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