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

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 $z\ne 0$.

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 $z\in D$.