% Level sets of density continuous functions 

 
\documentstyle[12pt]{article} \pagestyle{myheadings}
\markright{Density Continuous Functions}
 
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\title{Level sets of density continuous functions}
\author{Krzysztof Ciesielski \and Lee Larson }
\date{}
 
\newcommand{\C}{\mbox{$\cal C$}} \newcommand{\CD}{\mbox{$\C_D$}}
\newcommand{\SD}{\mbox{$S_D$}} \newcommand{\bstar}{\mbox{Baire*1}}
\newcommand{\ambig}{\mbox{$\Fs\cap\Gd$}}
\newcommand{\ind}{\mbox{$\Lambda$}} \newcommand{\cover}{\mbox{\cal S}}
 
\begin{document}
 
%\Timestwelve
 
\maketitle
 
 
 
\begin{abstract}
A function $f:\reals\into\reals$ is density continuous if it is
continuous when both its range and domain are endowed with the
density topology. The level sets of density continuous functions are
characterised as those sets which are density closed and ambiguous.
\footnote{AMS Subject Classification 26A21.}
\end{abstract}
 
\section{Introduction}
 
The density topology on the real line, \reals, consists of all sets,
$S$, such that each point of $S$ is a Lebesgue density point of $S$.
The density topology is a completely regular refinement of the ordinary
topology, which fails to be normal \cite{GNN:densitytop}. In this
paper, we consider some properties of functions $f:\reals\into\reals$
which are continuous when the density topology is applied to both the
range and domain. Such functions have been termed {\em density
continuous} \cite{KO:denscont}.
 
It has recently been shown that the density continuous functions do
not form a vector space and there are monotone, and even $C^\infty$
functions which are not density continuous \cite{CL:SpDensCont}. On
the other hand, all locally convex functions are density continuous
\cite{CL:SpDensCont} and density continuous functions are in the class
\bstar\/ \cite{CLO:DensContCont}.
 
In this paper, we answer a question posed by Krzysztof
Ostaszewski \cite{KO:SemiDensity} related to the properties of the
class of density continuous functions, \CD, when viewed as a
semigroup.  This question is:
\begin{description}
\item[Query]
Given a density
closed \Gd\/ subset $E$ of \reals, is there an $f\in\CD$ such that
$E=\inv{f}(0)$?
\end{description}
 
 
It turns out that this question can be answered negatively.
In Section \ref{charsec}, the level sets of density continuous functions are
characterized as consisting of the sets which are closed in the density
topology while being simultaneously \Fs\/ and \Gd.  This shows that
any density closed set $E$, which is \Gd, but not \Fs\/ is a
counterexample to Ostaszewski's query.
 
We use the following notation: \begin{description}
    \item $\reals$ -- the set of real numbers;
    \item $\nats$ -- the set of natural numbers;
    \item $|A|$ -- the Lebesgue measure of a
        measurable set $A\subset\reals$;
    \item $A^c$ -- the complement of the set $A$;
    \item \interior{A}--the interior of the set A;
    \item $\dist{P}{Q}=\inf\{ |p-q|:p\in P \mbox{ and } q\in Q\}$ is
the distance between sets $P$ and $Q$;
    \item $B(F,\e)=\{ x:\dist{F}{x}<\e\}$;
    \item ${\overline d}(A,x),\ {\underline d}(A,x),\
        d^+(A,x),\ d^-(A,x),\
        d(A,x)$ -- the upper, lower, right, left and ordinary
(respectively)
        densities of a set $A\subset\reals$ at a point
        $x\in\reals$;
    \item $L_f(\alpha)=\{x\in\reals:f(x)=\alpha\}$;
    \item \CD -- the set of all density continuous functions on \reals.
    \item Given $P\subset Q\subset\reals$, we say $P$ is a {\em
portion}
        of $Q$, if there exists an open interval $I$ such that
        $P=I\cap Q\ne\emptyset$. \end{description}
 
\section{Level Sets of Density Continuous Functions}\label{charsec}
 
The purpose of this section is to prove the characterization of
the level sets of functions in \CD. This is presented in Theorem
\ref{thm:LevelSets}. First, some preliminary results and definitions
must be presented. Several of these results are interesting in their
own right.
 
