Q:

Bounded holomorphic function on an unbounded region

Let $\Omega\subset\mathbb C$ is an unbounded region, with boundary $\Gamma$. Let $f$ be a function holomorphic in $\Omega$ and continuous on $\Omega\cup\Gamma$. If $f$ is bounded on $\Omega\cup\Gamma$, show that $\forall z\in \Omega$, $|f(z)|\le\max_{w\in\Gamma}|f(w)|$.  
This is from W. Rudin's Real and Complex Analysis.  And there is a hint: show that without loss of generality, we can assume that $\Omega$ doesn't intersect the (open) unit disk.   
Under this assumption, I can complete the proof using maximum modulus principle.   
My question is: why can we assume it?  If $\mathbb C\setminus\Omega$ has non-empty interior, I can see it is valid.  Can anyone give some advice? 

A:

Suppose that the interior of $\mathbb C\setminus\Omega$ is empty.  This means that for any $z\in\mathbb C\setminus\Omega$ and any $\varepsilon$, one can find $z'=z'(\varepsilon)\in \Omega$ such that $|z-z'|\lt\delta$. In other words, $\Omega$ is dense in $\mathbb C$.  In this case, $\Omega\cup\Gamma$ is the whole complex plane, and by Liouville's theorem, the assumption that $f$ is bounded on this sets implies that $f$ is constant.