% Ordinary and strong density continuity of complex analytic functions.

%%% For LaTeX 2e
\documentclass{rae}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{epsf}
%%%%%%

% Standard Spaces
\def\integer{\Bbb Z}
\def\natural{\Bbb N}
\def\rational{\Bbb Q}
\def\real{\Bbb R}
\def\complex{\Bbb C}

%End of proof marker
\def\qed{\hfill$\Box$}
%Proof
\def\proof{\noindent{\sc Proof.} }

\newcommand{\supp}{\operatorname{supp}}
\newcommand{\cl}{\operatorname{cl}}

%% Theorems, etc.

\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{problem}{Problem}

\newcommand{\thm}[2]{\begin{theorem}\label{#1}#2\end{theorem}}
\newcommand{\cor}[2]{\begin{corollary}\label{#1}#2\end{corollary}}
\newcommand{\examp}[2]{\begin{example}\label{#1}#2\end{example}}
\newcommand{\lem}[2]{\begin{lemma}\label{#1}#2\end{lemma}}
\newcommand{\prop}[2]{\begin{proposition}\label{#1}#2\end{proposition}}

\newcommand{\pr}[2]{\begin{problem}\label{#1}#2\end{problem}}

% Functions

\newcommand{\lm}[1]{\mbox{$\operatorname{\lambda}\left({#1}\right)$}}


\newcommand{\putillustration}[5]{
     \begin{figure}[htb]
     \[
          \vbox to #2in{\hrule width #1in height 0pt depth 0pt \vfill
          \special{illustration #3 scaled 250}}
     \]
     \caption{#4\label{#5}}
     \end{figure}
}
 
%%%%%%%%%%%%%%%%%%%%%   MISCELLANEOUS
 
\renewcommand{\P}{\mbox{$\cal P$}}
\newcommand{\B}{\mbox{$\cal B$}}
\newcommand{\K}{\mbox{$\cal K$}}
\newcommand{\U}{\mbox{$\cal U$}}
\newcommand{\V}{\mbox{$\cal V$}}
\newcommand{\e}{\varepsilon}
\renewcommand{\d}{\mbox{$\delta$}}

% characteristic function
     \newcommand{\charf}[1]{\mbox{\raise.48ex\hbox{$\chi$}$_{#1}$}}

%%%%%%%%%%%

\title{Ordinary and strong density continuity of complex analytic functions}

\author{Krzysztof Ciesielski,
Department of Mathematics,
West Virginia University,
Morgantown, WV 26506-6310, USA
(kcies@wvnvms.wvnet.edu)
}

\date{}

\markboth{K.~Ciesielski}{Density continuity of complex analytic functions}

\keywords{strong density continuity, complex analytic functions}
\MathReviews{26B05, 26A15, 30A99}


\begin{document}\maketitle

\begin{abstract}

In the paper we prove that the complex analytic functions are
(ordinarily) density continuous.
This stay in contrast 
with the fact that even such a simple function as $G\colon\real^2\to\real^2$, 
$G(x,y)=(x,y^3)$, is not density continuous 
\cite{CW:DensCTran}.
We will also characterize those analytic functions which are strongly
density continuous at the given point $a\in\complex$. From this we conclude that
a complex analytic function $f$
is strongly density continuous if and only if $f(z)= a + b z$, where
$a,b\in\complex$ and $b$ is either real or imaginary.

\end{abstract}
 
\section{Preliminaries} 

The notation used throughout this paper is standard. In particular, 
the complex plane $\complex$ will be identified with $\real^2$.
All sets considered in the paper will be Lebesgue measurable.
The two-dimensional Lebesgue measure of a set 
$A\subset\complex$ will be denoted by \lm{A}. 

