In the proof of trace zero theorem, we consider a sequence of $u_m\in C^\infty(\bar{\mathbb{R}}_+^n)$ such that
\begin{equation}
\Vert u_m-u\Vert_{W^{1,p}(\mathbb{R}_+^n)}\rightarrow 0
\end{equation}
where $u\in W^{1,p}(\mathbb{R}_+^n)$ and $Tu=0$ on $\partial \mathbb{R}_+^n$.
We can show that for $x_1>0$,
$$
\int_{\mathbb{R}^{n-1}}|u_m(x_1,z)|^pdz\leq C_p \int_{\mathbb{R}^{n-1}}\left(|u_m(x_1,z)|^{p}+x_1^{p-1}\int_0^{x_1}|u_{m,x_1}(\xi,z)|^pd\xi\right)dz.
$$
Then it says that by taking $m\rightarrow\infty$, we have
$$
\int_{\mathbb{R}^{n-1}}\left|u\left(x_{1}, z\right)\right|^{p} d z \leq C x_{1}^{p-1} \int_{0}^{x_{1}} \int_{\mathbb{R}^{n-1}}|D u(\xi, z)|^{p} d z d \xi.
$$
for a.e. $x_1>0$.
My question is about the left side, why for a.e. $x_1$,
$$
\lim\limits_{m\rightarrow\infty}\int_{\mathbb{R}^{n-1}}\left|u_m\left(x_{1}, z\right)\right|^{p} d z=\int_{\mathbb{R}^{n-1}}\left|u\left(x_{1}, z\right)\right|^{p} d z
$$
Is that kind of using Fubini's theorem?
Best Answer
I'm not too sure about the exact statement, but there's a sufficient simpler argument: let $D_m(x_1) = \int_{\mathbb{R}^{n-1}}{|u_m-u|^p(x_1,z)\,dz}$.
$D_m$ is a sequence of non-negative functions the integral of which goes to $0$: so there is a subsequence $u_{k_m}$ of $u_m$ such that $D_{k_m}$ goes a.e. to zero.
Then, for a.e. $x_1 > 0$, $\int_{\mathbb{R}^{n-1}}{|u_{k_m}|^p(x_1,z)\,dz} \rightarrow \int_{\mathbb{R}^{n-1}}{|u(x_1,z)|^p\,dz}$.