% Peano curve and \I-approximate differentiability

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\documentclass{rae}
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\firstpagenumber{608}
%\received{September 30, 1993}

\Volume{17}
\IssueNumber{2}
\Year{1991/92}

\markboth{Cieselski and Larson}{Peano Curve}

\keywords{$\cal I$-density, Peano curve, differentiable}
\MathReviews{26A21}



\title{The Peano curve and \I-approximate differentiability}

\author{Krzysztof Ciesielski, Department of Mathematics, West Virginia
University, Morgantown, WV 26506
\and
Lee Larson, Department of Mathematics,
University of Louisville, Louisville, KY 40292
\thanks{Received support from the University of Louisville Graduate School}}
%\email{Larson: lmlars01@ulkyvx.louisville.edu, 
%Ciesielski:\\ kcies@wvnvms.wvnet.edu}


\def\integer{\Bbb Z}
\def\natural{\Bbb N}
\def\rational{\Bbb Q}
\def\real{\Bbb R}



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%sets with the Baire property
     \newcommand{\BaireSets}{\mbox{$\cal B$}}
%the ideal of first category sets
     \newcommand{\I}{\ifmmode{\cal I}\else\mbox{$\cal I$}\fi}
     \newcommand{\T}{\ifmmode{\cal T}\else\mbox{$\cal T$}\fi}
%I-density topology
     \newcommand{\itopology}{\mbox{$\T_{\cal I}$}}
%deep I-density topology
     \newcommand{\deepitopology}{\mbox{$\T_{\cal D}$}}
 
%%%%%%%%%%%%%%%%%%%%%% FUNCTION OPERATIONS
 
%inverse of a function
     \newcommand{\inv}[1]{\mbox{$#1^{-1}$}}
%reals to reals
     \newcommand{\RtoR}{\mbox{$\colon\real\to\real$}}
 
%%%%%%%%%%%%%%%%%%%%%% SET OPERATIONS
 
%interior of a set
     \newcommand{\interior}[1]{\mbox{$\mbox{{\rm int}}\left(#1\right)$}}
%closure of a set
     \newcommand{\closure}[1]{\mbox{$\mbox{\rm cl}\left(#1\right)$}}
%complement of a set
     \renewcommand{\complement}[1]{\mbox{$#1^c$}}
% big union
     \renewcommand{\Cup}{\bigcup}
% characteristic function
     \newcommand{\charf}[1]{\mbox{\raise.48ex\hbox{$\chi$}$_{#1}$}}
%measure of a set
     \newcommand{\measure}[1]{\mbox{$\mbox{m}(#1)$}}
%density points
     \newcommand{\idensitypts}[1]{\mbox{$\Phi_{\cal I}(#1)$}}
     \newcommand{\deepidensitypts}[1]{\mbox{$\Phi_{\cal D}(#1)$}}

\newtheorem{theorem}{Theorem}%[section]
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{Lemma}
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#2\end{example}}
 
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     \newcommand{\iAppDer}{\mbox{D}^{(\I)}}
 
 
%%%%%%%%%%%%%%%%%%%%%   MISCELLANEOUS
 
\newcommand{\e}{\mbox{$\varepsilon$}}
\renewcommand{\d}{\mbox{$\delta$}}
\newcommand{\idens}{\I-den\-sity}
\newcommand{\idisp}{\I-dispersion}
\hyphenation{den-sity den-sities}

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\begin{document}


\maketitle
 
\section{Preliminaries}
 
A function $f\RtoR$ is {\em density continuous} ({\em \I-density continuous},
{\em deep-\I-den\-sity continuous\/}) at the point $x$ if it is
continuous at $x$ when the density topology (\I-density topology,
deep-\I-density  topology) is used on both the domain and the range 
\cite{CL:SpDensCont,CL:Examples,CLO:DiffDensCont,Wilczynski:CatAn}.
In \cite{CLO:DiffDensCont} it is proved that the first coordinate of
the classical Peano area-filling
curve is nowhere approximately differentiable, even though it is
continuous and density continuous. In this paper we generalize this result by
proving in Section 3 that the same function is also \I-density and
deep-\I-density continuous, even though it is nowhere \I-approximately
differentiable. To prove this it is shown that a point $x$ is a deep-\I-density
point of the Baire set $E$ if, and only if, $x$ is an \I-density point of 
the unique regular open set $\tilde E$ such that the symmetric difference
$E\triangle \tilde E$ is of the first category.
 
