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\begin{document}

\title{McMillan's area problem.}
\author{Michael D. O'Neill and Robert E. Thurman}



\address{ Department of Mathematics, UTEP\\El Paso, TX 79968}




\email{michael@math.utep.edu, bob@math.utep.edu}
\begin{abstract}
We answer  the  question asked by McMillan in 1970 concerning
distortion at the boundary by conformal mappings of the disk
which was left open in our earlier paper \cite{OT1}. 


\end{abstract}



\newtheorem{dfn}{Definition}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{claim}{Claim}
\newtheorem{theorem}{Theorem}
\newtheorem*{TheoremA}{Theorem A}
\newtheorem*{TheoremB}{Theorem B}
\newtheorem*{Case I}{Case I}
\newtheorem*{Case II}{Case II}
\newtheorem{ex}{Example}
\newtheorem{rmk}{Remark}
\newtheorem{prop}{Proposition}[section]
\def\pfclm{\noindent {\em Proof of claim}\quad}
\def\proof{\noindent{\em Proof.}\quad}
\def\pflm{\noindent {\em Proof of lemma.}\quad}
\def\pfthm{\noindent {\em Proof of theorem.}\quad}
\def\rse#1{\mbox{$^{\raisebox{.3ex}{$\scriptstyle{#1}$}}$}}

\maketitle
  
\begin{center}
1991 Mathematics subject classification: 30C25, 30C85
\end{center}

\section{Introduction}


Let $A$ denote the set of ideal accessible boundary points of  
a simply connected domain $\Omega$. Recall that these are the finite radial limit
 points of the Riemann map
from the unit disk onto $\Omega$ and that each radius along which the limit exists
gives a distinct ideal boundary point.
 In particular, distinct ideal accessible boundary points may have the 
same complex coordinate.
 Fix $w_0 \in \Omega$ and for each
$a\in A$ and $r<|w_0 -a|$ let $\gamma(a,r) \subset \{z:|z-a|=r\}$ be
the circular crosscut of $\Omega$ separating $a$ from $w_0$ which can
be joined to $a$ by a Jordan arc contained in $\Omega \cap \{z:|z-a| <
r\}$.
Throughout this paper we will refer to $\gamma(a,r)$ as the principle 
separating arc for $a$ of radius $r$.

Let $L(a,r)$ denote the Euclidean length of $\gamma(a,r)$ and let 
$$
A(a,r)= \int \limits _0 ^ r L(a, \rho) \, d\rho. 
$$
In \cite{McMillan1}, McMillan showed that
$$
\limsup_{r\to 0} \frac{A(a,r)}{\pi r^2} \ge \frac12 
$$
almost everywhere on $\partial \Omega$ with respect to harmonic
measure ( denoted by a.e.-$\omega$ hereafter).

The purpose of this paper is to prove

\begin{TheoremA}
$$
\liminf_{r\to 0} \frac{A(a,r)}{\pi r^2} \le \frac12 \qquad \mbox{a.e.-$\omega$}
$$
\end{TheoremA}
\noindent answering a question raised at the end of \cite{McMillan1}.
In an earlier paper, \cite{OT1}, we proved
\begin{TheoremB}
$$
\liminf_{r\to 0} \frac{L(a,r)}{2\pi r} \le \frac12 \qquad \mbox{a.e.-$\omega$}
$$
\end{TheoremB}
\noindent also in answer to the last paragraph of \cite{McMillan1}. Theorem A
implies Theorem B but the basic idea of the proof is the same as in
\cite{OT1}.
Let
$$
E_{m,k}= \{a\in A|\, A(a,r) > (\frac12 + \frac{1}{m})  \pi r^2\quad
\forall r<\frac{1}{k}\}
$$
and consider a Riemann map $f:\mathbb D  \to \Omega$ from the unit disk
to $\Omega$ such that $f(0)= w_0$. We will show that
 $f^{-1}(E_{m,k})$ has zero Lebesgue measure in the unit circle
$\mathbb T$ for each $m$ and $k$. We do this  by showing that if
$f^{-1}(E_{m,k})$ has a point of density for some $m,k$ then the image of
that point would be surrounded by a closed curve contained in
$\Omega$. Since the union of all such sets then has measure zero,
this completes the proof.

