%\documentstyle[12pt]{article}
\documentclass{rae}

\MathReviews{26A15}
\keywords{%Key words and phrases: 
symmetrically continuous, measurability, Baire property.}


\newcommand{\real}{{\bf R}}
\newcommand{\rat}{{\bf Q}}
\newcommand{\inte}{{\bf Z}}
\newcommand{\nat}{{\bf N}}
\newcommand{\qed}{$\#$} 
\newcommand{\cont}{{\bf c}} 
\title{Symmetrically continuous functions 
on various subsets of the real line}
\author{Marcin Szyszkowski%
\thanks{ The author wishes to thank professor K. Ciesielski for fruitful 
discussion and invaluable help with writing this paper.},
%{\footnotesize
Department of Mathematics, West Virginia University,
Morgantown, WV 26506-6310, 
and Department of Mathematics,
Gda\'nsk University, Wita Stwosza 57, Gda\'nsk, Poland. 
e-mail: {\tt Marcin@math.wvu.edu}} 
 
\newtheorem{theorem}{Theorem}%[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{problem}{Problem}%[section]
\newtheorem{example}{Example}%[section]
\newtheorem{factD}[theorem]{Fact}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{conjecture}{Conjecture}
 
\newcommand{\thm}[2]{\begin{theorem}\label{#1}#2\end{theorem}}
\newcommand{\cor}[2]{\begin{corollary}\label{#1}#2\end{corollary}}
\newcommand{\prop}[2]{\begin{proposition}\label{#1}#2\end{proposition}}
\newcommand{\lem}[2]{\begin{lemma}\label{#1}#2\end{lemma}}
\newcommand{\pr}[2]{\begin{problem}\label{#1}#2\end{problem}}
\newcommand{\ex}[2]{\begin{example}\label{#1}#2\end{example}}
\newcommand{\defi}[2]{\begin{definition}\label{#1}#2\end{definition}}
\newcommand{\conj}[2]{\begin{conjecture}\label{#1}#2\end{conjecture}}
\newcommand{\fact}[2]{\begin{factD}\label{#1}#2\end{factD}}

\begin{document} 
\maketitle 

%\footnotetext{Mathematical Reviews subject classification: 26A15} 
%\abstract{ 
\begin{abstract}
We define symmetric continuity for functions defined 
on arbitrary subsets of $\real$. The main result is that 
when a symmetrically continuous function is defined on a 
measurable set (a set with the Baire property) 
then it is continuous almost everywhere (on a residual set, respectively).  
This generalizes the known result for functions defined 
on the whole real line. 
\end{abstract}
%*****************************************************
%Intro
\section {Introduction} 
The study of symmetric functions and symmetrically continuous 
functions stems from the theory of trigonometric series. 
In theory of trigonometric series, we study functions defined 
on various subsets of $\real$, while  
symmetrically continuous functions have been studied 
so far only on the whole real line or on intervals. 
We examine basic properties of symmetrically continuous functions 
defined on arbitrary subsets of $\real$. 

We say that a function $f\colon \real\to\real$ is symmetrically 
continuous at a point $x\in\real$ if 
$$\forall {\varepsilon >0}~\exists {\delta >0}~
\forall {|h|<\delta}~~|f(x+h)-f(x-h)|<\varepsilon $$ 
or, equivalently 
$$ \lim_{h\to 0} f(x+h)-f(x-h) = 0.$$ 

We define symmetrical continuity for a function defined 
on any subset of reals in the following way. 
\defi{scd}{Let $A\subset\real$. A function 
$f\colon A\to\real$ is said to be symmetrically 
continuous at a point $x$ if
$$\forall {\varepsilon >0}~\exists {\delta >0}~\forall {|h|<\delta}~~
x\pm h\in A \Rightarrow |f(x+h)-f(x-h)|<\varepsilon .$$}
If $f$ is symmetrically continuous at every point of $A$ then we 
we say that $f$ is symmetrically continuous on $A$ or simply 
symmetrically continuous.

Because of the implication above it is possible (and in fact
 we will often consider such a situation) that $f$ is 
symmetrically continuous at $x$ only because for every 
$x+h\in A$ the number $x-h$ is not in $A$. If this happens 
then we say that $f$ is vacuously symmetrically continuous  
at this point. In Section 5 we address the issue when the 
domain $A$ is symmetric, that is,  $x+h\in A$ if and only 
if $x-h\in A$. Requiring additionally in 
Definition \ref{scd} that the function is defined on a 
symmetric domain would narrow our considerations 
and would not improve our results (at least when the
domain is measurable or has the Baire property). 

Note that we do not need a function $f$ to be defined at 
a point $x$ to talk about symmetric continuity  at $x$. 
However we will concentrate only on the points from 
the domain (except in Theorem \ref{mmarc}). 

Obviously ordinary continuity implies symmetric 
continuity, but not the other way around. 
Let us give some simple examples. 
The discontinuous function $f(x)=0$ for $x\not =0$ and $f(0)=1$ 
is symmetrically continuous everywhere. The function 
$\cos\left({1\over x}\right)$ is symmetrically continuous 
everywhere (including $0$) but $\sin\left({1\over x}\right)$ 
is not symmetrically continuous at $0$.

\ex{ex1}{Let $G\subset\real$ be an additive group and 
$f=\chi_G\colon\real\to\{0,1\}$ be the characteristic 
function of $G$. Then $f$ is symmetrically continuous on $G$.}
Indeed if $a\in G$ then for any point $s=a+h$ its symmetric 
reflection about $a$ is $a-h=2a-s$ and it belongs to $G$ if 
and only if $s$ does. Outside $G$, $f$ is symmetrically 
continuous only at points $x$ for which $2x\in G$. 
In particular if $G$ is divisible by 2 then $f$ is 
symmetrically discontinuous everywhere outside $G$.

\ex{ex2}{Let $G_1$ and $G_2$ be the additive subgroups 
in $\real$ with $G_1\cap G_2=\{0\}$ and let 
$f\colon G_1\cup G_2\to \{0,1\}=\chi_{G_1}$. Then 
$f$ is symmetrically continuous on its domain. } 
Just like above, for any $a\in G_i$, $i=1,2$, and 
$s\in\real$, we have $s\in G_i$ if and only if 
$2a-s\in G_i$. So $f$ is symmetrically continuous at $a$. 

\smallskip 
Using functions defined on domains smaller than $\real$ 
we may address an old problem of Marcus to characterize 
the set of points at which a function $f\colon \real\to\real$ 
may be symmetrically continuous. Without solving this 
problem we state the following theorem. 
\thm{mmarc}{ For any set $A\subset\real$ there is a set 
$X\subset\real$ and a function $f\colon X\to\{0,1\}$ that 
is symmetrically continuous (vacuously) at $x\in\real$ 
(not necessarily in $X$) if and only if $x\in A$.} 
This theorem follows immediately from \cite[lemma 4]{msz}.