In Section 4 we give an example of a bounded
\I-approximately continuous function that is not a derivative. 
 
 
The notation used throughout this paper is standard. In particular,
$\real$ stands for the set of real numbers and $\natural=\{1,2,3,\ldots\}$.
For $A,B\subset\real$ and $d\in\real$ the
complement of $A$ is denoted by \complement{A}, 
the symmetric difference of $A$ and $B$ is denoted by 
$A\triangle B=(A\cup B)\setminus(A\cap B)$,
while $B - d = \{x - d\in\real\colon x\in B\}$
and  $dB =\{dx\in\real\colon x\in B\}$.
The symbol \BaireSets\  stands for the family of subsets of $\real$ which
have the Baire property  and \I\ denotes the ideal of first category
subsets of $\real$.  A statement about a subset of $\real$ is true
\I-a.e. if the set on which it fails to be true is in \I. An open set
$E\subset\real$ is {\em regular }
if $E=\interior{\closure E}$. For a set $E\in\BaireSets$ we denote by $\tilde E$
the only regular open set $A$ for which $E\triangle A\in\I$. The Lebesgue
measure of a measurable set $A$ is denoted by $\measure{A}$.
 
 
If $A\in\BaireSets$, then $0$ is an {\em \idens\ point\/} of $A$ if
for every increasing sequence $\{n_m\}_{m\in\natural}$ of natural
numbers there exists a subsequence $\{n_{m_p}\}_{p\in\natural}$ such that
\[
\lim_{p\to\infty}\charf{n_{m_p}A\cap(-1,1)} = \charf{(-1,1)},\ \mbox{\I-a.e.},
\]
or, equivalently, that the set
\[
\liminf_{p\to\infty}n_{m_p}A =
\bigcup_{q\in\natural} \bigcap_{p\ge q} n_{m_p}A
\]
is residual in $(-1,1)$. 
We say that a point $a$ is an 
\idens\ point of $A\in\BaireSets$ if $0$ is an \idens\ point of $A-a$.
The set of all \idens\ points of $A\in\BaireSets$ is denoted by 
\idensitypts{A}. It is not difficult to see that 
$A\triangle\idensitypts{A}\in\I$ for every $A\in\BaireSets$
\cite[Theorem 3]{Wilczynski:CatAn} and that 
\begin{equation}\label{con_idens}
\idensitypts{A}=\idensitypts{B}\,\,\,
\mbox{ for every $A,B\in\BaireSets$ such that }\,\,\,  A\triangle B\in\I. 
\end{equation}
The family of sets
\[
\itopology=\{A\in\BaireSets\colon A\subset\idensitypts{A}\}
\]
forms a topology on $\real$ called
the {\em \idens\ topology\/} \cite{PWW:CatAn,Wilczynski:CatAn}.
 
A point $a\in\real$ is a {\em deep-\idens\ point\/}
\cite{Wilczynski:CatAn} of an $A\in\BaireSets$ if there exists a
closed set $F\subset A\cup \{a\}$ such that $a$ is an \idens\
point of $F$.
The set of all deep-\idens\ points of $A\in\BaireSets$ is denoted by 
\deepidensitypts{A}.
The family of sets
\[
\deepitopology=\{A\in\BaireSets\colon A\subset\deepidensitypts{A}\}
\]
forms a topology on $\real$ called the {\em deep-\idens\ topology\/} 
\cite{Lazarow:Coarsest,Wilczynski:CatAn}.
 
A point
$x$ is a {\em dispersion} ({\em \I-dispersion}, {\em deep-\I-dispersion})
point of $A$ if $x$ is a density (\idens, deep-\idens) point of
\complement{A}. 
 
A function $f\RtoR$ is said to be  
{\em \I-approximately differentiable at a point $x$} if there exists 
a number $\iAppDer f(x)$, called the {\em \I-approximate derivative}
of $f$ at $x$, such that for every
$\varepsilon >0$, $x$ is an \I-density point of some Baire subset of
\[
\left\{t\in\real\colon \frac{f(t)-f(x)}{t-x}\in
(\iAppDer f(x)-\varepsilon,\iAppDer f(x)+\varepsilon)\right\}.
\]
(Compare also \cite{LW:IAppDer:21A} and \cite[Definition 8]{Wilczynski:CatAn}.)
 
\bigskip
 
We also need the following easy fact
\cite[Lemma 4]{AversaWilczynski:Homeomorphisms}.
 