Recall that 
$0$ is a strong dispersion point of $A\subset\complex$ if
\begin{equation}\label{StrDisp}
\lim_{a\to 0^+,\, b\to 0^+}
\frac{\lm{A\cap[(-a,a)\times(-b,b)]}}{\lm{(-a,a)\times(-b,b)}}=0
\end{equation}
and it is an (ordinary) dispersion point of $A$ if 
in the above limit we replace rectangles with squares, i.e., 
we take $a=b$. It is also well known that the squares can be replaced 
with the balls $B(a)=\{z\in\complex\colon |z|<a\}$, i.e., that
$0$ is a dispersion point of $A\subset\complex$ if
\begin{equation}\label{Disp}
\lim_{r\to 0^+}
\frac{\lm{A\cap B(r)}}{\lm{B(r)}}=0.
\end{equation}
A point $z\in\complex$ is a dispersion (strong dispersion) 
point of $A\subset\complex$ if it is a dispersion (strong dispersion) point 
of $A-z$, and $z\in\complex$ is a density (strong density)
point of $A$ if it is a dispersion (strong dispersion)
point of the complement of $A$.
(Compare Saks
\cite[pages 106, 128]{Saks}.) 
The strong density topology ${\cal T}_{\cal S}$ on $\complex$ is defined as 
the family of all measurable subsets $A$ of $\complex$ such that every $z\in A$ is a 
strong density point of $A$ \cite{GNN}. 
Similarly we define the density topology
${\cal T}_{\cal N}$
on $\complex$ using the notion of ordinary density point on $\complex$. 
(Compare  \cite{GNN} and \cite{LMZ:FineTop}.)
Notice that the topologies ${\cal T}_{\cal N}$ and ${\cal T}_{\cal S}$ 
are invariant under translations and under multiplications
by positive real numbers. 

Function $f\colon\complex\to\complex$ is density continuous (strongly density
continuous) at $z\in\complex$ if it is continuous with the topology 
${\cal T}_{\cal N}$ (${\cal T}_{\cal S}$) used in the domain and the range.
In particular, it is easy to see that 
$f\colon\complex\to\complex$ is density continuous (strongly density
continuous) at $z\in\complex$ if and only if for every 
$A\subset\complex\setminus\{z\}$ if $z$ is not a dispersion (strong dispersion)
point of $A$ then $f(z)$ is not a dispersion (strong dispersion)
point of $f(A)$.

In what follows we will use also the following easy fact.
It will be left without proof.  

\lem{SubStr}{ Let $A\subset B(1)$ and $R_k\subset B(1)$, $k\in\natural$,
be a such that
\[
\frac{\lm{A\cap R_k}}{\lm{R_k}}>\delta\ \text{ for all }\ k\in\natural.
\]
If the sets $K^j\subset B(1)$ for $j<n$ are disjoint and
such that $\lm{\bigcup_{j<n} K^j}=\lm{B(1)}$ then there is a $j<n$ and
an increasing sequence $\{k_p\}$ such that
\[
\frac{\lm{A\cap K^j\cap R_{k_p}}}{\lm{K^j\cap R_{k_p}}}>\delta
\ \text{ for all }\ p\in\natural.
\]
}

We will also use the following version of the change of variables
formula.

\lem{ChangeVar}{ Let $F\colon\complex\to\complex$, $U\subset\complex$ 
be an open region and let $h\colon U\to\complex$ be analytic 
with analytic inverse. Then
\[
\int_{h(U)} F\, d\lambda=\int_U (F\circ h) \cdot |h^\prime(z)|^2\, d\lambda.
\]
}

\proof This immediately follows from the standard
change of variables formula \cite[Thm. 7.26]{Rudin}
if we notice that the Jacobian
of transformation $h$ is equal to
$|\det{h^\prime(x,y)}|=|h^\prime(z)|^2$
which, in turn, follows immediately from Cauchy-Riemann equations.
(Compare \cite[p. 250 Exercise 6]{Rudin}.)
\qed

In what follows we will also use the following notation 
for $\alpha\geq 0$ and $\e,r_0>0$
\[
K(\alpha,\e,r_0)=\{z=r e^{i\varphi}\in\complex\colon 0\leq r<r_0\ \&\ 
\alpha-\e<\varphi<\alpha+\e\}.
\]

\section{Functions $bz^n$.} 

We will start with the following equivalent form of the property that
$0$ is a strong dispersion point of $A\subset\complex$. 

