Every discrete effective divisor on a plane domain is the zero divisor of a holomorphic function

List the points of $A$ with each $a$ repeated $m(a)$ times. If the list is finite, the corresponding finite polynomial works (with $F\equiv 1$ for the empty list). If the list is infinite and $\mathrm{\Omega }=\mathbb{C}$, let $m_0$ be the multiplicity of $0$ and enumerate the remaining nonzero terms as $(a_n)$. Fact [L4] applied to $m_0$ and $(a_n)$ gives the required entire function. Thus assume $\mathbb{C}\setminus \mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$. Define $D:=\{z\in \mathrm{\Omega }:\mathrm{dist}(z,\mathbb{C}\setminus \mathrm{\Omega })<\frac{1}{\mathrm{\mid }z\mathrm{\mid }+1}\}\mathrm{.}$ Split the repeated list into $(b_j)$, consisting of the terms in $D$, and $(c_j)$, consisting of the remaining terms.

For each $b_j$, choose $p_j\in \mathbb{C}\setminus \mathrm{\Omega }$ with $\mathrm{\mid }{b}_j-p_j\mathrm{\mid }=\mathrm{dist}(b_j,\mathbb{C}\setminus \mathrm{\Omega })$. If $(b_j)$ is infinite, then $\mathrm{\mid }{b}_j-p_j\mathrm{\mid }\to 0$: otherwise a subsequence stays a fixed positive distance from the boundary, while the defining inequality for $D$ keeps that subsequence bounded, producing a limit point in $\mathrm{\Omega }$. Define $Q_j(z):=E_j\text{\hspace{0.17em}⁣}\left(\frac{b_j-p_j}{z-p_j}\right)\mathrm{.}$ Each $Q_j$ is holomorphic on $\mathrm{\Omega }$ and, by [L1], has its only zero at $z=b_j$.

If $(c_j)$ is infinite, then $\mathrm{\mid }{c}_j\mathrm{\mid }\to \mathrm{\infty }$. Indeed, a bounded subsequence would have a limit point; discreteness excludes a limit in $\mathrm{\Omega }$, while $c_j\mathrm{\notin }D$ gives $\mathrm{dist}(c_j,\mathbb{C}\setminus \mathrm{\Omega })\ge \frac{1}{\mathrm{\mid }{c}_j\mathrm{\mid }+1},$ which excludes a boundary limit. Let $m_0$ be the multiplicity of $0$ in this list and enumerate its nonzero terms as $(d_j)$. Then $\mathrm{\mid }{d}_j\mathrm{\mid }\to \mathrm{\infty }$, so [L4] gives an entire function $F_2(z)=z^{m_0}{\prod }_j{E}_{p_j}(z\mathrm{/}{d}_j)$ whose zeros, with repetition, are exactly the $c_j$. If the list is finite, take the corresponding finite polynomial, and if it is empty, take $F_2\equiv 1$.

Fix a compact set $K\subseteq \mathrm{\Omega }$ and put $d:=\mathrm{dist}(K,\mathbb{C}\setminus \mathrm{\Omega })>0$. For all sufficiently large $j$, ${\sup }_{z\in K}\mid \frac{b_j-p_j}{z-p_j}\mid \le \frac{\mathrm{\mid }{b}_j-p_j\mathrm{\mid }}{d}\le \frac{1}{2}\mathrm{.}$ Hence [L2] gives ${\sup }_{z\in K}\mathrm{\mid }1-Q_j(z)\mathrm{\mid }\le 2^{-j-1}$ after increasing the starting index if necessary. Thus ${\prod }_j{Q}_j$ is normally convergent on $\mathrm{\Omega }$, and [L3] gives a holomorphic function $F_1$ whose zeros, with repetition, are exactly the $b_j$.

The product $F:=F_1{F}_2$ is holomorphic on $\mathrm{\Omega }$. Steps 3.1 and 2.2 show that its zeros are exactly the original points $a\in A$, and repetition in the list gives each zero order $m(a)$.