Hi everyone: Let $ \omega $ be a bounded open set in $ \mathbb{R}^{q} $, $ q\geq 2 $, and $ E $ a subset of the boundary $ \partial\omega $ that has harmonic measure zero in $ \omega $. Let $ V $ be the interior of the closure of $ \omega $. We know that some points of $ E $ can be inside $ V $. If I take an open ball $ B $, with the closure of $ B $ inside V, can I say that $ \partial B\cap E $ has harmonic measure zero in $ B $?
1 Answer
No. Let $V$ be the disk $z<2$ in the complex plane, $E$ the arc $\{ e^{it}/2:t\leq1\}\subset V$. Now draw a simple arc $\gamma(t),\; 0\leq <1$, $\gamma(0)=1$, $\gamma\backslash\gamma(0)\subset V\backslash E$, and $\gamma$ is spiraling around $E$, so that the limit set of $\gamma(t)$ as $t\to 1$ equals $E$. And let $\omega=V\backslash\{\gamma\cup E\}=V\backslash\overline{\gamma}$. Then $V$ is the interior of the closure of $\omega$. Take $B=\{ z:z=1/2\}$. So $E\cap\partial B$ has non zero harmonic measure with respect to $B$.
On the other hand $E$ has zero harmonic measure with respect to $\omega$, simply because no point of $E$ is accessible from $\omega$, and the set of nonaccessible boundary points has zero harmonic measure. (This is completely evident if one recall the Brownian motion interpretation of the harmonic measure).

$\begingroup$ Thanks for your answer. But, there should be a superharmonic function $ v(x) $ that approaches $ +\infty $, as $ x $ apporaches any point of $ E $. Do you see such a function? $\endgroup$ Nov 25, 2016 at 20:55

$\begingroup$ I don't understand your second question. Superharmonic where, in which region? Why should it exist? $\endgroup$ Nov 26, 2016 at 0:41

$\begingroup$ Sorry, I am referring to a theorem that says: a set $ E\subset\partial \omega $ has harmonic measure zero, if and only if there is a positive superharmonic function $ u(x) $ on $ \omega $ that $ \rightarrow+\infty $, as $ x\rightarrow $ any point $ y\in E $ (I am using " classical potential theory", by D. Armitage and S. Gardiner, Theorem 6.5.2, pg 177) $\endgroup$ Nov 26, 2016 at 4:11

$\begingroup$ If they say so, such a function exists. Of course one cannot "explicitly" write it in this complicated domain. $\endgroup$ Nov 26, 2016 at 5:03

$\begingroup$ But it seems to me that this creates a problem (???). Because if we set $ w(x) $ equal to $ u(x) $ on $ V\setminus E $ and equal to $ +\infty $ on $ E\cap V $, then we obtain a superharmonic function on $ V $ that takes on the value $ +\infty $ on $ E\cap V $. That means $ E\cap V $ is a polar set and so its harmonic measure must be zero??? Does it make sense? $\endgroup$ Nov 26, 2016 at 5:39