# Harmonic measure

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 $B$?

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 non-accessible boundary points has zero harmonic measure. (This is completely evident if one recall the Brownian motion interpretation of the harmonic measure).
• 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? – M. Rahmat Nov 25 '16 at 20:55
• 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) – M. Rahmat Nov 26 '16 at 4:11
• 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? – M. Rahmat Nov 26 '16 at 5:39