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\section*{}{\begin{center}{\Large\bf\thesection.\quad#1}\end{center}}\par\bigskip}



\pagestyle{myheadings} 
\markright{CIESIELSKI, LARSON, OSTASZEWSKI}

\marginlabels{n}
 
\title{SEMIGROUPS OF \I-DENSITY CONTINUOUS FUNCTIONS}
 
 
\author{}
\date{}
 
\begin{document}
\thispagestyle{empty}

\vbox to 1.5in{}

\begin{center}
\LARGE SEMIGROUPS OF \I-DENSITY CONTINUOUS FUNCTIONS
\end{center}

\vskip1cm

\begin{center}
Krzysztof Ciesielski\footnote{This author was partially supported by a West
Virginia University Senate Research grant.}, Lee Larson$^2$, Krzysztof
Ostaszewski\footnote{These authors were partially supported by University of
Louisville research grants.} \end{center} 

\medskip

\begin{center}
{\small Communicated by Michael Mislove}

\end{center}

%\begin{abstract}
%This paper is concerned with the classes 
%of \I-density continuous functions and
%deep-\I-density continuous functions  as 
%semigroups with composition as the operation. We also analyze some of their 
%subsemigroups. It is shown that the groups of automorphisms of these semigroups
%and several of their subsemigroups have the inner automorphism property.
%\end{abstract}

%\pagebreak
 
\sgfsection{Preliminaries}

This paper is concerned with the classes of \idens\ continuous and deep-\idens\
continuous functions as semigroups with composition as the operation. Several
of their subsemigroups are also analyzed. In particular, it is shown that the
groups of automorphisms of these semigroups and several of their subsemigroups
have the inner automorphism property.

The notation used throughout this paper is standard. In particular,
\reals\ stands for the set of real numbers and $\nats=\{1,2,3,\ldots\}$.
The symmetric difference of sets $A$ and $B$ is denoted by $A\triangle B$
and the complement of a set $A\subset\reals$ by \complement{A}. 
The symbols \Lebesgue\ and
\BaireSets\  stand for the families of subsets of \reals\ which are
Lebesgue measurable and have the Baire property, respectively.
\N\ and \I\ denote the ideals of Lebesgue measure zero and 
of first category subsets of \reals.
If a statement is true everywhere except for those points of a set
belonging to \N, then it is true {\em almost everywhere 
 (a.e.).}
If it is true everywhere except for the points of a set belonging to \I,
then it is true {\em\I-almost everywhere 
 (\I-a.e.)}.
If $A\in\Lebesgue$, we denote its Lebesgue measure by \measure{A}.
 
The natural topology on \reals\ is denoted by \ordinarytop. The symbols
$\interior{A}$ and $\closure{A}$ stand for the interior and 
closure of $A\subset\reals$ with respect to \ordinarytop.
It is also easy to see that for any set
$B\in\BaireSets$ there is a unique open set $\tilde{B}$ such that
$\tilde{B}=\interior{\closure{\tilde{B}}}$ and $B=\tilde{B}\triangle I$ 
for some $I\in\I$ \cite{Oxtoby:MeasCat}. Any open set $G$ for which 
$G=\interior{\closure{G}}$ is called a {\em regular} open set.
In a sense, $\tilde{B}$ is the largest open set such that $B$ can
be written as $\tilde{B}\triangle I$ for some $I\in\I$.
 
Any of the sets $\bigcup_{n\in\smallnats}[a_n,b_n]$ 
or $\bigcup_{n\in\smallnats}(a_n,b_n)$
is a {\em right interval set\/} at $0$ if $0 < b_{n+1} < a_n < b_n$ for
$n\in\nats$ and $\lim_{n\to\infty}{a_n} = 0$. A {\em left interval set\/} at
$0$ is defined similarly. A set $E$ is a right (left) interval set at
$x\in\reals$ if $E - x$ is a right (left,
respectively) interval set at $0$.
 
For a topological space $(X, {\cal T})$ let \selfmaps{X}\, or 
 \selfmaps{(X,{\cal T})} be the semigroup of all
continuous selfmaps of $X$; i.e., continuous functions 
$f:X\to X$, with composition as the operation. 
If there is no topology defined on $X$, then
the discrete topology is used in the definition of \selfmaps{X}\ so
that $\selfmaps{X}=X^X$.
 
If $G$ is a semigroup, then $\aut{G}$ 
denotes the group of all automorphisms of $G$.
A subsemigroup $G$ of $\selfmaps{X}$ has the 
{\em inner automorphism property} if for every automorphism $\Y\in\aut{G}$
there is an $h\in G$ such that $\Y(g)=\conj{g}{h}$ for every $g\in G$.
 
For a topological space $X$ the property that $\selfmaps{X}$
has the inner automorphism property is strongly 
related to the following definitions and facts.

A collection of topological spaces is {\em S--admissible} if for
each pair of spaces $X$ and $Y$ from the collection, any isomorphism
$\Phi:\selfmaps{X}\to\selfmaps{Y}$ is induced by a
homeomorphism $h:X\to Y$; i.e., 
$\Phi(f)=\conj{f}{h}$ for all $f\in\selfmaps{X}$.  In other
words, within an S--admissible class of spaces, $X$ is homeomorphic
to $Y$ if, and only if, \selfmaps{X} is isomorphic to \selfmaps{Y},
so that the topological structure of the space $X$ is fully characterized
by the algebraic structure of \selfmaps{X}. It is easy to see that for 
a topological space $X$ a necessary and sufficient condition for belonging
to an S--admissible class is that the semigroup
$\selfmaps{X}$ has the inner automorphism property.
This gives a motivation for studying the topological spaces $X$ for which
$\selfmaps{X}$ has the inner automorphism property.
For more information on the subject see Magill \cite{KM:SemiSurvey}.
 
The basis for studying the inner automorphism property of subsemigroups $G$
of $\selfmaps{X}$ is given by the following theorem of 
Schreier \cite{Schreier:absMenge:37}.
 
