Riemann Mapping Theorem: Continuously Extending to Boundary I am interested in the following question.
Let $\Omega \subset \mathbb C$ be a bounded simply-connected domain and let $f: \Omega \rightarrow \Delta$ be a biholomorphism onto the unit disk. Does $f$ extend continuously to the boundary?
If this is difficult, consider the following example. Let $\Omega = \Delta \setminus [0, 1)$, and let $f: \Omega \rightarrow \Delta$ be a biholomorphism.  Is it true that for $x \in [0, 1)$ we have that
$$\lim_{\varepsilon \rightarrow 0^+}f(x + i \varepsilon) = \lim_{\varepsilon \rightarrow 0^+}f(x - i \varepsilon)? $$
 A: The answer for your question is very simple: 
$f: \Omega \to \Delta$ bi-holomorphism extends continuously to the boundary iff $\Omega$ is a Jordan domain 
(here we assume $\Omega$ bounded as in the question - otherwise the result holds only with obvious modifications to allow for the point at infinity to "complete" the boundary to a simple curve in $\hat C$ as the upper plane and the usual Mobius transform to the unit disc shows) 
Note that the converse is false and as noted in the comments, for bi-holomorphisms $g:\Delta \to \Omega$ we have continuous extension iff $\partial \Omega$ is locally connected and continuous injective extension (hence topological homeomorphism from the closed unit disc) extension iff $\Omega$ is a Jordan domain (and then lots of results about the curve bounding $\Omega$ if we impose stricter conditions on the boundary map like real differentiability or even Lipschitz, complex differentiability etc
The proof can be found for example in Conway, Functions of Complex Variables II, page 53-54 and it follows from three facts: extension of $f: \Omega \to \Delta$ bi-holomorphism to one boundary point implies the boundary point is simple and the extension is injective at simple boundary points; in our case the extension is then injective everywhere; the image of $\partial \Omega$ is a compact subset of the unit circle and as it follows that all points with radial limits for $f^{-1}$ on the unit circle are in the image, the image must be also surjective as those have full lebesgue measure. Hence the extension is a homeomorphism of the boundaries.
A: These properties are equivalent (if $\Omega$ is bounded and simply connected):


*

*A Riemann map $\tau\colon\Delta\to\Omega$ extends homeomorphically to a map $\bar\Delta\to\bar\Omega$

*$\partial\Omega\cong S^1$ (that is, a Jordan curve)

*Every point of $\partial\Omega$ is simple, which means that for every $\alpha_n\to\beta\in\partial\Omega$ there exist $\gamma\colon [0,1)\to\Omega$ and a sequence $t_n\in [0,1)$ such that $\gamma(t_n)=\alpha_n$.


A (partial) proof can be found in Rudin's Real and Complex Analysis. In addition, a theorem of Painlevé says that this extension is $C^\infty$ if the boundary of $\Omega$ has the same regularity. 