The details of the present argument are more complicated than in
\cite{OT1} so
it may be helpful to read \cite{OT1} first to get the main idea with 
fewer technicalities. It may also be helpful to take an early
glance at figures 1 and 2 near the end of the paper.
 For more detailed background on the problem,
one can also refer to \cite{McMillan1}, \cite{McMillan2} and \cite{O'Neill1}.
For the ideas from geometric function theory used here we refer to
\cite{Ahlfors}, \cite{Pommerenke2} and \cite{Garnett-Marshall}.















\section{Proof of Theorem A.}

  In order to construct a curve in $\Omega$ which will surround a
boundary point and thus give the contradiction which proves Theorem A,
 we will need to know that
centered at
almost every  point of $E_{m,k}$
there is a wide angled annular corridor whose thickness is bounded
from below. That such corridors exist will be a consequence of the
accumulation of $E_{m,k}$ near the image of a point of density of
 $f^{-1}(E_{m,k})$. In
fact,
the abundance of points of $E_{m,k}$ will allow us to construct a
chain of such corridors in $\Omega$ which will wrap around a boundary
point.

We will require the following lemma.
Let $\omega(z,E,\Omega)$ denote the harmonic measure of the set 
$E\subset \partial \Omega$ from the point $z \in \Omega$.
\begin{lemma}\label{crowdlemma}
Let $\Omega$ be a simply connected domain in $\mathbb C$ and let 
$f$ be a Riemann map $f:\mathbb D \to \Omega$.
Let $E\subset \partial \Omega$ be a Borel set such that
$f^{-1}(E)$ has a point of density. Then given $\delta >0$ there is a
point $w\in \Omega$ such that
$$
\omega(w,E,\partial \Omega) > 1- \delta.
$$
\end{lemma}
\begin{proof}
Let $\eta$ be a point of density of $f^{-1}(E)\subset \mathbb T$. For any interval
$I\subset \mathbb T$ centered at $\eta$ there is a unique
$r(I,\delta)$,
 $0<r(I,\delta)
< 1$  such that 
$$
\omega(r(I,\delta)\eta, I,\mathbb D) = 1 - \frac{\delta}2.
$$
Let $z_I = r(I,\delta)\eta$. Given any $\epsilon >0$ there is an
interval $I$ centered at $\eta$ such that
$$
|I\setminus f^{-1} (E)| < \epsilon |I|
$$
where $|\cdot|$ denotes linear measure.
Integrating the Poisson kernel at $z_I$ over $I\setminus f^{-1}(E)$
then gives
$$
\omega(z_I,I\setminus f^{-1}(E),\mathbb D) <\frac{\delta}2
$$
if $|I|$ is sufficiently small. Therefore
$$
\omega(z_I,I\cap f^{-1}(E),\mathbb D)> 1-\delta
$$
and taking $w= f(z_I)$ finishes the proof of the lemma.
\end{proof}

Let $d_f(z_I)$ denote the Euclidean distance from $f(z_I)$ to
$\partial \Omega$.
Actually, results of Beurling from \cite{Beurling} imply the existence
of a constant $K$ such that a disk of radius $K d_f(z_I)$ contains all the harmonic
measure of the set $E$ found in the lemma. (See \cite{Pommerenke2}, pg.142.) 


Let $w_0 = f(0)$ and assume that $\eta\in \mathbb T$ is a point of density of
$f^{-1}(E_{m,k})\subset \mathbb T$.
 The finite  number of steps required to get 
a contradiction in the construction to follow will only depend on the number $m$ in the definition of
$E_{m,k}$.  It will be clear from the construction that if $\delta >0$
is  sufficiently
small and $\omega(w_1,E_{m,k},\Omega) > 1-\delta$ for some point $w_1$
then the
required number of steps can be completed. Moreover, the choice of $\delta$
depends only on $m$. We choose $\delta$ to be this small and apply
Lemma \ref{crowdlemma} with $E= E_{m,k}$ thus obtaining the desired
point $w_1$.

 
Let $d_0$ be the Euclidean distance from $w_1$ to $\partial \Omega$
and let $x_0\in \partial \Omega$ be a point such that $|x_0 - w_1| =
d_0$.
Since $f(\eta)\in A$ we can assume  that
 $d_0 <<
\frac{1}{k}$ where $k$ is the integer in the definition of $E_{m,k}$.