\lem{lemma36}{If  $B = \bigcup_{n\in\natural} (a_n,b_n)$ is 
such that $\lim_{n\to\infty} b_n=0$
and there exists a positive number $c$ such that
\[
\frac{b_n - a_n}{b_n}>c,
\]
for every $n\in\natural$, then $0$ is not an \I-dispersion
point of $B$.}
 
\section{Basic Lemmas}
 
We start this section with the following lemma.
 
\lem{lem:regopen}{If $A$ is regular open, then 
$\idensitypts{A}=\deepidensitypts{A}$.}
 
Proof. The inclusion $\deepidensitypts{A}\subset\idensitypts{A}$ is obvious
from the definitions.
 
To prove the converse inclusion let $x\in\idensitypts{A}$. For simplicity we
assume that $x=0$. We must show that $0$ is a deep-\I-density point of $A$.
 
But $0$ is an \I-density point 
of $A$ if, and only if,  $0$ is an
\I-dispersion point of $A^c$.  Without using the full strength of Theorem 1 from
\cite{EL:BaireIApp} it follows that $0$ is an \I-density point of $A$ if, and
only if,
\begin{list}% 
{ }{\setlength{\rightmargin}{\leftmargin}}
\item for every nonempty interval $(a,b)\subset(-1,1)$ there exist
numbers $\varepsilon > 0$ and $n_0\in\natural$ such that for every $n\geq
n_0$ there is a nonempty interval $(c,d)\subset (a,b)$ with the properties:
\[
|d-c| > \varepsilon \mbox{\,\,\, and \,\,\,} (c,d)\cap n\widetilde{A^c} =
\emptyset.
\]
\end{list}
 
Notice that in the above it is
enough to consider only intervals $(a,b)$ with rational endpoints. Moreover,
by the regularity of $A$, 
$\closure{\widetilde{\complement{A}}}= \complement{A}$, 
and so 
\begin{eqnarray*}
(c,d)\cap n\widetilde{A^c} = \emptyset \, 
      & \Longleftrightarrow \,& (c,d)\cap n A^c = \emptyset \, \\
      & \Longleftrightarrow \,& \frac{1}{n}(c,d)\cap A^c = \emptyset \, \\
      & \Longleftrightarrow \,& \frac{1}{n}(c,d) \subset A.
\end{eqnarray*}
Hence, we can conclude that $0$ is an \I-density point of $A$ if, and
only if,
 \begin{list}%
{ {\bf ($\star$)}}{\setlength{\rightmargin}{\leftmargin}}
\item for every nonempty interval
$I=(a,b)\subset(-1,1)$ with rational endpoints there exists a
number $\varepsilon_I > 0$ and $n_I\in\natural$ such that for every $n \geq n_I$
there is a nonempty interval 
$(c,d)\subset [c,d]\subset (a,b)$ with the properties:
\[
|d-c| > \varepsilon_I \mbox{\,\,\, and \,\,\,} \frac{1}{n}[c,d]\subset A.
\]
\end{list}
 
Now we can construct a closed set $E\cup\{0\}\subset A\cup\{0\}$ which satisfies
($\star$). This will finish the proof.
 
Let $\{I_k\}_{k\in\natural}$ be an
enumeration of all nonempty subintervals of
$(-1,1)$ with rational endpoints and assume that 
$n_{I_k} \geq k$ for every $k\in\natural$.
For $k\in\natural$, let $E_k$ be a union of intervals $\frac{1}{n}[c_n,d_n]$
for $n \geq n_{I_k}$, where $[c_n,d_n]$ is a subset of $I_k$ satisfying
($\star$); i.e., such that
\[
|d_n-c_n| > \varepsilon_{I_k} \mbox{\,\,\, and \,\,\,}
\frac{1}{n}[c_n,d_n]\subset A.
\]
Let
\[
E = \bigcup_{k\in\natural} E_k.
\]
 
Notice that 
\[
E_k\subset (-\frac{1}{n_{I_k}},\frac{1}{n_{I_k}})
\subset (-\frac{1}{k},\frac{1}{k})
\]
so that  $E_k \setminus (-\frac{1}{n},\frac{1}{n})$ 
intersects only finitely many closed intervals forming $E_k$.
In particular, 
\[
E \setminus (-\frac{1}{k},\frac{1}{k}) = 
\bigcup_{i<k} E_i \setminus (-\frac{1}{k},\frac{1}{k}),
\]
where the right-hand side intersects only finitely many closed intervals from
the collection whose union forms
$E$. This implies that $E\cup\{0\}$ is closed.
This finishes the proof of Lemma \ref{lem:regopen}.
 