\proposition{prop:basic}{
Let $X$ be a set and let $G$ be a subsemigroup of $\selfmaps{X}$
such that every constant mapping is in $G$. Then, for every $\Y\in\aut{G}$
there exists a unique bijection $h$ of $X$ such that
$\Y(g)=\conj{g}{h}$ for every $g\in G$.}
 
Every semigroup considered in this paper contains all constant functions.
In particular, if $\Y$ is an automorphism of a semigroup $G\subset\selfmaps{X}$
containing all constant functions
then the unique bijection $h$ of $X$ for which 
$\Y(g)=\conj{g}{h}$ for all $g\in G$ will be called 
the {\em generating bijection} of \Y.
 
Following Magill \cite[Definition 2.2, p. 198]{KM:SemiSurvey}
we say that a topological space $X$ is {\em generated} if it is $T_1$ and
the collection of complements of level sets for its continuous selfmaps
$\{\left(f^{-1}(\{x\})\right)^c\colon x\in X \mbox{ and } f\in\selfmaps{X} \}$
forms a subbase for $X$. (Compare also \cite{JW:unique}.)
It is known that the class of all generated spaces is S--admissible 
\cite[Theorem 2.3, p. 198]{KM:SemiSurvey}. In particular, there is
the following.
 
\proposition{prop:generated}{If a topological space $X$ is generated then,
$\selfmaps{X}$ has the inner automorphism property.}
 
\sgfsection{Topologies}
 
This paper is concerned with the semigroups
$\idenscont=\selfmaps{(\reals,\itopology)}$ of \I-{\em density 
continuous functions}, 
$\deepidenscont=\selfmaps{(\reals,\deepitopology)}$ 
of {\em deep-\I-density continuous functions} and
$\denscont=\selfmaps{(\reals,\densitytop)}$ 
of {\em density continuous functions} and some of their subsemigroups.
We start with definitions of the topologies \densitytop,
\itopology\ and \deepitopology. For this, we first introduce the notions 
of a density point, \I-density point and deep-\I-density point.
(See \cite{Oxtoby:MeasCat,PWW:CatAn,Wilczynski:CatAn}. For discussion, 
compare also \cite{CL:Refinements}.)
 
Let $A\in\Lebesgue$. A number $x$, not necessarily in $A$, is a 
{\em density point\/} of $A$ if  
\begin{equation}\label{def_dens}
\lim_{h\to 0^+}\frac{\measure{A\cap(x-h,x+h)}}{2h}=1.
\end{equation}
The set of all density points of $A\in\Lebesgue$ we denote by \densitypts{A}.
It is a straightforward consequence of the Lebesgue density theorem that
for every $A\in\Lebesgue$, $\densitypts{A}\triangle A\in\N$. The family of
sets \[
\densitytop=\{A\in\Lebesgue\colon A\subset\densitypts{A}\}
\]
forms a topology on \reals\ \cite{Oxtoby:MeasCat,GW:AppContTrans} called
the {\em density topology\/}.
 
If $A\in\BaireSets$, we say that $0$ is
an {\em \I-density point\/} of $A$ if
for every increasing sequence $\{n_m\}_{m\in\smallnats}$ of natural
numbers there exists a subsequence $\{n_{m_p}\}_{p\in\smallnats}$ such that
\begin{equation}\label{def_idens1}
\lim_{p\to\infty}\charf{n_{m_p}A\cap(-1,1)} = \charf{(-1,1)}
\ \mbox{\I-a.e.}
\end{equation}
We say that a point $a$ is an 
\I-density point of $A\in\BaireSets$ if $0$ is an \I-density point of $A-a$.
The set of all \I-density points of $A\in\BaireSets$ we denote by 
\idensitypts{A}. 
Similar to the case with Lebesgue density, as noted above, 
$A\triangle\idensitypts{A}\in\I$ for every $A\in\BaireSets$
\cite[Theorem 3]{Wilczynski:CatAn} and 
\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 \reals\ \cite{PWW:CatAn,Wilczynski:CatAn} called
the {\em \I-density topology\/}.
 
Finally, we say that 
a point $a\in\reals$ is a {\em deep-\I-density 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 \I-density
point of $F$.
The set of all deep-\I-density points of $A\in\BaireSets$ is denoted 
\deepidensitypts{A}.
The family of sets
\[
\deepitopology=\{A\in\BaireSets\colon A\subset\deepidensitypts{A}\}
\]
forms a topology on \reals\ \cite{Lazarow:Coarsest,Wilczynski:CatAn} called the
{\em deep-\I-density topology\/}.
 
The inclusion relations between the topologies
\ordinarytop, \densitytop, \itopology\ and \deepitopology\ are given by the 
following theorem \cite[Theorem 2.5]{CL:Examples}.
 
\theorem{th:densvsidens}{If ${\cal P}(\reals)$ stands for 
the discrete topology on \reals\ then
\[
\begin{array}{ccccccc}
\ordinarytop\cap\densitytop & \subset & \deepitopology\cap\densitytop &
\subset & \itopology\cap\densitytop & \subset & \densitytop \nonumber \\
 
\| & & \cap & & \cap & & \cap \nonumber \\
 
\ordinarytop & \subset & \deepitopology &
\subset &   \itopology & \subset & {\cal P}(\reals) \nonumber
\end{array}
\]
Moreover, all the inclusions are proper.
}
 
We also need the following fact.
For the \I-density case it can be found in \cite[Lemma 2.4]{CL:Baire},
\cite[Theorem 1]{PWW:RemIDens} or \cite[Theorem 2]{Wilczynski:CatAn}.
 
\proposition{sequence}{
If $E = \bigcup_{n\in\smallnats}[a_n,b_n]$ is a right interval set such that
\begin{description}
\item[(i)] $\lim_{n\to\infty}{(b_n - a_n)/a_n} = 0$; and
\item[(ii)] $\lim_{n\to\infty}b_{n + 1}/a_n = 0,$
\end{description}
then $0$ is a density and deep-\I-density point of $\complement{E}$. 
In particular, $E^c\in\densitytop\cap\deepitopology$.
}
 
\sgfsection{Density Continuous Functions}

Let $\Diff$ be the class of all differentiable functions from \reals\ to \reals.
Moreover, let \AppDiff\ stand for the class of all approximately
differentiable functions (\cite{Bruckner:DiffReal}, see also 
\cite{KO:SemiDensity})
and let \aeAppDiff\ be the class of all almost everywhere
approximately differentiable functions.
Often used is the group of homeomorphisms
 $f:(\reals,\ordinarytop)\to(\reals,\ordinarytop)$, 
which is written as $\cal H$.
 