 We will introduce positive constants $c_0, c_1,c_2,\dots$ and $C_1,
C_2,\dots$. Their values will be determined in the discussion to
follow and will either be purely numerical or depend only on $m$
(in the definition of $E_{m,k}$).
For any $w\in \mathbb C$ and $r>0$, let $D(w,r)$ denote the set
$$
\{z\in \mathbb C:|z-w| < r\}.
$$

Let $N$ be a large integer which will be determined later.
We will see that it can be chosen so that $N\le (\mbox{const. } m^{\frac32})$.
Since $x_0$ is a boundary point nearest to $w_1$ we may choose $R_0$ so that
$D(w_1,d_0) \cap D(x_0,2^NR_0)$ has area greater than $(\frac12 -
\frac{1}{8m}) \pi (2^N R_0) ^2$. Choose $c_0$ so that if $x_0^*$ is any point
in $D(x_0,c_0R_0)$ then the area of $D(w_1,d_0) \cap D(x_0^*,R_0)$ is greater than $(\frac12 -
\frac{1}{4m}) \pi R_0 ^2$. Later, we will also need
$c_0 << \frac{1}{\sqrt{2m}}$. It is clear that $R_0$ is proportional to $ d_0$ in a
ratio
 depending only on $m$.

 If $\delta>0$ is sufficiently small, there is
a set of points of $E_{m,k}$ of positive harmonic measure contained in
 $D(x_0,c_0R_0)$. In fact,
the circular arc $\partial D(x_0,c_0 R_0) \cap D(w_1,d_0)$ extends to a 
circular crosscut of $\Omega$ which determines a unique subdomain, $U_0$, of
$\Omega$ not containing $w_1$. The midpoint,  $w^*$, of the circular
arc $\partial D(x_0,\frac{c_0R_0}2) \cap D(w_1,d_0)$ is contained in $U_0$.
By the comparison principle for harmonic
measure and the Beurling projection theorem there is a constant
$C_1>0$
such that 
$$
\omega(w^*, \partial U_0 \cap \partial \Omega \cap D(x_0, c_0R_0),
\Omega)\ge C_1 >0
$$
and by repeated application of Harnack's inequality in $D(w_1,d_0) \cup
U_0$, there is then a
constant $C_2$ such that 
$$
\omega(w_1, \partial U_0 \cap \partial \Omega \cap D(x_0, c_0R_0),
\Omega)\ge C_2 >0.
$$
By lemma \ref{crowdlemma}, if $\delta$
is sufficiently small then
\begin{equation}\label{inequality}
\omega(w_1, \partial U_0 \cap \partial \Omega \cap D(x_0, c_0R_0)
\cap E_{m,k},
\Omega)\ge \frac{C_2}2 >0,
\end{equation}
as claimed.

Let $x_0^*$ be an element of $\partial U_0 \cap D(x_0,c_0 R_0) \cap
E_{m,k}$. 
Note that because $x_0^* \in E_{m,k}$ we have
$$
\int \limits _ {\frac{R_0}{\sqrt{2m}}}^{R_0}L(x_0^*,\rho)\, d\rho \ge (\frac12 +
\frac{1}{2m}) \pi r^2
$$
and by the choice of $c_0$, the area of 
$$
\{z \in \mathbb C:\frac{R_0}{\sqrt{2m}} \le |z- x_0^*| \le R_0\} \cap D(w_1,d_0)
$$
is greater than $(\frac12 - \frac{1}{2m})\pi R_0^2$. If 
$$
\gamma(x_0^*,r)\cap D(w_1,d_0) = \emptyset
$$
for each $r\in [\frac{R_0}{\sqrt{2m}},R_0]$ then the area of the
annulus
$$
\{z \in \mathbb C:\frac{R_0}{\sqrt{2m}} \le |z- x_0^*| \le
R_0\}
$$
is greater than
$$
(\frac12 + \frac{1}{2m})\pi R_0^2 + (\frac12 - \frac{1}{2m}) \pi R_0^2
= \pi R_0^2.
$$
This contradiction shows that there  exists $r\in
[\frac{R_0}{\sqrt{2m}},R_0]$ such that
$\gamma(x_0^*,r)\cap D(w_1,d_0) \ne \emptyset $.  Simple topological
considerations show  that circular crosscuts of smaller radius centered at $x_0^*$
which intersect $D(w_1,d_0)$ must be principle separating arcs for
$x_0^*$. Let  $c_1 =
\frac{1}{\sqrt{2m}}$. Thus by shrinking $R_0$ by a factor no
smaller than $\frac{c_1}3$, we may assume that for each $r\le 3R_0$
we have $\gamma(x_0^*,r)\cap D(w_1,d_0)\ne \emptyset$.
 It follows that 
for each $ r\le 2R_0$ we have $\gamma(x_0,r)\cap D(w_1,d_0) \ne \emptyset$.