\lem{lem:homeom}{ Let $f\colon\real\to\real$ be such
that $f^{-1}(E)\in\I$  for every $E\in\I$. Then $f$ is deep-\I-density
continuous if, and only if, $f$ is \I-density continuous. 
}
 
Proof. It is known that every \I-density continuous function is deep-\I-density
continuous \cite[Theorem 4.1(iv)]{CL:Examples}. 
To prove the converse implication 
choose a deep-\I-density continuous function $f$ satisfying the assumption and
let $f(x)$ be an \idens\ point of  $E\in\BaireSets$ with
$f(x)\in E$. Then, by (\ref{con_idens}), $f(x)$ is an \idens\ point 
of $\tilde E$ and, by the
regularity of $\tilde E$ and Lemma \ref{lem:regopen},
$f(x)$ is also a deep-\idens\ point of
$\tilde E$. Thus, $x$ is a deep-\idens\ point of $f^{-1}(\tilde E)$.
Moreover, by  the assumption,
\[
f^{-1}(\tilde E)\triangle f^{-1}(E) = f^{-1}(\tilde E\triangle E)\in\I.
\]
So, by (\ref{con_idens}), 
$x$ is an \idens\ point of $f^{-1}(E)$.
 
\bigskip
 
We will need also the following characterization of a deep-\I-density point.
 
\lem{lem:Zajicek}{The following conditions are equivalent:
\begin{description}
\item[(i)]    $x$ is a deep-\I-density point of $A$;
\item[(ii)]   given $s\in(0,1)$, there exists $D_s>0$ and $R_s\in(0,1)$ such
              that whenever $0<D<D_s$ and 
              $(y-\delta,y+\delta)\subset(x-D,x+D)\setminus\{x\}$ with
              $2\delta /D>s$, then there is an interval
              $J\subset(y-\delta,y+\delta)\cap A$ with 
              $\measure{J}/2\delta>R_s$.
\end{description}
}
 
Proof. Without any loss of generality it may be assumed that $x=0$. In 
\cite[Theorem (5)]{Zajicek:AlternativeDefinition} Zaj\'{\i}\v{c}ek proves
that $0$ is an \I-density point of $A$ if, and only if, (ii) is satisfied,
where the inclusion $J\subset(y-\delta,y+\delta)\cap A$ is replaced by
\begin{equation}\label{con_Zai}
\left[(y-\delta,y+\delta)\cap A\right]\setminus J\in \I.
\end{equation}
But we can assume that $A$ is open, since there is a closed set 
$F\subset A\cup\{0\}$ for which $0$ is an \I-density point and then,
by Lemma \ref{lem:regopen}, $0$ is also an \I-density point of
$\tilde F=\interior F$. But then (\ref{con_Zai}) implies that 
$J\subset(y-\delta,y+\delta)\cap A$. Lemma \ref{lem:Zajicek} is proved.
 
\section{Main Theorem}
 
\thm{nowhereIappDiff}{There exists a continuous,
density continuous and \I-density continuous 
function $f$ which is nowhere approximately and
\I-approximately differentiable. }
 
Proof. 
We begin by defining a version of the classical Peano area-filling
curve\index{Peano curve} $P:[0,1]\to [0,1]\times [0,1]$. To do this, a sequence
of continuous paths
$P_n:[0,1]\to [0,1]\times [0,1]$ for $n=0,1,\ldots$, are defined which
converge uniformly to $P$.
This definition is facilitated by the following basic construction,
which will be referred to as BCP.
 
Given a square $[a,b]\times[c,d]$ with one of its diagonals a
parametrized constant speed path
$\lambda:[\alpha,\beta]\to[a,b]\times[c,d]$, we construct a new
path $\lambda_1:[\alpha,\beta]\to[a,b]\times[c,d]$ as shown in Figure
\ref{PeanoIterate}, 
where the speed of the new path is constant and three times the speed of
$\lambda$. 
 