O'Malley \cite{RO:Baire*1} defined the class \bstar\/ to consist of all
functions $f:\reals\to\reals$ such that for every perfect set $P$,
there exists a portion $Q$ of $P$ such that $f|_Q$ is continuous. The
following theorem is known \cite{CLO:DensContCont}.
 
\begin{theorem}\lab{thm:DensContB*}
    $\CD\subset\bstar$. \end{theorem}
 
\begin{theorem}\lab{thm:lattice}
    \CD\/ is a lattice. \end{theorem}
 
Proof. Let $f,g\in\CD$, $h=\max\{f,g\}$ and $x_0\in\reals$. Assume
first that $h(x_0)=f(x_0)>g(x_0)$ and that $m\in (g(x_0),f(x_0))$. If
$G$ is a density neighborhood of $f(x_0)$ contained in $(m,\infty)$,
then  $H=\inv{f}(G)\cap\inv{g}((-\infty,m))$ is a density neighborhood
of $x_0$ with the property that $h|H=f|H$.  This implies that $h$ is
density continuous at $x_0$.  A symmetrical argument handles the case
when $h(x_0)=g(x_0)>f(x_0)$.
 
Now, assume $f(x_0)=g(x_0)=h(x_0)$ and $G$ is a density neighborhood
of $h(x_0)$. Both $\inv{f}(G)$ and $\inv{g}(G)$ are density
neighborhoods of $x_0$, so $H=\inv{f}(G)\cap \inv{g}(G)$ is also a
density neighborhood of $x_0$. If $x\in H$, then $f(x)\in G$ and
$g(x)\in G$, so $h(x)=\max (f(x),g(x))\in G$. From this, it follows
that $h^{-1}(G)\supset H$ and $h$ is density continuous at $x_0$.
 
Therefore \CD\/ is closed under the operation of taking the maximum of
two functions. Since $\min(f(x),g(x))=-\max(-f(x),-g(x))$, we see
\CD\/ is closed under the minimization operation also. These two
statements prove that \CD\/ is a lattice.
 
The following lemma was proved by the present authors  \cite[Corollary
1]{CLO:DiffDensCont}.
 
\begin{lemma}\lab{lem:distClosed} If $F$ is a closed set, then
$f(x)=\dist{x}{F}\in\CD$. \end{lemma}
 
This theorem can be proved from a known result characterising the
associated sets of \bstar\/ functions due to Pu and Pu
\cite{PP:AssBaire*}.  Since it is in the spirit of what follows, we
include the following shorter proof.
 
\begin{theorem}\lab{thm:B*FsGd} A set $A$ is a level set for a
function $f\in\bstar$  if and only if $A\in\ambig$.
\end{theorem}
 
Proof. Suppose $f\in\bstar$ and $\alpha\in\reals$. Since \bstar\/ is
contained in Baire 1, it follows that $L_f(\alpha)\in\Gd$.  To prove
that $L_f(\alpha)$ is also in \Fs\/, we use induction on
$\zeta<\omega_1$ to define sequences $K_\zeta$ and $F_\zeta$ such that
for every $\zeta<\omega_1$ \begin{description} \item[(a)]
$K_\zeta=\cl{L_f(\alpha)\setminus\Cup_{\eta<\zeta}F_\eta}$, \item[(b)]
$F_\zeta\subset L_f(\alpha)\cap K_\zeta$, \item[(c)] $F_\zeta$ is
relatively open in $K_\zeta$, and \item[(d)] if $K_\zeta\ne\emptyset$,
then $K_{\zeta+1}$ is a proper subset of $K_\zeta$. \end{description}
 
By (a), we have defined $K_\zeta$, provided $F_\eta$ are defined for
all  $\eta<\zeta$.  In particular, $K_0=\cl{L_f(\alpha)}$.
 
Thus, let us assume that $K_\zeta$ is defined.  If
$K_\zeta=\emptyset$, then define $F_\zeta=\emptyset$.  If
$K_\zeta\ne\emptyset$, we can find, using the fact that $f\in\bstar$,
a portion $F_\zeta\subset K_\zeta$ such that $f|F_\zeta$ is
continuous.  But, by (a), $L_f(\alpha)\cap F_\zeta$ is dense in
$F_\zeta$, so $f|F_\zeta\equiv\alpha$ and consequently,
$F_\zeta\subset L_f(\alpha)$.  This implies (b) and (c),  and,
together with (a), also (d).  The construction is finished.
 
