Thinness and polarity Let $D$ be a bounded open set in $\mathbb{R}^{n}$ with $n\geq2$ and $E$ a subset of the boundary $\partial D$ of $D$. $D$ is said to be thin at a point $y\in D$ if there is a superharmonic function $u$ on a neighborhood $U$ of $y$ such that 
$$\liminf u(x)>u(y)$$ as $x\to y$ form inside $D\cap U$.
Suppose $E$ is Borel and  $D$ is thin at each point of $E$. 
1) Does it imply that $E$ is polar?
2) What if $\overline{D}\setminus E$ is thin at each point of $E$? Can we conclude that $E$ is polar? ($\overline{D}$ is the closure of $D$)
 A: Here is an example to show that 1) does not hold (when $n=2$) : consider the measure $\mu=\sum n^{-2}\delta_{\alpha_{n}}$ where the sum runs over all rational numbers of $[-1,1]$, and denote by $U^{\mu}$ the associated logarithmic potential. Since $U^{\mu}$ is finite except on a polar set, the set 
$$A_{n}=\{z\in[-1,1],~U^{\mu}(z)<n\},\qquad n\text{ large enough},$$ 
is of positive capacity. Moreover, the set 
$$S_{n}=\{z\in\mathbb{C},~U^{\mu}(z)>n\}$$ 
is open and non-empty, it contains $\mathbb{Q}\cap[-1,1]$, and $A_{n}\subset \overline S_{n}\setminus S_{n}$ because each point of $A_{n}$ is a limit point of $\mathbb{Q}\cap[-1,1]$ and $U^{\mu}$ equals $\infty$ at those points. Finally, $S_{n}$ is thin at each point $\zeta$ of $A_{n}$ since
$$
\liminf_{z\to\zeta,~z\in S_{n}}U^{\mu}(z)\geq n>U^{\mu}(\zeta).
$$
This example shows that 2) is also false, because on $[-1,1]$, the points $z$ of 
$\overline S_{n}\setminus A_{n}$ obviously satisfy $U^{\mu}(z)\geq n$, and outside of $[-1,1]$, the potential is continuous and thus the points of $\overline S_{n}$ also satisfy $U^{\mu}(z)\geq n$. Thus $\overline S_{n}\setminus A_{n}$ is thin at $A_{n}$.
A: Yes, according to Proposition 7 on page 112 of Markov Processes, Brownian Motion, and Time Symmetry by Kai Lai Chung and John B. Walsh,

A polar set is very thin; a very thin set is thin; a thin set is semi-polar.

They also mention that semi-polar and polar are equivalent for Brownian motion in any dimension, by a deep result called the Kellogg-Evans theorem.
