I stuck when reading the following proof of the Poincare inequality (Calculus of variations, Jurgen Jost & Xianqing Li-jost, Page 177-178):
Theorem (Poincare inequality)
Let $\Omega\subset\Bbb R^d$ be open and bounded. For any $u\in\pmb H_0^{1,p}(\Omega )$,
$$\Vert u \Vert_{\pmb L^p(\Omega)}\le \left(\frac{meas\Omega}{\omega_d}\right)^{1/d}\Vert Du \Vert_{\pmb L^p(\Omega)}$$
where $\omega_d$ is the Lebesgue measure of the unit ball in $\Bbb R^d$.
Proof
Since $\pmb C_0^1(\Omega)$ is dense in $\pmb H_0^{1,p}(\Omega )$, we may assume $u\in\pmb C_0^{1}(\Omega )$. Put $u(x)=0$ for all $x \in \Bbb R^d\setminus\Omega$. For $\nu\in\Bbb R^d $ with $\lvert \nu \rvert =1$, we have
$$u(x)=-\int_{0}^\infty \frac{\partial}{\partial r}u(x+r\nu)dr$$ Integration w.r.t. $\nu$ yields
$$\lvert u(x) \rvert= \lvert -\frac{1}{d\omega_d}\int_{0}^{\infty}\int_{\lvert \nu \rvert =1}\frac{\partial}{\partial r}u(x+r\nu)d\nu dr \rvert$$
$$\le\frac{1}{d\omega_d}\int_{\Omega}\frac{1}{\lvert x-y\rvert ^{d-1}}\lvert Du(y)\rvert dy$$
(I wonder why can we get this inequality.)
Therefore
$$\left( \int_{\Omega}\lvert u(x) \rvert ^p dx\right)^{1/p}$$
$$\le \frac{1}{d\omega_d}\left( \int_{\Omega} \left(\int_{\Omega} \frac{1}{\lvert x-y \rvert ^{d-1}}\lvert Du(y)\rvert dy\right)^p\right)^{1/p}$$
$$\le \frac{1}{d\omega_d} \left(\int_{\Omega} \left(\int_{\Omega} \frac{1}{\lvert x-y \rvert ^{d-1}}\lvert Du(y)\rvert^p dy\right)\left(\int_{\Omega} \frac{1}{\lvert x-y \rvert ^{d-1}}dy\right)^{p-1}dx\right)^{1/p}$$ (By holder's inequality)
$$=\frac{1}{d\omega_d}\left( \int_{\Omega}\lvert Du(y) \rvert ^p dy\right)^{1/p} \left(\int_{\Omega} \frac{1}{\lvert x-y \rvert ^{d-1}}dx\right)$$
using Fubini's theorem to exchange the order of integration in the first factor. (How??)
In order to control
$$\int_{\Omega} \frac{1}{\lvert x-y \rvert ^{d-1}}dx$$
we choose $R$ with
$$meas\Omega = meas B(R) =\omega_d R^d$$
($B(R):=\{z\in \Bbb R ^d|\lvert z \rvert \le R\}$)
Since
$$\frac{1}{\lvert x-y \rvert ^{d-1}} \le \frac{1}{R^{d-1}}, \lvert x-y \rvert \le R$$
$$\frac{1}{\lvert x-y \rvert ^{d-1}} \ge \frac{1}{R^{d-1}}, \lvert x-y \rvert \ge R$$
we have
$$\int_{\Omega} \frac{1}{\lvert x-y \rvert ^{d-1}}dx \le \int_{B(R)} \frac{1}{\lvert x-y \rvert ^{d-1}}dx = d\omega_dR = dw_d^{1-1/d}(meas\omega)^{1/d}$$
(I don know how we get the first inequality since there is no inclusion between $\Omega$ and $B(R)$, and how is the value of that integral on $B(R)$ becomes $d\omega_dR$.)
Q.E.D.
Any help would be appreciated!