As the sequence $\{ K_\zeta :\zeta<\omega_1\}$ of closed sets cannot be
strictly decreasing, (d) implies that there exists an $\eta<\omega_1$
such that $K_\eta=\emptyset$; i.e., that \[
L_f(\alpha)=\Cup_{\zeta<\eta}F_\zeta. \] But, (c) implies that each
$F_\zeta\in\Fs$, which immediately yields $L_f(\alpha)\in\Fs$.
 
Next, suppose that $A\in\ambig$, $f(x)=\charf{A}(x)$ and $P$ is
a perfect set.   By supposition, both $A$ and $\comp{A}$ are \Gd\/
sets, so the Baire category theorem shows they both cannot be dense in
$P$. Therefore, there must exist a portion $Q\subset P$ such that
either $Q\subset A$ or $Q\subset \comp{A}$.  In either case, $f|Q$ is
constant and therefore continuous.  This shows that $f\in\bstar$ with
$A=L_f(1)$.
 
 
\begin{lemma}\lab{lem:openclosed} Let $A$ be a density closed set from
$\Fs\cap\Gd$ such that $A$ does not contain any infinite interval.
Then there is a bounded open set $G$ in \reals\/ such that $A\cap G$
is a nonempty compact set. \end{lemma}
 
Proof. First, suppose that $A$ contains no open intervals.  Assume the
lemma is not true in this case.  We claim that with this supposition,
both $A$ and \comp{A} are dense \Gd\/ subsets of \cl{A}.  This claim
is clear for $A$.  To prove it for \comp{A}, let $I$ be any open
interval such that $I\cap A\ne\emptyset$. Since $A$ contains no open
intervals, there must exist a closed interval $[c,d]\subset I$ such
that $[c,d]\cap A=(c,d)\cap A$ is nonempty and not closed.  It follows
easily from this that  \[ I\cap\cl{A}\setminus A\supset
[c,d]\cap\cl{A}\setminus A\ne\emptyset, \] which implies the claim.
But, this obviously contradicts the Baire category theorem.
Therefore, the lemma must be true in this case.
 
Now, returning to the general case for $A$, we introduce an equivalence
relation, $\sim$, on $A$ by $a\sim b\Longleftrightarrow [a,b]\subset
A$; i. e., all the closed and connected components of $A$ are the
equivalence classes of $A/\!\!\sim$. There is obviously a set
$B\subset\reals$ which is homeomorphic to $A/\!\!\sim$; i.e., there is
a nondecreasing function $h:A\into B$ such that
$h(a)=h(b)\Longleftrightarrow a\sim b$. $B$ does not contain any
interval, $B\in\Fs\cap\Gd$ and $B$ is density closed. If $B\cap (a,b)$
is compact, then $\inv{h}((a,b))$ satisfies the conclusion of the
lemma.
 
The following lemma is an easy consequence of the preceding lemma.
 
\begin{lemma}\lab{lem:MaxFam} If $A$ is a nonempty density closed set
from $\Fs\cap\Gd$ such that $A$  does not contain any infinite
interval, then there is a maximal family $\{ U_n:n\in\nats\}$ of
pairwise disjoint bounded open sets such that $A\cap U_n$ is compact
and nonempty. \end{lemma}
 
\begin{lemma}\lab{lem:decomp} Let $A$ be a density closed set from
$\Fs\cap\Gd$ that does not contain any infinite interval.  There is a
countable decomposition $\{ F_\gamma\}_{\gamma\in\Gamma}$ of $A$ into
compact sets and a family $\{ \e_\gamma\}_{\gamma\in\Gamma}\subset
(0,\infty)$ such that \[ B_x=\{ \gamma\in\Gamma : x\in
B(F_\gamma,\e_\gamma) \} \] is finite for all $x\in\reals$ and \[ \{
B(F_\gamma,\e_\gamma)\setminus F_\gamma : \gamma\in\Gamma\} \] is
locally finite in the density topology. \end{lemma}
 
