The harmonic measure on the unit disc is absolutely continuous with respect to length

Question

I have read some pages from the book Conformally Invariant Processes in the Plane by Lawler, and found there the following definition of a harmonic measure: $$\text{hm}(z,D;V) = \mathbb{P}^z\left(B_{\tau_D}\in V\right)$$ (that is, the probability that a $d$-dimensional Brownian motion starting at $z$ is in $V$ the moment it hits the complement of $D$).

Later on, the book says that for the unit disc $\mathbb{D}$, it is obvious that the harmonic measure $\text{hm}(z,\mathbb{D};\cdot)$ is absolutely continuous (with respect to length).

Could you explain why is it so?

Attempt

A possible interpretation of the claim is as follows: starting a Brownian motion at $z$, which is somewhere in the open disc $D$, the probability of exiting the disc via a set $A\subseteq\partial D$ whose (Lebesgue) measure is 0, is 0. That makes sense - when the "gate" is so small, one wouldn't expect to exit through it "by chance".

So the attempt is as follows: suppose otherwise. Let $A$ be a set with Lebesgue measure $0$, and let $a>0$ be its harmonic measure (for $z=0$). Now, turn the disc with angle $\theta$ such that $A$, and the turned $A^\theta$ are disjoint. (this is the part I'm not sure about - is that always possible?). Then, the harmonic measure of $A^\theta$ is also $a$. Now, do that enough ($k$) times so that $ka>1$, and that will lead to a contradiction.

Update

The attempt above is wrong, as I may not necessarily have disjoint rotations of a 0-measured set. However, for $z=0$ it's not difficult to show that the harmonic measure of a set on the boundary is simply its normalized Lebesgue measure.

It's only left to see why it follows for $z\ne 0$.

Answer 1

The density of the harmonic measure with respect to surface measure on the boundary is given by the Poisson kernel. So the result is true, but is there an easier argument that shows that the result is "obvious"?

Here's my attempt.

Let $A$ be a subset of the boundary $\partial D$ of the disk $D=\{z: \|z\|< 1\}$ with zero Lebesgue (surface) measure, let $\tau=\inf\{t\geq 0: \|B_t\|\geq1\}$ be the hitting time of $D^c$ and define $h(z)=\mathbb{P}_z(B_\tau\in A)$ for $z\in D$. Then $h$ is harmonic on $D$ and (as you've noted) $h(0)=0$. But by the mean value property of harmonic functions $$0=h(0)={1\over \omega_d}\int_D h(z)\,dz.$$ Because $h$ is non-negative, this means $h(z)=0$ almost surely on $D$ with respect to Lebesgue measure, and by continuity $h(z)=0$ for all $z\in D$.