\lem{wycinek}{ The point $0$ is a strong dispersion point of $A\subset\complex$
if and only if for every $\alpha= m\frac{\pi}{2}$, $m\in\natural$, 
and every parameter $\e\in(0,\pi/4)$ (that might depend of $r$)
\begin{equation}\label{StrDispWycinek}
\lim_{r\to 0^+}
\frac{\lm{A\cap K(\alpha,\e,r)}}{\lm{K(\alpha,\e,r)}}=0.
\end{equation}
}

\proof Fix $A\subset\complex$. 

By way of contradiction assume first that (\ref{StrDispWycinek}) is false
for some $\alpha= m\frac{\pi}{2}$, $m\in\natural$.
Then, there exists $\delta>0$ and sequences $\e_k\in(0,\pi/4)$ and $r_k>0$ such that
$r_k$ converges to $0$ and
\[
\frac{\lm{A\cap K(\alpha,\e_k,r_k)}}{\lm{K(\alpha,\e_k,r_k)}}>\delta
\ \text{ for all }\ k\in\natural.
\]
For convenience we will assume $\alpha=0$, the other cases being similar.

\protect\begin{center}\leavevmode\epsfbox{pic1.ps}\\
\medskip
Figure 1.\end{center}


Let $a_k=r_k$ and $b_k=2\,\e_k r_k$. Then, $b_k\geq r_k \tan \e_k$, since
$2> \frac{\tan (\pi/4)}{\pi/4}> \frac{\tan \e_k}{\e_k}$. In particular, 
$K(\alpha,\e_k,r_k)\subset(-a_k,a_k)\times(-b_k,b_k)$. (See Figure 1.)
Moreover, 
$\lm{(-a_k,a_k)\times(-b_k,b_k)}=8 r_k^2 \e_k = 8\,\lm{K(\alpha,\e_k,r_k)}$.
Hence
\[
\frac{\lm{A\cap[(-a_k,a_k)\times(-b_k,b_k)]}}{\lm{(-a_k,a_k)\times(-b_k,b_k)}}
\geq
\frac{\lm{A\cap K(\alpha,\e_k,r_k)}}{8\,\lm{K(\alpha,\e_k,r_k)}}>
\frac{\delta}{8}
\]
for all $k\in\natural$, contradicting (\ref{StrDisp}).

Conversely, assume that (\ref{StrDisp}) is false, i.e., that 
there exists $\delta>0$ and sequences $a_k$, $b_k$ converging to $0$ 
such that
\[
\frac{\lm{A\cap[(-a_k,a_k)\times(-b_k,b_k)]}}{\lm{(-a_k,a_k)\times(-b_k,b_k)}}
>\delta\ \text{ for all }\ k\in\natural.
\]
Then, by Lemma \ref{SubStr} used with sets 
$K^j=\{r e^{i\varphi}\in B(1)\colon j\frac{\pi}{2}<\varphi<(j+1)\frac{\pi}{2}\}$
for $j\in\{0,1,2,3\}$,
we can assume that 
the similar property holds when in the above limit the sequence
of rectangles 
$\{(-a_k,a_k)\times(-b_k,b_k)\}$ is replaced by one of the following 
four sequences: $\{(0,a_k)\times(0,b_k)\}$, $\{(0,a_k)\times(-b_k,0)\}$,
$\{(-a_k,0)\times(0,b_k)\}$, or $\{(-a_k,0)\times(-b_k,0)\}$.
For convenience we will assume that this is the case for the first of these
sequences, i.e., that
\[
\frac{\lm{A\cap[(0,a_k)\times(0,b_k)]}}{a_k b_k}
>\delta\ \text{ for all }\ k\in\natural.
\]
Furthermore, choosing subsequence, if necessary, we can assume that
either $a_k\leq b_k$ for all $k$ or $b_k\leq a_k$ for all $k$.
We will assume that
\[
b_k\leq a_k\ \text{ for all }\ k\in\natural.
\]
Let $r_k=\sqrt{a_k^2+b_k^2}$ and $\e_k=\arctan{\frac{b_k}{a_k\delta/2}}$.
Then $r_k\leq 2\, a_k$ and 
\[
(a_k\delta/2,a_k)\times(0,b_k)\subset K(0,\e_k,r_k)\subset 
(0,2\, a_k)\times(-\frac{4}{\delta}b_k,\frac{4}{\delta}b_k).
\]
(See Figure 2.) 

