QUESTION: So I don't understand how he gets the last line of the proof. Namely, how we go from $$u_m(x',x_n)\leq u_m(x',0)+\int_0^{x_n}u_{m,n}(x',t)dt$$ to $$\int_{{\mathbb{R}}^{n-1}}u_m(x',x_n)\leq \int_{{\mathbb{R}}^{n-1}}u_m(x',0)+x_n^{p-1}\int_0^{x_n}\int_{{\mathbb{R}}^{n-1}}u_{m,n}(x',t)dt$$
The idea is just to raise both sides of
$$|u_m(x',x_n)| \le |u_m(x',0)|+\int_0^{x_n}|u_{m,n}(x',t)|dt$$
to the power $p$, then integrate over $\mathbb R^{n-1}$. First, we have $(a+b)^p \le C_p (a^p + b^p)$, so
$$\frac 1 C |u_m(x',x_n)|^p \le |u_m(x',0)|^p + \left(\int_0^{x_n}|u_{m,n}(x',t)|dt\right)^p.$$
Now we treat the integral using Hölder's inequality: writing $|u_{m,n}| = |u_{m,n}| \times 1$ we have
$$\int |u_{m,n}|\le\left(\int|u_{m,n}|^p\right)^{1/p}\left(\int1^{p/(p-1)}\right)^{(p-1)/p}$$ and thus $$\left(\int |u_{m,n}| \right)^p \le\left(\int|u_{m,n}|^p \right)x_n^{p-1},$$ since the domain of integration $[0,x_n]$ has measure $x_n$. Putting this together and integrating over $\mathbb R^{n-1}$ we get
$$\frac 1 C\int_{{\mathbb{R}}^{n-1}}|u_m(x',x_n)|^p dx' \leq \int_{{\mathbb{R}}^{n-1}}|u_m(x',0)|^p dx'+x_n^{p-1}\int_0^{x_n}\int_{{\mathbb{R}}^{n-1}}u_{m,n}(x',t)dt dx'. $$
Switching the order of integration and using $|u_{m,n}| \le |Du_m|$ finishes the estimate.
