% spellchecked:pdh 11-10-98 % last edit::pdh 11-10-98 jch 10/22/98 % gallies sent: % gallies corrected: % set in production style:10/22/98 % Section Inroads from Omalley \documentclass{rae} \usepackage{amsmath,amsthm,amssymb} %\coverauthor{Eric S. Key} %\covertitle{Symmetric Measure-Preserving Systems} \received{July 28, 1997} \MathReviews{Primary: 28D05. Secondary: 47A35} \keywords{Measure-preserving, symmetric tent map} \firstpagenumber{1} \markboth{Eric S. Key}{Symmetric Measure-Preserving Systems} \author{Eric S. Key, Department of Mathematical Sciences, University of Wisconsin-Milwaukee, Milwaukee, Wisconsin 53201, USA, e-mail: {\tt ericskey@csd.uwm.edu}} \title{SYMMETRIC MEASURE-PRESERVING SYSTEMS} %%%%%%Put Author's Definitions Below Here %%%%%%%%%%% \def\pf{\noindent{\sc Proof.} } \newcommand{\allconsecutive} {\newtheorem{lemma}{Lemma} \newtheorem{prop}[lemma]{Proposition} \newtheorem{theorem}[lemma]{Theorem} \newtheorem{corollary}[lemma]{Corollary} } \newenvironment{myitem}{\begin{list}{\arabic{enumi}.}{\usecounter{enumi}\itemsep0.15cm\topsep0.15cm}}{\end{list}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \allconsecutive \begin{document} \maketitle \begin{abstract} A symmetric measure-preserving system is one where the measure $\Pr$ is preserved by two maps $T$ and $R$ where $R$ is self-inverse and \hbox{$T\circ R = T$}. We discuss the existence of such systems and some consequences, including when unimodal maps are conjugate to the symmetric tent map. \end{abstract} \bigskip \section{Introduction} A continuous map $T:[0,1]\rightarrow[0,1]$ is called {\it unimodal with turning point} $m$ if $m\in (0,1)$ and $T$ is continuous, strictly increasing on $[0,m]$ and strictly decreasing on $[m,1]$. For the moment, let us call a unimodal map {\it two-to-one} if $T(0) = T(1) = 0$ and $T(m) = 1$. To each two-to-one map we can associate a unique continuous map $R:[0,1]\rightarrow[0,1]$ such that $R$ is not the identity and $T\circ R = T$. The most well-known such pair of maps is $\tau(x) = \min(2x,2(1-x))$ and $\rho(x) = 1-x$. For each probability measure $\Pr$ on the Borel subsets of $[0,1]$ we may define the function $F:[0,1]\rightarrow[0,1]$ defined by $F(t)=\Pr([0,t])$. We will call $F$ the {\it distribution function associated with} $\Pr$. Given a two-to-one map $T$ one problem of interest is to characterize the probability measures $\Pr$ which are preserved by $T$. It is well-known that such measures exist. In the case of $\tau$ we know that Lebesgue measure on $[0,1]$ is one such probability measure. We also note that this measure is preserved by~$\rho$. Suppose for a moment that given a two-to-one map $T$ with turning point $m$ and its associated map $R$ that we can find a probability measure $\Pr$ which is preserved by both $T$ and $R$. Let $F$ be the distribution function associated with $\Pr$. Since $R$ preserves $\Pr$ we have $$F(R(t)) = 1 - F(t) = \rho(F(t)),$$ and since $T$ preserves $\Pr$, for $x\in[0,m]$ we have $$F(T(x)) = F(x) + 1 - F(R(x)) = 2F(x) = \tau(F(x)),$$ while for $x\in[m,1]$ we have $$F(T(x)) = F(T(R(x))) = \tau(F(R(x)) = \tau(\rho(F(x))) = \tau(F(x)).