By a slight strengthening of the above argument it is  clear that  there are 
constants $ c_2>0,c_3 >0$ such that if $0< R < \frac1{k}$ and $a\in E_{m,k}$
then 
\begin{equation}
\label{*}\bigl|\{r\in [c_1R,R]: L(a,r) > (1 +
\frac{c_2}{m})\pi r  \} \bigl | \ge c_3 R.
\end{equation}



We will now assume without loss of generality that $x_0$ is the origin
and that $w_1$ is on the positive  imaginary axis.
Let 
$$
A_0 = \{z:R_0 <|z|< 2 R_0\}
$$
Let 
$$
\theta _0 = \inf\{\theta \in (-\frac{\pi}2,\pi):J_{\theta}\cap
\partial \Omega \ne \emptyset\}
$$
where 
$$
J_{\theta}= \{z: \arg (z) = -\theta, R_0 <|z| < 2R_0 \}.
$$
Let
$$
S_0 = \{z:R_0 <|z|< 2R_0,-\theta_0\le \arg (z) <\frac{\pi}2\}.
$$

See figure 2.

Choose $x_1 \in J_{\theta_0} \cap \partial \Omega$.
Let $R_1=\frac{|x_1 |}2$ and consider the annulus
 $A_1= \{z: c_1 R_ 1 \le | z- x_1| \le R_1\}$. Any circular arc $K$
centered at $x_1$ in $A_1$ with an angle of at least 
$(1+ \frac{c_2}{m})\pi$ is divided into two or three  subarcs by the
ray $\{z: \arg z = - \theta _0\}$. At least two of the arcs
have an angle larger than $\frac{c_2 \pi}{ 2 m}$. If $\alpha >0$ is
sufficiently small then the ray $L_1 = \{z: \arg z = - (\theta_0 +
\alpha)\}$ also divides $K$ into two or three subarcs, at least two of
which have an angle larger than $\frac{c_2 \pi}{4 m}$.
 The same angle 
$\alpha$ will be used in each step of the construction. It is
determined so that in each new step, newly constructed annular
corridors centered at points $x_{j+1}$ with $\arg x_{j+1} = - \theta_j$
will cross the ray $\{z: \arg z = - (\theta_j + \alpha )\}$. The angle
$\alpha$ does not depend on the size of $R_1$ (or $R_j$ for later $j$)
but only on $c_1$ and $c_2$. Specifically, choose 
$\alpha <
\alpha ^*$ where $\alpha ^*$ is found by solving the triangle with
sides $A=1$, $B=\frac{c_1}2$ and $C$ and angles $\angle AB = \pi - \frac{c_2
\pi}{2m}$, $\angle CA = \alpha ^*$ and $\angle BC$.
The choice $\alpha = \frac{c_1 c_2 \pi}{32 m}$ is sufficient for our
purposes.

We can further choose a sufficiently small constant $c_4 >0$ so that
any circular arc centered at $a\in D(x_1, c_4 R_1)$ with an angle of
at least $(1+ \frac{c_2}{m}) \pi$ and with radius between $c_1 R_1$
and $R_1$ will also be divided by the ray $L_1$ into at least two
subarcs with angle larger than $\frac{c_2 \pi} {8m}$. Notice that
$c_4$ depends only on $c_1$ and $c_2$ and not on $R_1$.
We will use the same constant $c_4$ in subsequent similar steps of the
construction with different radii $R_j$.