The classes $\Diff$, $\AppDiff$, \aeAppDiff\ are connected to the 
class \denscont\ 
of density continuous functions by the following theorem.
 
\theorem{th:DensityDiff}{ If \measureF\ stands for the class of measurable
functions, then
\[
\begin{array}{ccccccc}
\Diff & \subset & \AppDiff 
& \subset & \aeAppDiff & \subset & \measureF \nonumber \\
\cup & & \cup & & \cup & & \cup \nonumber \\
\Diff\cap\denscont & \subset & \AppDiff\cap\denscont 
& \subset & \aeAppDiff\cap\denscont & \subset & \denscont \nonumber \\
\end{array}
\]
and all the inclusions are proper.
}
 
Proof. The inclusions are obvious.
The vertical inclusions are proper because there is a \cinfinity\
function which is not density continuous 
 \cite[Example 1]{CL:SpDensCont}.
 
$\denscont\not\subset\aeAppDiff$ follows immediately from 
\cite[Theorem 4]{CLO:DiffDensCont}.
 
$\aeAppDiff\cap\denscont\not\subset\AppDiff$ 
is proved by the function $h(x)=|x|$.
 
$\AppDiff\cap\denscont\not\subset\Diff$ is easily justified by the function
$f$ defined by $f(x)=0$ for $x\in E^c$ and 
$f(x)=(x-a_n)^2(x-b_n)^2$ for $x\in [a_n,b_n]$ and $n\in\nats$,
where $E=\UoverN [a_n,b_n]$ is a right interval set 
from Proposition \ref{sequence}.
 
\bigskip
 
The motivation to study 
 the inner automorphism property of the
classes from Theorem \ref{th:DensityDiff}
arises from the following theorem of Magill \cite{Magill:Diff}.
 
\theorem{th:innerdiff}{The semigroup \Diff\ of all differentiable functions
has the inner automorphism property.}
 
Following this path Ostaszewski proved the following \cite{KO:SemiDensity}.
 
\theorem{th:innerdens}{The semigroups \denscont, $\aeAppDiff\cap\denscont$,
$\AppDiff\cap\denscont$ and $\Diff\cap\denscont$ have the inner automorphism 
property.}
 
Notice that the above theorem cannot be deduced from Proposition
\ref{prop:generated}, because the density topology is not generated
\cite{CL:Generated}. In fact, the real line equipped with the density topology is
the only known example of a completely regular not generated topological space
for which the semigroup of continuous selfmaps has the inner automorphism
property.  (Also note the remark after Theorem \ref{theorem18}.)
 
Theorem \ref{th:innerdens} is proved by showing the following facts:
for every generating bijection $h$ of
$\Y\in\aut{\denscont}\cup\aut{\aeAppDiff\cap\denscont}$
we have $h,\inv{h}\in\cal H\cap\denscont\cap\AppDiff$ and 
for every generating bijection $h$ of
$\Y\in\aut{\AppDiff\cap\denscont}\cup\aut{\Diff\cap\denscont}$
we have $h,\inv{h}\in\cal H\cap\denscont\cap\Diff$. 
In fact, if we identify the group \aut{H}\ with the set of
generating bijections, we obtain
 
\corollary{cor:densauto}{The following relations hold:
\[
\aut{\Diff\cap\denscont}=\aut{\AppDiff\cap\denscont}\subset
\aut{\aeAppDiff\cap\denscont}=\aut{\denscont}
\]
and the inclusion is proper.}
 
Theorems \ref{th:innerdiff} and \ref{th:innerdens} prove
that the classes
\Diff, $\Diff\cap\denscont$, $\AppDiff\cap\denscont$,
$\aeAppDiff\cap\denscont$ and 
 \denscont\ have the inner automorphism
property.  For the remaining three classes the question about their inner
automorphism property is moot as those classes do not form
semigroups. For the class \AppDiff, the proof is given in Example
\ref{ex:closureiAppDiff}. For the two other classes the result follows from the
example below.
 
\example{ex:notclosed1}{There exist almost everywhere differentiable
functions $f$ and $g$ such that $g\circ f$ is not measurable. In particular,
the classes \aeAppDiff\ and \measureF\ are not closed under composition.}
 
Proof. Let $P\subset\reals$ be any perfect nowhere dense set with positive 
measure,
$S\subset P$ be nonmeasurable and let $C$ be a Cantor set of measure $0$.
Let $f$ be a homeomorphism such that $\inv{f}(C)=P$ and let 
$g=\charf{S}\circ\inv{f}$. 
Then, $f$ is differentiable almost everywhere, as a
homeomorphism and $g$ is differentiable almost everywhere, as
$g=\charf{f(S)}$ and $f(S)\subset C$ has measure $0$.
On the other hand, $g\circ f=\charf{S}\circ\inv{f}\circ f=\charf{S}$
is not measurable, because $S$ is nonmeasurable. 
 
\sgfsection{\I-approximate Derivative}
 
In this section we introduce the notion of the \I-approximate
derivative as category analogue of the approximate derivative. Let us recall
that a function $f\RtoR$ is \I-approximately continuous 
(deep-\I-approximately continuous) if $f$ is continuous with respect to the
ordinary topology on the range and \I-density (deep-\I-density) topology on the
domain. It is well known that the classes of \I-approximately continuous
functions and  deep-\I-approximately continuous functions
coincide  \cite[Theorem 2]{Lazarow:Coarsest}.
 
We start with the following definitions. A function $f\RtoR$ is
 said to be %CHANGE DONE FOR SPACEING 
{\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
\begin{equation}\label{DEF1}
\left\{t\in\reals\colon \frac{f(t)-f(x)}{t-x}\in
(\iAppDer f(x)-\varepsilon,\iAppDer f(x)+\varepsilon)\right\}.\end{equation}
(Compare also \cite{LW:IAppDer:21A} and \cite[Definition 8]{Wilczynski:CatAn}.)
 