%\putpicture{4.13325}{1.522}{PeanoIterate.eps}{Basic
%construction BCP.}{PeanoIterate}  

\putepsf{4.13325}{1.522}{PeanoIterate.ps}{Basic
construction BCP.}{PeanoIterate}  
 
Using symmetries, BCP can be applied to
either of the two diagonals of any square with either path orientation. 
Also, if $\|G\|_\infty=\sup_x|G(x)|$, then it is clear that 
\begin{equation}\label{PeanoDistance}
\|\lambda-\lambda^\prime\|_\infty\le\sqrt{(b-a)^2+(d-c)^2}
\end{equation}
for every $\lambda^\prime:[\alpha,\beta]\to[a,b]\times[c,d]$.
 
To construct the Peano curve, let $P_0(t)=(t,t)$ and define $P_1$ by
applying BCP to $P_0$. The image of $P_1$ consists of a diagonal from
each of the nine squares
\[
\left[ \frac{i}{3},\frac{i+1}{3}\right]\times
\left[ \frac{j}{3},\frac{j+1}{3}\right],\quad i,j=0,1,2.
\] 
(See Figure \ref{PeanoIterate}
with $a=c=0$ and $b=d=1$.)
Construct $P_2$ by applying BCP to each 
of the diagonals of these squares as shown in Figure
\ref{PeanoIterate2}.

%\putpicture{3.94}{1.522}{PeanoIterate2.eps}{Construction of 
%$P_2$.}{PeanoIterate2} 

\putepsf{3.94}{1.522}{PeanoIterate2.ps}{Construction of 
$P_2$.}{PeanoIterate2} 

 
This process can be continued inductively in the obvious way to form
the sequence $P_n$, $n\in\natural$. From (\ref{PeanoDistance}) it follows
that
\[
\| P_n-P_m\|_\infty\le\sqrt{2}\,3^{-\min(n,m)}.
\]
This shows that $P_n$ converges uniformly to $P$. It is also easy to
see that the image of $P$ is a dense, compact subset of
$[0,1]\times[0,1]$, so $P$ is an area-filling curve.
 
If $P=(p_1,p_2)$, where $p_i:[0,1]\to [0,1]$, $i=1,2$ are the
coordinate functions for $P$, then we claim $f=p_1$ is a function
satisfying the conditions of Theorem \ref{nowhereIappDiff}.
 
To see this, it might be helpful to see how $f$ can be defined directly as a
uniformly convergent sequence of continuous functions
$f_n:[0,1]\to[0,1]$, where each $f_n$ is the first coordinate of $P_n$. The
first coordinate of BCP can be represented by the construction shown
in Figure \ref{PeanoX}. A similar construction can be done with either 
diagonal via an obvious
reflection. This construction is denoted BCX.
 
%\putillustration{4.7}{3.02}{PeanoX.cill}{Basic construction
%BCX.}{PeanoX}

\putepsf{4.7}{3.02}{PeanoX.ps}{Basic construction
BCX.}{PeanoX}


 
Notice that Figure \ref{PeanoX}(B) also represents 
$f_1:[0,1]\to[0,1]$, if we take $a=\alpha=0$ and $b=\beta=1$. To form
$f_2$ it is enough to apply
BCX to each linear segment of $f_1$. Then, apply BCX to each linear
segment of $f_2$ to arrive at $f_3$, etc. 
 
Evidently, $f$ is continuous, as the first coordinate of the continuous function
$P$.  Also, as proved in \cite{CLO:DiffDensCont},
it is density continuous and nowhere approximately differentiable.
 
In the rest of the proof, we will need the following easy
observations.
 
The function $P$ is self-similar in the sense that for every
$n\in\natural$ and every $i=0,1,\ldots,9^n-1$, there exist 
$l(i),r(i)\in\{0,1,\ldots,3^n-1\}$ such that the image
\begin{equation}\label{Peano1a}
P\left(\left[\frac{i}{9^n},\frac{i+1}{9^n}\right]\right)
=\left[\frac{l(i)}{3^n},\frac{l(i)+1}{3^n}\right]
\times
\left[\frac{r(i)}{3^n},\frac{r(i)+1}{3^n}\right],
\end{equation}
and the path followed is a scaled and reflected copy of the entire path
of $P$ in $[0,1]\times[0,1]$.
Since $f$ is the first coordinate of $P$, condition (\ref{Peano1a}) implies 
also that for each integer
$i\in\{0,1,\ldots,9^n-1\}$, there is an integer
$l(i)\in\{0,1,\ldots,3^n-1\}$ such that \begin{equation}\label{Peano1}
f\left(\left[\frac{i}{9^n},\frac{i+1}{9^n}\right]\right)
=\left[\frac{l(i)}{3^n},\frac{l(i)+1}{3^n}\right].
\end{equation}
 
Also notice the following easy geometrical fact.
 