\medskip
In Section 2 we state the main results about continuity of 
symmetrically continuous functions. Section 3 consists 
only of the proofs of theorems from Section 2. 
In  Section 4 we investigate the extension properties 
of symmetrically continuous functions to bigger 
domains and the last section is devoted to 
functions on symmetric domains. 

Our notation is standard and follows \cite{Ci} and \cite{t}. 
By an interval we always mean a nontrivial interval (i.e., 
containing more than one point). For different $a,b\in\real$ 
we write $[a,b]$ for the interval $[\min(a,b),\max(a,b)]$. 




We sumarize the properties of symmetrically continuous functions 
defined on subsets of $\real$ in the chart below. 
We assume here that the function $f:A\to\real$ is symmetrically continuous. 
The property ``$f$ is extendable'' means that $f$ has 
a symmetrically continuous extension, and ``$f$ is almost extendable'' 
means that there is a symmetrically continuous function 
$F$ defined on a set containing $A$ such that 
$\{x\in A\colon f(x)\not = F(x)\}$ has measure zero. 
The property ``$f$ can be extended" indicates that there 
is an extension of $f$ which is symmetrically continuous in 
every point of $A$. 

The symbol ``+'' means that the function $f$ has the property 
in the left column, ``-'' means that it does not. 
Directly before the table we include some short explanations. 
%Marks ``(i..)'' refer to short explanations below the table. 
We also included the appropriate examples and theorems 
for fast reference.




%\bigskip 
%\newpage





\begin{description} 
\item[(i)] A function $f$ from $A\subset\real$ to $\real$ 
is measurable (or has the Baire property) 
if $f$ is the restriction of some measurable $F\colon\real\to\real$ 
(having the Baire property) to the set $A$. This is the same 
as saying that the preimages under $f$ of open sets are intersections 
of measurable sets (sets with the Baire property) with $A$. 
If $f$ is as in Example \ref{ex2} and 
$G_1$ and $G_2$ are of the full outer measure (nowhere meager) 
then $f$ is nonmeasurable (does not have the Baire property). 

\item[(ii)] The function $\chi_{(0,\infty)}$ on 
$\real\setminus\{0\}$ is an example here. 

\item[(iii)] Take $f$ is as in Example \ref{ex2} with
$G_1$ and $G_2$ nowhere meager and 
of the full outer measure. 

\item[(iv)] Extend $f$ putting $0$ on $\real\setminus A$.

\item[(v)] Define $f$ on the intervals 
$(-{1\over {2n}},-{1\over {2n+1}})$ and 
$({1\over {2n}},{1\over {2n-1}})$
by assigning value $1$ on $({1\over {2n}},{1\over {2n-1}})$
and $0$ otherwise.
%We do not 
%know whether any $f$ can be almost extended to almost whole~$\real$. 
\end{description} 



\noindent 
\begin{tabular}{|c|c|c|c|c|}\hline 
Domain $A$ is : $\rightarrow$ & $\real$ & any set & 
measurable or & symmetric \\ 
  & & & with the Baire & \\ 
 & & & property &  \\ \cline{1-1} 
$\downarrow ~f$ is : $\downarrow$ & & & & \\ \hline 
measurable or has & Thm. \ref{pp}& (i)& 
Cor. \ref{mmd}\ \& \ref{mbd}  & (i) Ex.\ref{exsd} \\ 
the Baire property & + & -- & + & -- \\ \hline 
continuous almost &Cor. \ref{be}&Ex. \ref{ex3} & 
Cor. \ref{mmd}\ \& \ref{mbd}& Ex. \ref{exsd} \\ 
everywhere on $A$ & + & -- & + & --  \\ \hline 
  & & (ii) & (ii) & Ex. \ref{exsd}\\ 
extendable to $\real$ & N/A & -- & -- & -- \\ \hline 
extendable to a & &Ex. \ref{ex3} & & Ex. \ref{exsd} \\ 
measurable set or & & & & \\ 
a set with the & & & & \\ 
Baire property & N/A & -- & N/A & -- \\ \hline 
almost extendable & & (ii)& (ii)& Ex. \ref{exsd}\\ 
to $\real$ & N/A & -- & -- & -- \\ \hline 
almost extendable & &(iii) &Cor. \ref{mmd}\ \& \ref{mbd} & Ex. \ref{exsd} \\ 
to a residual set & N/A & -- & + & -- \\ \hline 
\multicolumn{5}{|c|}{extensions preserving only original 
points of symmetric continuity} \\ \hline 
can be extended & & Thm. \ref{cne}&Ex. \ref{ex2} & (iv) \\ 
to $\real$ & N/A & -- & -- & + \\ \hline 
can be extended & & Thm. \ref{cne}&Ex. \ref{ex2} & (iv)\\ 
to a residual set & N/A & -- & -- & + \\ \hline 
can be almost & & (v)&Fact \ref{cb} & (iv) \\ 
extended to $\real$ & N/A & -- & + & + \\ \hline 
\end {tabular} 





\bigskip 



%***********************************************************
%continuity 
\section{Continuity of symmetrically continuous functions}

It is always a fundamental question how some other type of continuity 
is related to the ordinary continuity. Although symmetric continuity 
is weaker than continuity, symmetrically continuous functions 
on ``nice" domains must be continuous at many points. 
In this section we are going to investigate the set of 
points of continuity of symmetrically continuous functions 
defined on arbitrary subsets of $\real$. Our main results here 
are Theorem \ref{mu} and Corollaries \ref{mmd} and \ref{mbd}, 
which are strengthenings of earlier known theorems, 
quoted below, for functions from $\real$ to $\real$. 
The proofs have been shifted to the next section as they 
are long and technical.

\smallskip
The first result that we state was published by Stein and 
Zygmund in 1960.
\thm {se}{{\rm (Stein, Zygmund \cite{stz})} 
If $f\colon\real\to\real$ is symmetrically 
continuous and measurable then it is continuous everywhere  
except on a set of measure zero and first category.}
Later on Pesin and Preiss showed that the measurability 
assumption may be dropped.
\thm{pp}{{\rm (Pesin \cite{pes}, Preiss \cite{pr})} 
If $f\colon\real\to\real$ is symmetrically 
continuous then $f$ is measurable.}
So the obvious corollary says that symmetric continuity implies 
continuity almost everywhere (in the sense of measure and category).