\protect\begin{center}\leavevmode\epsfbox{pic2at35.ps}\\
\medskip
Figure 2.
\bigskip
\end{center}

In particular, 
$\lm{K(0,\e_k,r_k)}\leq 2\, a_k \, 2\, \frac{4}{\delta}b_k=\frac{16}{\delta} a_k b_k$
and
\begin{eqnarray*}
\lefteqn{\frac{\lm{A\cap K(0,\e_k,r_k)}}{\lm{K(0,\e_k,r_k)}}}\\
& \text{\hspace*{1.0in}} \geq &  
\frac{\lm{A\cap[(a_k\delta/2,a_k)\times(0,b_k)]}}{\frac{16}{\delta} a_k b_k}\\
& \text{\hspace*{1.0in}} \geq &  
\frac{\delta}{16}
\frac{\lm{A\cap[(0,a_k)\times(0,b_k)]}-\lm{(0,a_k\delta/2]\times(0,b_k)}}{a_k b_k}\\
& \text{\hspace*{1.0in}} > &  
\frac{\delta}{16}\left(\delta-\frac{\delta}{2}\right) = \frac{\delta^2}{32},
\end{eqnarray*}
for all $k\in\natural$, 
which contradicts (\ref{StrDispWycinek}). \qed

In what follows we will need also the following inequality.

\lem{MeasuresEst}{ Let $f(z)=e^{i\varphi}z^n$, $A\subset\complex$,
$\alpha,\beta\geq 0$ and  $\e,r>0$. If 
\begin{equation}\label{EqMesEstAss}
A\cup K(\alpha,\e,r)\subset K(\beta,\frac{\pi}{n},1)
\end{equation}
then for every $d\in(0,1)$
\[
\lm{f(A)\cap K(n\alpha+\varphi,n\e,r^n)}\geq 
n^2 (d\, r)^{2n-2} [\lm{A\cap K(\alpha,\e,r)} - d^2 \e r^2].
\]
}

\proof By (\ref{EqMesEstAss}) we can restrict $f$ to $K(\beta,\frac{\pi}{n},1)$.
Then, $f$ is one-to-one and has an analytic inverse $f^{-1}$. Hence, using 
Lemma \ref{ChangeVar}, we obtain
\begin{eqnarray*}
\lefteqn{\lm{f(A)\cap K(n\alpha+\varphi,n\e,r^n)} }\\
& \text{\hspace*{1.2in}} = &  
\lm{f(A)\cap f(K(\alpha,\e,r))}\\
& \text{\hspace*{1.2in}} = &  
\int_{f(K(\alpha,\e,r))} \charf{f(A)}\, d\lambda\\
& \text{\hspace*{1.2in}} = &  
\int_{f(K(\alpha,\e,r))} \charf{A}\circ f^{-1}\, d\lambda\\
& \text{\hspace*{1.2in}} = &  
\int_{K(\alpha,\e,r)} \charf{A}\circ f^{-1}\circ f(z)\cdot |f^\prime(z)|^2\, d\lambda\\
& \text{\hspace*{1.2in}} = &  
\int_{K(\alpha,\e,r)} \charf{A}(z)\cdot n^2|z|^{2n-2}\, d\lambda\\
& \text{\hspace*{1.2in}} \geq &  
\int_{K(\alpha,\e,r)\setminus K(\alpha,\e,d r)} \charf{A}(z)\cdot n^2|z|^{2n-2}\, d\lambda\\
& \text{\hspace*{1.2in}} \geq &  
\int_{K(\alpha,\e,r)\setminus K(\alpha,\e,d r)} 
                                       \charf{A}(z)\cdot n^2(d\, r)^{2n-2}\, d\lambda\\
& \text{\hspace*{1.2in}} = &  
n^2(d\, r)^{2n-2}\left[
\int_{K(\alpha,\e,r)}\charf{A}\, d\lambda - \int_{K(\alpha,\e,d r)}\charf{A}\, d\lambda\right]\\
& \text{\hspace*{1.2in}} \geq &  
n^2(d\, r)^{2n-2}\left[
\lm{A\cap K(\alpha,\e,r)} - \lm{K(\alpha,\e,d r)}\right]\\
& \text{\hspace*{1.2in}} = & 
n^2 (d\, r)^{2n-2} [\lm{A\cap K(\alpha,\e,r)} - d^2 \e r^2].
\end{eqnarray*}
\qed

Now, we are ready for the proof of the main lemma.

