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Note: Here $\mathcal H^k$ denotes the $k$-dimensional Hausdorff measure.

Let $f \in C^1 (\mathbb \Omega)$ for some open, connected, bounded subset $\Omega$ of $\mathbb R^n$. We consider for each $t \in \mathbb R$ the level set $E_t := \{x \in \Omega \, | \, f(x) = t\}$.

Question: Is it true that for Lebesgue almost every $t \in \mathbb R$, we have $Df(x) \neq 0$ for $\mathcal H^{n-1}$-almost every $x \in E_t$?

Comments:

  1. It is interesting to contrast the desired result to the fact that we have $Df = 0$, $\mathcal H^n$-a.e. on every level set $E_t$. Note that this is not a contradiction!

  2. The desired result is true for $f \in C^n (\Omega)$ by Sard's theorem.

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  • $\begingroup$ For more smooth functions Sard's theorem says that for almost all $t$ we have $Df\ne 0$ on $E_t$ $\endgroup$
    – Fedor Petrov
    Commented May 20 at 9:45
  • $\begingroup$ @FedorPetrov Indeed, in a later edit I added that additional remark. $\endgroup$
    – Nate River
    Commented May 20 at 9:49
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    $\begingroup$ Isn't it true that if $t$ is a regular value, then $Df$ is never zero on $E_t$; and $E_t$ has a positive $\mathcal{H}^{n-1}$ measure because it is a $C^1$ submanifold of dimension $n-1$? $\endgroup$
    – KhashF
    Commented May 20 at 14:24
  • $\begingroup$ @KhashF The first statement is indeed true, and Sard’s theorem says that almost every $t$ is a regular point. However Sard’s only applies for smooth enough functions, $C^n$ in this case. The second statement is not true as stated, though something similar does hold. The keywords are $C^1$ Sard’s lemma. $\endgroup$
    – Nate River
    Commented May 20 at 22:10
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    $\begingroup$ @GiorgioMetafune Oh, I think I misread his second statement as saying something else. It is true. I guess he wanted to use it to conclude a contradiction by Sard’s theorem. $\endgroup$
    – Nate River
    Commented May 23 at 14:02

1 Answer 1

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It follows from co-area formula: $$ \int_A |Df|=\int_{\mathbb R} \mathcal H^{n-1} (\{f=t\}\cap A). $$ By taking $A=\{|Df|= 0\}$ one gets that $$ \int_{\mathbb R} \mathcal H^{n-1} (\{f=t\}\cap \{|Df|=0\})=0 $$ and thus ,for a.e. $t$, $$ \mathcal H^{n-1} (\{f=t\}\cap \{|Df|=0\})=0 $$

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  • $\begingroup$ Oh dang that is nifty. Thanks! $\endgroup$
    – Nate River
    Commented May 23 at 11:04

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