 
The circular arc $\partial D(x_1,c_4 R_0) \cap S_0$ extends to a
crosscut of $\Omega$ which determines a subdomain $U_1$ not containing
$w_1$. Because the width of $S_0$
is greater than $(const.)  d_0$, we may argue as before using
Harnack's inequality and  the Beurling projection theorem in
$D(w_1,d_0) \cup S_0\cup U_1$  to
find a constant $C_3>0$ depending only on $m$ such that
\begin{equation}
\omega(w_1,\partial U_1 \cap \partial \Omega \cap D(x_1,c_4R_0), \Omega) > C_3 >0.
\end{equation}
Therefore
\begin{equation}\label{**}
\omega(w_1,\partial U_1 \cap \partial \Omega \cap D(x_1,c_4R_0) \cap
E_{m,k}, \Omega) > \frac{C_3}2 >0
\end{equation}
by lemma \ref{crowdlemma} with a sufficiently small initial
 choice of $\delta>0$.


For each point $a\in E_{m,k} \cap \partial U_1 \cap D(x_1,c_4R_0)$
let $F_a \subset [c_1R_1,R_1]$ denote the set of $r$ such that
$L(a,r) > (1+ \frac{c_2}{m})\pi r$. By (\ref{*}), the set $F_a$ has $|F_a| > c_3
R_1$ and for each $r\in F_a$, $\gamma(a,r)$ intersects the ray $L_1$. Let $x$ denote the orthogonal
projection of $x_1$ on the line $L_1$. For points $z,w$ in the plane,
let $\overline{zw}$ denote the line segment with endpoints $z$ and $w$. Then $L_1 = \overline{x_0x}
\cup \overline{x\{\infty\}}$ and we write $F_a = F_a^+ \cup F_a^-$
where $F_a^+$ (respectively, $F_a^-$) is the set of $r\in F_a$ such
that  $\overline{x\{\infty\}}$
(respectively, $\overline{x_0x}$)   divides $\gamma(a,r)$ into two subarcs,
the smaller of which has an angle at least $\frac{c_2 \pi}{ 8 m}$.
Then  either $|F_a^+| \ge \frac{c_3}2  R_1$ or
$|F_a^-| \ge \frac{c_3}2  R_1$. Making a choice of $+$ or $-$ so that
the previous inequality holds, we rename the chosen set $F_a^*$. Let 
$L_1^*$ denote the corresponding side of $L_1$ with respect to the
point $x$ and let
$$
G_a= \{L_1^* \cap \gamma(a,r): r \in F_a^*\}.
$$
 
By (\ref{**}) and the pigeonhole principal we find $a_1$ and $a_1^*$ in 
$E_{m,k} \cap \partial U_1 \cap D(x_1,c_4R_0)$ and constants $c_5>0$
and $c_6>0$  so that $\frac{c_5 R_0}2 < | a_1 - a_1^* | < c_5 R_0$
 and so that $|G_{a_1}\cap G_{a_1^*}| > c_6 R_1$. Note that here,
$c_5<<c_4$. In fact it will be seen in the following paragraph
that $c_5$ should be chosen to be small compared to the angle
$\frac{c_2 \pi}{8 m}$.
 
There are now two cases to consider.
\begin{Case I}
{\em For each $\rho$ such that $c_1 R_1 \le \rho \le R_1$ we have
$\gamma(a_1,\rho)\cap S_0 \ne \emptyset$.}
\end{Case I}
\begin{Case II}
{\em There is some radius $\rho$ with $c_1 R_1 \le \rho \le R_1$, such
that
$\gamma(a_1,\rho)\cap S_0 = \emptyset$.}
\end{Case II}
%%In Case II, the arc which intersects $S_0$ cannot be a separating arc at all
%%since it can be reached from $a_1$ by a Jordan curve lying in $S_0
%%\cup U_1$.