A function $f\RtoR$ is {\em  \I-approximately differentiable}
if it is \I-approximately differentiable 
at every point. The class of all \I-approximately differentiable
functions is denoted by \iAppDiff.
 
A function $f\RtoR$ is said to be {\em  \I-approximately differentiable
\I-almost everywhere} if there exists a set $A\in\I$
such that $f$ is \I-approximately differentiable at every point of $A^c$.
The class of all functions \I-approximately differentiable 
\I-almost everywhere is denoted by \IaeiAppDiff.
 
Similarly, the classes of functions that are either 
deep-\I-approximately  or \I-approximately continuous
\I-almost everywhere may be defined.
 
It is easy to introduce the idea of a function $f\RtoR$ 
being deep-\I-approximately
differentiable at point $x$ by requiring that $x$ is a deep-\I-density point
of a set from (\ref{DEF1}). However it can be proved that for the 
\I-approximately continuous functions the notions of 
\I-approximate differentiability and deep-\I-approximate differentiability
at a point coincide. Thus, there is no good reason to pursue this notion.
 
We start our investigation 
with the following propositions.
 
\proposition{pr:idiff1}{If a function $f\RtoR$ is \I-approximately 
differentiable
at $x$ then $f$ is \I-approximately continuous at $x$.}
 
Proof. Choose $\e > 0$. 
We must prove that $x$ is an \I-density point of the set 
\[
\{t\in\reals\colon |f(t)-f(x)|\in(-\varepsilon,\varepsilon)\}.
\] 
But, if we choose $\delta >0$ such that 
$\delta \, (\iAppDer f(x)-\varepsilon,\iAppDer f(x)+\varepsilon)
\subset(-\varepsilon,\varepsilon)$, then
\[
\{t\in\reals\colon |f(t)-f(x)|\in(-\varepsilon,\varepsilon)\}\supset
\]
\[
\{t\in\reals\colon |t-x|<\delta\mbox{ \& }
|f(t)-f(x)|\in\delta \, 
(\iAppDer f(x)-\varepsilon,\iAppDer f(x)+\varepsilon)\}\supset
\]
\[
\{t\in\reals\colon |t-x|<\delta\mbox{ \& }
|f(t)-f(x)|\in |t-x| \, 
(\iAppDer f(x)-\varepsilon,\iAppDer f(x)+\varepsilon)\}=
\]
\[
\left\{t\in\reals\colon \frac{|f(t)-f(x)|}{|t-x|} - \iAppDer f(x)\in
(-\varepsilon,\varepsilon)\right\}\cap (x-\delta,x+\delta)
\]
and, by assumption, $x$ is an \I-density point of some Baire 
subset of the last set. This finishes the proof.
 
\proposition{pr:idiff2}{If $f\RtoR$ is \I-approximately differentiable
\I-almost everywhere, then $f$ is a Baire function.}
 
Proof. It was shown by Poreda, Wagner-Bojakowska and Wilczy\'{n}ski that a
function $f\RtoR$ is \I-approximately continuous \I-almost everywhere if, and
only if, $f$ has the Baire property \cite[Theorem 7]{PWW:CatAn}. Use of 
Proposition \ref{pr:idiff1} completes the proof.
 
\bigskip
 
In the sequel we also need the following two examples.
 
\example{ex:hunch}{For every $a<b\leq c<d$ there exists a differentiable 
\I-density continuous and density continuous function 
$f\colon\reals\to[0,1]$ such that $f(x)=0$ for $x\in (a,d)^c$ and $f(x)=1$
for $x\in[b,c]$.}
 
Proof. We have to define $f$ only on $(a,b)$ and $(c,d)$. So, define
\[
f(x)=\frac{(x-a)^2 (x-2b+a)^2}{(b-a)^4}
\]
for $x\in(a,b)$ and 
\[
f(x)=\frac{(x-d)^2 (x-2c+d)^2}{(c-d)^4}
\]
for $x\in(c,d)$. It is easy to see that
$f$ is differentiable and, because it is everywhere unilaterally a
polynomial, it is also density and \I-density continuous \cite[Corollary
3]{CL:SpDensCont},  \cite{CL:Analytic}.
 
\bigskip
 
The previous example easily implies the existence of the following example.
 
\example{ex:appDiffNotCont}{There exists an approximately and 
\I-approximately differentiable, density continuous and \I-density
continuous function which is not continuous. In particular,
\[
\AppDiff\cap\iAppDiff\cap\denscont\cap\idenscont\not\subset{\cal C}.
\]
}
 
Proof. Let  $E=\UoverN [a_n,d_n]$ be an interval set as in 
Proposition \ref{sequence}. For each $n$, let 
$f_n$ be as in Example \ref{ex:hunch} with $[a,d]=[a_n,d_n]$.
Define $f(x)=0$ on $E^c$ and $f=f_n$ on $[a_n,d_n]$. The rest is obvious.
 
\bigskip
 
Now we are ready to prove the next theorem.
 
\theorem{th:semigrinclusions}{ If \BaireF\ stands for the Baire
functions, then
\[
\begin{array}{ccccccc}
\Diff & \subset & \iAppDiff & \subset & 
\IaeiAppDiff & \subset & \BaireF \nonumber \\
\cup & & \cup & & \cup & & \cup \nonumber \\
\Diff\cap\deepidenscont & \subset & \iAppDiff\cap\deepidenscont & \subset & 
\IaeiAppDiff\cap\deepidenscont & \subseteq & \deepidenscont \nonumber \\
\cup & & \cup & & \cup & & \cup \nonumber \\
\Diff\cap\idenscont & \subset & \iAppDiff\cap\idenscont & \subset & 
\IaeiAppDiff\cap\idenscont & \subseteq & \idenscont \nonumber \\
\end{array}
\]
and all the inclusions with the symbol $\subset$ are proper.
}
 
Proof. The inclusion $\idenscont\subset\deepidenscont$  can be found in 
\cite[Theorem 4.1(iv)]{CL:Examples}.
The other inclusions are obvious.
 
