|\;f(x)=0\text{ and }\nabla f(x)\ne 0\}$. Can we find $f$ such that $E_f$ has positive $n$-dimensional Lebesgue measure?

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    $\begingroup$ Use the implicit function theorem to show that locally, $E_f$ has measure zero. $\endgroup$
    – copper.hat
    Aug 28, 2018 at 23:24
  • $\begingroup$ Nice, thanks a lot! $\endgroup$
    – Mohan Swaminathan
    Aug 28, 2018 at 23:38
  • $\begingroup$ @copper.hat Does your comment(answer) applies also for non-analytic smooth functions like flat functions? $\endgroup$
    – Joako
    Jul 23 at 22:03
  • $\begingroup$ @Joako It applies to any $C^1$ function. Roughly it says that locally $E_f$ is $n-1$ dimensional. The definition of $E_f$ rules out flat spots (at least open sets on which $f$ is constant). $\endgroup$
    – copper.hat
    Jul 23 at 22:06
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    $\begingroup$ I am not sure what your point is. The definition of $E_f$ requires that the point is not critical, but it could easily be 'flat' along some direction (for example, if $f$ is affine) . In your example, $E_f = \emptyset$. $\endgroup$
    – copper.hat
    Jul 23 at 22:36

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