\lem{zTOn}{ Let $b\in\complex$ and $n\in\natural$, $n\geq 1$. 
Then the function $f(z)=b z^n$ is density continuous at $0$. Moreover,
it is strongly density continuous at $0$ if and only if
$b$ is either real or imaginary number. }

\proof If $b=0$ then the lemma is certainly true. So, assume that
$b\neq 0$. The topologies 
${\cal T}_{\cal N}$ and ${\cal T}_{\cal S}$ 
are invariant under multiplications
by positive real numbers. So, without loss of generality we can
assume that $|b|=1$, i.e., that $b=e^{i\varphi}$ for some $\varphi\geq 0$.

To prove that $f$ is density continuous at $0$ let $A\subset B(1)$ be
such that $0$ is not a dispersion point of $A$. We will show that $0=f(0)$
is not a dispersion point of $f(A)$. 
To this order first notice that, by (\ref{Disp}),
there exists a sequence $r_k\in (0,1)$ converging to $0$
and $\delta\in(0,1)$ such that
\[
\frac{\lm{A\cap B(r_k)}}{\lm{B(r_k)}}>\delta\ \text{ for all }\ k\in\natural.
\]
Then, by Lemma \ref{SubStr} used with sets 
$K^j=K(\frac{j\pi}{2n},\frac{\pi}{4n},1)$, $j<4n$, 
we can assume that for some $\alpha=\frac{j\pi}{2n}$
we have $A\subset K(\alpha,\frac{\pi}{4n},1)$
and 
\[
\frac{\lm{A\cap K(\alpha,\frac{\pi}{4n},r_k)}}{\lm{K(\alpha,\frac{\pi}{4n},r_k)}}
>\delta\ 
\text{ for all }\ k\in\natural.
\]
Then, by Lemma \ref{MeasuresEst} used with $d=\delta/2$, and the above we have
\begin{eqnarray*}
\frac{\lm{f(A)\cap B(r_k^n)}}{\lm{B(r_k^n)}}
& \geq &
\frac{\lm{f(A)\cap K(n\alpha+\varphi,\frac{\pi}{4},r_k^n)}}
                                       {\pi r_k^{2n}}\\
& \geq &
\frac{n^2 (\frac{\delta}{2}\, r_k)^{2n-2} 
    [\lm{ A\cap K(\alpha,\frac{\pi}{4n},r_k) } - \frac{\delta^2}{4}\frac{\pi}{4n} r_k^2]}
{4n\frac{\pi}{4n}r_k^{2n}}\\
& = & \frac{n}{4} \left(\frac{\delta}{2}\right)^{2n-2}\left[
\frac{ \lm{ A\cap K(\alpha,\frac{\pi}{4n},r_k) } }{\lm{K(\alpha,\frac{\pi}{4n},r_k)}}
- \frac{\frac{\delta^2}{4}\frac{\pi}{4n} r_k^2}{\frac{\pi}{4n} r_k^2}
\right]\\
& > & \frac{n}{4} \left(\frac{\delta}{2}\right)^{2n-2}
\left[\delta - \frac{\delta^2}{4} \right]>0
\end{eqnarray*}
for every $k\in\natural$. 
Therefore, by (\ref{Disp}), $0$ is not a dispersion point of $f(A)$.
We proved that $f$ is density continuous at $0$.

To prove the second part, assume first that $b=e^{i\varphi}$ is real or imaginary.
Thus, $\varphi=m\frac{\pi}{2}$ for some $m\in\natural$. 
Let $A\subset B(1)$ be
such that $0$ is not a strong dispersion point of $A$. 
We will show that $0=f(0)$
is not a strong dispersion point of $f(A)$. 