$$ Thus we have \begin{align*} F\circ R & = \rho\circ F\\ F\circ T & = \tau\circ F \end{align*} Note that to this point we only use symmetry. Suppose in addition we know that $F$ is strictly increasing and continuous. Such would be the case if the probability measure was non-atomic and assigned positive probability to all sub-intervals of $[0,1]$. In this case we have \begin{align*} R & = F^{-1}\circ\rho\circ F\\ T & = F^{-1}\circ\tau\circ F, \end{align*} and we would have proven that $T$ and $\tau$ (and $R$ and $\rho$) are topologically conjugate. Conversely, suppose that $T$ is two-to-one with turning point $m$, $R$ is the associated map with $T\circ R = T$, and for some homeomorphism $F:[0,1]\rightarrow[0,1]$ we have $F\circ T = \tau\circ F$ and $F\circ R = \rho\circ F$. Then $F$ is the distribution function of a probability measure preserved by both $T$ and $R$. To see why, let $\Pr$ be the probability measure on the Borel subsets of $[0,1]$ whose distribution function is $F$. Such $\Pr$ exists by the Carath\'{e}odory extension theorem. It is sufficient to check that the measures of intervals of the form $[0,y]$ are preserved. Since $T$ is two-to-one there is a unique $x\in[0,m]$ with $T(x) = y$, and we have $T^{-1}([0,y]) = [0,x]\cup[R(x),1]$, and $R^{-1}([0,y]) = [R(y),1]$. Note that $R(m) = m$ so that $F(m) = F(R(m)) = \rho(F(m)) = 1-F(m)$ so $F(m) = 1/2$. Hence \begin{align*} \Pr([0,x]\cup[R(x),1]) & = F(x) + 1-F(R(x)) = F(x) + \rho(F(R(x))\\ & = 2F(x) = \tau(F(x)) = F(T(x)) = \Pr([0,y]), \end{align*} and \begin{align*} \Pr([R(y),1]) & = 1 - F(R(y)) = \rho(F(R(y)) = F(y) = \Pr([0,y]). \end{align*} In this paper we \begin{enumerate} \item Generalize the idea of two-to-one maps to abstract measure spaces. \item In the case where the measure space is a compact metric space, show that there are non-atomic probability measures preserved by two-to-one maps (suitably defined) which are also preserved by the reflection map~$R$. \item In the case where the compact metric space is $[0,1]$, give conditions on two-to-one maps which ensure that this measure will give positive probability to any subinterval of $[0,1]$. \item In the case of $[0,1]$, look at what happens if we have non-atomic probability measures which give probability $0$ to some subintervals of $[0,1]$. \end{enumerate} \section{Some Additional Definitions and Examples} We will call the quintuple $(\Omega,{\cal F}, \Pr, T, R)$ a {\it symmetric measure-preserving system} if \begin{description} \item[P0:] $(\Omega,{\cal F}, \Pr)$ is a probability space; \item[P1:] $(\Omega,{\cal F}, \Pr, T)$ is a measure-preserving system; \item[P2:] $(\Omega,{\cal F}, \Pr, R)$ is a measure-preserving system; \item[P3:] $\{\omega \in \Omega: R(\omega) \neq \omega\} \in {\cal F}$ and $\Pr(\{\omega \in \Omega: R(\omega) \neq \omega\}) > 0$; \item[P4:] $R(R(\omega)) = \omega$ for all $\omega\in\Omega$; \item[P5:] $T\circ R = T$. \end{description} We shall call a measurable map $R$ of $(\Omega,{\cal F},\Pr)$ a {\bf reflection} of $(\Omega,{\cal F},\Pr)$ if it satisfies (P3) and (P4). If we have no measure in mind, we shall call a measurable map $R$ of $(\Omega,{\cal F})$ a {\bf reflection} of $(\Omega,{\cal F})$ if $R$ is not the identity map and $R\circ R$ is the identity map. Two examples of symmetric measure-preserving systems are \begin{itemize} \item $\Omega = [0,1]$; \item $\cal F =$ the Borel subsets of $[0,1]$; \item $\Pr(E) =$ the ordinary Lebesgue measure of $E$; \item $T(x) = \min(2x,2(1-x))$; \item $R(x) = 1-x$; \end{itemize} and \begin{itemize} \item $\Omega = [0,1]$; \item $\cal F =$ the Borel subsets of $[0,1]$; \item $\displaystyle{\Pr(E) = \int_E\frac{1}{\pi\sqrt{x-x^2}}\,dx}$'; \item $T(x) = 4x(1-x)$; \item $R(x) = 1-x$. \end{itemize} Note that the probability measure in the second example is non-atomic and gives positive probability to all subintervals of $[0,1]$. This provides one example of the situation discussed in the previous section. We now proceed to generalize our earlier idea of two-to-one. Note that we drop the requirement that the map be onto. Suppose that $\cal F$ is a $\sigma$-algebra on the set $\Omega$ and that $T$ is a measurable map from $\Omega$ to $\Omega$. We shall say that $T$ is {\bf two-to-one} if there are measurable sets $\Omega_l$ and $\Omega_r$ and a reflection $R$ of $(\Omega,\cal F)$ with the properties that \begin{itemize} \item $\Omega = \Omega_l \cup \Omega_r$; \item $\Omega_l \cap \Omega_r$ is the set of fixed points of $R$; \item $T\circ R = T$; \item The restriction of $T$ to each of $\Omega_l$ and $\Omega_r$ is one-to-one; \item If $F\in{\cal F}$ then $T(F\cap\Omega_l)\in{\cal F}$ and $T(F\cap\Omega_r)\in{\cal F}$. \end{itemize} Since we can show that there is exactly one such $R$ for any two-to-one map $T$, we will refer to $R$ as {\bf the reflection associated with} $T$. Also note that the sets $\Omega_l$ and $\Omega_r$ cannot be empty and that $R$ maps each of these sets onto the other. Two-to-one maps are a natural generalization of unimodal maps. We have seen examples of two-to-one maps on $[0,1]$. Here are some examples on the closed unit disk and on the unit circle in the complex plane. Suppose that $a$ and $b$ are complex numbers with $|a|^2 = |b|^2 + 1$. The fractional linear transformation $f(z) = (az + b)/(\overline{b}z + \overline{a})$ maps the unit disk onto itself and maps the unit circle onto itself. The map $T(z) = (f(z))^2$ maps the unit circle onto itself and maps the unit disk onto itself. In each case $T$ is two-to-one. To see why, take $R(z) = f^{-1}(-f(z))$. $R$ is a fractional linear transformation which maps the unit circle to the unit circle and the unit disk to the unit disk. As a map of the unit disk to itself, $R$ has exactly one fixed point at $z = -b/a$, and this fixed point does not lie on the unit circle. What is interesting about this example is that as a map of the unit circle to itself, $R$ has no fixed points, in contrast with the examples on $[0,1]$. \section{Constructing Symmetric Measures} In this section we assume that $T$ is two-to-one and that $R$ is the reflection associated with~$T$. As we shall not consider more than one two-to-one map at a time, this should cause no confusion. We will give conditions on $T$ which assure the existence of a probability measure $\Pr$ such that the system $(\Omega,{\cal F},\Pr,T,R)$ is a symmetric measure-preserving system. Let ${\cal I}_T$ denote the invariant $\sigma$-algebra of $T$ and let ${\cal I}'_T = \{G\in {\cal I}_T:\break T(G) = G\}$. In some cases, Theorem~\ref{t7.1} below can be used to show that ${\cal I}'_T$ only contains the empty set, as we shall see in the next section. \begin{lemma}\label{lemma8.3} Suppose that $G\in {\cal I}'_T$ and $G\neq \emptyset$. Let $\mu$ be a probability measure on $(\Omega, {\cal F})$ and suppose that $\mu(G) = 1$. Then the set function $\nu$ defined on $\cal F$ by \begin{displaymath} \nu(E) = \frac{1}{2}\mu(T(E\cap G \cap \Omega_l)) + \frac{1}{2}\mu(T(E\cap G \cap \Omega_r)) \end{displaymath} is a probability measure on $(\Omega,{\cal F})$ with $\nu(G) = 1$, $\nu\circ T^{-1} = \mu$ and $\nu\circ R^{-1} = \nu$. \end{lemma} \pf It is clear that $\nu$ is well-defined and non-negative, since $T$ carries elements of $\cal F$ to elements of $\cal F$. Next note that $T(G\cap \Omega_l) = T(G\cap \Omega_r) = G$, so $\nu(G) = 1$, and that since the restriction of $T$ to each of $\Omega_l$ and $\Omega_r$ is one-to-one, $\nu$ is countably additive. Hence $\nu$ is a probability measure on $\cal F$. Note that $R(G) = R^{-1}(G) = R^{-1}(T^{-1}(G)) = (T\circ R)^{-1}(G) = T^{-1}(G) = G$, and $R(\Omega_l) = \Omega_r$, so $\nu\circ R^{-1} = \nu$. Finally we show that $\nu\circ T^{-1} = \mu$. First observe that for any set $E\in{\cal F}$ we have \begin{displaymath} T(T^{-1}(G\cap E) \cap \Omega_l) = G\cap E = T(T^{-1}(G\cap E) \cap \Omega_r). \end{displaymath} To see why, recall that $T$ maps $G$ onto $G$. Therefore \begin{displaymath} T(T^{-1}(G\cap E) ) = G\cap E. \end{displaymath} Therefore, $g\in G\cap E$ if and only if there is some $g'\in T^{-1}(G\cap E)$ such that $T(g') = g$. Now, $g'\in T^{-1}(G\cap E)$ if and only if $R(g') \in T^{-1}(G\cap E)$. Since either $g'\in\Omega_l$ and $R(g')\in\Omega_r$ or vice versa, $T$ maps both $T^{-1}(G\cap E) \cap \Omega_l$ and $T^{-1}(G\cap E) \cap \Omega_r$ onto $G\cap E$, as claimed. Therefore, for any $E\in \cal F$, \begin{align*} 2 \nu(T^{-1}(E)) & = 2 \nu(T^{-1}(E)\cap G)\\ & = 2 \nu(T^{-1}(E\cap G))\\ & = \mu(T(T^{-1}(G\cap E) \cap \Omega_l)) + \mu(T(T^{-1}(G\cap E) \cap \Omega_r))\\ & = 2\mu(G\cap E)\\ & = 2\mu(E), \end{align*} which finishes the proof of the lemma.\qed \begin{lemma}\label{prop10.2} Suppose that $G\in {\cal I}'_T$ and $G\neq \emptyset$. Let $\mu$ be a probability measure on $(\Omega, {\cal F})$ and suppose that $\mu(G) = 1$. There is a sequence $\mu_n$ of $R$-invariant probability measures on $(\Omega,{\cal F})$ such that $\mu_n(G) = 1$ and $\mu_n\circ T^{-1} = \mu_{n-1}$ for $n = 1, 2, \dots$. \end{lemma} \pf We give a recursive construction. Put $\mu_0 = (\mu + \mu\circ R^{-1})/2$. Since $R\circ R$ is the identity map on $\Omega$, $\mu_0$ is $R$-invariant. Since $R^{-1}(G) = G$ we have $\mu_0(G) = 1$. Suppose now that $n$ is a positive integer and $\mu_0,\dots,\mu_{n-1}$ have been constructed to satisfy Lemma~\ref{prop10.2}. Define $\mu_n$ by \begin{displaymath} \mu_n(E) = \frac{1}{2}\mu_{n-1}(T(E\cap G \cap \Omega_l)) + \frac{1}{2}\mu_{n-1}(T(E\cap G \cap \Omega_r)). \end{displaymath} Then Lemma~\ref{lemma8.3} shows that $\mu_n$ satisfies the conditions of Lemma~\ref{prop10.2} as well.