\smallskip
The category version of that corollary is an old theorem of Fried 
from 1937 (which uses the method of Charzy\'nski).
\thm{fri}{{\rm (Fried \cite{fr})} 
Let $f\colon\real\to\real$ be symmetrically continuous 
on a residual set. Then $f$ is continuous at 
every point except on a set of the first category.} 

Stein and Zygmund had also a stronger result saying 
that measurability and symmetric continuity only on a 
measurable set also implies continuity almost 
everywhere on that set.
\thm{sms}{{\rm (Stein, Zygmund \cite{stz})} 
If $f\colon\real\to\real$ is measurable and symmetrically 
continuous on a measurable set $E\subset\real$ then 
$f$ is continuous almost everywhere (in the sense of measure) on $E$.}
One may suspect that here, as before, the measurability 
assumption is not necessary. Indeed this is the case. It 
follows from the following theorem of Belna from 1983.
\thm{uh}{{\rm (Belna \cite{bel})} Let $f\colon\real\to\real$ be 
an arbitrary function. Then the set of points at which $f$ 
is symmetrically continuous but not continuous has inner 
measure zero and contains no second category set having the 
Baire property.} 
Since for every function from $\real\to\real$ the set of 
points of continuity is a $G_\delta$ set, which is both 
measurable and has the Baire property, we have the 
following corollary. 
\cor{be}{{\rm (Belna \cite{bel})} Let $f\colon\real\to\real$ be 
symmetrically continuous on a set $E$ which is measurable 
(has the Baire property). Then $f$ is continuous at almost 
every point of $E$ (on a residual subset of $E$).}

In the corollary above the function $f$ may of course fail to be 
even measurable on the set $\real\setminus E$. This raises 
a question whether we need at all the assumption that 
the function is defined on the set $\real\setminus E$. 
The answer is not obvious since the  proofs of 
all the theorems mentioned above use heavily the fact 
that the function is defined everywhere on $\real$. 
However, it is indeed enough that the function $f$ is 
defined only on the set $E$. It follows from the next 
theorem, which can be regarded as a strengthening of 
the Belna's result. 
\thm{mu}{Let $f\colon A\to\real$, where $A$ is an arbitrary 
subset of $\real$. Then the set of points at which $f$ is 
symmetrically continuous but not continuous has the inner measure 
zero and contains no second category set having the Baire property.}

Two corollaries below are strenghenings of Corollary \ref{be}. 
\cor{mmd}{Let $E\subset\real$ be measurable and 
$f\colon E\to\real$ be symmetrically continuous. 
Then $f$ is continuous almost everywhere on $E$.}
\cor{mbd}{Let $E\subset\real$ have the Baire property and 
$f\colon E\to\real$ be symmetrically continuous. Then $f$ is 
continuous at every point of $E$ except on a meager set.}

In these corollaries the assumption that $E$ is measurable 
(has the Baire property) cannot be dropped, as the following example shows.
\ex{ex3}{Let $G_1$ be a group of full outer measure and let $G_2=\rat$. 
Then $f\colon G_1\cup G_2\to\{0,1\}$ defined like in Example \ref{ex2} 
is symmetrically continuous (on its domain) but 
discontinuous everywhere on $G_1\cup G_2$. The same is true 
if $G_1$ is a group of second category.}
Indeed, since both $G_1$ and $\rat$ are dense in $\real$ 
then $f$ must be discontinuous everywhere. 
%**************************************************************
%Proofs 
\section {Proof of the main result}
We give the proof of Theorem \ref{mu}. Theorems \ref{m1} 
and \ref{m2} are technical and lengthy, but then the 
proof of Theorem \ref{mu} is short and easy.

We introduce two useful (though technical) definitions. 
\defi{scr}{A symmetric covering relation on a set $E\subset\real$ is a 
family $V$ of closed intervals $[a,b]$ such that ${a+b\over2}\in  E$.}
\defi{5scr}{If $V$ is a symmetric cover relation on $E\subset\real$ 
then  $V^5$ is a family of closed intervals $[a,b]$ such 
that there is a sequence of points $a=x_0,x_1,\ldots ,x_5=b$ 
such that $[x_i,x_{i+1}]$ belong to $V$ for $i=0,1,2,3,4$. 
We call the points $x_1,\ldots ,x_4$ the intermediate endpoints.} 
Uher \cite{uh} proved the following theorems (which have been reformulated 
by Thomson, and we give here Thomson's versions). 
\thm{u1}{{\rm (Uher \cite[thm. 3.25]{t})} Let $E\subset\real$ have the 
Baire property and suppose $V$ is a symmetric cover relation 
on $E$ having the property that for every $x\in  E$ there 
is a positive number $\delta(x)$ so that for every $t\in\real$ 
$$0<t<\delta(x) \Rightarrow [x-t,x+t]\in V.$$
Then there is an open set $G$ such that $E\setminus G$ 
is of the first category and every interval $[x-t,x+t]$ 
contained in G belongs to~$V^5$.}
\thm{u2}{{\rm (Uher \cite[thm. 3.26]{t})} Let $E\subset\real$ be measurable 
and suppose $V$ is a symmetric cover relation on $E$ having the 
property that for every $x\in E$ there is a positive number 
$\delta(x)$ so that for every $t\in\real$ 
$$0<t<\delta(x) \Rightarrow [x-t,x+t]\in V.$$
Then, for almost all points $x\in E$ (in the sense of measure), there is 
a neighborhood $U_x$ of $x$ such that  for each $x+t\in U_x$ the interval  
$[x,x+t]$ belongs to $V^5$.} 

We will prove two theorems that are strengthened versions of Uher 
theorems.
\thm{m1}{Let $E\subset\real$ have the Baire property and suppose 
$V$ is a symmetric cover relation on $E$ having the property that 
for every $x\in E$ there is a positive number $\delta(x)$ so that 
for every $t\in\real$ 
$$0<t<\delta(x) \Rightarrow [x-t,x+t]\in V.$$
Then there is an open set $G$ such that $E\setminus G$ is 
of the first category and every interval $[x-t,x+t]$ 
contained in $G$ is in $V^5$. Moreover, the intermediate 
endpoints may be chosen from the set $E$.}
\thm{m2}{Let $E\subset\real$ be measurable and suppose $V$ is a 
symmetric cover relation on $E$ having the property that for 
every $x\in E$ there is a positive number $\delta(x)$ so that 
for every $t\in\real$  
$$0<t<\delta(x) \Rightarrow [x-t,x+t]\in V.$$
Then, for almost all points $x\in E$ (in the sense of measure), 
there is a neighborhood $U_x$ of $x$ such that whenever 
$x+t\in U_x$ then the interval $[x,x+t]$ belongs to $V^5$. 
Moreover, the intermediate endpoints may be chosen from the set $E$.}

As we see, only the last parts of Theorems \ref{m1} and 
\ref{m2} make them stronger then their Uher/Thomson's counterparts. 
The structure of our proof is essentially identical to that 
 presented in \cite{t} (attributed to Humke and Laczkovich). 
However, our proof is considerably more complicated from technical 
point of view. 

