Line segment is removable for continuous functions
Let's begin with a concrete result: If $L$ is a line, then $L\cap \Omega$ is removable for $\mathscr X=C(\Omega)$. Let $f\in C(\Omega)$ be holomorphic on $\Omega\setminus L$. By Morera's theorem it suffices to prove that the integral of $f$ along the boundary of every triangle $T\subset \Omega$ is zero. Consider three cases:
- $T$ does not meet $L$. This case is clear.
- $T$ has a side that lies on $L$. Let $T_n=T+\delta_n$ where $\delta_n$ are complex numbers such that $\delta_n\to 0$ and $T_n\cap L=\varnothing$. In other words, we translate $T$ by a small amount to move it off the line $L$. Using the uniform continuity of $f$ on compact subsets of $\Omega$, we find that $$\int_{\partial T}f(z)\,dz=\lim_{n\to\infty}\int_{\partial T_n}f(z)\,dz=0$$
- $T$ has points on both sides of $L$. Then $L$ divides $T$ into a triangle and another polygon (triangle or quadrangle). The quadrangle can be further divided into two triangles. The integral of $f$ over the boundary of each resulting triangle is $0$ by the above. Add these integrals together to obtain $\int_{\partial T}f(z)\,dz=0$.
Line segment is not removable for bounded functions
To complement the above result: a line segment is not removable for $\mathscr X=L^{\infty}(\mathbb C)$. Indeed, the function $f=z+z^{-1}$ maps the unit disk $\mathbb D$ bijectively onto $\mathbb C\setminus [-2,2]$. The inverse $g=f^{-1}$ is holomorphic in $\mathbb C\setminus [-2,2]$ and is bounded by $1$, but has no holomorphic extension to $\mathbb C$. (For one thing, $g(z)$ approaches both $i$ and $-i$ as $z\to 0$. For another, a holomorphic extension would have to be constant by Liouville's theorem.)
General sets
The case $\mathscr X=C^{\alpha}(\Omega)$, $0\le \alpha<1$, was settled by E. P. Dolzhenko in 1963: a compact set $E$ is removable if and only if its $(1+\alpha)$-dimensional Hausdorff measure is zero.
As often happens, the integer Hölder exponent turned out to be harder than fractional. It waited under 1979, when Nguyen Xuan Uy proved that a compact set $E$ is removable for holomorphic functions of class $\mathrm{Lip}(\Omega)$ if and only if $E$ has 2-dimensional measure zero.
The two remaining cases, $L^\infty$ and $C$, are much more involved. Two classical results are:
- A set of $1$-dimensional measure $0$ is removable for $L^\infty$ (hence also for $C$).
- A set of Hausdorff dimension greater than $1$ is not removable for $C$ (hence also for $L^\infty$).
But the removable sets for these classes admit no complete characterization in terms of Hausdorff measures. Instead, they are studied via the concepts of analytic capacity $\gamma$ and continuous analytic capacity $\alpha$ (the latter is defined similarly to $\gamma$ but using $C$ instead of $L^{\infty}$.) Removability is easily shown to be equivalent to the vanishing of the appropriate capacity. One then proceeds to investigate necessary and sufficient conditions for the vanishing of analytic capacity, its invariance under classes of maps, continuity under nested unions/intersections, etc... The webpage of Xavier Tolsa has a wealth of information on this topic.
If $f\colon\Omega\to\mathbb{D}$ is biholomorphic and has an analytic continuation across $\partial\Omega$, it must map $\partial\Omega$ to $\partial\mathbb{D}$. Since $f'$ can have only isolated zeros on $\partial\Omega$, $f$ is locally biholomorphic at almost all points of $\partial\Omega$, and thus $\partial\Omega\cap V = f^{-1}(\partial\mathbb{D}\cap U)$ must be an analytic arc. So that $\partial\Omega$ consists of analytic arcs is essential for the existence of a continuation across the boundary at least if $f$ is injective and the image of $f$ has a boundary consisting of analytic arcs. Sometimes it is of course possible to continue $f$ across the boundary if $\partial\Omega$ is less regular, but a general theorem asserting the existence of a continuation across the boundary needs strong premises.
It seems to me that non-extendability happens at unbounded points
Not only there. If $(\alpha_n)$ is a sequence in $\mathbb{D}\setminus\{0\}$ with
$$\sum_{n=1}^\infty 1 - \lvert\alpha_n\rvert < \infty,$$
the Blaschke product
$$B_{k,\alpha}(z) = z^k\cdot\prod_{n=1}^\infty \frac{\lvert \alpha_n\rvert}{\alpha_n}\frac{z-\alpha_n}{1-\overline{\alpha_n}z}$$
is, for every $k\in\mathbb{N}$ a bounded holomorphic function on $\mathbb{D}$, and it cannot be analytically continued across every part of the boundary containing a limit point of $(\alpha_n)$. Choosing $(\alpha_n)$ so that all points on $\partial\mathbb{D}$ are limit points, you get a bounded holomorphic function whose natural boundary is $\partial\mathbb{D}$.
Best Answer
Answer: yes. I am going to formalize where you said "$f/g$ is always finite" to mean: "for any $P \in \Omega$, if $g$ has a zero at $P$, then $f$ has a zero of at least the same order."
Recall that we define the order of a zero to be smallest $n \geq 0$ such that $f^{(n)}(P) \neq 0$ but $f^{(n-1)}(P) = 0$. That this exists (when the function is not identically the zero function) is a consequence of holomorphicity, which is obvious if you already know that holomorphic functions are analytic.
Given that $f$ and $g$ are holomorphic, this is the only hypothesis needed to ensure that $f/g$ admits a holomorphic continuation to a domain including $P$. Moreover, this condition is necessary. This is seen easily by Taylor expanding at $P$.
In particular, if a continuous extension exists, it is automatically holomorphic.