# Continuous extension of a Bounded Holomorphic Function on $\mathbb{C}\setminus K$

Let $f:\mathbb{C}\setminus K\rightarrow\mathbb{D}$ be a holomorphic map, where $K$ is a compact set with empty interior. My question:

Prove or disprove that: $f$ extends continuously on $\mathbb{C}.$

Remark: Observe that if $K$ is discrete then by the Riemann Removable Singularity Theorem we know that infact there is a holomorphic extension.

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Here is an example from Which sets are removable for holomorphic functions?

The function $f=z+z^{-1}$ maps the punctured unit disk $\mathbb D$ bijectively onto $\mathbb C\setminus [-2,2]$. The inverse $g=f^{-1}$ is holomorphic in $\mathbb C\setminus [-2,2]$ and is bounded by $1$, but has no holomorphic (or even continuous) extension to $\mathbb C$. Indeed, $g(z)$ approaches both $i$ and $-i$ as $z\to 0$.

A compact set $K$ is removable for bounded holomorphic functions if and only if its analytic capacity is zero. A simple sufficient condition was given by Painlevé: if the $1$-dimensional Hausdorff measure of $K$ is zero, then it’s removable for bounded holomorphic functions. That is, every bounded holomorphic function on $\mathbb C\setminus K$ extends to a holomorphic function on $\mathbb C$ (which is necessarily constant by Liouville’s theorem).

Even for compact subsets of the real line, this is not true. More precisely, suppose that $K \subseteq \mathbb{R}$ is compact. Then $K$ is removable (for bounded holomorphic functions) if and only if $m(K)=0$, where $m$ is the one-dimensional Lebesgue measure.

See e.g. my answer to this question

Hence just take your favorite Cantor set in the real line of positive one-dimensional Lebesgue measure, and it will be non-removable. One can even give an explicit expression for a bounded holomorphic function $f$ on $\mathbb{C}\setminus K$ which is not constant. See the above link.