By Lemma \ref{wycinek} we can find $\delta\in(0,1)$, 
$\beta= p\frac{\pi}{2}$, where $p\in\natural$, and sequences 
$\e_k\in(0,\pi/4)$ and $r_k\in(0,1)$ such that $r_k$ converges to $0$ and 
\[
\frac{\lm{A\cap K(\beta,\e_k,r_k)}}{\lm{K(\beta,\e_k,r_k)}}>\delta
\ \text{ for all }\ k\in\natural.
\]
By Lemma \ref{SubStr} used with sets 
$K^j=K(\frac{j\pi}{2n},\frac{\pi}{4n},1)$, $j<4n$, 
we can assume that for some $\alpha=\frac{j\pi}{2n}$
\[
\frac{\lm{A\cap K(\beta,\e_k,r_k)\cap K(\alpha,\frac{\pi}{4n},1)}}
{\lm{K(\beta,\e_k,r_k)\cap K(\alpha,\frac{\pi}{4n},1)}}
>\delta\ 
\text{ for all }\ k\in\natural.
\]
We can also assume that either 
$\e_k> \frac{\pi}{4n}$ for all $k$ or 
$\e_k\leq \frac{\pi}{4n}$ for all $k$.
However, the first case implies that $0$ is not an ordinary dispersion
point of $A$, since in this case we would have
\[
\frac{\lm{A\cap B(r_k)}}{\lm{B(r_k)}}
\geq
\frac{\lm{A\cap K(\beta,\e_k,r_k)}}{4n\, \lm{K(\beta,\e_k,r_k)}}>\frac{\delta}{4n}
\]
for all $k\in\natural$. 
Thus, by the first part of the Lemma, $0$ is not
a (strong) dispersion point of $f(A)$.

So, assume that $\e_k\leq \frac{\pi}{4n}$ for all $k$.
But then, $\alpha=\beta$ since otherwise 
$K(\beta,\e_k,r_k)\cap K(\alpha,\frac{\pi}{4n},1)=\emptyset$.
Hence,
$K(\beta,\e_k,r_k)\cap K(\alpha,\frac{\pi}{4n},1)=K(\alpha,\e_k,r_k)$
and
\[
\frac{\lm{A\cap K(\alpha,\e_k,r_k)}}{\lm{K(\alpha,\e_k,r_k)}}
>\delta\ 
\text{ for all }\ k\in\natural.
\]
We can also assume that $A\subset K(\alpha,\frac{\pi}{4n},1)$.
Then, by Lemma \ref{MeasuresEst} used with $d=\delta/2$, and the above we have
\begin{eqnarray*}
\lefteqn{\frac{\lm{f(A)\cap K(n\alpha+\varphi,n\e_k,r_k^n)}}
                                   {\lm{K(n\alpha+\varphi,n\e_k,r_k^n)}} }\\
& \text{\hspace*{1.5in}} \geq &
\frac{n^2 (\frac{\delta}{2}\, r_k)^{2n-2} 
    [\lm{ A\cap K(\alpha,\e_k,r_k) } - \frac{\delta^2}{4}\e_k r_k^2]}
   {n \e_k r_k^{2n}}\\
& \text{\hspace*{1.5in}} = & n \left(\frac{\delta}{2}\right)^{2n-2}\left[
\frac{ \lm{ A\cap K(\alpha,\e_k,r_k) } }{\lm{K(\alpha,\e_k,r_k)}}
          - \frac{\frac{\delta^2}{4}\e_k r_k^2}{\e_k r_k^2}
                                                          \right]\\
& \text{\hspace*{1.5in}} > & 
n \left(\frac{\delta}{2}\right)^{2n-2}
\left[\delta - \frac{\delta^2}{4} \right]>0
\end{eqnarray*}
for every $k\in\natural$. 
But notice that $n\alpha+\varphi=(j+m)\frac{\pi}{2}$.
Therefore, by Lemma \ref{wycinek}, $0$ is not a strong dispersion point of $f(A)$.
We proved that $f$ is strong density continuous at $0$.

