Suppose $W$ is a bounded open subset of $\mathbb{R}^{n}$ and $n\geq2$. Let $V$ be the interior of the closure of $W$ and $E$ a subset of the boundary of $V$. If $\omega(x,W)(E)=0$ ($\omega(x,W)$ is the harmonic measure of $W$ at $x\in W$), can we conclude that $\omega(x,V)(E)=0?$
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Not really. Take one of the usual counter-examples in potential theory: $$W = (0,1) \times (0,1) \setminus \biggl(\bigcup_{n=2}^\infty\{\tfrac{1}{n}\} \times (\tfrac{1}{n}, 1-\tfrac{1}{n})\biggr).$$ Then $V = (0,1) \times (0,1)$. If $E = \{0\} \times (0, 1)$, then $\omega(x, W)(E) = 0$, but of course $\omega(x, V)(E) > 0$.
answered Oct 22, 2019 at 21:37
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$\begingroup$ In Gardiner-Armitage book, pg183, Theorem 6.6.10., they say that if $W$ is a subset of $V$ and $E$ is a subset of $\partial V$ and outside the closure of $V\setminus W$, then $\omega(x,V)(E)=0$ iff $\omega(x,W)(E)=0$. Your example is not in contradiction with this theorem? $\endgroup$ Commented Oct 23, 2019 at 18:28
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$\begingroup$ No, it is not: the closure of $V \setminus W$ clearly contains $\{0\} \times [-1, 1]$. $\endgroup$ Commented Oct 23, 2019 at 18:37
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$\begingroup$ Do you know any other conditions ( besides the theorem I cited from Armitage -Garfiner) that guarantees $(\omega(x,W)(E)=0)$ implies $(\omega(x,V)=0$? $\endgroup$ Commented Oct 23, 2019 at 19:56
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$\begingroup$ If $W$ is not equal to $V$ in a neighbourhood of $E$, this gets rather delicate. I do not think I have ever seen any results on that, but mind that I am not a specialist on classical potential theory. Do you have a particular example of $V$ and $W$ in mind? The claim is clearly true if $V$ and $W$ are smooth enough, and I think it should also be true if $V$ and $W$ are assumed to be Lipshitz. But I do not know if this goes in the direction that you are interested in. $\endgroup$ Commented Oct 23, 2019 at 20:38
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$\begingroup$ For example, take for $V$ a bounded open set and a subharmonic function $u$ on a neighborhood of the closure of $V$. Suppose for example that $u<c$ ($c$ is a constant) on $V$ and take $E$ the set of points in the boundary of $V$ where $u>c$. Then $E$ is negligible for $V$. Can we say that it is also negligible for $W$, the interior of the closure of $V$? $\endgroup$ Commented Oct 23, 2019 at 22:57