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


*

*subharmonic on $\mathbb{C}$

*harmonic on $\mathbb{C}\setminus K$

*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). 