The vertical inclusions are proper because 
$\Diff\cap\deepidenscont\not\subset\idenscont$ 
\cite[Example 6.7]{CL:Examples}
and $\Diff\not\subset\deepidenscont$ 
\cite[Example 5.7]{CL:Examples}.
 
The fact that the horizontal inclusions of the form
$\subset$ are proper follows from 
$\iAppDiff\cap\idenscont\not\subset\Diff$ 
(Example \ref{ex:appDiffNotCont}) and 
$\IaeiAppDiff\cap\idenscont\not\subset\iAppDiff$ which is proved by
$f(x)=|x|$.
 
This finishes the proof.
 
\bigskip
 
It remains open whether the inclusions of the form
$\subseteq$ in the previous
theorem are proper. In particular, a positive answer to this question
would follow from a positive answer for the following problem.
 
\problem{pr:nowheridiffer}{Is there an \I-density continuous function
which is nowhere \I-approximately differentiable?}
 
To prove the next theorem, the following two lemmas are needed.
The first one is a version of the chain rule. The analogous theorem for 
approximate derivatives can be found in \cite{KO:denscont2}.
 
\lemma{ichainrule}{If $f,g\RtoR$ are such that
$\iAppDiff f(x_0)$ and $\iAppDiff g(f(x_0))$ exist
then $\iAppDiff (g\circ f)(x_0)$ also exists and
\begin{equation}\label{eq22}
\iAppDiff (g\circ f)(x_0)=\iAppDiff g(f(x_0))\,\iAppDiff f(x_0).
\end{equation}
}
 
Proof. First assume that $\iAppDiff g(f(x_0))>0$ and 
$\iAppDiff f(x_0)>0$. Put 
$z_0=f(x_0)$ and let $u_1,u_2,v_1,v_2$ be arbitrary positive numbers 
such that $0<u_1<\iAppDiff f(x_0)<u_2$ and $0<v_1<\iAppDiff g(z_0)<v_2$.
Then, there exist sets $U,V\in\itopology$, $x_0\in U$, $z_0\in V$, 
such that
\begin{equation}\label{eq:f}
u_1 \le \frac{f(x)-f(x_0)}{x-x_0} \le u_2
\end{equation}
for all $x\in U$ and
\begin{equation}\label{eq:g}
v_1 \le \frac{g(z)-g(z_0)}{z-z_0} \le v_2
\end{equation}
for all $z\in V$.
Since, by Proposition \ref{pr:idiff1},
$f$ is \idens\ continuous at $x_0$, there exists $W\in\itopology$
such that $x_0\in W\subset\inv{f}(V)$. In particular, by (\ref{eq:g}),
for $x\in W$ we have $z=f(x)\in V$ and
\begin{equation}\label{eq:f2}
v_1 \le \frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)} \le v_2.
\end{equation}
Multiplying (\ref{eq:f}) by (\ref{eq:f2}) we obtain
\[
u_1v_1 <
\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}\,\frac{f(x)-f(x_0)}{x-x_0}
\le u_2v_2;
\]
i.e.,
\[
u_1v_1 < \frac{g(f(x))-g(f(x_0))}{x-x_0} \le u_2v_2,
\]
for every $x\in U\cap W\in\itopology$. 
But, 
\[
u_1v_1 < \iAppDiff g(f(x_0))\,\iAppDiff f(x_0) < u_2v_2
\]
where the 
numbers $u_1v_1$ and $u_2v_2$ 
can be choosen as close to the number \[\iAppDiff g(f(x_0))\,\iAppDiff f(x_0)\]
as we wish. This implies (\ref{eq22}).
 
The remaining cases, when 
$\iAppDiff g(f(x_0))\leq 0$ or $\iAppDiff f(x_0)\leq 0$, are very similar,
modulo some little technical problems with the signs of the
inequalities. This finishes the proof.

 
\lemma{lem:iapproxdiffae}{
If $f\in\idenscont\cap\IaeiAppDiff$ 
and $g\in\IaeiAppDiff$, then $g\circ f\in\IaeiAppDiff$. }
 
Proof. Let $A,B\in\I$ be such that $f$ is \I-approximately differentiable
on $A^c$ and $g$ is \I-approximately differentiable on $B^c$.
 
By Lemma \ref{ichainrule} the function $g\circ f$ is 
\I-approximately differentiable on the set $D=A^c\cap\inv{f} (B^c)$.
Moreover, for every $x$, if $E_x=\inv{f} (x)$ then $g\circ f$ 
is \I-approximately differentiable on the set 
$E_x\cap\widetilde E_x\in\itopology$, as 
$g\circ f$  is constant on this set.
Thus, if $E=\Cup_{x\in\reals} (E_x\cap\widetilde E_x)$, then $g\circ f$ is 
\I-approximately differentiable on $D\cup E$.
 
To finish the proof it is enough to prove that $(D\cup E)^c\in\I$; in other
words
that every $x$ is an \I-density point of $D\cup E$.
 
So, let $x\in\reals$. Then, $f(x)$ is an \I-density point of 
$B^c\cup \{f(x)\}$, because $B$ is discrete in \itopology. So, 
since $f$ is \I-density continuous, $x$ is an \I-density point of 
\begin{eqnarray*}
\inv{f} (B^c\cup \{f(x)\}) & = & \inv{f} (B^c) \cup \inv{f} (f(x))\\
& = & \inv{f} (B^c) \cup E_{f(x)}\\
& = & \inv{f} (B^c) \cup (E_{f(x)}\cap\widetilde E_{f(x)}) \cup 
(E_{f(x)}\setminus\widetilde E_{f(x)}).
\end{eqnarray*}
But $E_{f(x)}\setminus\widetilde E_{f(x)}\in\I$ and $A\in\I$. Hence, 
$x$ is an \I-density point of 
$A^c\cap(\inv{f} (B^c) \cup (E_{f(x)}\cap\widetilde E_{f(x)}))
\subset D\cup E$. This finishes the proof.
 
