Let $S\subseteq \mathbb D:=\{z\in \mathbb C:|z|<1\}$. Suppose that $\overline S$, the Euclidean closure of $S$
 meets $\mathbb T=\{z\in\mathbb C:|z|=1\}$. A point $\xi=e^{i\theta}\in \overline S\cap \mathbb T$ is called a non-tangential boundary
 point, if there is a cone 
 $$S\_\alpha(\theta):=\{z\in \mathbb D: |\arg(1-e^{-i\theta}z)|<\alpha, \;{\rm Re}\,(e^{-i\theta}z)\geq 0\},$$
 and a sequence $(z\_n)$ in $S\cap S\_\alpha$ such that $\lim z\_n=\xi$. My question is,
 whether the set $S\_{nt}$ of all non-tangential boundary points of $S$ is Lebesgue measurable 
 in $\mathbb T$. I would be surprised if this were the case, but could not come up with a counter example. This is connected to the Brown-Shields-Zeller theorem.



Measurability of the set of non-tangential boundary points
For given $\alpha$ the corresponding set of nontangential boundary points is a $G\_{\delta}$ set, since the cone must contain points of $S$ in annuli arbitrarily close to the unit circle. Here $G\_n$ consists of boundary points $\xi$ where the cone $S\_\alpha(\theta)$ intersects $\{z: 1-1/n<|z|<1\} \cap S$. Thus
 the set of all non tangential boundary points is a $G\_{\delta \sigma }$ set and in particular a Borel set. 