\qed \begin{theorem}\label{t7.1} Suppose that $\Omega$ is a compact metric space, that $\cal F$ is the Borel sigma algebra and that $T$ and $R$ are continuous. Suppose that $G\in {\cal I}'_T$ and $G\neq \emptyset$. Then there is a probability measure $\Pr$ on $(\Omega,{\cal F})$ having $\Pr(G) = 1$ which is invariant under both $T$ and~$R$. Furthermore, if $R$ has at most one fixed point and $T$ and $R$ have no fixed points in common, then $\Pr$ is non-atomic. \end{theorem} \pf Let $\mu_n$ be the sequence of measures constructed in Lemma~\ref{prop10.2}. Put $\sigma_n = n^{-1}(\mu_0 + \cdots + \mu_{n-1})$ for $n = 1, 2,\dots$. Each $\sigma_n$ is invariant under $R$ and $R$ is continuous, so any limit point of the sequence $\sigma_n$ will also be invariant under $R$. Since \begin{displaymath} \sigma_n = n^{-1}(\mu_{n}\circ T^{-n} + \mu_n\circ T^{-n + 1} + \cdots + \mu_n\circ T^{-1}) \end{displaymath} it is easy to show that any limit point of the sequence $\sigma_n$ will also be $T$ invariant. (See Theorem~6.9 of Walters [1982] for the case of Borel measures on $[0,1]$.) Now suppose that $R$ has at most one fixed point and $R$ and $T$ have no fixed points in common. We first show that no periodic point of $T$ may be an atom of $\Pr$. Suppose that $\omega$ is a periodic point of $T$ with period $n$. Let $p = \Pr(\{\omega\}) > 0$. Note that the inverse image of an atom under $T$ is never empty, and therefore, contains either $1$ or $2$ points. Observe that \begin{enumerate} \item $\omega\in T^{-n}(\omega)$; \item $T^{-n}(\omega)$ contains $\omega$ and at least one other point, and has probability $p$. \item Each element of $T^{-n}(\omega)$ is an atom, and these atoms each have a probability which is less than $p$. \end{enumerate} Therefore $p > 0$ is not possible, meaning there are no periodic atoms. Now we show that no non-periodic point may be an atom either. Begin with the purported atom $\omega$. For each positive integer $n$ the elements of $T^{-n}(\omega)$ are atoms, and since no atom is a periodic point, the sets $T^{-n}(\omega)$, $n= 1, 2, \dots$ are disjoint. Since these sets all have the same probability, they must have probability $0$ which contradicts our assumption that $\omega$ is an atom.\qed \begin{corollary}\label{c7.1} Suppose that $T:[0,1]\rightarrow[0,1]$ is a continuous, onto, unimodal map with $T(0) = 0 = T(1)$. Then there is a non-atomic probability measure on the Borel sets of $[0,1]$ and a continuous reflection $R$ of $[0,1]$ so that $([0,1],{\cal B},\Pr,T,R)$ is a symmetric measure-preserving system. \end{corollary} \section{Applications} Next we will show how Theorem~\ref{t7.1} can be used to analyze the behavior of some symmetric unimodal maps of $[0,1]$ to itself. \begin{lemma}\label{lemma11.1} Let $\cal I$ be a closed bounded interval, let $a$ be the left endpoint of ${\cal I}$ and let $b$ be in the interior of $\cal I$. Suppose $f:{\cal I}\rightarrow{\cal I}$ \begin{myitem} \item is continuous; \item satisfies $f(x) > x$ on $(a,b]$; \item satisfies $f(f(b)) > a$. \end{myitem} Then for each $y\in(a,b]$ there is some integer $k \geq 2$ for which $f^{(k)}(b) > y$. \end{lemma} \pf Suppose not. Then for each positive integer $k$ we have $y\geq f^{(k+1)}(b)\! = f(f^{(k)}(b)) > f^{(k)}(b)$ so $p \equiv \lim_{k\rightarrow\infty}f^{(k)}(b) \in (f^{(2)}(b),y] \subset (-a, b]$ is a fixed point of $f$. This contradicts our assumption that $f$ has no fixed points in $(a, b]$.