# 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 $$D$$)

• 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. – Josiah Park Nov 5 '19 at 6:57

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 $$A_{n}=\{z\in[-1,1],~U^{\mu}(z) is of positive capacity. Moreover, the set $$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}$$.

• Thanks. But can you please explain why $A_{n}$, for $n$ large enough, is of positive capacity? – M. Rahmat Dec 16 '19 at 8:18
• The subset of $[-1,1]$ where $U^\mu$ is finite is of positive capacity. It is the countable union of the $A_n$. – user111 Dec 16 '19 at 9:01
• 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? – M. Rahmat Dec 16 '19 at 17:09

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.

• 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? – M. Rahmat Nov 5 '19 at 14:54