\bigskip
 
As a corollary we obtain
 
\theorem{th:Isemigroups}{The classes 
$\Diff$,
$\idenscont$,
$\deepidenscont$,
$\Diff\cap\idenscont$,
$\Diff\cap\deepidenscont$,
$\iAppDiff\cap\idenscont$,
$\iAppDiff\cap\deepidenscont$ and 
$\IaeiAppDiff\cap\idenscont$
are closed under composition. In particular, they form semigroups.}
 
Proof. The classes $\Diff$, $\idenscont$ and $\deepidenscont$ are 
obviously closed under composition. Therefore the same is true of
the classes $\Diff\cap\idenscont$ and $\Diff\cap\deepidenscont$.
 
For the classes $\iAppDiff\cap\idenscont$ and $\iAppDiff\cap\deepidenscont$
the conclusion follows immediately from Lemma \ref{ichainrule}, while for
the class $\IaeiAppDiff\cap\idenscont$ 
it follows from Lemma \ref{lem:iapproxdiffae}.
 
\problem{pr:closure}{Are the classes $\IaeiAppDiff\cap\deepidenscont$
and \IaeiAppDiff\ closed under composition? If so,
do they have the inner automorphism property?}
 
We finish this section by showing that the remaining classes 
\iAppDiff\ and \BaireF\ of Theorem \ref{th:semigrinclusions} are not
closed under composition.
 
\example{ex:closureiAppDiff}{There exists a differentiable function
$f$ and an approximately and \I-approximately differentiable function
$g$ such that $g\circ f$ is neither approximately 
nor \I-approximately continuous. In particular, the classes
\AppDiff\ and \iAppDiff\ are not closed under composition.}
 
Proof. Let $D=\UoverN (p_n,q_n)=(0,\infty)\setminus E$ where $E$ is from 
Proposition \ref{sequence}. Let us define
$h\RtoR$ by putting $h(x)=0$ for $x\in D^c$ and 
\[
h(x)=u_n (x-p_n)(x-q_n)
\]
for $x\in(p_n,q_n)$, where $u_n>0$. Define $f$ by $f(x)=\int_0^x h(y)\,dy$.
It is easy to see that $f$ is differentiable and constant
on each interval contained in $D^c$. Decreasing the constants $u_n$, if
necessary, we may assume that
$\lim_{n\to\infty} f(p_{n+1})/f(p_n) = 0$.
 
Let us choose intervals $(a_n,b_n)$ centered at $f(p_n)$ such that
the assumptions from Proposition \ref{sequence} are satisfied and
let $E=\UoverN [a_n,b_n]$. Then, $E^c\in\densitytop\cap\deepitopology$.
Let us define $g(x)=0$ for $x\in E^c$ and 
\[
g(x)=v_n (x-a_n)^2 (x-b_n)^2
\]
for $x\in(a_n,b_n)$,
where the $v_n$ are choosen in such a way that $g(f(p_n))=1$.
Then $g$ is differentiable at every point $\neq 0$ and $g$ is constantly
equal $0$ on $E^c\in\itopology\cap\densitytop$. Hence,
$h\in\AppDiff\cap\iAppDiff$. On the other hand, $g\circ f=1$ on
the set $D^c$ while, $g\circ f(0)=0$. Thus, 
$g\circ f\not\in\AppDiff\cup\iAppDiff$.
 
\example{ex:closureBaireF}{The class \BaireF\ is 
not closed under composition.}
 
Proof. Let $h$ be an embedding of the irrationals
$\reals\setminus\rationals$ into the Cantor set $C\subset[0,1]$. Put 
$f(q)=2$ for $q\in\rationals$ and $f(x)=h(x)$ for $x\in\reals\setminus\rationals
$.
Then, $f\in\BaireF$. Moreover, let $S\subset\reals\setminus\rationals$ be a set
without the Baire property and let $g$ be the characteristic
function of $f(S)$; i.e.,
$g=\charf{f(S)}$. Then $g\in\BaireF$, since $f(S)\subset C$ and 
$g\circ f=\charf{S}\not\in\BaireF$. 
 
 
\sgfsection{Semigroups of \I-density Continuous Functions}
 
For the reminder of this paper let \E\ stand for any of the
semigroups of Theorem \ref{th:Isemigroups}; e.g., one of the classes
$\idenscont$,
$\deepidenscont$,
$\Diff\cap\idenscont$,
$\Diff\cap\deepidenscont$,
$\iAppDiff\cap\idenscont$,
$\iAppDiff\cap\deepidenscont$ or 
$\IaeiAppDiff\cap\idenscont$. 
In particular, $\Diff\cap\idenscont\subset\E\subset\deepidenscont$.
It will be shown that these semigroups, 
with the possible exception of $\IaeiAppDiff\cap\idenscont$, have the inner
automorphism property.
 
We start with the following lemma.
 
\lemma{lemma23}{
If $h$ is a generating bijection of an automorphism \Y\/ of \E, then $h$ is
\iapp\/ continuous.
}
 
Proof.
Let $r\in\reals$ and $\e>0$ be arbitrary.  We will show that $h$ is
\iapp\/ continuous at $r$. 
 
Let $f\RtoR$ be as in Example \ref{ex:hunch} for
$a=h(r)-\e$, $b=c=h(r)$ and $d=h(r)+\e$.
Let $g\RtoR$ with $g(x)=f(x)+h(r)$.  It is easy to see that 
$g\in\Diff\cap\idenscont\subset\E$.
 
\Y\/ is an automorphism of \E, thus there is an 
$\alpha\in\E\subset\deepidenscont$ such that $\Y(\alpha)=g$.
We have $h\circ\alpha=g\circ h.$
Note that
$h(\alpha(r))=g(h(r))=f(h(r))+h(r)=1+h(r)$.  
Therefore, $h(\alpha(r))\ne h(r)$. 
But, $h$ is a bijection, so $\alpha(r)\ne r$.  
 
The function $\alpha$ is deep-\I-density continuous.  
Thus, there exists a set $U\in\deepitopology$ such that
$\alpha(x)\ne r$ for all $x\in U$.  
Then, for all $x\in U$, 
$f(h(x))+h(r)=g(h(x))=h(\alpha(x))\ne h(r)$, 
which implies $f(h(x))\ne 0$; i.e., $h(x)\in (a,d)=(h(r)-\e,h(r)+\e)$.
So $|h(x)-h(r)|<\e$ for $x\in U$.  
This means precisely that $h$ is \iapp\/ continuous.
 
