currently I am reading Greg Lawler's book conformally invariant processes and I am already having some experience with harmonic measures and calculating explicit Poisson kernels and related exit distributions of Brownian motion.
My question is more about the following fact that Greg Lawler states in his book:
Terminology: We denote as follows.
- We take a domain $D \subseteq \mathbb C$ that has at least one regular point.
- We say that $\partial D$ is locally analytic at $z$ if there exists some conformal function (analytic and one-to-one) $f \colon \mathbb D \to \mathbb C$ such that $$ f(\mathbb D) \cap D = f\left(\{z \in \mathbb D \colon \Im(z) > 0\}\right) $$
- Finally we call $\partial D$ piecewise analytic if $\partial D$ is locally analytic except at a finite amount of points.
Claim: Suppose that $D \subseteq \mathbb C$ be a domain that has at least one regular point. Then if $D$ is piecewise analytic then for fixed $z \in \mathbb D$ the measure $\text{hm}(z,D;\cdot)$ is absolutly continuous with respect to the $1$-dimensional Lebesgue measure given by the arclength.
The case for the unit disk: Let us start with the easiest case, which is given by the unit disk $\mathbb D$. Fix any set $B \subseteq \partial D$ and denote by $h(\cdot) = \text{hm}(\cdot,\mathbb D;B)$ then $h$ is harmonic, since it can be explicitly calculated via the following IVP: $$ \begin{cases} \Delta h(z) = 0 \quad& \text{ if } z \in \mathbb D \\ h(z) = 1 \quad& \text{ if } z \in A \\ h(z) = 0 \quad& \text{ if } z \in \partial D \setminus A. \end{cases} $$ Since we know that harmonic functions satisfy the mean value property this gives $$ 0 = h(0) = \frac{1}{2\pi} \int_{\mathbb D} h(z) d\lambda(z), $$ and since $h$ is positive it must be that $h(z) = 0$ for all $z \in \mathbb D$. But this is enough since then $$ \liminf_{\epsilon \downarrow 0} \frac{ \text{hm}\left(z,\mathbb D;I_{\epsilon}(w)\right)}{\lambda(I_{\epsilon}(w))} = 0, $$ where $I_{\epsilon} = \{\tilde{w} \in \partial \mathbb D \colon \big| \arg(\tilde{w}-w) \big| \leq \epsilon\}$. Hence by measure theory $\text{hm}(z,\mathbb D;\cdot)$ is absolutly continuous with respect to the Lebesgue measure.
General locally analytic domains: Fix any $w \in D$ we want to show that $\text{hm}(w,D;\cdot)$ is absolutly continuous with respect to $|dw|$. Fix any $w^{\prime} \in \partial D$ and denote by $B(w^{\prime},\epsilon)$ the $\epsilon$ ball around $w^{\prime}$ then it sufficies to show that $$ \liminf_{\epsilon \downarrow 0} \frac{\text{hm}\left(w,D; B(w^{\prime},\epsilon) \cap \partial D\right)}{\lambda(B(w^{\prime},\epsilon) \cap \partial D)} \leq C < \infty. $$ Since $\partial D$ is piecewise analytic we can find a conformal map $f \colon \mathbb D \to D$ such that $f(0) = w^{\prime}$ but now I am stuck. The idea should be to push forward the Lebesgue measure $|dw| = |d(f \circ z)|$ while also pushing forward the harmonic measure locally around a neighborhood of $w^{\prime}$ such that $\text{hm}(w,D,\cdot) = C(w,D) \text{hm}(z,\mathbb D,\cdot)$ but I am not sure about this since the conformal map can be pretty rough in terms of its boundary values.
Thanks you in advance!