For every $t\in\natural$, $t>1$, and nonempty interval $(a,b)\subset[0,1]$
there are $i,n\in\natural$ such that 
\begin{equation}\label{Peano2}
K=\left[\frac{i}{t^n},\frac{i+1}{t^n}\right]\subset(a,b)\ \ 
\mbox{ and }\ \  \frac{\measure{K}}{b-a}\geq\frac{1}{2t}
\end{equation}
 
To see this, let $n$ be the smallest natural number such that
$$1/t^n<(b-a)/2.$$ Thus, $2/t^{n-1}\geq(b-a)$ and there exists $i$ such
that $i/t^n\in(a,(b+a)/2)$. Hence,
$K=[i/t^n,(i+1)/t^n]\subset(a,b)$ and 
$\measure{K}/(b-a)\geq(1/t^n)/(2/t^{n-1})=1/2t$. This finishes the proof
of (\ref{Peano2}).
 
Notice that (\ref{Peano1}) implies
$\inv{f}(E)$ is nowhere dense for every nowhere dense set $E$.
 So,
\[
\inv{f}(E)\in\I\ \mbox{ for every }\ E\in\I.
\]
Thus, by Lemma \ref{lem:homeom}, to show that $f$ is \I-density continuous
it is enough to prove that $f$ is
deep-\I-density continuous. 
 
Let $x\in[0,1]$ and let 
$A\subset\real\setminus\{f(x)\}$ be a set such that
$f(x)$ is a deep-\I-density point of $A$. It must be shown
that $x$ is a deep-\I-density point of $\inv{f}(A)$.
This will be done with the aid of Lemma \ref{lem:Zajicek}.
 
Let $s=1/9^k\in(0,1)$. We must find 
$D_s>0$ and $R_s\in(0,1)$ such that whenever $0<D<D_s$ and an interval
$I\subset(x-D,x+D)\setminus\{x\}$ with
$\measure{I}/D>s$, then there is an interval
$J\subset I\cap\inv{f}(A)$ with 
\begin{equation}\label{PeanoAA}
\frac{\measure{J}}{\measure{I}}>R_s.
\end{equation}
 
Let $s^\prime=s/9^3$. Using Lemma \ref{lem:Zajicek} with
$A$ and $f(x)$, there exists 
$D_{s^\prime}>0$ and $R_{s^\prime}=1/3^l\in(0,1)$ such that 
\begin{itemize}
\item{} 
whenever 
$0<D^\prime\leq D_{s^\prime}$ and an interval
\[ I^\prime\subset(f(x)-D^\prime,f(x)+D^\prime)\setminus\{f(x)\} \] 
with
$\measure{I^\prime}/D^\prime\geq s^\prime$, then there is an interval
$J^\prime\subset I^\prime\cap A$ 
with 
\begin{equation}\label{PeanoBB}
\measure{J^\prime}/\measure{I^\prime}>R_{s^\prime}
\end{equation}
\end{itemize}
 
Let $D_s>0$ be such that 
\begin{equation}\label{PeanoCC}
|f(x)-f(y)|<D_{s^\prime}\ \mbox{ for }\ |x-y|<D_s
\end{equation}
and let
$R_s=1/9^{l+5}$. Let $0<D<D_s$ and choose an interval
$I\subset(x-D,x+D)\setminus\{x\}$ with $\measure{I}/D>s$. We will find an
interval $J\subset I\cap\inv{f}(A)$ with $\measure{J}/\measure{I}>R_s$.
 
Assume that $I\subset(x,x+D)$. The other case is similar.
 
Using (\ref{Peano2}), we can find $I_0=[j/9^{n-1},(j+1)/9^{n-1}]\subset I$
such that 
\begin{equation}\label{PeanoDD}
\frac{\measure{I_0}}{\measure{I}}\geq\frac{1}{18}.
\end{equation}
Moreover, using (\ref{Peano1a}),
it is easy to find $I_1=[i/9^n,(i+1)/9^n]\subset I_0$ such that
$f(x)\not\in f(I_1)$. Thus, 
\[
\frac{\measure{I_1}}{D}=\frac{1}{9}\frac{\measure{I_0}}{D}\geq
\frac{1}{9}\frac{1}{18}\frac{\measure{I}}{D}>\frac{s}{9^3}=s^\prime.
\]
In particular, there exist $p=\inv{(s^\prime)}$ contiguous
intervals $I^1,I^2,\ldots,I^p$ of length $1/9^n$,
one of which is $I_1$ and such that 
$x\in I^1\cup I^2\cup\ldots\cup I^p$.
 