\corollary{cor:generhomeom}{If $h$ is a generating bijection 
of an automorphism \Y\/ of \E, then $h$ is a homeomorphism.}
 
Proof. $h$ is an \I-approximately continuous bijection of \reals.
Since $h$ is also a Darboux Baire one function, it must be a homeomorphism.
 
\bigskip
 
As a next step we need the following
 
\lemma{lemma32}{
Let $h\RtoR$ be an increasing homeomorphism which is not \idens\/
continuous at $0$.  Then there exists a function 
$f\in\Diff\cap\idenscont\subset\E$ 
such that $\conj{f}{h}$ is not deep-\idens\/ continuous.
}
 
Proof. Without any loss of generality we may assume that $h(0)=0$. 
 
It is very easy to verify that any deep-\I-density
continuous homeomorphism  is also \I-density continuous. (Compare e.g.
\cite[Theorem 23]{Wilczynski:CatAn}.) Thus, $h$ is also not deep-\I-density
continuous at $0$. Again, without any loss of generality we may further assume
that $h$ is not right deep-\I-density continuous at $0$. The left-hand side
argument is essentially the same.
 
The sets $E\cup (-E)\cup \{0\}$, where $E$ is an open right interval set, form 
basis for $\deepitopology$ at $0$ \cite{Lazarow:Coarsest}.
Thus, there exists a right interval set
$E=\UoverN[a_n,b_n]$ such that $0$ is an \I-density point of 
$\complement{E}$, while $0$ is not a right \I-density point of the complement of
\[
D=\UoverN[\inv{h}(a_n),\inv{h}(b_n)].
\]
 
Using the definition of deep-\I-density point we can easily choose the sequences
$\{\alpha_n\}$ and $\{\beta_n\}$ such that
$\inv{h}(a_n)<\alpha_n<\beta_n<\inv{h}(b_n)$ for $n\in\nats$
and $0$ is still not a right \I-density point of the complement of
\[
\UoverN[\alpha_n,\beta_n]
\]
while, evidently, $0$ is an \I-density point of the complement of
\[
\UoverN [h(\alpha_n),h(\beta_n)]\subset\UoverN [a_n,b_n].
\]
 
For each $n\in\nats$, let $f_n\colon\reals\to[0,1]$ be a function from Example
\ref{ex:hunch} for $\inv{h}(a_n)<\alpha_n<\beta_n<\inv{h}(b_n)$.
Define $f(x)=0$ for $x\in E^c$ and $f(x)=c_n f_n(x)$ for $x\in[a_n,b_n]$ and
$n\in\nats$, where $c_n>0$ are choosen in such a way that 
\begin{equation}\label{cond:limit}
\lim_{n\to\infty}\frac{h(c_{n+1})}{h(c_n)}=0.
\end{equation}
Then it is easy to see that $f$ is differentiable and
\I-density continuous.
 
On the other hand
\[
T=(\conj{f}{h})\left(\UoverN (\alpha_n,\beta_n)\right)
=
\UoverN h\left(c_n\right).
\]
Now, by (\ref{cond:limit}) and Proposition \ref{sequence}, 
it is easy to see that there is a right interval set $S\supset T$ such that
$0$ is a deep-\I-density point of $\complement{S}$, while $0$ is not an
\I-density point of $\left(\UoverN [\alpha_n,\beta_n]\right)^c$.
Hence, \conj{f}{h}\/ is not deep-\idens\/ continuous.
 
\corollary{cor:genIdens}{If $h$ is a generating bijection of an automorphism
of \E, then $h$ and \inv{h}\/ are
\idens\/ continuous homeomorphisms of \reals.}
 
Proof. By Corollary \ref{cor:generhomeom}, $h$ is a homeomorphism of \reals.
So, by Lemma \ref{lemma32} it must be \I-density continuous.
The statement for \inv{h}\/ follows from the fact that \inv{h}\/ is
a generating bijection for \inv{\Y}.
 
\bigskip
 
As an immediate corollary we obtain the following.
 
\theorem{theorem18}{
The semigroups \idenscont\/ and \deepidenscont\/  
have the inner automorphism property. Moreover,
\[
\aut{\idenscont}=\aut{\deepidenscont}.
\]
}
 

Let us notice that we can also deduce that \deepidenscont\/ has 
the inner automorphism property by using Proposition \ref{prop:generated},
as the deep-\I-density topology \deepitopology\ is generated.
This follows immediately from the fact that the function which demonstrates
the complete regularity of the \deepitopology\ \cite[Theorem 5]{Lazarow:Coarsest}
is also deep-\I-density  continuous\footnote
{If $0\in A\in\deepitopology$, then there exist right interval sets 
$E\subset V\subset A$ composed of closed and open intervals, respectively,
such that $0$ is a deep-\I-density point of a closed set
$P=\closure{(-E)\cup E}$ which is a subset of $U=(-V)\cup V\cup\{0\}$.
\L azarow's function is defined by $f(0)=1$ and 
$f(x)=\left(\dist{x}{U^c}\right)/\left(\dist{x}{U^c}+\dist{x}{P}\right)$
for $x\neq 0$. 
Because $f$ is piecewise linear on a neighborhood of every point $x\ne0$, it
must be density continuous at $x$. It is  deep-\I-density continuous at $0$
because $f$ is constant on $U\in\deepitopology$. }. However, Proposition
\ref{prop:generated} cannot be used in the case of the \I-density topology,
because \itopology\ is not generated. This follows from the fact that for
\itopology\ to be generated,
 the
family $\F = \{\left(f^{-1}(\{x\})\right)^c\colon x\in\reals 
\mbox{ and } f\colon(\reals,\itopology)\to(\reals,\itopology)\}$ would have to
form a subbase of \itopology. This is not the case since even the family
$\{\left(f^{-1}(\{1\})\right)^c\colon  
f\colon(\reals,\itopology)\to[0,1]\}\subset\F$ does not form
a subbase for \itopology, because \itopology\ is not regular
\cite[Theorem 5]{PWW:CatAn}.
 