\baselineskip15pt \noindent 
{\bf Proof of Theorem \ref{m1}}. Let $\cal I$ denote the family of all open
intervals $I$ with the property that for every $x,y\in I$ there is a chain 
$x=x_0,x_1,\ldots ,x_5=y$ satisfying 
\begin{center}\hfill $\displaystyle
[x_i,x_{i+1}]\in V ~{\rm for}~i=0,1,2,3,4~~
{\rm and}~x_i\in E~~{\rm for}~i=1,2,3,4.$\hfill $(*)$ \end{center} 
We will show that 
\begin{equation} \label{1} 
\hspace{-3pt} \!
\hbox{if $E$ is residual in $(a,b)$ then there is a 
subinterval of $(a,b)$ belonging to $\cal I$.}
\end{equation} 
The statement of the theorem then follows by taking for the set $G$ 
a union of a maximal pairwise disjoint subfamily of $\cal I$. 
Indeed any interval contained in $G$ is also contained in some 
interval from $\cal I$ so it satisfies the assertion of the theorem. 
To see that $E\setminus G$ is first category note that 
the set $E$ is residual in some nonempty open set $U$ 
for which $E\setminus U$ is meager. But then $U$ consists of 
disjoint intervals $(a,b)$ and $E$ is residual in each of them. 
So $G$ is dense in $U$ and, since it is open, it is residual in $U$. Thus 
$E\setminus G\subset (E\setminus U)\cup (U\setminus G)$ is meager as claimed. 

To show (1) assume that $E$ is residual in $(a,b)$. 
For any positive $n\in\omega$ let
$$E_n=\left\{ x\in E\colon  0<t<{1\over n}~\Rightarrow~[x-t,x+t]\in
V\right\}.$$ Then $E=\bigcup_{n=1}^\infty E_n$ and there is a number $n\in
\omega$ so  that $E_n$ is of second category in $(a,b)$. Thus there is an
interval 
$I\subset (a,b)$ such that $E_n$ is of second category in every 
subinterval of $I$. Assume additionally that the length $|I|$ of 
$I$ is less than $1\over n$. Take the interval $I^\prime$ 
concentric with $I$ and of the length ${1\over 10}|I|$. This is 
this interval $I^\prime$ that, as we shall show, belongs to $\cal I$. 

Let $x,y\in I^\prime,~ x<y$. For the sake of simplicity assume that $x=0$.  
For $x\not=0$ we repeat the construction below with the set $E$ replaced 
by $E-x$. Note that for any numbers $t$ with 
$|t|<{3\over 2}|I^\prime|={3\over 20}|I|$ 
and $r$ with ${1\over 2}\leq |r|\leq 2$, 
the set  
\begin{equation} rE_n+t~~\mbox{is second category in every subinterval 
of $I^\prime$}. \label{2} \end{equation} 
To see this note that since $0\in I^\prime$,  
$[-0.9|I|,0.9|I|]\subset I$. Since $|r|\geq {1\over 2}$ we have 
$[-0.45|I|,0.45|I|]\subset rI$ and since 
$|t|\leq {3\over 2}|I^\prime|=0.15|I|$, $[-0.3|I|,0.3|I|]\subset rI+t$. 
We see that the interval $I^\prime$ is contained in every interval 
$rI+t$. Since $E_n$ is of second category in 
every subinterval of $I$ then so is $rE_n+t$ in $rI+t$.  

Similarly, the sets   
\begin{equation}\pm {1\over 2}E+t~~
\mbox{are residual in $I^\prime$ for $|t|<{3\over 2}|I^\prime|$.} 
\label{3} \end{equation} 
We want to find a 5-element chain $0=x_0,x_1,\ldots ,x_5=y$ 
satisfying $(*)$. To do this we will find an interval 
$J\subset\left( {9\over 10}y,y\right)$ such that for every $z\in J$ there 
is a four element chain $0=x_0,x_1,\ldots ,x_4=z$ satisfying 
\begin{center}\hfill $\displaystyle 
[x_i,x_{i+1}]\in V ~{\rm for}~ i=0,1,2,3~~
{\rm and}~x_i\in E~~{\rm for}~ i=1,2,3.$\hfill $(**)$ \end{center} 
This will finish the proof since then we select $z\in J$ such that 
$z$ is also in the set 
$Z=\left\{ z\in\left( {9\over 10}y,y\right) 
\colon z\in E~{\rm and}~[z,y]\in V\right\}$. 
The intersection of $J$ and $Z$ is nonempty since 
$(2E_n-y)\cap E\cap\left( {9\over 10}y,y\right)\subset Z$ 
(as $z\in 2E_n-y$ implies ${{z+y}\over 2}\in E_n$ so $[z,y]\in V$) 
and, by (\ref{2}), $2E_n-y$ is second category in every subinterval 
of $I^\prime$. 

In the interval $\left( {9\over 10}y,y\right)$ the set $E$ 
is residual and, by (\ref{2}), $E_n+{3\over 4}y$ is second 
category since ${3\over 4} y<{3\over 2}|I^\prime|$. 
Hence $E\cap\left( E_n+{3\over 4}y\right)$ 
is also second category in  $\left( {9\over 10}y,y\right)$. 
It follows, from the fact that $E=\bigcup_{n=1}^\infty E_n$, 
that there is an integer $m$ and an open interval 
$J\subset\left( {9\over 10}y,y\right)$ so that the set 
$E_m\cap\left( E_n+{3\over 4}y\right)$ is second category in 
every subinterval of $J$. We may assume that $|J|<{1\over m}$. 

To show that $J$ satisfies ($**$) pick $z\in J$. 
By (\ref{3}) the set ${1\over 2}E+{3\over 4}y$ 
is residual in $J$ and, by (\ref{2}), the set 
${1\over 2}E_n+{1\over 4}z+{3\over 8}y$ is second category in $J$ since 
${3\over 4}y,{1\over 4}z+{3\over 8}y<y<|I^\prime|$. 
Therefore we may choose a point 
$$t\in \left( {1\over 2}E+{3\over 4}y \right)\cap 
\left( {1\over 2}E_n+{1\over 4}z+{3\over 8}y\right)\cap J. $$
Since $t\in {1\over 2}E+{3\over 4}y$, we have  
\begin{equation} q=2t-{3\over 2}y\in E.\label{4} \end{equation} 

Now we choose a point $s$ satisfying 
$$ s\in \left( {1\over 2}E+{3\over 4}y\right)\cap 
\left( -{1\over 2}E+2t-{3\over 4}y\right)\cap 
\left( {1\over 2}E+{1\over 2}z\right) 
\cap \left[ E_m\cap\left( E_n+{3\over 4}y\right)\right] \cap J $$ 
and $|s-t|<{1\over 2}\delta (q)$. 

