Holomorphy of continuous function on $(A\cup B)^C$ Let $f:\mathbb C\to \mathbb C$ be a continuous function and $A,B\subseteq \mathbb C$ two open connected sets with $\overline{A \cup B}=\mathbb C$. Further, we know that $f\mid_A$ and $f\mid_B$ are holomorphic. Does this also imply holomorphy of $f$? If not can you give precise conditions on when we get holomorphy?
I think in the case where $A\cap B\ne \emptyset$ we should get holomorphy.
I'm in particular interested in the case where $A$ is the upper half plane $\mathbb H$  and $B$ is the lower half plane.
 A: In general, $f$ need not be holomorphic on the entire plane. Let $K\subset \mathbb{C}$ be a compact set with empty interior and positive Lebesgue measure, such that its complement is connected, e.g. a product of two fat Cantor sets. Define
$$f(z) = \int_K \frac{d\overline{\zeta}\wedge d\zeta}{\zeta-z}.$$
Then $f$ is holomorphic on $\mathbb{C}\setminus K$, and continuous on all of $\mathbb{C}$.
But, if $C$ is a circle containing $K$ in its interior, then we have
\begin{align}
\int_C f(z)\,dz &= \int_C \int_K \frac{d\overline{\zeta}\wedge d\zeta}{\zeta-z}\,dz\tag{Fubini}\\
&= \int_K \left(\int_C \frac{dz}{\zeta-z}\right) d\overline{\zeta}\wedge d\zeta\\
&= \int_K -2\pi i\, d\overline{\zeta}\wedge d\zeta\\
&= 4\pi \lambda(K),
\end{align}
where $\lambda$ is the Lebesgue measure. Thus $\int_C f(z)\,dz \neq 0$, and by Cauchy's integral theorem it follows that $f$ is not holomorphic on all of $\mathbb{C}$.
Now pick e.g. $A = B = \mathbb{C}\setminus K$, or $A,B$ such that $A\cup B = \mathbb{C}\setminus (K \cup L)$, where $L$ is a straight line, if you want $A$ and $B$ different, even disjoint.

In the case of particular interest, where $A$ is the upper half-plane, and $B$ the lower half-plane, however, it follows that $f$ is entire.
That is most conveniently seen by using Morera's theorem.