To discuss the inner automorphism properties of the other semigroups
we need the following two lemmas.
 
\lemma{diffVSidiffhom}{The notions of \I-approximate derivative,
approximate derivative
and ordinary derivative coincide on the class of all homeomorphisms. In
other words, if $f\RtoR$ is a homeomorphism and $x\in\reals$ then 
$f'(x)=\iAppDiff f(x)=\AppDiff f(x)$ whenever any of these derivatives
exists.}
 
Proof. See {\L}azarow and Wilczy\'nski \cite{LW:IAppDer:21A}, also Wilczy\'nski
\cite[Theorem 38]{Wilczynski:CatAn}.
 
\lemma{lemma34}{
Let \Y\/ be an automorphism of 
$\Diff\cap\idenscont$,
$\Diff\cap\deepidenscont$,
$\iAppDiff\cap\idenscont$ or 
$\iAppDiff\cap\deepidenscont$ and let $h$ be its generating
bijection. Then $h$ is differentiable; i.e., $h\in\Diff$.
}
 
Proof. It is already known that $h$ is a homeomorphism. 
Thus $h$ is differentiable almost
everywhere \cite{Bruckner:DiffReal}.
Let $x_0$ be a point of differentiability of $h$ and let $x\in\reals$ be any 
other point.
Define $f(t)=t+(x-x_0)$. 
Then $f\in\Diff\cap\idenscont$, so it belongs to all of the
semigroups under consideration.  For any positive \d\/ we have
\begin{eqnarray*}
\frac{h(x+\d)-h(x)}{\d} & = & \frac{h\circ f(x_0+\d)-h\circ f(x_0)}{\d}\\
& = & \frac{\Y(f)\circ h(x_0+\d)-\Y(f)\circ h(x_0)}{\d}.\label{equation27}
\end{eqnarray*}
 
If \Y\/ is an automorphism of $\Diff\cap\idenscont$ 
or $\Diff\cap\deepidenscont$ then $\Y(f)$ is differentiable, and since $h$
is differentiable at $x_0$, the quotient in (\ref{equation27}) converges to
$\Y(f)'(h(x_0))\,h'(x_0)$.
 
If \Y\/ is an automorphism of $\iAppDiff\cap\idenscont$ or 
$\iAppDiff\cap\deepidenscont$ then $\Y(f)$ is \iapp\/ differentiable. 
So, by Lemma \ref{ichainrule},
$\Y(f)\circ h$ is \iapp\/ differentiable at $x_0$.
Hence, condition (\ref{equation27}) guarantees that $\iAppDer h(x)$ exists and
is equal $\iAppDer (\Y(f)\circ h)(x_0)$. But, $h$ is a homeomorphism. 
Thus, by Lemma \ref{diffVSidiffhom},
$h$ must be differentiable at $x$ as well.  This ends the proof.
 
\bigskip
 
As an immediate corollary we obtain
 
\theorem{theorem19}{
The semigroups 
$\Diff\cap\idenscont$,
$\Diff\cap\deepidenscont$,
$\iAppDiff\cap\idenscont$ and 
$\iAppDiff\cap\deepidenscont$ 
have the inner automorphism property. Moreover,
\[
\aut{\Diff\cap\idenscont}=\aut{\Diff\cap\deepidenscont}=
\aut{\iAppDiff\cap\idenscont}=\aut{\iAppDiff\cap\deepidenscont}
\]
\[
\subset\aut{\idenscont}=\aut{\deepidenscont}
\]
and the inclusion is proper.
}
 
We are not able to prove or disprove that the semigroup 
$\IaeiAppDiff\cap\idenscont$ has the inner automorphism
property.  We proved that each automorphism of 
$\IaeiAppDiff\cap\idenscont$ is generated by a
homeomorphism $h$ of the real line which must be differentiable almost
everywhere.  However, to be an element of 
$\IaeiAppDiff\cap\idenscont$, $h$ would need to be
differentiable outside of a set of first category. In general, a homeomorphism 
of the real line need not be \I-a.e. differentiable.  
In fact, Belna, Cargo, Evans
and Humke \cite{BelnaCargoEvansHumke:5} show that there exists a strictly
increasing homeomorphism $h:[0,1]\to[0,1]$ and $\tau\in(0,1)$ such that
\[
\llDini{h}(x)=\lrDini{h}(x)=\tau
\mbox{ and }
\ulDini{h}(x)=\urDini{h}(x)=\infty
\]
for a residual set of points $x\in [0,1]$. 
On the other hand, there are certain restrictions on the ``severity'' 
of the nondifferentiability of $h$ on sets of the second category.  
Neugebauer \cite{Neugebauer:Derivatives} shows that a continuous
function $f$ has $\llDini{f}(x)=\urDini{f}(x)$ and 
$\ulDini{f}(x)=\urDini{f}(x)$ on a residual set.
 
We are not able to find a satisfactory answer to our question.  Therefore, the
following remains.
 
\problem{problem1}{
Let $h$ be a homeomorphism of \reals\/ such that for every \idens\/ continuous,
\I-a.e. \iapp\/ differentiable $f$, \conj{f}{h}\/  is also \idens\/ continuous,
and \I-a.e. \iapp\/ differentiable.  Is $h$ differentiable on a residual
set? }
 
Clearly, an affirmative answer to this problem is equivalent to 
the fact that $\IaeiAppDiff\cap\idenscont$ has the
inner automorphism property.
 
\bigskip
 
\bibliographystyle{plain}
%\bibliography{semigroups}

\begin{thebibliography}{10}

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\end{thebibliography}


\bigskip

\begin{flushleft}
Krzysztof Ciesielski, Department of Mathematics, West Virginia University,
Morgantown, WV 26506\\
Lee Larson and Krzysztof Ostaszewski, Department of Mathematics, University of
Louisville, Louisville, KY 40292\\
{\bigskip}
Received October 11, 1990\\
and in the final form May 21, 1990
\end{flushleft}
 
\end{document}