Proof. We will use induction to define sets $K_\zeta$, $F(\zeta,\eta)$
and $U(\zeta,\eta)$ for $\eta\in\nats$ and $\zeta<\omega_1$ satisfying
\begin{description} \item[(a)] $F(\zeta,\eta)=K_\zeta\cap
U(\zeta,\eta)$, \item[(b)] $F(\zeta,\eta)$ is compact, \item[(c)]
$U(\zeta,\eta)$ are open and bounded, \item[(d)]
$K_\zeta=A\setminus\Cup_{\alpha<\zeta}\Cup_\eta U(\alpha,\eta)$,
\item[(e)] for fixed $\zeta$, the collection $\{
U(\zeta,\eta):\eta\in\nats\}$ is pairwise disjoint, and \item[(f)] if
$K_\zeta\ne\emptyset$, then there is an $\eta\in\nats$ such that
$K_\zeta\cap U(\zeta,\eta)\ne\emptyset$. \end{description}
 
It follows from (d) that $K_0=A$.   Properties (c) and (d) imply that
every $K_\zeta$ satisfies the conditions of Lemma \ref{lem:MaxFam}.
Inductive application of Lemma \ref{lem:MaxFam} easily establishes
(a)--(f).  Moreover, $K_\zeta$ is a decreasing sequence of sets which
are relatively closed in $A$, so there is an $\alpha<\omega_1$ such
that for all $\beta>\alpha$, $K_\beta =K_\alpha$. Conditions (c), (d)
and (f) guarantee that $K_\alpha=\emptyset$. Thus, putting \[
\Gamma=\{ (\zeta,\eta):\zeta<\alpha, \eta\in\nats\}, \] we can define
$F_\gamma=F(\zeta,\eta)$ and $U_\gamma=U(\zeta,\eta)$ for all
$\gamma=(\zeta,\eta)\in\Gamma$.
 
Decreasing $U_\gamma$ does not change the above properties as long as
$F_\gamma\subset U_\gamma$ and $U_\gamma$ is open.  Combining this
with the fact that $F_\gamma$ is compact, we may assume that  for
every $\gamma$ there is a $\d_\gamma>0$ such that
$U_\gamma=B(F_\gamma,\d_\gamma)$ and $\sum_\gamma\d_\gamma<1$. Let
$a_\gamma>0$ be such that $\sum_\gamma a_\gamma<1$ and choose
$\e_\gamma\in (0,\d_\gamma)$ such that
\begin{equation}\label{eq:epsfraction}
\frac{|B(F_\gamma,\e_\gamma)\setminus F_\gamma|}{\d_\gamma}<a_\gamma.
\end{equation}
 
First notice that for every $x\in\reals$, the set  $U_x=\{\gamma:x\in
U_\gamma\}$ is finite. Otherwise, there is an infinite sequence,
$\gamma_k=(\zeta_k,\eta_k)$ such that $x\in U_{\gamma_k}$.  It follows
from (e) that $\zeta_n\ne\zeta_m$ whenever $n\ne m$.  By passing to a
subsequence, if necessary, we may assume that $\zeta_k$ is an
increasing sequence.
 
Fix  \[ \d=\dist{K_{\zeta_1}}{x}\ge\dist{K_{\zeta_0+1}}{x}>0, \] where
the first inequality follows from the fact that $K_\zeta$ is a
decreasing sequence and the second from (d). Since
$\lim_k\d_{\gamma_k}=0$, there is an $n>1$ such that
$0<\d_{\gamma_m}<\d$ for all $m\ge n$. In particular,  \[
\d_{\gamma_n}<\d=\dist{K_{\zeta_1}}{x}\le\dist{F_{\gamma_n}}{x}; \]
i.e., $x\notin B(F_{\gamma_n},\d_{\gamma_n})=U_{\gamma_n}$.  This
contradiction establishes that $U_x$ is finite.  This implies that
$B_x$ is also finite, which establishes the first part of the lemma.
 
Let \[ \Gamma_1=\Gamma\setminus B_x, \] \[ S=\Cap_{\gamma\in
B_x}B(F_\gamma,\e_\gamma)\setminus F_\gamma, \] \[
T=\Cup_{\gamma\in\Gamma_1}B(F_\gamma,\e_\gamma)\setminus F_\gamma, \]
and \[ U=S\setminus T. \]
 
Evidently, $x\in U$, $S$ is open and $U$ intersects only finitely many
of the $B(F_\gamma,\e_\gamma)\setminus F_\gamma$. To finish the proof,
it is enough to show that $U$ has full density at $x$. This is done by
establishing that $d(T,x)=0$.
 