Such a point may be found since, by (3), the first three sets are residual 
in $J$, and the fourth one is of second category in every subinterval of 
$J$ so their intersection is dense in $J$. 
Denote by $p$ the point $p=s-{3\over 4}y$ and note that 
\begin{equation}
|q-2p|=2|s-t|<\delta (q) ~{\rm and}~ p\in E_n.\label{5}\end{equation} 

Now define our points $x_i$:
\begin{itemize}
\item[] $x_0=0=x$,
\item[] $x_1=2p=2s-{3\over 2}y~\in E$ ~since~ 
$2\left( {1\over 2}E+{3\over 4}y\right) -{3\over 2}y=E$, 
\item[] $x_2=2q-2p=4t-{3\over 2}y-2s~\in E$ ~since~ 
$4t-{3\over 2}y -2\left( -{1\over 2}E+2t-{3\over 4}y\right) =E$, 
\item[] $x_3=2s-z~\in E$ ~since~ 
$2\left( {1\over 2}E+{1\over 2}z\right) -z=E$, 
\item[] $x_4=z$. 
\end{itemize}
So the second part of ($**$) is satisfied. 

The centers of $[x_i,x_{i+1}]$ are: \begin{itemize} 
\item[] ${{x_0+x_1}\over 2}=p\in E_n$ by (5), 
\item[] ${{x_1+x_2}\over 2}=q\in E$ by (4), 
\item[] ${{x_2+x_3}\over 2}=2t-{3\over 4}y-{1\over 2}z~\in E_n$ 
~since~ $t\in {1\over 2}E_n+{1\over 4}z+{3\over 8}y$,  
\item[] ${{x_3+x_4}\over 2}=s\in E_m$. 
\end{itemize} 
To finish the proof we only need to verify that each $[x_i,x_{i+1}]$ 
is in $V$. But this can be seen as follows.
Intervals $[x_0,x_1]$ and $[x_2,x_3]$ are in $V$ since their 
centers are in $E_n$ and each $x_i$ is in the interval $I$ 
which has the length $<{1\over n}$. The interval $[x_1,x_2]\in V$ 
since, by (5), $|x_1-{{x_1+x_2}\over 2}|=|2p-q|<\delta (q)$. 
And finally $[x_3,x_4]$ belongs to $V$ because 
$|x_4-s|=|z-s|<{1\over m}$ as both $s,z\in J$ which has 
length less than ${1\over m}$. 
\qed

\bigskip
In the proof of the next theorem we use the following notions and facts.
By $|A|$ we denote the outer (Lebesgue) measure of the set $A$, 
$d(A)$ denotes the set of points of (outer) density of $A$, i.e., the set 
$\left\{ x\in\real\colon \lim_{h\to 0^+}~{|A\cap[x-h,x+h]|\over
{2h}}=1\right\}$.  The well known theorem of Lebesgue (see e.g. \cite[thm.
A.11]{t})  says that for any set $A$ almost all points of $A$ are its points 
of density. $d(A)$ is always a measurable set (even for nonmeasurable $A$) 
and $d(d(A))=d(A)$. If $x\in d(A)$ then $ax+b\in d(aA+b)$ for 
$a,b\in\real,~a\not=0$. In particular if $A$ consists only 
of its density points then so does $aA+b$ for 
$a,b\in\real,~a\not=0$. We also use the following lemma. 
\lem{a10}{{\rm \cite[thm. A.11]{t}} If sets $A_n\subset\real$, where 
$n\in \{1,2,\ldots ,k\}$, are all measurable except at most 
one of them and $z$ is a point of density of each of them then 
$z$ is also a point of density of their intersection. In particular 
in every neighborhood of $z$ there is a point of $\bigcap_{n=1}^k A_n$.} 

\medskip
\noindent
{\bf Proof of Theorem \ref{m2}}. The proof follows the pattern  
of the proof of Thm.~\ref{m1}; we will refer to the conditions 
$(*)$ and $(**)$ from that proof. 

We assume throughout the proof 
that every measurable set consists only of its density points. Let 
$$E_n=\{ x\in E\colon  0<t<{1\over n}~\Rightarrow~[x-t,x+t]\in V\}.$$ 
We show that every point that belongs to the set 
$\bigcup_{n=1}^\infty~d(E_n)$ has a neighborhood satisfying 
the assertion of the theorem. Thus almost every point of $E$ 
satisfies it since $E\setminus\bigcup_{n=1}^\infty~d(E_n)$ is of measure 0.

Let $x$ be a point of density of $E_n$. Assume for notational convenience 
that $x=0$ (for if $x\not=0$ then replace the set $E$ by $E-x$). 
There is a number $\eta ,~0<\eta <{1\over n}$, such that for every 
$y$ with $|y|<\eta$ and every $t$ with $|t|<2|y|$ each of the sets 
\begin{equation} 2E_n+t,~\pm {1\over 2}E_n+t,~{3\over 2}E_n+t,
~{\rm and}~ E\cap (E_n+t)\label{6}\end{equation} 
intersects the interval $({9\over 10}y,y)$ in a set 
of outer measure greater than ${9\over 100}|y|$. 
We may take $\eta ={1\over 6}\delta$, where $\delta$ is such that 
for every $0<h<\delta$ we have 
${|E_n\cap[-h,h]|\over {2h}}\geq {999\over {1000}}$. 
To see that this $\eta$ works take for example the set 
${1\over 2} E_n+t$ and note that 
$\left( {1\over 2} E_n+t\right)\cap (0.9y,y)={1\over 2}\left[ E_n\cap 
(1.8y-2t,2y-2t)\right]+t$. Numbers $|1.8y-2t|$ and $|2y-2t|$ are less then 
$6|y|<\delta$, so 
\newpage
\begin{eqnarray*} 
|E_n\cap (1.8y-2t,2y-2t)| & = & 
|E_n\cap (-6y,6y)\\
& &\setminus \left[ E_n\cap [(-6y,6y)
\setminus (1.8y-2t,2y-2t)]\right]|\\
& \geq & 
{999\over 1000}12|y|-(12|y|-0.2|y|)\\
& = & {188\over 1000}|y|>2{9\over 100}|y|. 
\end{eqnarray*}  
The calculations for other sets in (6) are similar. 

Fix any $y\in (0,\eta)$. We prove that there is a chain 
$0=x_0,x_1,\ldots ,x_5=y$ satisfying ($*$). 
A similar argument would work for $y<0$.

Let $Z=\left\{ z\in \left( {9\over 10}y,y\right) \colon [z,y]\in V\right\}$. 
Note that $(2E_n-y)\cap ({9\over 10}y,y)\subset Z$ and so, 
by (\ref{6}), $|Z|\geq {9\over 100}y$. To prove 
the theorem it is enough to show that there is a measurable set 
$F\subset\left( {9\over 10}y,y\right)\cap E$ such that 
$|F|>{1\over 100}y$ and that for every $z\in F$ there  
is a four element chain $0=x_0,\ldots ,x_4=z$ satisfying ($**$). 
Indeed if this were true then $F\cap Z\not=\emptyset$ and 
choosing any $z\in F\cap Z$ we would obtain a four element 
chain which expanded by $x_5=y$ would give us our desired chain.

