Is my proof correct? (a generalization of the Laurent expansion in an annulus) I want to see if my solution to the following problem in Ahlfors' Complex Analysis text is correct. The problem reads:

Let $\Omega$ be a doubly connected region whose complement consists of the components $E_1, E_2$.   Prove that every analytic function $f(z)$ in $\Omega$ can be written in the form $f_1(z)+f_2(z)$ where $f_1(z)$ is analytic outside of $E_1$ and $f_2(z)$ is analytic outside of $E_2$.

(Note that this is a generalization of the familiar Laurent expansion in an annulus.)
Here is my attempt at the proof:
Firstly, we will prove the statement for a bounded region $\Omega$:
It is given that $\Omega^c=E_1 \cup E_2$, where the complement is taken in the Riemann sphere. Exactly one of these components contains $\infty$, and the other one is hence bounded. Suppose WLOG that $E_1$ is bounded and $\infty \in E_2$.
The boundary of $\Omega$ is $$\partial \Omega=\partial (\Omega^c)=\partial \left(E_1 \cup E_2 \right)=\overline{E_1 \cup E_2} \setminus \text{int}(E_1 \cup E_2)=E_1 \cup E_2 \setminus (\text{int}E_1 \cup \text{int} E_2)=(E_1 \cup E_2) \cap [(\text{int} E_1)^c \cap (\text{int} E_2)^c]\\=(E_1 \cap (\text{int} E_1)^c \cap (\text{int} E_2)^c) \cup (E_1 \cap (\text{int} E_1)^c \cap (\text{int} E_2)^c) \\ =\partial E_1 \cap (\text{int} E_2)^c \bigcup \partial E_2 \cap (\text{int} E_1)^c=\partial E_1 \cup \partial E_2.$$
where I have used information from this question.
Next, given $\delta>0$ we cover the plane by the net of squares of side $\delta$, induced by the lines $x=m \delta,y=n \delta$, and we denote by $Q_j, j \in J$ the closed squares in the next which are contained entirely in $\Omega$; because $\Omega$ is bounded $J$ is finite, and if $\delta$ is sufficiently small $J$ is not empty.
Denote the distance between $\partial E_1,\partial E_2$ by $\rho$. It is clear that $\rho>0$, and therefore for $\delta<\frac{\rho}{\sqrt{2}}$ the closed squares of the net are partitioned into three groups:


*

*The squares contained entirely in $\Omega$, $Q=\{Q_j: j \in J \}$.

*The squares which meet $E_1$ (exclusively), $R=\{R_k: k \in K \}$.

*The squares which meet $E_2$ (exclusively), $S=\{S_l: l \in L \}$.


Furthermore, if $\delta<\frac{\rho}{2\sqrt{2}}$ two squares from $R$ and $S$ can't be adjacent to each other. It is proven in the text that $$ f(z)=\frac{1}{2 \pi i} \oint_{\partial Q} \frac{f(\zeta)}{\zeta-z} \mathrm{d} \zeta,$$ for all $z \in \text{int} Q$. We can prove once more that $\partial Q=\partial R \cup \partial S$, so that
$$f(z)=\frac{1}{2\pi i}\oint_{\partial S} \frac{f(\zeta)}{\zeta-z} \mathrm{d} \zeta+\frac{1}{2 \pi i}\oint_{\partial R} \frac{f(\zeta)}{\zeta-z} \mathrm{d} \zeta=:f_1(z)+f_2(z)$$ with the suitable orientation. This will work for any $z \in \Omega$ provided that we take $\delta$ sufficiently small, such that $z \in \text{int} Q$.
Finally, if $\Omega$ is not bounded, and $z \in \Omega$, repeat this proof in the restricted domain $\Omega'=\Omega \cap \{\zeta: |\zeta|<|z|+1 \}$.
Is this correct?
 A: Your proof has the right idea, but it does some things that are not needed (e.g., $\partial \Omega=\partial E_1\cup \partial E_2$) and 
skips more relevant ones (checking the holomorphicity of $f_1$ and $f_2$ in appropriate sets). Also, 
it's possible to work with both bounded and unbounded $\Omega$ at once if you don't focus on covering $\Omega$ itself. 
As you said, one of two components of the complement is bounded; let it be $E_1$. Pick $\delta>0$ such that 
$\operatorname{dist}(E_1,E_2)>10\,\delta$. Cover the plane by the net $\mathcal N$ of squares of sidelength $\delta$. Let 
$$\mathcal A=\{Q\in\mathcal N : Q\cap E_1\ne\varnothing\}$$
$$\mathcal B=\{Q\in\mathcal N\setminus \mathcal A : Q \cap Q'\ne\varnothing \text{ for some  }Q'\in \mathcal A\}$$
Since $E_1$ is bounded, $\mathcal A$ is finite. Hence $\mathcal B$ is also finite. Neither collection is empty. All squares
in either $\mathcal A$ or $\mathcal B$ lie in $\Omega$. 
Let $\gamma = \sum_{Q\in\mathcal B} \partial Q$, understood as the sum of cycles (with cancellation). The edges present in $\gamma$ are of two kinds: 


*

*an edge between a square in $\mathcal B$ and a square in $\mathcal A$

*an edge between a square in $\mathcal B$ and a square not in $\mathcal A\cup \mathcal B$


Write $\gamma=\gamma_1+\gamma_2$ accordingly, and observe that $\gamma_1$ and $\gamma_2$ are disjoint. Let 
$$f_k(z)= \frac{1}{2\pi i}\int_{\gamma_k} \frac{f(\zeta)}{z-\zeta}\,d\zeta,\quad k = 1,2  $$
Each function $f_k$ is holomorphic in the complement of the support of $\gamma_k$. Hence, $f_1$ is holomorphic in $\mathbb C \setminus \bigcup \mathcal A$, 
and $f_2$ is holomorphic in the interior of $\bigcup \mathcal A\cup \bigcup \mathcal B$.   
For points $z$ in the interior of $Q_0\in \mathcal B$ and for any square $Q\in  \mathcal B$ we have 
$$
\frac{1}{2\pi i}\int_{\partial Q} \frac{f(\zeta)}{z-\zeta}\,d\zeta =\begin{cases} f(z),  \quad &Q=Q_0 \\ 0, \quad & Q\ne Q_0 \end{cases} 
$$
Summing over $Q\in\mathcal B$, we conclude that 
$$f(z)=f_1(z)+f_2(z)\tag1$$
holds in the interior of any square in $\mathcal B$. By the continuity of both sides,  (1) also holds in the interior of $\bigcup\mathcal B$. 
Since $f-f_1$ is holomorphic in  $\Omega \setminus \bigcup \mathcal A$, the function $f_2$ has a holomorphic extension to  $\mathbb C\setminus E_2$. And since $f-f_2$ is holomorphic in the interior of $\Omega\cap (\bigcup \mathcal A\cup \bigcup \mathcal B)$,   the function $f_1$ has a holomorphic extension to  $\mathbb C\setminus E_1$. 
