% Peano curve and \I-approximate differentiability %\documentstyle[rae12mod]{article} %\input{epsf.tex} \documentclass{rae} \usepackage{amsmath} \usepackage{amssymb} \usepackage{epsf} \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} \newcommand{\putepsf}[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} } %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] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}%[theorem] {Lemma} \newtheorem{corollary}[theorem]{Theorem} \newtheorem{definition}[theorem]{Definition} \newtheorem{problem}{Problem} \newtheorem{example}{Example} \newcommand{\thm}[2]{\begin{theorem}\label{#1}$\!\!\!${\bf.} #2\end{theorem}} \newcommand{\lem}[2]{\begin{lemma}\label{#1}$\!\!\!${\bf.} #2\end{lemma}} \newcommand{\examp}[2]{\begin{example}\label{#1}$\!\!\!${\bf.} #2\end{example}} \newcommand{\urDini}[1]{\mbox{$\overline{D}^+#1$}} \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} %%%%%% % % Picture Macros %General picture Displayer % #1=width, #2=height, #3=name, #4=caption, #5=label \newcommand{\putpicture}[5]{ \begin{figure}[htb] \[ \vbox to #2in{\hrule width #1in height 0pt depth 0pt \vfill \special{pictfile #3 scaled 250}} \] \caption{#4\label{#5}} \end{figure} } \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} } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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_{i0$ and $R_s\in(0,1)$ such that whenever $0s$, 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 $0s$, 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 $0R_{s^\prime} \end{equation} \end{itemize} Let $D_s>0$ be such that \begin{equation}\label{PeanoCC} |f(x)-f(y)|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 $D^\primeR_{s^\prime}$. 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\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*} \begin{thebibliography}{10} \bibitem{AversaWilczynski:Homeomorphisms} V.~Aversa and W.~Wilczy\'nski. \newblock Homeomorphisms preserving {$\cal {I}$}-density points. \newblock {\em Boll. Un. Mat. Ital.}, B(7)1:275--285, 1987. \bibitem{CL:SpDensCont} Krzysztof Ciesielski and Lee Larson. \newblock The space of density continuous functions. \newblock {\em Acta Math. 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