# Continuous extension of conformal map to accessible point

Let $$G \subset \mathbb{C}$$ be a simply connected bounded domain. By the Riemann mapping theorem there is a holomorphic bijection $$f : G \to \mathbb{D}$$.

An accessible point of $$\partial G$$ is an equivalence class of jordan half intervals $$\gamma : [0,1) \to G$$ where two jordan half intervals $$\gamma_1, \gamma_2$$ are said to be equal if $$\lim_{t \to 1^{-}} \gamma_1(t) = \lim_{t \to 1^{-}} \gamma_2(t) = \zeta$$ for some $$\zeta \in \partial G$$ and there is a jordan half interval $$\gamma_3$$ such that for any neighbourhood $$N$$ of $$\zeta$$ in $$G$$ $$\gamma_3$$ intersects $$\gamma_1$$ and $$\gamma_2$$ in $$N$$. Abusing notation I will identify an accessible point with its final point $$\zeta$$ ( note $$\zeta$$ may correspond to many (even uncountably many) accessible points).

The point of this is that $$f$$ admits a continuous extension at an accessible point. Precisely,

For every accessible point $$\zeta$$, $$f$$ admits a continuous extension to $$\zeta$$ such that $$f(\zeta) \in \partial \mathbb{D}$$ and distinct accessible points are sent to distinct points on $$\partial \mathbb{D}$$.

Given an accessible point $$\zeta$$ one defines $$f(\zeta) = \lim_{t \to 1^{-}} f(\gamma(t))$$ for any $$\gamma$$ in the accessible point. I have no problem showing that $$f$$ is well-defined and that distinct accessible points are sent to distinct boundary points, but am stuck on showing that $$f$$ is continuous at $$\zeta$$. To me it seems that for this theorem to even make sense one would have to provide a topology for $$G$$ together with its set of accessible points. Any ideas on how to proceed or a reference to a proof? Many thanks!