%\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 $t\in\real$ $$02{9\over 100}|y|. \end{eqnarray*} The calculations for other sets in (6) are similar. Fix any $y\in (0,\eta)$. We prove that there is a chain $0=x_0,x_1,\ldots ,x_5=y$ satisfying ($*$). A similar argument would work for $y<0$. Let $Z=\left\{ z\in \left( {9\over 10}y,y\right) \colon [z,y]\in V\right\}$. Note that $(2E_n-y)\cap ({9\over 10}y,y)\subset Z$ and so, by (\ref{6}), $|Z|\geq {9\over 100}y$. To prove the theorem it is enough to show that there is a measurable set $F\subset\left( {9\over 10}y,y\right)\cap E$ such that $|F|>{1\over 100}y$ and that for every $z\in F$ there is a four element chain $0=x_0,\ldots ,x_4=z$ satisfying ($**$). Indeed if this were true then $F\cap Z\not=\emptyset$ and choosing any $z\in F\cap Z$ we would obtain a four element chain which expanded by $x_5=y$ would give us our desired chain. Recall that, by (\ref{6}), $|E\cap\left( E_n+{3\over 4}y)\right)\cap \left( {9\over 10}y,y\right) |>{9\over 100}y$, so from the fact that $E$ is an increasing sum of $E_m$'s, there is an integer $m$ so that \begin{equation} \left|E_m\cap\left( E_n+{3\over 4}y\right)\cap \left( {9\over 10}y,y\right)\right| >{8\over 100}y.\label{7} \end{equation} Denote $A_m= E_m\cap\left( E_n+{3\over 4}y\right)\cap \left( {9\over 10}y,y\right)$ and write $$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. 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