\qed \begin{theorem}\label{th45.1} Suppose that $([0,1],{\cal B},\Pr,T,R)$ is a symmetric measure--preserving system and that \begin{enumerate} \item $\Pr$ has no atoms; \item $T$ is unimodal with turning point $m$; \item $T(x) > x$ on $(0,m]$; \item $T(0) = T(1) = 0$. \end{enumerate} Then for any $a\in[0,1]$, if $T(a) < 1$ then $\Pr([T(a),1]) > 0$. \end{theorem} \pf Suppose not. Then $\Pr([0,T(a)]) = 1$. We will use Lemma~\ref{lemma11.1} to derive a contradiction. It is sufficient to examine the case $a\in(0,m]$. Note that since $\Pr$ and $T$ are both invariant under $R$ and $R$ is self-inverse, $\Pr(A) = 0$ implies $\Pr(T(A)) = 0$. Since $T$ is continuous and maps both $0$ and $1$ to $0$, and $\Pr([T(a),1]) = 0$, for every $k \geq 1$ we have $\Pr([0,T^{(k+1)}(a)]) = 0$. From Lemma~\ref{lemma11.1} for some such $k$ we have $T^{(k+1)}(a) > a$ so $[0,a] \subset [0,T^{(k+1)}(a)]$. This implies $\Pr([0,a]) = 0$, which in turn implies $\Pr([0,T(a)]) = 0$, which is our contradiction.\qed \begin{corollary} Suppose that $m\in (0,1)$ and \begin{enumerate} \item $T:[0,1]\rightarrow[0,1]$ is unimodal with turning point $m$; \item $T(x) > x$ on $(0,m]$; \item $T(0) = T(1) = 0$; \item $T(m) < 1$, $T(T(m)) > 0$. \end{enumerate} Then ${\cal I}_T'$ contains only the empty set. \end{corollary} \pf Suppose not. We will now apply Theorem~\ref{t7.1}. Let $\Pr$ be the probability measure on the Borel subsets of $[0,1]$ which is preserved by both $R$ and $T$. Note that $\Pr$ is not atomic as $m$ is the only fixed point of $R$ and $m$ is not a fixed point of $T$. Hence $\Pr([T(m),1]) = \Pr(T^{-1}([T(m),1])) = \Pr(\{m\}) = 0$. This contradicts Theorem~\ref{th45.1}.\qed Next we consider the question of when symmetric measure-preserving systems are isomorphic. Following Walters [1982] we say that two symmetric measure-preserving systems $(\Omega_i, {\cal F}_i, \Pr_i, T_i, R_i)$, $i = 1, 2$ are {\bf isomorphic} if there exist $M_i\in {\cal F}_i$ with $\Pr_i(M_i) = 1$ for $i = 1, 2$ such that \begin{description} \item[(a)] $T_i(M_i)\subset M_i$ for $i= 1, 2$; \item[(b)] There is an invertible measure-preserving transformation $\Phi:M_1\rightarrow M_2$ with \begin{align*} \Phi(T_1(\omega)) & = T_2(\Phi(\omega))\\ \Phi(R_1(\omega)) & = R_2(\Phi(\omega)) \end{align*} for all $\omega \in M_1$. \end{description} Recall the symmetric tent map system, $([0,1],{\cal B}([0,1]),\lambda,\tau,\rho)$, defined in the introduction. Here is a formalization of the situation described in the introduction. \begin{theorem}\label{isomorphism} Suppose that $T:[0,1]\rightarrow[0,1]$ is a continuous unimodal map with turning point $m$, $T(0) = T(1) = 0$, $T(m) = 1$, and reflection $R$. Suppose that $([0,1], {\cal B},\Pr,T,R)$ is a symmetric measure-preserving system and the distribution function of $\Pr$ is a homeomorphism of $[0,1]$ onto $[0,1]$. Then $([0,1], {\cal B},\Pr,T,R)$ is isomorphic to $([0,1],{\cal B},\lambda,\tau,\rho)$. \end{theorem} Since the key