Recall that, by (\ref{6}), $|E\cap\left( E_n+{3\over 4}y)\right)\cap 
\left( {9\over 10}y,y\right) |>{9\over 100}y$, 
so from the fact that $E$ is an increasing sum of $E_m$'s,  
there is an integer $m$ so that \begin{equation} 
\left|E_m\cap\left( E_n+{3\over 4}y\right)\cap 
\left( {9\over 10}y,y\right)\right| >{8\over 100}y.\label{7}
\end{equation} 
Denote $A_m= E_m\cap\left( E_n+{3\over 4}y\right)\cap 
\left( {9\over 10}y,y\right)$ and write 
$$F=\left( {1\over 2}E+{3\over 4}y\right)\cap 
d\left( {2\over 3}E_n+{1\over 2}y\right)\cap d(A_m)\cap E.$$
We show that all points of $F$ have the property claimed for it.

First note that, by (6) and (7), 
$$|F|>{1\over 10}y-{1\over 100}y-{1\over 100}y-{2\over 100}y-
{1\over 100}y>{5\over 100}y.$$ 
Let $z\in F$. So $z$ is a density point of $F$. 
Since $z\in d\left( {2\over 3}E_n+{1\over 2}y\right)$ it follows that  
$z$ belongs to $d\left( {1\over 2}E_n+{1\over 4}z+{3\over 8}y\right)$ 
(indeed ${3\over 4}z\in 
d\left( {3\over 4}\left( {2\over 3}E_n+{1\over 2}y\right)\right)$ so 
${1\over 4}z+{3\over 4}z\in d
\left( {3\over 4}\left( {2\over 3}E_n+{1\over 2}y\right) +
{1\over 4}z\right)$), and since $z\in E,~z\in{1\over 2}E+{1\over 2}z$. 

We may pick a point $t$ so that 
$$t\in \left( {1\over 2}E_n+{1\over 4}z+{3\over 8}y\right)\cap
\left( {1\over 2}E+{1\over 2}z\right)\cap
\left( {1\over 2}E+{3\over 4}y\right)\cap d(A_m)$$ 
and $|z-t|<{1\over m}$. 

This is possible since by Lemma \ref{a10} since all sets above, 
except possibly the first one, are measurable and the point $z$ 
is a density point of each of them. 

We define $q=2t-{3\over 2}y$ and note that $q\in E$ 
since $2\left( {1\over 2}E+{3\over 4}y\right) -{3\over 2}y=E$. 
Note also that $t$ is a density point of the set 
$-{1\over 2}E+2t-{3\over 4}y$ as $t\in {1\over 2}E+{3\over 2}y$. 

We now pick a point $s$. We want $s$ to satisfy:
$$ s\in\left({1\over 2}E+{1\over 2}z\right)\cap\left( 
{1\over 2}E+{3\over 4}y\right)\cap\left( 
-{1\over 2}E+2t-{3\over 4}y\right)\cap A_m, $$ 
$$|s-t|<{1\over 2}\delta(q)~~{\rm and}~~|s-z|<{1\over m}.$$ 

Once again by Lemma \ref{a10} there exists a point $s$ being in 
all four sets and arbitrary close to $t$ ($t$ is a density point 
of each of the above set). Taking $s$ close enough to $t$ we 
guarantee the last condition since $|s-z|<|z-t|+|t-s|$ and 
$|z-t|<{1\over m}$. 

Define the point $p=s-{3\over 4}y$ and note that 
$|q-2p|=2|s-t|<\delta(q)$. 
The points $x_0,x_1,\ldots ,x_4$ satisfying ($**$) are defined as follows:
\begin{itemize} 
\item[] $x_0=x=0$, 
\item[] $x_1=2p=2s-{3\over 2}y\in E$ ~since~ 
$2\left( {1\over 2}E+{3\over 4}y\right)-{3\over 2}y=E$, 
\item[] $x_2=2q-2p=4t-2s-{3\over 2}y\in E$ ~since~ 
$4t-{3\over 2}y-2\left( -{1\over 2}E+2t-{3\over 4}y\right) =E$, 
\item[] $x_3=2s-z\in E$ ~since~ 
$2\left( {1\over 2}E+{1\over 2}z\right) -z=E$, 
\item[] $x_4=z$.  \end{itemize}

The centers of the intervals $[x_i,x_{i+1}]$ are:
\begin{itemize} 
\item[] ${{x_0+x_1}\over 2}=p=s-{3\over 4}y\in E_n$ since 
$s\in A_m\subset E_n+{3\over 4}y$. 
\item[] ${{x_1+x_2}\over 2}=q=2t-{3\over 2}y\in E$ since 
$t\in {1\over 2}E+{3\over 4}y$. 
\item[] ${{x_2+x_3}\over 2}=2t-{3\over 4}y-{1\over 2}z~\in E_n$ since 
$t\in {1\over 2}E_n+{1\over 4}z+{3\over 8}y$. 
\item[] ${{x_3+x_4}\over 2}=s~\in E_m$ since 
$s\in A_m\subset E_m$.  \end{itemize} 

Intervals $[x_0,x_1]$ and $[x_2,x_3]$ are in $V$ since their 
centers are in $E_n$ and each $x_i$ is in the interval $I$ 
which has the length $<{1\over n}$. The interval $[x_1,x_2]\in V$ 
since $|x_1-{{x_1+x_2}\over 2}|=|x_1-q|=2|s-t|<\delta (q)$. 
And finally $[x_3,x_4]$ belongs to $V$ because 
$|x_4-s|=|z-s|<{1\over m}$. 
This way we see that each interval $[x_i,x_{i+1}]$ belongs to 
$V$ and the proof is finished.
\qed 

\medskip \noindent
{\bf Proof of Theorem \ref{mu}.} Let us denote by $SC_f$ the set of 
points where $f$ is symmetrically continuous and by $C_f$ the set of 
continuity points. We prove only that $SC_f\setminus C_f$ contains no 
measurable set of positive measure. The proof that $SC_f\setminus C_f$ 
contains no second category set having the Baire property is essentially 
the same. 

Assume that $E\subset SC_f$ is measurable. We will 
show that almost all points in $E$ are also in $C_f$. 
This shows that the only measurable subsets of 
$SC_f\setminus C_f$ are of measure zero so 
$SC_f\setminus C_f$ has inner measure zero.

