$\textbf{Background for problem statement}$:
Let $B \subset \mathbb{C}$ be a bounded domain, and $g_{B}(z,z_0) = g$ its Green's function with pole at $z_0 \in B$, so $g$ is harmonic in $B \setminus \{z_0 \}$, equal to zero on the boundary of $B$, and near $z_0$, $g + \log|z-z_0| = \alpha$ is a bounded harmonic function. Set $\alpha(z_0) = R_0$.
Set $\psi(r) = \mu r^{\nu}$ for some fixed positive constants $\mu,\nu$. For a fixed $r > 0$, set $\psi = \psi(r)$.
Define, for any fixed $r > 0$,and $\delta \neq 0$, $h(z) = \delta \beta_{r} g(z)$, where $\beta_{r} = -(\log \psi)^{-1}[ 1 + (\log \psi)^{-1} R_0]$.
According to a textbook I am reading ("Condenser Capacities and Symmetrization in Geometric Function Theory" by Vladimir N. Dubinin), the following is true (here $U(z_0,\psi(2r))$ is the open disk center $z_0$ and radius $\psi(2r)$):
$\textbf{Problem statement}$
For $r > 0$ sufficiently small (and fixed), the boundary of $S = U(z_0, \psi(2r)) \cap \{h(z) > \delta \}$ is precisely $\partial S = U(z_0,\psi(2r)) \cap \{h(z) = \delta \}$. But no explanation is provided.
I fail to see why this is true, and if something significant in the definition of $h$ is needed to establish this. As far as I can see, $h$ is the same as $g$ up to a factor, and thus near $z_0$ , $h$ has a logarithmic singularity, so it should be the case that in an entire closed disk of sufficiently small radius the intersection is wholly empty, or $S$ is the entire open disk $U(z_0, \psi(2r))$.
This book is not well known and there may well be a typo here, but I believe that the statement being made here is at least close to one which is correct, either that or I am failing to comprehend what the author is really trying to say.
My suspicion (based on a rough calculation that I needed to do manually) is that on $\partial U(z_0, \psi(2r))$, $\beta_r g = 1 + \frac{\nu \log 2 + o(1)}{\log \mu + \nu\log r} + O((\frac{1}{\log(r)})^2) < 1$ for $r>0$ small. Therefore $h = \delta \beta_{r} g < \delta$ on the boundary of $U(z_0, \psi(2r))$ for $r > 0$ small, which implies the desired result.
However I hesitate to believe this just yet, and the calculation itself is messy, so I leave the question open in case there is an easier solution.