To do this, let $\e>0$. Choose $\Gamma_2\subset\Gamma_1$ such that
$\Gamma_1\setminus\Gamma_2$ is finite and
$\sum_{\gamma\in\Gamma_2}a_\gamma<\e$.  Since $\e_\gamma<\d_\gamma$,
$k_0$ can be chosen such that for $k\ge k_0$, \[ (x,x+1/k)\cap \left(
B(F_\gamma,\e_\gamma)\setminus F_\gamma\right)= \emptyset,\ \
\forall\gamma\in\Gamma_1\setminus\Gamma_2. \] Thus,
\begin{eqnarray}\label{eq:bigineq}
\lefteqn{\left|\frac{(x,x+1/k)\cap\Cup_{\gamma\in\Gamma_1}
(B(F_\gamma,\e_\gamma)\setminus F_\gamma )}{1/k}\right|
 \le}\nonumber \\  & & k\sum_{\gamma\in\Gamma_2}| (x,x+1/k)\cap
\left( B(F_\gamma,\e_\gamma)\setminus F_\gamma\right) | \\ & & \le
k\sum_{\gamma\in\Gamma_2}a_\gamma /k < \e.\nonumber \end{eqnarray} The
first inequality in (\ref{eq:bigineq}) is obvious. The second
inequality in (\ref{eq:bigineq}) is handled in two cases. If
$\d_\gamma\le 1/k$, then by (\ref{eq:epsfraction}),  \[ |(x,x+1/k)\cap
(B(F_\gamma,\e_\gamma)\setminus F_\gamma)| \le
|B(F_\gamma,\e_\gamma)\setminus F_\gamma| \le a_\gamma\d_\gamma \le
a_\gamma/k \] If $\d_\gamma >1/k$, then  \[
1/k<\d_\gamma=\dist{F_\gamma}{\comp{U_\gamma}} \le \dist{F_\gamma}{x}
\] by the definitions of $F_\gamma$ and $U_\gamma$ with $x\notin
U_\gamma$. Hence, \[ |(x,x+1/k)\cap (B(F_\gamma,\e_\gamma)\setminus
F_\gamma)|= \] \[ =
|(x,x+1/k)\cap(B(F_\gamma\cap(-\infty,x],\e_\gamma)\setminus
F_\gamma)|+ \] \[ +|(x,x+1/k)\cap B(F_\gamma\cap
[x,\infty),\e_\gamma)\setminus F_\gamma|. \] The first term on the
right-hand side is the empty set because
$\e_\gamma<\d_\gamma<\dist{x}{F_\gamma}$ and the assumption that
$1/k<\d_\gamma$.   Letting $y=\inf(F_\gamma\cap [x,\infty))$, we see \[
|(x,x+1/k)\cap (B(F_\gamma,\e_\gamma)\setminus F_\gamma)| =
|(x,x+1/k)\cap (y-\e_\gamma,y)|. \] In case $y=\infty$, the right-hand
side vanishes and the conclusion is obvious.   Assume $y\in\reals$.
Notice that $y>x+1/k$, $y-\d_\gamma>x$ and $y-\d_k+1/k<y$ because \[
\dist{x}{y}\ge\dist{x}{F_\gamma}\ge\d_\gamma>1/k. \] Hence,
\begin{eqnarray*} |(x,x+1/k)\cap (y-\e_\gamma,y)| &\le&
|(y-\d_\gamma,y-\d_\gamma+1/k)\cap (y-\e_\gamma,y)|\\ &=&
\frac{|(y-\d_\gamma,y-\d_\gamma+1/k)\cap (y-\e_\gamma,y)|}
        {k|(y-\d_\gamma,y-\d_\gamma+1/k)|}\\ &\le&
\frac{|(y-\d_\gamma,y)\cap(y-\e_\gamma,y)|}{k|(y-\d_\gamma,y)|}\\ &=&
\frac{\e_\gamma}{k\d_\gamma}\\ &\le&
\frac{|B(F_\gamma,\e_\gamma)\setminus F_\gamma|}{k\d_\gamma}\\ &\le&
a_\gamma/k. \end{eqnarray*} The second inequality in the expression
written above is justified by noting that if $0<a<b$ and $c>0$, then
$a/b<(a+c)/(b+c)$.
 
Therefore $d^+(T,x)=0$.
 
A similar argument can be presented for the left density.  The lemma
follows from these densities.
 