in this theorem is having the distribution function of $\Pr$ be an increasing function, the following corollary to Theorem~\ref{th45.1} is of interest. \begin{corollary}\label{cor45.2} Suppose that $([0,1],{\cal B},\Pr,T,R)$ is a symmetric measure--preserving system and \begin{enumerate} \item $\Pr$ has no atoms; \item $T$ is unimodal with turning point $m$; \item $T(m) = 1$ and $T(0) = T(1) = 0$; \item For every interval $I\subset [0,1]$ there is some positive integer $k$ so that $m\in T^{(k)}(I)$. \end{enumerate} Then the distribution function of $\Pr$ is a homeomorphism of $[0,1]$ onto $[0,1]$. In particular, $([0,1],{\cal B},\Pr,T,R)$ is isomorphic to $([0,1],{\cal B},\lambda,\tau,\rho)$. \end{corollary} \pf Suppose not. Note that we must have $T(x) > x$ on $(0,m]$. Let $F$ denote the distribution function of $\Pr$. Then for some $0 \leq a < b \leq 1$ we have $F(a) = F(b)$, so $\Pr([a,b]) = 0$. Let $I$ denote $[a,b]$, and choose $k$ so that $m\in T^{(k)}(I)\equiv I_k$. Note that since $T$ is continuous $I_k$ is a closed interval, and $I_k$ has probability $0$. It is also clear that $I_k$ has non-empty interior. Let $J_k = I_k \cup R(I_k)$. $J_k$ is a closed interval with probability $0$ which contains $m$ in its interior. Hence $T(J_k)$ is an interval of probability $0$ with right endpoint $1$ and non-empty interior. This contradicts Theorem~\ref{th45.1}.\qed We are, however, in a position to assert the existence of symmetric measure-preserving systems. Using Corollary~\ref{c7.1} and the idea of the proof of Corollary~\ref{cor45.2}, it is easy to see \begin{theorem}\label{isom} Suppose \begin{enumerate} \item $T:[0,1]\rightarrow[0,1]$ is continuous; \item $T$ is unimodal with turning point $m$; \item $T(m) = 1$, $T(0) = T(1) = 0$; \item For every interval $I\subset [0,1]$ there is some positive integer $k$ so that $m\in T^{(k)}(I)$. \end{enumerate} Then there is a continuous reflection of $[0,1]$, denote it by $R$, and non-atomic probability measure $\Pr$ on $\cal B$ which assigns positive probability to all intervals, such that $([0,1],{\cal B},\Pr,T,R)$ is a symmetric measure-preserving system which is isomorphic to $([0,1],{\cal B},\lambda,\tau,\rho)$. \end{theorem} Condition 4 in the theorem is satisfied in many cases. See the discussion of homtervals and stable periodic orbits in Collet and Eckmann [1980]. \section{Symmetry in $([0,1],{\cal B},\Pr)$} Suppose that we are given a probability measure $\Pr$ on the Borel subsets, $\cal B$, of $[0,1]$. We would like to construct transformations $T$ and $R$ so that $(\Omega,{\cal B}, \Pr, T, R)$ is symmetric. We have seen that this is easily done if the distribution function of $\Pr$ is continuous and strictly increasing. Suppose then we only require that it be continuous. \begin{theorem}\label{theorem9.1} If $\Pr$ is a non-atomic probability measure on the Borel sets of $[0,1]$ then there is a symmetric measure-preserving system $([0,1],{\cal B},\Pr,T,R)$. \end{theorem} The proof is presented as a series of lemmas. As before, put $F(t) = \Pr([0,t])$. Put $F^{-1}(y) = \sup\{x: F(x) \leq y\}$ and $R(t) = F^{-1}(1-F(t))$ for all $t\in[0,1]$. Then we have: \begin{lemma}\label{prop9.1} There