 Fix an $\varepsilon >0$ and define a symmetric covering relation on $E$: 
$$V=\{[x-t,x+t]\colon ~x\pm t\in E\Rightarrow |f(x-t)-f(x+t)|<\varepsilon\}.$$

This relation satisfies assumption of Theorem \ref{m2}. 
So for almost every $x\in E$ there is a neighborhood 
$U_x$ so that $x+t\in U_x\Rightarrow [x,x+t]\in V^5$ 
and intermediate endpoints are in $E$, that is there are 
points $x_0=x,x_1,\ldots ,x_5=x+t$ such that $x_i\in E$ for 
$i=1,2,3,4$ and $[x_i,x_{i+1}]\in V$ for $i=0,\ldots ,4$.
Therefore  
$|f(x)-f(x+t)|<5\varepsilon$. Since 
$\varepsilon$ has been chosen arbitrarily, $f$ is 
continuous in almost all points of $E$.     
\qed 
%**************************************************************
%extensions
\section {Extensions of continuous and symmetrically 
continuous functions}
\baselineskip 13pt
In this section we look at the extension properties of 
symmetrically continuous functions and compare them with 
these properties for continuous functions.

We know that every continuous function $f\colon A\to\real$ 
defined on a subset of $\real$ can be extended to 
a continuous function defined on a dense 
$G_\delta$ set. This set can be of course residual. 
Note that continuity distinguishes between 
measure and category as it is possible to find function 
that can be only extended to a measure zero $G_\delta$ set. 
We will see in Theorem \ref{cne} that some 
symmetrically continuous functions cannot be extended 
beyond measure zero and first category domain.

If we look at Example \ref{ex3} we see that 
symmetrically continuous functions defined there is 
discontinuous on its domain so on a full outer measure 
or second category set. Thus, by Corollaries \ref{mmd} and 
\ref{mbd}, $f$ cannot be extended to a symmetrically continuous 
function defined on a measurable set or on a set having the Baire 
property. 

For an example of a similar function $f$ defined on a symmetric 
domain see Section 5. In Theorem \ref{cne} we give much 
stronger example in which domain may be chosen to be 
of measure zero and first category.

\medskip
A different question is whether having a function $f\colon A\to\real$ 
that is continuous (or symmetrically continuous) we may  extend it to 
the whole real line leaving it continuous (or symmetrically continuous) 
on the set $A$. For continuous functions as the next theorem 
shows this is true. 
\fact {cb}{If $f\colon A\to\real$ is continuous then there is an 
extension $\hat f$ of $f$ defined on the whole $\real$ with 
$A\subset C_{\hat f}$, where $C_{\hat f}$ is the set of points 
of continuity of $\hat f$.}
\baselineskip 15pt 
{\bf Proof.} If $A$ is not dense in $\real$ then 
$\real\setminus {\rm cl}(A)$ 
consists of disjoint intervals $(a,b)$ and we extend our 
function by defining $f$ on each of them as a continuous 
function joining $\sup_{x\to a, x\in A} f(x)$ with 
$\sup_{x\to b, x\in A} f(x)$ (even if one of them or both 
are infinity). 
So we may assume that $A$ is dense in $\real$. 

Assume first that $f$ is bounded. In this case 
we may assume that $f\colon A\to [0,1]$. 
Take a closure cl$(f)$ of the function $f$ in $\real ^2$ 
($f$ is a set of pairs thus a subset of $\real ^2$). 
Since $A$ is dense, the projection $\pi ($cl$(f))$ of cl$(f)$ 
onto the first coordinate is the whole real line. 
This is so, as for every finite interval $[a,b]$, the set 
${\rm cl}(f|_{[a,b]})$ is compact and its projection is also 
compact, thus contains $A\cap [a,b]$ --- a dense set in $[a,b]$. 

Let $I_x=\{ y\colon \langle x,y\rangle\in {\rm cl}(f)\}$. 
Then every $I_x$ is nonempty. 
Also, for every $x\in A$ the set $I_x$ is the one element set 
$\{f(x)\}$. Indeed, for every $\{\langle x_n,f(x_n)\rangle\}$ 
in $f$ with $\{x_n\}$ converging to $x$, we have 
$\lim f(x_n)=f(x)$ by the continuity of $f$.
Take an indexed selector $S$ of the family $\{I_x\colon x\in\real\}$ 
(i.e., $S$ is a set of pairs $\langle x,y\rangle$, where $y\in I_x$). 
Then $S$ is a function from $\real$ to $[0,1]$ and 
$S$ agrees with $f$ on every $x\in A$. 
It is easy to check that $S$ is continuous in every point of $A$.  

\smallskip
If $f$ is unbounded then take the function 
$g=\arctan \circ f$, which is bounded. By the continuity of 
$\arctan$ we have that $g$ is continuous in every point 
where $f$ is continuous. 

Having an extension $\hat g$ of $g$, we come back to 
$f$ taking $\hat f=\tan\circ\hat g$, where we define 
$\hat f(\pm{\pi\over 2})=0$. Once again by the continuity of the 
function $\tan$ we preserve points of continuity of $\hat g$.
\qed

\medskip
Once again the translation of the above theorem into 
symmetrical language fails to hold.
\thm {cne}{There is a function $f\colon  A\to \{0,1\}$ that is 
symmetrically continuous everywhere on $A$ with the following property: 
if $F$ is an extension of $f$ to a domain $D$ such that $D$ has 
positive measure or $D$ is of second category with the Baire 
property, then $F$ is not symmetrically continuous at some point 
on $A$.} 
{\bf Proof.} We will construct inductively (using 
transfinite induction) a set $A$ and will define 
function $f$ on $A$. Function $f$ will have two properties:  
for every perfect set $P$, there is a 
point $x\in P$ such that there is a pair 
of sequences $\{y_n\}$ and $\{z_n\}$ in $A$ converging to $x$ 
and symmetric about $x$, (i.e. ${{y_n+z_n}\over 2}=x$ for all $n$); 
$f$ will have value $1$ for all $y_n's$ and value $0$ 
for all $z_n's$. Then, of course, $x$ cannot 
belong to the domain of any symmetrically continuous extension of 
$f$ and, since any measurable set of positive measure or Baire 
measurable set of second category contains a perfect set, 
$f$ cannot be extended to any of these sets.

Let $\{P_\alpha \colon \alpha <\cont\}$ be a list of all perfect 
subsets of $\real$. Take any point $x_0\in P_0$ and define 
$y_n^0=x_0-{1\over n}$ and $z_n^0=x_0+{1\over n}$. Put also 
$f_0\colon \{y_n^0,z_n^0\colon n\in \omega\}\to\{0,1\}$, $f_0(y_n^0)=1$, 
$f_0(z_n^0)=0$. (We do not define $f_0(x_0)$.) 

Since $x_0$ is the only accumulation point of the set 
$\{y_n^0,z_n^0\colon n\in \omega\}$, the function $f_0$ is 
symmetrically continuous (vacuously) on its domain.
Denote by $T_0$ the linear space over $\rat$ spanned by the set 
$\{y_n^0,z_n^0\colon n\in \omega\}$ and note that $x_0\in T_0$. 

