Every meromorphic function on a plane domain is a quotient of holomorphic functions

Let $P$ be the pole set of $f$, with multiplicities equal to its positive finite pole orders. Each $x\in \mathrm{\Omega }\setminus P$ has a holomorphy neighbourhood disjoint from $P$, and each $a\in P$ has an isolated-pole neighbourhood meeting $P$ only at $a$. Hence every point of $\mathrm{\Omega }$ has a neighbourhood meeting $P$ in finitely many points. For a compact $K\subseteq \mathrm{\Omega }$, the family of all such open neighbourhoods covers $K$; a finite subcover shows that $K\cap P$ is finite. Thus [L2] applies. Choose a holomorphic function $h$ on $\mathrm{\Omega }$ whose zero divisor is exactly $P$.

On $\mathrm{\Omega }\setminus P$, define $g:=fh$. Near a pole $a\in P$, the zero of $h$ has exactly the same order as the pole of $f$, so $g$ is locally bounded on a punctured neighbourhood of $a$. By [L3], $g$ extends holomorphically across every point of $P$.

Away from $P$, one has $f=g\mathrm{/}h$. Since both sides are meromorphic and agree on the dense open set $\mathrm{\Omega }\setminus P$, this quotient represents $f$ on all of $\mathrm{\Omega }$.