In the book [1], it seems to me that the proof of Theorem 8.12 subtly uses the following result.
My question: Is the following result indeed true? And, is my proof of this result correct? (If anybody is interested in supplying an easier proof, that would also be interesting).
Proposition: Let $I \subset \mathbb{R}$ be an open interval, $1 \leq p \leq \infty$ and $u \in W^{1,p}(I)$. Let $S = \{ x \in I : u(x) \neq 0 \}$ (here we have fixed a choice of representative for $u$). Then $u' = u' \chi_S$ a.e. on $I$.
Proof: We first consider the $1 \leq p < \infty$ case. Let $G \in C^1(\mathbb{R})$ such that $|G(x)| \leq |x|$ for all $x \in \mathbb{R}$, and
$G(x) = \begin{cases} 0 & \quad \textrm{if } |x| \leq 1, \\ x & \quad \textrm{if } |x| > 2. \end{cases}$
Set $G_n(x) = (1/n) G(nx)$ and $u_n(x) = G_n(u(x))$ for each $n \geq 1$.
By Corollary 8.11 of [1] we have that $u_n \in W^{1,p}(I)$ with $u_n'(x) = G_n'(u(x)) u'(x) = G'(n u(x)) u'(x)$. Also, we see that $u_n(x) \rightarrow u(x)$ and $u_n'(x) \rightarrow u'(x) \chi_S (x)$ pointwise for every $x \in \mathbb{R}$. An easy application of dominated convergence therefore yields that $u_n \rightarrow u$ in $L^p(I)$ and $u_n' \rightarrow u' \chi_S$ in $L^p(I)$. Since $u_n \in W^{1,p}(I)$ for all $n$, this implies that $u' = u' \chi_S$ a.e. on $I$. This completes the proof for $1 \leq p < \infty$.
Lastly, consider the $p = \infty$ case. Assume that $u \in W^{1, \infty}(I)$. Let $J \subset I$ be a bounded interval. Then $u \in W^{1,p}(J)$ for any $1 \leq p < \infty$. From our work above it follows that $u' = u' \chi_S$ a.e. on $J$. We conclude that $u' = u' \chi_S$ a.e. on $I$ since $J$ was any arbitrary bounded interval.
[1]: "Functional Analysis, Sobolev Spaces and PDEs" by Haim Brezis