 
 
\begin{theorem}\lab{thm:LevelSets}
    The level sets of density continuous functions are precisely the
    density closed sets which are in \ambig. \end{theorem}
 
Proof. In light of Theorem \ref{thm:B*FsGd} and Theorem
\ref{thm:DensContB*} we see that the level sets of any density
continuous function are in $\Fs\cap\Gd$ and are density closed.  To
finish the proof of the theorem, it suffices to show that given a
density closed set $A\in\Fs\cap\Gd$, there is a function $f\in\CD$
such that $A=L_f(0)$.
 
First assume that $A$ does not contain any infinite interval. Let
$\Gamma$, $B_x$, $F_\gamma$ and $\e_\gamma$  be as in Lemma
\ref{lem:decomp}. Define \[ h_\gamma (x)=\dist{x}{F_\gamma}/\e_\gamma,
\] so that $h_\gamma\equiv 0$ on $F_\gamma$ and $h_\gamma\ge 1$ on
$B\comp{(F_\gamma,\e_\gamma)}$. According to Lemma \ref{lem:decomp},
$B_x$ is finite for all $x$, so that a function $f$ may be defined as
\[ f(x)=\min_{\gamma\in\Gamma}\{ 1,h_\gamma (x)\}. \] Evidently,  in
light of Lemma \ref{lem:distClosed},  $h_\gamma$ and 1 are density
continuous.  Also,  $\inv{f}(0)=\Cup_{\gamma\in\Gamma}F\gamma =A$. It
suffices to prove that $f$ is density continuous.
 
Let $x\in\reals$. According to Lemma \ref{lem:decomp} there is a
density open set $G$ containing $x$ such that \[ G_x=\{\gamma : G\cap
B(F_\gamma,\d_\gamma)\setminus F_\gamma\ne\emptyset\} \] is finite.
We consider two cases.
 
Case 1. Assume $x\notin A$.   Then, Theorem \ref{thm:lattice} shows
that \[ f|G=\min_{\alpha\in G_x}\{ 1,h_\alpha\} \] is density
continuous on $G$.
 
Case 2. Let $x\in A$. If $g=\min_{\gamma\in G_x}\{1,h_\gamma\}$, then,
as in Case 1, $g$ is density continuous. Suppose $D$ is a density
neighborhood of $f(x)=0$.  Then \begin{eqnarray*} \inv{f}(D) & = &
\left[\inv{f}(D)\cap\comp{A}\right]\cup\left[\inv{f}(D)\cap A\right] \\
& \supset & \left[\inv{f}(D)\cap G\cap \comp{A}\right]\cup A \\ & = &
\left[\inv{g}(D)\cap G\cap\comp{A}\right]\cup A\\ & \supset &
\inv{g}(D)\cap G. \end{eqnarray*} This case is now clear because
$\inv{g}(D)\cap G$ is a density neighborhood of $x$.
 
Finally, if $A$ does contain an infinite interval, let  $-\infty\le
a<b\le\infty$ be such that $[a,b]\cap A$ does not contain an infinite
interval and such that $\reals\setminus (a,b)\subset A$. Then, if $f$
is defined for $A\cap [a,b]$ as above, the function \[ f_0(x)=\min\{
f(x), \dist{x}{\reals\setminus (a,b)}\} \] satisfies the conclusion of
the theorem.
 
\begin{corollary}\lab{cor:asssets} The associated sets of density
continuous functions are precisely the density open sets which are in
$\Fs\cap\Gd$. \end{corollary}
 
Proof. If $f$ is any real function and $\alpha\in\reals$, define
$g=\max\{ f,\alpha\}$.  Since $\{
x:f(x)>\alpha\}=\comp{(L_g(\alpha))}$, this corollary follows from
Theorems \ref{thm:lattice} and \ref{thm:LevelSets}.
 
\bibliographystyle{plain} 

%\bibliography{levelsets}

\begin{thebibliography}{1}

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\bibitem{PP:AssBaire*}
H.~W. Pu and H.~H. Pu.
\newblock Associated sets of {B}aire* 1 functions.
\newblock {\em Real Anal. Exch.}, 8(2):479--485, 1982-83.

\end{thebibliography}

 
\vspace{.125in}
 
{\noindent\footnotesize Lee Larson, Department of Mathematics,
University of Louisville, Louisville, KY 40292 \linebreak Krzysztof
Ciesielski, Department of Mathematics, West Virginia University,
Morgantown, WV 26506}
 
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