IHow to apply the density of $C_0^1(\Omega)$ in $W_0^{1,1}(\Omega)$ to derive the identity in the following? I'm reading the book by Gilbarg and Trudinger.
In Lemma 7.14, the authors proved that if $u \in C_0^1(\Omega)$, then 
$$u(x)=\frac{1}{nw_n}\int_{\Omega}\frac{(x_i-y_i)D_iu(y)}{|x-y|^n}dy.$$
Then right after this the authors claim that if $u \in W_0^{1,1}(\Omega)$, then using the fact that $C_0^1(\Omega)$ is dense in $W_0^{1,1}(\Omega)$, one can obtain 
$$u(x)=\frac{1}{nw_n}\int_{\Omega}\frac{(x_i-y_i)D_iu(y)}{|x-y|^n}dy \quad a.e. $$
I can't see why here, because $\frac{(x_i-y_i)}{|x-y|^n}$ is singular.
Can anyone give me some ideas? Thanks a lot!
 A: I would argue as follows. Assume that $n\ge 2$ and let 
$$
T_i f (x)=\int_{\mathbb{R}^n} f(y)\frac{x_i-y_i}{\lvert x-y\rvert^{n}}\chi_\Omega(y)\, dy.$$
Here $\chi_\Omega(y)$ is $1$ if $y\in \Omega$ and $0$ otherwise. The kernel of this operator can be estimated in absolute value by the kernel of the Riesz potential:
$$
\left\lvert \frac{x_i-y_i}{\lvert x-y\rvert^{n}} \right\rvert\le C\frac1{\lvert x-y\rvert^{n-1}}.$$
It is a consequence of this that $T_i$ is continuous from $L^1(\Omega)$ to $L^{\frac n{n-1}}(\mathbb{R}^n)$, because 
$$
\left\lVert T_if\right\rVert_{L^\frac{n}{n-1}(\mathbb{R}^n)}\le C_1\left\lVert I_1(\lvert f\rvert\chi_\Omega)\right\rVert_{L^\frac{n}{n-1}(\mathbb{R}^n)}\le C_2\lVert f\rVert_{L^1(\Omega)}, $$
the second inequality being Hardy-Littlewood-Sobolev's one$^{[1]}$. 
In the problem at hand, if $u\in W^{1,1}_0(\Omega)$ we can approximate it with a sequence $u_k\in C^1_0(\Omega)$ in such a way that 
$$u_k\to u\qquad \text{in }W^{1,1}(\Omega).$$
We have that $u_k=T_i \left(\frac{\partial u_k}{\partial x_i}\right)$ for all $k$. Here the left hand side is $L^1$ convergent to $u$ and the right hand side is $L^{\frac n{n-1}}$ convergent to $T_i\left(\frac{\partial u}{\partial x_i}\right)$. Therefore the limits must be equal almost everywhere, because both modes of convergence imply the pointwise convergence of a subsequence at almost all points.

$[1]$ If $\Omega$ is bounded, one might probably avoid using this inequality and resort to the conceptually simpler Young's inequality for convolutions with $p=1$ and $r=q\in \left[1, \frac{n}{n-1}\right)$ (notations taken from Wikipedia's page). This works in dimension $1$ too. If needed I can fill in the details. 