Define 
\[
D^\prime=\max\{|f(x)-f(i/9^n)|,|f(x)-f((i+1)/9^n)|\}>0
\]
and $I^\prime=f(I_1)$. By 
(\ref{PeanoCC})
we see that
$D^\prime<D_{s\prime}$ 
and, by (\ref{Peano1a}), $f(i/9^n)$ and $f((i+1)/9^n)$ are the end points
of $I^\prime$ so that
$I^\prime\subset[f(x)-D^\prime,f(x)+D^\prime]\setminus\{f(x)\}$. Moreover,
since $x,i/9^n,(i+1)/9^n\in I^1\cup I^2\cup\ldots\cup I^p$ then, by
(\ref{Peano1}), we have
\[
D^\prime\leq
\mbox{m}\left( f\left( \Cup_{j=1}^{p}I^j \right) \right)\leq
\sum_{j=1}^{p}\measure{f(I^j)}=p\measure{I^\prime}.
\]
Hence,
\[
\frac{\measure{I^\prime}}{D^\prime}\geq
\frac{\measure{I^\prime}}{p\measure{I^\prime}}=\inv{p}=s^\prime.
\]
Thus, 
by (\ref{PeanoBB}),
there is an interval 
$J^\prime\subset I^\prime\cap A$ 
such that $\measure{J^\prime}/\measure{I^\prime}>R_{s^\prime}$.
 
Using (\ref{Peano2}), we can find an interval 
\[J^\prime_1=[j_0/3^m,(j_0+1)/3^m]\subset J^\prime\] such that
$\measure{J^\prime_1}/\measure{J^\prime}\geq 1/6>1/9$. Hence,
\[
\frac{\measure{J^\prime_1}}{\measure{f(I_1)}}=
\frac{\measure{J^\prime_1}}{\measure{I^\prime}}=
\frac{\measure{J^\prime_1}}{\measure{J^\prime}}
                                  \frac{\measure{J^\prime}}{\measure{I^\prime}}
>\frac{1}{9}R_{s^\prime}=\frac{1}{3^{l+2}}
\]
and 
$J^\prime_1=[j_0/3^m,(j_0+1)/3^m]\subset f(I_1)=f([i/9^n,(i+1)/9^n])$.
But now condition (\ref{Peano1a}) implies easily that there exists
an interval 
$$J=[j/9^m,(j+1)/9^m]\subset I_1=[i/9^n,(i+1)/9^n]$$ such that
$f(J)=J^\prime_1$ and
\[
\frac{\measure{J}}{\measure{I_1}}>\left(\frac{1}{3^{l+2}}\right)^2
=\frac{1}{9^{l+2}}.
\]
Hence, 
by (\ref{PeanoDD}),
\[
\frac{\measure{J}}{\measure{I}}\geq
\frac{\measure{J}}{18\measure{I_0}}
=\frac{1}{9}\frac{\measure{J}}{18\measure{I_1}}
>\frac{1}{9^3}\frac{1}{9^{l+2}}=R_s.
\] 
Condition (\ref{PeanoAA}) is proved.
This finishes the proof that $f$ is \I-density continuous.
 
\smallskip
 
To see that $f$ is not \I-approximately differentiable at a point $x\in[0,1]$
let us do the following construction for each $n\in\natural$.
Choose $i\in\natural$ such that $x\in[i/9^n,(i+1)/9^n]$. Then, by (\ref{Peano1}),
$f([i/9^n,(i+1)/9^n])=[j/3^n,(j+1)/3^n]$ for some $j\in\natural$. 
It is also not
difficult to see that condition (\ref{Peano1a}) implies that 
\[
\left\{f\left(\left[\frac{9i}{9^{n+1}},\frac{9i+1}{9^{n+1}}\right]\right),
f\left(\left[\frac{9i+8}{9^{n+1}},\frac{9i+9}{9^{n+1}}\right]\right)\right\}
\]
\[
=\left\{
\left[\frac{3j}{3^{n+1}},\frac{3j+1}{3^{n+1}}\right],
\left[\frac{3j+2}{3^{n+1}},\frac{3j+3}{3^{n+1}}\right]
\right\}.
\]
 