Assume that for $\beta <\alpha$ we have defined points
$x_\beta\in P_\beta$ and sequences $\{y_n^\beta\}$ and $\{z_n^\beta\}$ 
converging to $x_\beta$ and ${{y_n^\beta +z_n^\beta }\over 2}=x_\beta$. 
Assume also that $x_\beta ,y_n^\beta ,z_n^\beta \not\in T_\beta$ where 
$T_\beta =$linear space over $\rat$ spanned by the set 
$\{x_\gamma ,y_n^\gamma ,z_n^\gamma \colon n\in \omega ,\gamma <\beta \}$ 
(i.e. the points $x_\beta ,y_n^\beta ,z_n^\beta$ are independent 
of the previous ones). 

Functions $f_\beta \colon \{y_n^\beta ,z_n^\beta \colon n\in
\omega\}\to\{0,1\}$  are defined like $f_0$, that is, $f_\beta (y_n^\beta
)=1$ and 
$f_\beta (z_n^\beta )=0$. (We leave $f_\beta (x_\beta )$ not defined.) 

Let $T_\alpha$ be the linear space over $\rat$ spanned by the set 
$\{x_\beta ,y_n^\beta ,z_n^\beta \colon  n\in \omega ,\beta <\alpha \}$. 
The space $T_\alpha$ has cardinality $|\alpha |\times\omega <\cont$ 
and we may find a point $x_\alpha\in P_\alpha\setminus T_\alpha$ and 
sequences $\{y_n^\alpha\}$, $\{z_n^\alpha\}$ disjoint with $T_\alpha$, 
converging to $x_\alpha$, and with 
${{x_n^\alpha +z_n^\alpha}\over 2}=x_\alpha$ for all $n\in\omega$. 
Like before we define $f_\alpha (y_n^\alpha)=0$ and 
$f_\alpha (z_n^\alpha)=1$. Again $f_\alpha$ is (vacuously) 
symmetrically continuous on its domain. 

Having our construction done for all $\alpha <\cont$ we 
define $f=\bigcup_{\alpha <\cont}f_\alpha$. Since points 
$y_n^\alpha$ and $z_n^\alpha$ are linearly independent of 
the points $y_n^\beta$ and $z_n^\beta$ for $\alpha\not =\beta$, 
the function $f$ is vacuously symmetrically continuous. 
Putting $A={\rm dom}(f)=
\{y_n^\alpha,z_n^\alpha;n\in \omega , \alpha <\cont\}$ we 
have constructed the desired set and the function defined
 on it.  
\qed

\smallskip
Is is worth to add that we may strengthen the theorem 
above by adding that $A$ may be of measure zero and 
first category. (We may select points $y_n^\alpha$ and $z_n^\alpha$ 
from the set $C$ constructed in \cite[Col.10]{msz}.) 
%***********************************************************
%Functions on symmertic domains
\section {Functions on symmetric domains}
Definition \ref{scd} may seem a bit strange since we 
do not require both of the points $x+h$ and $x-h$ to be 
in the domain. Most of our counterexamples 
have nonsymmetric domain and the functions are 
vacuously symmetrically continuous. Here we try to justify 
our definition by showing that adding symmetry to 
the domain does not give us any stronger properties 
that functions on nonsymmetric domains do not possess.

We say that a set $A\subset\real$ is symmetric if 
$$\forall_{x\in A}\forall_h ~~x+h\in A \iff~x-h\in A.$$
For example every additive group in $\real$ is symmetric.

\ex{g3}{ Let $A=\left\{{k\over 3^n}\colon k\in\inte ,n\in\omega\right\}$ 
and $f\colon A\to \{0,1\}$ be defined by 
$$ f\left({k\over 3^n}\right)= \left\{
  \matrix{
  0~~~ & \hbox{ when k is even} \cr
  1~~~ & \hbox { when k is odd} .   \cr }
  \right. $$
Then $f$ is symmetrically continuous and $A$ is symmetric 
yet $f$ is discontinuous everywhere on its domain.} 
{\bf Proof.} $A$ is a group so is obviously symmetric. 
It is immediate that the symmetric reflection
 of any point ${k\over 3^n}$ about a point ${l\over 3^m}$ 
preserves the parity of the numerator 
$\left(2{l\over 3^m} -{k\over 3^n}={{2l3^n-k3^m}\over{3^{m+n}}}\right).$ 
Since both $f^{-1}(0)$ and $f^{-1}(1)$ are dense, $f$ is 
discontinuous everywhere. 
\qed

\ex{exsd}{ Let $H$ be a Hamel base containing 1 and let 
$S$ be linear space over $\rat$ spanned by $H\setminus \{1\}$. 
Define $A=\{x+k\colon x\in S,~k\in\inte\}$ and $f\colon A\to\{0,1\}$ by 
$$ f(x+k)= \left\{
  \matrix{
  0~~~ & \hbox{ when $k$ is even} \cr
  1~~~ & \hbox { when $k$ is odd}.   \cr }
  \right. $$
Then $f$ is symmetrically continuous on $A$, $A$ is 
symmetric, and yet $f$ is discontinuous everywhere on $A$. 
Moreover $f$ cannot be extended to a symmetrically continuous function 
defined on any measurable set or a set with the Baire property.} 
{\bf Proof.} Just like in the previous example $A$ is a group, 
thus symmetric, and $f$ is symmetrically continuous (we 
preserve the parity of $k$ in reflections). Both 
$f^{-1}(1)$ and $f^{-1}(0)$ are dense in $\real$ 
so $f$ is discontinuous. Moreover $f$ cannot be extended 
to any measurable (or having the Baire property) domain being 
symmetrically continuous on the greater domain as that 
would imply continuity almost everywhere (on a residual 
set) on its domain (Corollaries \ref{mmd} and \ref{mbd}).
\qed

The function $f$ above cannot be even ``almost extended" 
(in a sense that the extension differs from $f$ only on a set of measure 0) 
to a symmetrically continuous function. This follows from the fact 
that the sets $f^{-1}(1)$ and $f^{-1}(0)$ are 
nonmeasurable. Note that all rational translations of 
$f^{-1}(1)$ and $f^{-1}(0)$ cover $\real$ so these sets have 
positive outer measure. So even upon removing sets of measure zero 
from them they remain dense in $\real$. 

\medskip
As we see, symmetric domains when nonmeasurable do not 
improve regularity of symmetrically continuous functions 
defined on them. This is a strong argument in favor of our 
definition (Definition \ref{scd}) since we do not need to 
distinguish between symmetric and nonsymmetric domains and we 
are still able to get positive results 
(e.g. Corollaries \ref{mmd} and \ref{mbd}).  

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\end{document}