Riesz measure of a log modulus records the holomorphic zeros

Let $a\in \mathrm{\Omega }$ with $f(a)=0$ and let $m:={\mathrm{ord}}_a(f)\ge 1$ by [F1]. By [F1] there are a disc $D(a,r)\subseteq \mathrm{\Omega }$ and a holomorphic zero-free $g$ on it with $f(z)=(z-a{)}^{m}g(z)$, so $f(z)\mathrm{\ne }0$ for $0<\mathrm{\mid }z-a\mathrm{\mid }<r$: every zero of $f$ is isolated.

Fix $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$. If $\phi =0$ the identity is immediate; otherwise put $K_0:=\mathrm{supp}\phi$. For each $x\in K_0$ there are concentric relatively compact discs $V⋐D⋐\mathrm{\Omega }$ with $x\in V$ and $\bar{D}$ containing at most one zero of $f$, since zeros are isolated. These smaller discs form an open cover of $K_0$; compactness gives finitely many pairs $(V_i,D_i)$ with the $V_i$ covering $K_0$. By [F6], choose ${\beta }_i\in C_c^{\mathrm{\infty }}(D_i)$ with ${\beta }_i\ge 0$ and ${\beta }_i=1$ on a neighbourhood of $\overline{V_i}$. On $W:={\bigcup }_i{V}_i$ the sum $S:={\sum }_i{\beta }_i$ is positive. Define ${\phi }_i:=\phi {\beta }_i\mathrm{/}S$ on $W$ and zero off $W$; since $K_0⋐W$, each ${\phi }_i\in C_c^{\mathrm{\infty }}(D_i)$, and ${\sum }_i{\phi }_i=\phi$.

Let $K\subseteq \mathrm{\Omega }$ be compact and suppose it contained infinitely many distinct zeros of $f$. By [F5] the infinite bounded set $Z(f)\cap K$ has an accumulation point $z_{\ast }\in K\subseteq \mathrm{\Omega }$; continuity gives $f(z_{\ast })=0$, contradicting the isolation of zeros from step 1.1. Thus each compact subset meets $Z(f)$ finitely. Since $\mathrm{\Omega }$ is second-countable and every zero is isolated, $Z(f)$ is at most countable: assign each zero the least element of a fixed enumerated basis that contains it and no other zero; distinct zeros receive distinct basis elements. The countable sum $\lambda :={\sum }_{a\in Z(f)}{\mathrm{ord}}_a(f){\delta }_a$ is a positive Borel measure by [F5], locally finite by the compact finiteness just proved. The open set $\mathrm{\Omega }$ is second-countable and locally compact Hausdorff by [F8]; Dependent Choice supplies the Countable Choice of [F7], so $\lambda$ is Radon.

For each $D_i$ from step 1.2, either $f$ has no zero, or it has exactly one zero $a_i$ of order $m_i$. In the latter case the local factorization [F1] extends $F_i(z):=f(z)\mathrm{/}(z-a_i{)}^{m_i}$ holomorphically and without zeros throughout $D_i$; in the zero-free case set $F_i:=f$ and $m_i:=0$. Then $u=m_i\log \mathrm{\mid }z-a_i\mathrm{\mid }+\log \mathrm{\mid }{F}_i\mathrm{\mid }$ on $D_i$ when $m_i>0$, and $u=\log \mathrm{\mid }{F}_i\mathrm{\mid }$ when $m_i=0$. By [F2], $\log \mathrm{\mid }{F}_i\mathrm{\mid }$ is harmonic, so its Riesz functional vanishes by [F4]; the point-mass normalization [F3] therefore gives ${\mu }_u({\phi }_i)=m_i{\phi }_i(a_i)=\int {\phi }_{i}d\lambda$ when $m_i>0$, and both sides are zero when $m_i=0$.

Summing the local identities of step 2.2 over the finite partition $\phi ={\sum }_i{\phi }_i$ from step 1.2 gives ${\mu }_u(\phi )={\sum }_i{\mu }_u({\phi }_i)={\sum }_i\int {\phi }_{i}d\lambda =\int \phi d\lambda$ for every $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$.

Since $\lambda$ is a positive Radon measure by step 2.1 that represents ${\mu }_u$ on all smooth compactly supported tests, the uniqueness clause of [F3] gives ${\mu }_u=\lambda ={\sum }_{a\in Z(f)}{\mathrm{ord}}_a(f){\delta }_a$, which is the asserted formula; in particular every compact subset of $\mathrm{\Omega }\setminus Z(f)$ carries no $\lambda$-mass, so $\log \mathrm{\mid }f\mathrm{\mid }$ has no Riesz mass off the zero set.