Let $\mathbb{D}$ denote the open unit disk of the complex plane and let $f : \mathbb{D} \rightarrow \mathbb{C}$ be an analytic function. Assume that $f$ has non-tangential limits on a dense set on $\partial\mathbb{D}$ such that the values can be extended to a continuous function on $\partial\mathbb{D}$. Is it enough to deduce that $f$ can be extended to a continuous function on $\overline{\mathbb{D}}$ that is analytic on its interior?
Thank you very much!
Here's a counterexample: Let $E\subset \partial\mathbb D$ be a dense subset of measure $0.$ Then there exists a positive harmonic function $u$ on $\mathbb D$ such that
$$\tag 1 \lim_{z\to \zeta,z\in \mathbb D}u(z)=\infty$$
for each $\zeta\in E.$*
Using $(1),$ let $v$ be a harmonic conjugate of $u.$ Set $f=e^{-(u+iv)}.$ Then
$$\lim_{z\to \zeta,z\in \mathbb D}f(z)=0$$
for each $\zeta\in E.$ Note that these limits allow full approach within the disc, not just nontangential approach.
I'll leave the verification that this $f$ is indeed a counterexample. But ask if you have questions.
*See Exercise 18, Chapter 6, Harmonic Function Theory by Axler, Bourdon, and Ramey https://www.axler.net/HFT.pdf