Do the level sets of a subharmonic function have measure zero?

Question

Suppose $u$ is subharmonic, real valued and continuous in the open disk $D\subset\mathbb{C}$ and is non constant on any open set in $D$. By adding a constant to $u$ if necessary, I want to know whether the set $$ \{z\in D: u(z)=0\} $$ has measure zero?

Answer 1

No. Let $K\subset\mathbb{C}$ be a compact set with positive measure, empty interior, and connected complement (e.g., a simple curve of positive area or a "thick" Cantor-type set). Let $\mu$ be the equilibrium measure of $K$, i.e., the measure that maximizes the energy $$ I(\mu) = \iint \log|z-w|\,d\mu(z)\,d\mu(w) $$ among all probability measures supported on $K$. Such a measure exists: see, e.g., section 3.3 of Potential Theory in the Complex Plane by Ransford or these lecture notes by Saff. Moreover (from the same sources), the potential of $\mu$, defined by $$ P_\mu(z) = \int \log|z-w|\,d\mu(w) $$ is

1. subharmonic on $\mathbb{C}$
2. harmonic on $\mathbb{C}\setminus K$
3. equal to $I(\mu)$ a.e. on $K$

Property 2 implies $P_\mu$ is not constant on any open set, since such a set would overlap $\mathbb{C}\setminus K$, and a harmonic function can't be constant on an open subset of a connected open set unless it's identically constant (recall that harmonic functions are real-analytic).