This implies, in particular, that for every
$y\in[9i/9^{n+1},(9i+1)/9^{n+1}]$ and 
$y^\prime\in[(9i+8)/9^{n+1},(9i+9)/9^{n+1}]$ we have
\[
\frac{|f(y)-f(y^\prime)|}{|y-y^\prime|}\geq
\frac{1/3^{n+1}}{1/9^n}=3^{n-1}.
\]
Hence, an easy geometrical argument implies that for one of the intervals
$[9i/9^{n+1},(9i+1)/9^{n+1}]$ or $[(9i+8)/9^{n+1},(9i+9)/9^{n+1}]$,
which we denote by $[a_n,b_n]$, we have $x\not\in[a_n,b_n]$ and
\[
\frac{|f(y)-f(x)|}{|y-x|}\geq 3^{n-1} \ \mbox{ for every }\ y\in[a_n,b_n].
\]
But, by Lemma \ref{lemma36}, $x$ is not an \I-dispersion point of
$\bigcup_{n\in\natural} [a_n,b_n]$. Thus, for every \I-density open set
$U$ containing $x$, for every $\e>0$ and $n\in\natural$ there is an 
$y\in(x-\e,x+\e)\cap U\cap\bigcup_{m>n} [a_m,b_m]$ for which 
\[
\frac{|f(y)-f(x)|}{|y-x|}\geq 3^n.
\]
This implies that $f$ is not \I-approximately differentiable.
Also notice that the construction of the intervals $[a_n,b_n]$
given above also implies that $f$ is not approximately
differentiable. This finishes the proof of Theorem \ref{nowhereIappDiff}.
 
\section{Derivatives and \I-approximate continuity}
 
In this section we show that the well-known fact that every 
bounded approximately countinuous function is a derivative is not 
true for the bounded \I-approximately continuous functions.
 
\examp{ex:iapproxnotder}{
There exists a bounded \I-density continuous function
which is not a derivative.}
 
Proof.
Let $P\subset(0,1]$ be a nowhere dense closed
set with positive measure. Choose a sequence 
$\{n_k\}_{k\in\natural}$ of natural numbers
satisfying $\lim_{k\to\infty} n_k/n_{k+1}=0$ and define 
\[
A=\Cup_{k\in\natural}\frac{1}{n_k}P.
\]
Then, by \cite[Lemma 2.4]{CL:Examples}, $0$ is a deep-\I-dispersion point of
$A$. Hence, there exists a closed set $B\cup\{0\}\subset\complement A$ such that
$0$ is an \I-density point of $B$. Moreover, it can be assumed that
\[
B=\Cup_{k\in\natural}[a_k,b_k]\cup[c_k,d_k],
\]
where $a_k<b_k<a_{k+1}<0<d_{k+1}<c_k<d_k$ \cite{Lazarow:Coarsest,PW:34}.
 
On the other hand, for all $k\in\natural$, 
\begin{equation}\label{dfract}
\frac{\measure{\complement{B}\cap(0,1/n_k)}}{1/n_k}\geq
\frac{\measure{A\cap(0,1/n_k)}}{1/n_k}>\measure{P}>0,
\end{equation}
so $0$ is not a dispersion point of \complement{B}.
 
Define the function $f$ on $A\cup B$ by
\[
f(x)=\left\{\begin{array}{ll}
1 & x\in A\\
0 & x\in B
\end{array}\right.
\]
and extend $f$ on elsewhere 
in such a way that it is piecewise linear on
$(0,\infty)$ and bounded by $1$. Since $0$ is an \idisp\ point of \complement{B},
it is apparent that $f$ is \I-density continuous. 
On the other
hand, $f$ cannot be a derivative. To see this, suppose $F$ is any
primitive function for $f$ and define
\[
G(x)=\int_0^x f.
\]
Since $f$ is continuous on $\real\setminus\{0\}$, we see that $F-G$
must be constant on both $(-\infty,0)$ and $(0,\infty)$. Since both
$F$ and $G$ are continuous, this implies that $F-G$ is constant on
$\real$ and therefore $G$ is differentiable on $\real$.
But, this is impossible 
since, by (\ref{dfract}),
\begin{eqnarray*}
D^-G(0) & = &
0<\measure{P}\\
& \le &\liminf_{k\to\infty}\frac{\measure{E\cap(0,1/n_k)}}{1/n_k}\\
& \le &\limsup_{k\to\infty}\frac{G(1/n_k)}{1/n_k}\le\urDini{G}(0). 
\end{eqnarray*}
 
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\end{document}