\protect\begin{center}\leavevmode\epsfbox{pic3.ps}\\
\medskip
Figure 3.
\medskip
\end{center}

To finish the proof let us assume that $b=e^{i\varphi}$ is neither real nor imaginary.
Thus, $\varphi=p\frac{\pi}{2}$ for some $p>0$, $p\not\in\natural$. 
Let $A=\{(x,y)\colon x>0\ \&\ -x^2<y<x^2\}$. 
It is easy to see that
$0$ is not a strong dispersion point of $A$.
On the other hand, $f(A)$
does not contain any axis. (See Figure 3.)
Using this fact it is not difficult to argue that
$0$ is a strong dispersion point of $f(A)$. \qed

\section{General case}

We will need the following fact.

\lem{GenAppr}{ Let $f$ be analytic on a neighborhood of 
$0$ and assume that $f(z)=\sum_{k=n}^\infty a_k z^k$
where $n>0$ and $a_n\neq 0$. If $g(z)=a_n z^n$ then there exists $r>0$ such that
for every $A\subset B(r)$
\begin{equation}\label{GenAprCon}
\frac{1}{4n}\lm{g(A)}\leq \lm{f(A)}\leq 4\lm{g(A)}.
\end{equation}
In particular, $f$ is density (strongly density) continuous at $0$
if and only if $g$ is density (strongly density) continuous at $0$.
}

\proof Notice that condition (\ref{GenAprCon}) imples that $0$ is a dispersion
(strong dispersion) point of $f(A)$ if and only if
$0$ is a dispersion
(strong dispersion) point of $g(A)$. Thus, the
additional part follows immediately from (\ref{GenAprCon}).

To prove (\ref{GenAprCon}) let us first choose
$r_0>0$ such that $f^\prime$ does not have any zeros in 
$B(r_0)\setminus\{0\}$.

Notice that it is enough to prove that for every $j<n$ we can find $r>0$
such that 
\[
\frac{1}{4}\lm{g(A)}\leq \lm{f(A)}\leq 4\lm{g(A)}.
\]
holds for all
$A\subset K^j=K(\alpha,\frac{\pi}{n},r)$, where $\alpha=\frac{2\pi j}{n}$.
So, choose $\alpha=\frac{2\pi j}{n}$
and consider functions $f$ and $g$ as restricted to $K^j$. Then,
$g$ is one-to-one and has an analytic inverse $g^{-1}\colon g(K^j)\to K^j$. 

Put $h=f\circ g^{-1}$. Then, $h$ is analytic and
\begin{eqnarray*}
h^\prime(z)
& = &
f^\prime(g^{-1}(z)) (g^{-1})^\prime(z)\\
& = & 
\left(\sum_{k=n}^\infty k a_k (g^{-1}(z))^{k-1}\right)
\frac{1}{n a_n(g^{-1}(z))^{n-1}}\\
& = & 
\sum_{k=n}^\infty \frac{k a_k}{n a_n} (g^{-1}(z))^{k-n}
\end{eqnarray*}
Thus, we can pick $r\in(0,r_0)$ such that
$|h^\prime(z)-1|=|h^\prime(z)-h^\prime(0)|<\frac{1}{2}$, i.e., that
\begin{equation}\label{ConAAA}
\frac{1}{2}<|h^\prime(z)|<\frac{3}{2}<2
\end{equation}
for all $z\in g(K^j)$ with $|z|<r$. 
We will show that this choice of $r$ implies (\ref{GenAprCon}).

