Singular boundary points and natural boundaries of function elements

Let $(f,U)$ be a function element and let $\zeta \in \mathrm{\partial }U$.

The boundary point $\zeta$ is regular for $(f,U)$ when there is a function element $(g,V)$ with $\zeta \in V$ such that

$g=f\text{on }U\cap V\mathrm{.}$

Since $V$ is an open neighbourhood of the boundary point $\zeta$, the intersection $U\cap V$ is nonempty. So a regular boundary point is one across which $(f,U)$ extends holomorphically on a full neighbourhood, not merely one where some remote overlap carries a direct analytic continuation in the sense of Function elements and direct analytic continuation.

The point $\zeta$ is a singular boundary point of $(f,U)$ when it is not regular.

If $B\subseteq \mathrm{\partial }U$ and every point of $B$ is singular for $(f,U)$, then $B$ is a natural boundary for $(f,U)$. In particular, saying that the whole boundary $\mathrm{\partial }U$ is natural means that $(f,U)$ admits no holomorphic extension across any boundary point.