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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 $D$)

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    $\begingroup$ For the second question, the following result is a consequence of Wiener's criterion which characterizes when a set is thin at a point in $\mathbb{C}$ (and for sets in $\mathbb{R}^n$ more generally). The result below can be found as Theorem 5.4.3 in Ransford's book (if you are looking for generalizations to higher dimensions Ransford will not do). Theorem Let $F$ be an $F_\sigma$ subset of $\mathbb{C}$. If $F$ is thin at $0$, then $E=\{ e^{i\theta}|\ r_{n}e^{i\theta}\in F, \text{for some sequence }r_n\rightarrow 0\}$ is a polar set. $\endgroup$ – Josiah Park Nov 5 '19 at 6:57
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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 $A_{n}$.

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  • $\begingroup$ Thanks. But can you please explain why $A_{n}$, for $n$ large enough, is of positive capacity? $\endgroup$ – M. Rahmat Dec 16 '19 at 8:18
  • $\begingroup$ The subset of $[-1,1]$ where $U^\mu$ is finite is of positive capacity. It is the countable union of the $A_n$. $\endgroup$ – user111 Dec 16 '19 at 9:01
  • $\begingroup$ With this line of reasoning you need to prove that all super harmonic functions on that neighborhood is finite on $A_{n}$. If one superharmonic function is finite at each point of a set that does not imply that this set is not polar. Take a single point $x_{0}$ where $U^{\mu}$ is finite. Steel $\{x_{0}\}$ is polar since it is an isolate point. No? $\endgroup$ – M. Rahmat Dec 16 '19 at 17:09
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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.

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  • $\begingroup$ Thanks. I am not familier with Brownian motion, is the definition of a thin set in potential theory the same as in Brownian Motion theory? I mean does the theorem you cited applies correctly to my case? $\endgroup$ – M. Rahmat Nov 5 '19 at 14:54

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