I don't completely understand the proof of the following result, which is also called Green's third identity:
Let $D\subset{\mathbb R^m}$ be a bounded domain of class $C^1$ and let $u\in C^2(\bar{D})$ be harmonic in $D$. Then $u(x)=\int_{\partial D}\bigg\{\frac{\partial u}{\partial\nu}(y)\Phi(x,y)-u(y)\frac{\partial \Phi(x,y)}{\partial\nu}\bigg\}ds(y),\quad x\in D$ where $\Phi(x,y):= \begin{cases} \dfrac{1}{2\pi}\ln\dfrac{1}{|x-y|},\quad m=2\\ \dfrac{1}{4\pi}\dfrac{1}{|x-y|},\quad m=3. \end{cases}$
Proof. For $x\in D$ choose a sphere $\Omega(x;r):=\{y\in{\mathbb R}^m:|y-x|=r\}$ of radius $r$ such that $\Omega(x;r)\subset D$ and direct the unit normal $\nu$ to $\Omega(x;r)$ into the interior of $\Omega(x;r)$. Apply Green's second identity to the harmonic functions $u$ and $\Phi(x,\cdot)$ in the domain $\{y\in D:|y-x|>r\}$ to obtain $ \int_{\partial D\cup\Omega(x;r)}\bigg\{u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}-\frac{\partial u}{\partial\nu}(y)\Phi(x,y)\bigg\}ds(y)=0. $ Now we have $ \int_{\Omega(x;r)}\bigg\{u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}-\frac{\partial u}{\partial\nu}(y)\Phi(x,y)\bigg\}ds(y)=\int_{\partial D}\bigg\{\frac{\partial u}{\partial\nu}(y)\Phi(x,y)-u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}\bigg\}ds(y) $
Here is my question:
How can I show that $ \lim_{r\to 0}\int_{\Omega(x;r)}\bigg\{u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}-\frac{\partial u}{\partial\nu}(y)\Phi(x,y)\bigg\}ds(y)=u(x)? $