TITLE: Continuous extension of a Bounded Holomorphic Function on $\mathbb{C}\setminus K$
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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.