Differentiability of absolute function Let $U\subset\mathbb{R}^n$ be open set and $u\in C^1(U)$. Is it true that the set of all points in $U$ such that $|u|$ is not differentiable is a set of measure zero? Or maybe, is there a counterexample?
 A: Thanks to Dominique who pointed out a serious flaw in my previous argument.
Let $f(x) = |u(x)|$.
The map $x \mapsto |x|$ differentiable except at $x=0$.
Let $Z= u^{-1} ( \{0 \} )$, and $C = (Du)^{-1} ( \{0 \} )$.
If $x \in Z \cap C$, then $f$ is differentiable at $x$.
If $x \in Z \setminus C$, the implicit function shows that there
is a neighbourhood $U$ of $x$ such that $U \cap Z$ has measure zero.
Since $\mathbb{R}^n$ is second countable, it follows that $Z \setminus C$ has
measure zero.
A: The absolute value function $| \cdot | : \mathbb{R} \to \mathbb{R}$ is differentiable everywhere except at $0$. By the chain rule, if $x \in U$ is a point such that $u(x) \neq 0$, then $|u|$ is differentiable at $x$. 
Now suppose that $x$ is such that $u(x)=0$. Then $|u|$ may or may not be differentiable at $x$. However, if $du (x) =0$ (where $du$ is the Fréchet derivative), then $|u|$ is differentiable at $x$ and its derivative is also $0$. Indeed,
\begin{align}
\frac{\big| \; |u(x+h)|-|u(x)| \; \big|}{||h ||} 
= \frac{|u(x+h)|}{|| h||} = \frac{|u(x+h)-u(x)|}{||h||} \longrightarrow 0 \qquad \text{as} \quad h \longrightarrow 0
\end{align}
This implies that the set of points where $|u|$ is not differentiable is included in the set of points where $u(x)=0$ and $du(x) \neq 0$. Denoting the latter set by $E$, it suffices to show that $E$ has measure $0$. Let $R$ be the set of regular points of $u$, that is the set of points where $du \neq 0$. We have $E= u^{-1}(0) \cap R = (u|_{R}) ^{-1} (0)$.
Now, $R$ is an open set and the restriction $u|_{R} : R \to \mathbb{R}$ is $C^1$. Since $0$ is a regular value of $u |_{R}$, the preimage $(u|_R)^{-1}(0)$ is a smooth submanifold of codimension $1$ of $\mathbb{R}^n$ by the regular value theorem. Since smooth submanifolds of $\mathbb{R}^n$ of positive codimension have Lebesgue measure $0$, we conclude that $E$ has measure $0$.