Since $f^\prime$ does not have any zeros in
$Z=K^j\cap B(r)\setminus\{0\}$, the set $Z$ can be covered by open sets
$S\subset Z$ such that $f$ has an inverse 
$f^{-1}\colon f(S)\to S$. 
So, pick $S\subset Z$ with this property.
Since Lebesgue measure is countable additive, we may assume that $A\subset S$.
Then, function $h$ restricted to $g(S)$ 
is one-to-one and, by Lemma \ref{ChangeVar},
\begin{eqnarray*}
\lm{f(A)}
& = &
\int_{f(S)} \charf{f(A)}\, d\lambda\\
& = & 
\int_{h(g(S))} \charf{A}\circ f^{-1}\, d\lambda\\
& = & 
\int_{g(S)} (\charf{A}\circ f^{-1}\circ f\circ g^{-1}(z))\cdot|h^\prime(z)|^2\, d\lambda\\
& = & 
\int_{g(S)} \charf{g(A)}(z)\cdot|h^\prime(z)|^2\, d\lambda
\end{eqnarray*}
But, by (\ref{ConAAA}), $\frac{1}{4}<|h^\prime(z)|^2<4$ for all 
$z\in g(S)\subset g(K^j)$.
Thus
\[
\lm{f(A)}=\int_{g(S)} \charf{g(A)}(z)\cdot|h^\prime(z)|^2\, d\lambda
\geq \int_{g(S)} \charf{g(A)}(z)\cdot\frac{1}{4}\, d\lambda
=\frac{1}{4} \lm{g(A)}
\]
and
\[
\lm{f(A)}=\int_{g(S)} \charf{g(A)}(z)\cdot|h^\prime(z)|^2\, d\lambda
\leq \int_{g(S)} \charf{g(A)}(z)\cdot 4\, d\lambda
=4 \lm{g(A)}.
\]
\qed


Now, we are ready for the main theorems. 

\thm{main1}{ Every analytic function is density continuous. }

\proof It follows immediately from Lemmas \ref{zTOn}, \ref{GenAppr}
and the fact that the topology ${\cal T}_{\cal N}$
is invariant under translations. \qed

\thm{main2}{ A non-constant analytic function $f$ is strongly density continuous
at $z$ if and only if $f^{(n)}(z)$ is either real or imaginary
number, where $n=\min\{k>0\colon f^{(k)}(z)\neq 0\}$.
}

\proof Since topology ${\cal T}_{\cal S}$
is invariant under translations we can assume that $z=0$.
Then, Theorem follows immediately from Lemmas \ref{zTOn} and \ref{GenAppr}.
\qed

\thm{mainCor}{ An analytic function $f$ is strongly density continuous
on its domain
if and only if $f(z)=a + bz$ where $a,b\in\complex$ and 
$b$ is either real or imaginary number.
}

\proof First assume that $f(z)=a + bz$ where $a,b\in\complex$.
If $f$ is constant, then evidently $f$ is 
strongly density continuous. So, assume $f$ is not constant
and pick an arbitrary $z\in\complex$.
Then
$\min\{k>0\colon f^{(k)}(z)\neq 0\}=1$ and $f^{(1)}(z)=b$.
So, 
by Theorem \ref{main2}, $f$ is strong density continuous
at $z$ if and only if $b$ is either real or imaginary number.

So, assume that $f$ is not linear. Then, $f^\prime$ is analytic and not constant.
Therefore, there exists $z$ in the domain of $f$ such that 
$f^\prime(z)$ is neither  real nor imaginary number.
Thus, 
by Theorem \ref{main2}, $f$ is not strongly density continuous
at $z$. \qed


\begin{thebibliography}{22}

\bibitem{CW:DensCTran} Krzysztof Ciesielski, W\l adys\l aw Wilczynski,
Density continuous transformations on $\real^2$, Real Analysis Exchange, 
to appear.

\bibitem{GNN} C. Goffman, C.J. Neugebauer, T. Nishiura, 
Density topology and approximate continuity, 
{\sl Duke Math. J. 28} (1961), 497--506.

\bibitem{LMZ:FineTop} Jaroslav Luke\v{s}, Jan Mal\'{y},
 Lud\v{e}k Zaj\'{\i}\v{c}ek,
Fine Topology Methods in Real Analysis and Potential Theory,
{\sl Lecture Notes in Mathematics 1189}, Springer-Verlag 1986.

\bibitem{Rudin} W. Rudin, {\sl Real and Complex Analysis}, 
McGraw-Hill, Inc., 1976.

\bibitem{Saks} S. Saks, {\sl Theory of the integral}, 
Monografie Matematyczne, Warsaw 1937.

\end{thebibliography}


\end{document}