Assume that we are in Case I.
Given $a$ and $b$ in $G_{a_1} \cap G_{a_1^*}$, let $S(a,b) \subset
\Omega$
 be the subdomain of $\Omega$ between the crosscuts $\gamma(a_1,|a_1 -
a|)$
and $\gamma(a_1,|a_1-b|)$. Let $S^*(a,b)$ denote the annular corridor
bounded by $\gamma(a_1,|a_1-a|)$, $\gamma(a_1,|a_1-b|)$,
 $\overline{ab}$, and $\partial S_0$. 
We claim that there is a constant $c_{7} > 0$ and 
there are points $a$ and $b$ in $G_{a_1}\cap G_{a_1^*}$
such that $|a-b| > c_7 R_1$ and such that 
$S^*(a,b)$
contains no point of $\partial \Omega$.
In fact, if $|a-b|< c_7^* R_1$  and
if there is  a point $\tau \in \partial \Omega$ contained in $S^*(a,b)$ then some
piece of $\partial \Omega$ must connect $\tau$ to $\overline{ab}$
and then must extend past $L_1$ through an angle of at least
$\frac{c_2 \pi}{8 m}$  in $S(a,b)$. Since $c_5$ is very small compared
to $\frac{c_2 \pi}{8 m}$ and since $|a_1^* - a_1| \ge \frac{c_5 R_0}2$, simple geometric
considerations show that if $c_7^*$ is sufficiently small,
then one of the arcs $\gamma (a_1^*, |a_1^* - b| )$ or $ \gamma (a_1^*,|a_1^*
- a| )$ would intersect $\partial \Omega$ at a point too close to
$L_1$ for the points $a$ and $b$ to be contained in $G_{a_1^*}$ (see
Figure \ref{fig1}).
As $| G_{a_1} \cap G_{a_1^*} | > c_6 R_1$ and
 $\mbox{diam }(G_{a_1} \cap G_{a_1^*}) < (1-c_1) R_1$ we find the desired constant $c_7$ with  $c_7^* > c_7 >0$ and the
points
$a$ and $b$ with $c_7 R_1<|a-b|\le c_7^* R_1 $.
Note that the constant $c_7$ only depends on previously introduced
constants and therefore only on $m$.
We rename the above annular corridor $S^*(a,b)\subset \Omega$ as $ S_0^*$.

\begin{figure}
%\centerline{\psfig{figure=figure1.ps,height=4in}}
\includegraphics*{figure1.eps}
\caption{$S^*(a,b)$ can contain no point of $\partial \Omega$.}
\label{fig1}
\end{figure}

Now, still assuming Case I, let
$$
J_{\theta} = \{z: \arg(z) = -\theta, \quad | a | < | z |
< | b| \}
$$ 
and let 
$$
S_1 = \{z: |  a| < | z | < | b|, \quad -\theta_1 \le
\arg z \le -(\theta_0 + \alpha) \}
$$
where
$$\theta_1 = \inf \{ \theta \in ((\theta_0 + \alpha), \pi): J_{\theta}
\cap \partial \Omega \neq \emptyset\}.
$$
See Figure \ref{fig2}.

\begin{figure}
%\centerline{\psfig{figure=figure2.ps,height=4in}}
\includegraphics*{figure2.eps}
\caption{Step one of the construction.}
\label{fig2}
\end{figure}


Choose $x_2 \in J_{\theta _1} \cap \partial \Omega$.
Let $R_2 = \frac{|x_2|}2$ and let $L_2$ be the ray $\{z: \arg z
= - (\theta _1 + \alpha)\}$. The arc $\partial D(x_2, c_4 R_2)
\cap S_0 \cup S_0^* \cup S_1 $ defines a subdomain $U_2$ not
containing
$w_1$. Arguing
as before with  
 Harnack's inequality, the comparison principle and the
Beurling projection theorem  but now in 
$D(w_1,d_0) \cup S_0 \cup S_0^* \cup S_1 \cup U_2$ we find,
using Lemma \ref{crowdlemma} with a sufficiently small choice of
$\delta >0$, a constant  $C_4 >0$
such that 

$$
\omega(w_1, \partial U_2 \cap \partial \Omega \cap D(x_2, c_4 R_2)
\cap E_{m,k},
\Omega)\ge C_4 >0.
$$

  As in the
previous step we  find points $a_2$ and $a_2^* $ in $D(x_2,c_{4}R_2) \cap
\partial U_2 \cap E_{m,k}$ and  sets
$G_{a_2},G_{a_2^*} \subset L_2$ with the same properties as before.  We then have again
\begin{Case I}
{\em For each $\rho$ such that $c_1 R_2 \le \rho \le R_2$
we have  $\gamma(a_2,\rho)\cap S_0 \cup S_0^* \cup S_1 \ne \emptyset$.}
\end{Case I}
\noindent and the complementary Case II.

