Trace-zero functions in $W^{1,p}$ This is an excerpt of a textbook's proof for a theorem (Trace-zero functions in $W^{1,p}$), from PDE Evans, 2nd edition, page 275.

Next let $\zeta \in C^\infty(\mathbb{R}_+)$ satisfy $$\zeta \equiv 1 \text{ on } [0,1], \zeta \equiv 0 \text{ on } \mathbb{R}_+ - [0,2], 0 \le \zeta \le 1$$ and write
  \begin{cases}
\zeta_m(x) := \zeta(mx_n) \quad (x \in \mathbb{R}_+^n) \\
w_m := u(x)(1-\zeta_m).
\end{cases}
  Then
  \begin{cases}
w_{m,x_n} := u_{m,x_n}(1-\zeta_m) \\
D_{x'}w_m = D_{x'}u(1-\zeta_m).
\end{cases}
  Consequently
  \begin{align}\int_{\mathbb{R}_+^n} |Dw_n-Du|^p \, dx &\le C \int_{\mathbb{R}^n} |\zeta_m|^p |Du|^p \, dx + Cm^p \int_0^{2/m} \int_{\mathbb{R}^{n-1}} |u|^p \, dx' dt \\
&=:A+B \tag{10} \end{align}
  Now $$A \rightarrow 0 \quad \text{as } m \rightarrow \infty, \tag{11}$$ since $\zeta_m \not=0$ only if $0 \le x_n \le 2/m$. To estimate the term $B$, we utilize the inequality $\text{(9)}$:
  \begin{align}
B &\le Cm^p\left( \int_0^{2/m} t^{p-1} \, dt \right) \left(\int_0^{2/m} \int_{\mathbb{R}^{n-1}} |Du|^p \, dx' dx_n \right) \\
&\le C \int_0^{2/m} \int_{\mathbb{R}^{n-1}} |Du|^p \, dx' dx_n \rightarrow 0 \quad \text{as }m \rightarrow \infty \tag{12}
\end{align}

My question is how I can fill in the details for $\text{(10)}$ and $\text{(12)}$? So far I've done some work and tried to resemble as close as possible to the textbooks results in some way.
For $\text{(10)}$, I did so far:
$$\int_{\mathbb{R}_+^n} |Dw_m-Du|^p \, dx = \int_{\mathbb{R}_+^n} |Du \cdot (1-\zeta_m) - Du|^p \, dx = \int_{\mathbb{R}_+^n} |\zeta_m|^p |Du|^p \, dx$$
and for $\text{(12)}$ I also did (for $p > 0$ at least),
\begin{align}
B &= Cm^p \int_0^{2/m} \int_{\mathbb{R}^{n-1}} |u|^p \, dx' dt \\
&= Cm^p \int_0^{2/m} \, dt \int_{\mathbb{R}^{n-1}} |u|^p \, dx' \\
&\le Cm^p \int_0^{2/m} t^{p-1} \, dt \int_{\mathbb{R}^{n-1}} |u|^p \, dx' 
\end{align}
 A: Since $w_m$ is a product, its gradient consists of two terms: $Du  (1-\zeta_m)$, and $u(D\zeta_m)$. So, the difference $|Dw_m-Du|$ is estimated by $|\zeta_m| |Du|+ |u||D\zeta_m|$. Next we want to show that this sum tends to zero in $L^p$. 


*

*The  term $|\zeta_m| |Du|$ is handled by a general fact about integrable functions (absolute continuity with respect to the set of integration): if $\int |f|<\infty$, then for every $\epsilon>0$ there is $\delta>0$ such that $\int_E |f|<\epsilon$ whenever the measure of $E$ is less than $\delta$. Here the set $E$ is the intersection of support of $\zeta_m$ (which is contained in the slab $0\le x_n\le 2/m$) which the support of $|Du|$ (which is compact). The Lebesgue measure of this intersection tends to zero. 

*The term $|u||D\zeta_m|$ is a bit trickier. To begin with, $|D\zeta|$ is bounded, and when we scale it to become $\zeta_m$, the gradient is multiplied by $m$. So we have to deal with $m^p\int |u|^p$, where the integral is over $0\le x_n\le 2/m$. Luckily the inequality (9) is available: it estimates the integral of $|u|^p$ over any slice (hyperplane) $u_n=t$. So we use Fubini to integrate over each such hyperplane first, and then integrate over   $0\le t\le 2/m$. At the end we again use absolute continuity, as for the first term: the integral of $|Du|^p$ over a small set is small.
