# A set of zero harmonic measure

We know that a set may be of zero harmonic measure without its Lebesgue measure being zero (see Armitage and Gardiner, classical potential theory, pg 178).

Now, consider the following problem. Let $$V$$ be a bounded open set in $$\mathbb{R}^{m}$$, $$m\geq 2$$, and $$W$$ the interior of the closure of $$V$$. Let $$E$$ be a subset of $$\partial V\cap W$$ ($$\partial$$ means boundary). We suppose $$E$$ has the following property: for every $$x\in E$$ and every $$r>0$$ such that the ball $$B(x,r)$$ is relatively compact in $$W$$, the boundary $$\partial B(x,r)$$ of $$B(x,r)$$ contains at least one point of $$E$$. My question is: can the harmonic measure $$\omega(E,x,V)$$ of $$E$$ be zero?

Intuitively (Brownian motion interpretation of harmonic measure) the answer seems negative. (?)

• Yes. In $R^3$, $V$ is a ball with removed radius, and $E$ is this radius. – Alexandre Eremenko Dec 10 '19 at 13:29
• Your assumption does not imply that $E$ is connected. Neither it implies that Lebesgue measure of $E$ is positive. – Alexandre Eremenko Dec 10 '19 at 14:00
• Yes you are right. I edited the post. Thanks. – M. Rahmat Dec 11 '19 at 4:26

It is easy to construct such an example in $$R^n$$ for $$n\geq 3$$ (see my comment). In $$R^2$$ let $$V$$ be a rectangle with removed segments: $$\{ x+iy:|x|<1,0 and $$E=(0,i)$$.
Similar example will work in $$R^n$$ if you want the $$n-1$$ Lebesgue measure positive. In fact you can even have $$n$$-dimensional Hausdorff measure positive, if you take something widely curved for $$E$$ instead of a straight thing.