Existence of holomorphic functions The theorem below is discussed in the book 'Selected Problems on Exceptional Sets' written by Carleson (page 73) but I would like to generalize it to certain subharmonic functions.  
Theorem : 
 Let $E\subset\mathbb C$ be a compact polar set, $D=\mathbb C\setminus E$ and $f\in\mathcal O(D)$ be holomorphic. If $$\int_D |f|^2 dxdy < \infty \mbox{ then } f\equiv 0.$$
Can anyone help me with a detailed proof or some references that may help me to find a detailed proof of the theorem? Any help will be greatly appreciated. Thanks
 A: I will prove that if $u\geq0$ is subharmonic on $D$ with $\int_D u(z) dx\,dy<\infty$, then $u=0$ on $D$. 
Let $\phi$ be a subharmonic function such, that $\phi|_E=-\infty$.
Let $M=\int_D u(z) dx\,dy$. Consider $u_\epsilon = u+\epsilon\phi$ for $\epsilon>0$. Those functions are subharmonic on $\mathbb{C}$. Choose $w\in D$. Pick $\delta>0$ and choose $R$ such that $\pi R^2>2M/\delta$. Let $B_R(w)$ be a disk with the center at $w$ and the radius $R$. Let $\epsilon_0>0$ be so small that $\epsilon_0 \int_{B_R(w)} \phi(z)dx\,dy<M$. Then for $0<\epsilon<\epsilon_0$ we have $\int_{B_R(w)}u_\epsilon dx\,dy<2M$, thus
$$u_\epsilon (w)\leq \frac{1}{\pi R^2}\int_{B_R(w)} u_\epsilon(z) dx\,dw<\delta\;.$$ 
This imply that $u(w)\leq \delta$. 
This proof obviously works for $\mathbb{R}^n$. 
A: I found this proof in a book but I did not understand the inequalities. Any explanation will be appreciated.
Let $D_1\subset D_2\subset\cdots\subset D,$ where $D_m$ is bounded by a finite number of analytic curves. Let $g_m$ be the Green's function of $D_m$  with pole at $\infty$ and let $h_m$ be its conjugate function. it is well known  that $E$ is polar means $g_m(z)$  converges uniformly to $\infty,$ $z\in D,$  on inside domains. Hence if we introduce the coordinates $\zeta=\xi+\eta=g_m+i h_m$ in $D_m$ and   consider the domain $\Omega_m=\{z, 0<g_m<1\}$ we have for $f\in H^2(D),$  $$\int\int_{\Omega_m}|f(z)|^2dxdy=\int_0^1d\xi\int_0^{2\pi}\frac{|f|^2d\eta}{(\frac{\partial g_m}{\partial n})^2}=\epsilon_m\rightarrow 0,$$
 On the other hand, the following estimate holds :
$$\int_0^1d\xi\left(\int_{g_m=\xi}|f(z)||dz|\right)^2\leq\int_0^1d\xi\left\{\int_{g_m=\xi}\frac{\partial g_v}{\partial n}|dz|.\int_{g_m=\xi}\frac{|f|^2}{\frac{\partial g_m}{\partial n}}|dz|\right\}$$
 $$=2\pi\int_0^1\int_0^{2\pi}\frac{|f|^2}{(\frac{\partial g_m}{\partial n})^2}d\eta d\xi=2\pi\epsilon_m.$$
Hence $\xi_m,$ $0<\xi_m<1,$ exists so that $$  \int_{g_m=\xi_m}|f(z)||dz|\leq \sqrt{2\pi\epsilon_m}  $$ 
The curves $g_m=\xi_m$ approach $E$ uniformly as $m\rightarrow \infty.$ Let $z_0$ be fixed in $D.$ Then for $m>m_0$ since  $f(\infty)=0$ $$|f(z_0)|=|\frac{1}{2\pi i}\int_{g_m=\xi_m}\frac{f(z)dz}{z-z_0}|\leq const. \sqrt{2\pi\epsilon}\rightarrow 0.$$ This implies $f(z)\equiv 0$ and concludes the proof. 
