Example 3 Sobolev spaces Evans PDE In the following example p.260 Evans. 
Can someone explain me the part that says if $\alpha + 1 < n, |Du(x)| \in L^1(U)$
And how do i get the middle inequality in the next line.
Thanks in advance
 A: It is an integrability condition, in particular it is the $n$-dimensional analogue to the statement  $\int_0^1\frac{1}{x^p}\,dx < \infty$ if $p < 1$. This condition is readily obtained by the Coarea formula: $$\int_{B(0,1)}\frac{1}{|x|^p}\,dx = \int_0^1\int_{\partial B(0,r)}\frac{1}{r^p}\,dS\,dr = \int_0^1\frac{1}{r^p}C(n)r^{n-1}\,dr = C(n)\int_0^1r^{-(p +1-n)}\,dr.$$ This proves that $|x|^{-p}$ is in $L^1(B(0,1))$ provided $p < n$. Hence, since $|Du(x)| = \alpha|x|^{-(\alpha + 1)}$, we obtain the condition $\alpha + 1 < n$.
Notice that, by definition, $u$ restricted to $\partial B(0,\epsilon)$ is constant and equal to $\epsilon^{-\alpha}$. Moreover $|\nu^i| \le |\nu| = 1$ since $\nu$ is the unit inward pointing normal to $\partial B(0,\epsilon)$. Hence one has
\begin{align}
\Big|\int_{\partial B(0,\epsilon)}u\phi\nu^i\,dS\Big| \le &\ \int_{\partial B(0,\epsilon)}|u\phi\nu^i|\,dS \le \int_{\partial B(0,\epsilon)}\|\phi\|_{\infty}|\nu||u|\,dS \\
= &\ \|\phi\|_{\infty}\epsilon^{-\alpha}\int_{\partial B(0,\epsilon)}\,dS.
\end{align}