Assume we are again in  Case I. We repeat the argument made for the
point $x_1$ at the new  point $x_2$ and find two  annular sectors. First
$S_1^*$  is found by
the pigeon hole argument  in the same way that $S_0^*$ was found in
the previous step. The new annular corridor $S_1^*$ is centered at the
point $a_2$ near $x_2$ and  ends on the  ray $L_2$ after having
passed through the additional angle of $ \alpha$ clockwise around $x_0$.
Now  $S_2$ is obtained in the same that $S_1$  was previously. That is,
$S_2$
is  centered at $x_0$, begins where $S_1^*$ ends on $L_2$  and is stopped in its clockwise
course around $x_0$ by a point $x_3 \in J_{\theta_2} \cap \partial
\Omega$.
In the $j^{\mbox{th}}$ subsequent step a point $x_j$ is found at the
end of $S_{j-1}$ and nearby points $a_j,a_j^*\in
E_{m,k}$ are found as before.  Case I at the $j^{\mbox{th}}$ step  means that
every  principle separating arc for $a_j$ with radius $\rho$ between $c_1 R_j$ and $R_j$
 intersects the
union of the previously constructed annular corridors
$S_0,S_0^*,S_1,S_1^*,
\dots,S_{j-1}$. The new annular corridors $S_{j-1}^*$ and $S_j$ are now
found as in previous steps. Note that after the $j^{\mbox{th}}$ step, the
union of annular corridors so far constructed has turned through an
angle of at least $j\alpha$ clockwise from the horizontal through $x_0$.
A sufficiently small initial choice of $\delta >0$ ensures that there
is an abundance of points of $E_{m,k}$ near the point
$x_j$
at the end of $S_{j-1}$ so that the construction may
continue to the $(j+1)^{\mbox{st}}$ step.


Assuming that we only encounter Case I in each step, a sufficiently
small choice of $\delta$ at the beginning of the proof allows us to
repeat the argument $N= [\frac{2\pi}{\alpha}]$ times and this
determines the choice of $N$ at the beginning of the construction.
 Since the union
of constructed corridors turns by an additional angle  of  at least
$\alpha$  with each step, we will have
constructed a connected union of annular corridors $\mathcal C$ in $\Omega$
contained in the annulus
$$
\{z: 2^{-N} R_0 < | z- x_0| < 2^N R_0 \}.
$$
 The union of $\mathcal C$ with  $D(w_1, d_0)$ contains a closed
curve in $\Omega$ surrounding the boundary point $x_0$.

If case II occurs at any step $n$ before the $N^{\mbox{th}}$
then there is a principle separating arc for $a_n$
of radius $\rho$, $c_1 R_n \le \rho \le R_n$, which does not intersect
$S_0 \cup S_0^* \cup S_1\cup S_1^* \cup \dots \cup S_{n-1}$. It
follows that the circular crosscut centered at $a_n$ of radius $\rho$
which does intersect $S_0 \cup S_0^* \cup S_1\cup S_1^* \cup \dots
\cup S_{n-1}$ cannot be a separating arc for $a_n$ at all. 
This means that $w_0$ is located in $\Omega$ on the concave side of this arc
but on the convex side of the arcs which make up $S_{n-1}$.
We then continue the construction at the $(n+1)^{\mbox{st}}$ step with
the original
annulus $A_0$ centered at $x_0$ but
now turning
in the counterclockwise direction. Since we have found case II in the
clockwise direction,  we
cannot find case II in the counterclockwise direction without repeating
the situation of $w_0$ being on the concave side of the last non-separating
circular arc but on the convex side of the arcs in the last $S_{n-1}$ from Case I. 
Simple topological considerations rule out this possibility and we
therefore find a closed curve in $\Omega$ surrounding $x_0$ in at most
$N$ more steps.

It follows that there can be no point of density of $f^{-1}(E_{m,k})$
and that  the harmonic measure of $E_{m,k}$ is therefore zero.
The theorem is proved.





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\end{thebibliography}



\end{document}



