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I'm reading Morse-Sard lemma on Francesco Maggi's book, that is the following

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Question: Does $|u(E)|=0$ imply that also the measure of the boundary of the set $\{u>t\}$ is zero? In particular, does $|u(E)|=0$ imply $|\partial \{u>t\} |=0$?

I would like to have $|\partial \{u>t\} |=0$ and my professor told me as a hint to use Morse-Sard lemma. However, although I think I understand the statement, I can' t see this implication easily. As I realize it, $|u(E)|=0$ yields that the image of the level sets $\{u=t\}$ has zero measure. Moreover, if $\{u=t\}=\partial \{u>t\}$ (I am not quite sure if this is true), then it follows that the image of the boundary of the sets $\{u>t\}$ has zero measure. But this in turn implies (cf. Boundary of the image is the image of the boundary that the boundary of the image of the sets $\{u>t\}$ has zero measure. And that is not exactly what I needed. What am I missing here?

Arctic Char
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kaithkolesidou
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  • What do you mean by "the image of the level sets ${u=t}$? Also, is $t$ a regular value? – Arctic Char Oct 19 '21 at 07:51
  • @ArcticChar $u(E)$ should be the image of the set of all critical points of $u$. I assumed that this is equivalent to saying the image of the level sets ${u=t}$ but maybe I'm wrong... What do you mean by regular here? – kaithkolesidou Oct 19 '21 at 07:54
  • You claim that "the image of the level sets ${u=t}$ has zero measure". I am asking what is the definition of "the image of the level sets ${u=t}$".... by regular I mean ${u=t} \cap E = \emptyset$. – Arctic Char Oct 19 '21 at 07:57
  • @ArcticChar Oh yes, it is a regular point. Well, this definition is a bit vague to me... I just try to "read" the lemma in terms of level sets so I can pass to boundaries somehow, but there is a high chance I make logical errors... – kaithkolesidou Oct 19 '21 at 08:00

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Your question has not much to do with Morse Sard theorem.

It is assumed that $t$ is a regular value. That is, $\nabla u (x) \neq 0$ for all $x\in \{u=t\}$. Then the regular value theorem implies that $\{u=t\}$ is a smooth submanifold in $\mathbb R^n$ with dimension $n-1$. Since it is of one less dimension, $|\{u = t\}| = 0$.

Now it suffices to show that $\partial \{ u >t\} = \{ u =t\}$. It is clear that the closure of $\{ u>t\}$ is inside $\{ u\ge t\}$. On the other hand, if $y\in \{u=t\}$, since $\nabla u (y) \neq 0$, the function

$$ g(s)= u(y+ s\nabla u(y))$$

has $'g(0) = |\nabla u (y)|^2 >0$ at $s=0$ and $g(0) = t$. Thus $g(s) >t$ for all $s>0$ and thus $y$ is in the closure of $\{ u> t\}$.

Thus $\overline{\{ u>t\}} = \{ u\ge t\}$. Argue similarly using $g$ as above, the interior of $\{ u\ge t\}$ is $\{u>t\}$. Thus

$$\partial \{u > t\} = \{ u =t\}.$$

Remark: This is all under the assumption that $t$ is a regular value. If not then $\{u>t\}$ might have boundary with positive measure. For example, let $A$ be a fat Cantor set (which has positive measure). Then there is a smooth function $u: \mathbb R \to [0,\infty)$ so that $u^{-1}(0) = A$. In this case, $\partial \{ u>0\} = A$ has positive measure.

Arctic Char
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  • Thank you very much for the answer! It helps quite a lot. I have a question though. The fact that $t$ is regular doesn't follow from sard's lemma? Starting with a smooth function $u$, isn't this what the lemma implies? Moreover, do you have any book reference for the regular value theorem? – kaithkolesidou Oct 19 '21 at 08:41
  • The sard theorem says that $t$ is a regular value for a.e. $t\in \mathbb R$, but not for all of $t\in \mathbb R$. For the regular value theorem in this setting, I guess you can find it in Spivak's Calculus of manifold, or Rudin's Principle of mathematical analysis. @kaithkolesidou – Arctic Char Oct 19 '21 at 09:34
  • Great! Many thanks again! – kaithkolesidou Oct 19 '21 at 09:39