exists $\Omega_0 \subset {\cal B}$ with $\Pr(\Omega_0) = 1$ such that $R(R(\omega)) = \omega$ for all $\omega\in \Omega_0$. \end{lemma} \pf We shall take $\Omega_0$ to be the complement of the union of all intervals where $F$ is constant. Precisely, we define \begin{align*} {\cal J} = \{&[a,b]\subset[0,1]: a < b, F(a) = F(b),\\ & x < a < b < y {\rm\;implies\;} F(x) < F(a) < F(y)\} \end{align*} Since the elements of $\cal J$ are disjoint closed subintervals of $[0,1]$ of positive length, $\cal J$ is countable, and the union of its elements is not $[0,1]$ since each element of $\cal J$ has probability $0$. Let $\Omega_0$ be the complement of the union of the elements of $\cal J$. It is clear that $\Pr(\Omega_0) = 1$ and that $F$ is strictly increasing on~$\Omega_0$. It is easy to see that for all $x\in[0,1]$ we have $F(F^{-1}(x)) = x$. What we need to know is that if $x\in \Omega_0$ then $F^{-1}(F(x)) = x$. To see this, observe that for all $x$ we have $x \leq F^{-1}(F(x))$, so we suppose that $x\in\Omega_0$ and $x < F^{-1}(F(x))$. However, since $F(x) = F(F^{-1}(F(x)))$, this would imply that both $x$ and $F^{-1}(F(x))$ were in $\Omega_0^c$, a contradiction. Now it is a simple matter to check that if $x\in \Omega_0$ then $R(R(x)) = x$.\qed \begin{lemma}\label{lemma9.1} Suppose that $g:[a,b]\rightarrow[0,1]$ is monotone and continuous. Let $h = F^{-1}\circ g$. For $z\in(a,b]$ put $c_z = h(z^-)$ and put $c_a = a$. For $z\in[a,b)$ put $d_z = h(z^+)$ and put $d_b=b$. Then for any $z\in[a,b]$, we have $F(c_z) = F(d_z)$. \end{lemma} \pf Simply observe that since $F$ and $g$ are continuous, $F(c_z) = g(z) = F(d_z)$.\qed First we apply Lemma~\ref{lemma9.1} to prove: \begin{lemma}\label{prop9.2} $R$ preserves $\Pr$. \end{lemma} \pf It is sufficient to prove that for any $b\in[0,1]$, $\Pr([b,1]) =\break \Pr(R^{-1}([b,1]))$. First notice that $F^{-1}$ is strictly increasing and continuous from the right. Since $F$ itself is non-decreasing and continuous we conclude that $R$ is non-increasing and continuous from the left. Let $b\in[0,1]$ be given and put $t_b = \sup(\{x:R(x) \geq b\})$. It is straightforward to check that $R(t_b) \geq b$ and that $R^{-1}([b,1]) = [0,t_b]$. Since $\Pr([b,1]) = 1-F(b)$ and $\Pr(R^{-1}([b,1])) = \Pr([0,t_b]) = F(t_b)$, it is sufficient to show that $1-F(b) = F(t_b)$. This is easily done by applying Lemma~\ref{lemma9.1} with $g(x) = 1 - F(x)$ and $z = t_b$, and observing that $R(t_b^+) \leq b \leq R(t_b) = R(t_b^-)$.\qed We now focus our attention on constructing $T$ which preserves $\Pr$ and which satisfies $T = T\circ R$. We omit the straightforward proof of the following: \begin{lemma}\label{prop9.3} $m \equiv F^{-1}(1/2)$ is the unique fixed point of $R$. \end{lemma} Define the function $T$ as follows: \begin{displaymath} T(x) = \left\{\begin{array}{cc} F^{-1}(2F(x)) & {\rm if}\;x\in[0,m]\\ F^{-1}(2(1-F(x))) & {\rm if}\;x\in[m,1] \end{array} \right. \end{displaymath} \begin{lemma}\label{prop9.4} $T = T\circ R$ \end{lemma} \pf It is easy to check that for any $x\in[0,1]$ that $F(x) = 1 - F(R(x))$. Suppose that $x\in[0,m]$. Then $R(x) \geq R(m) = m$ so $R(x) \in [m,1]$. So, $T(x) = F^{-1}(2F(x)) = F^{-1}(2(1-F(R(x)))) = T(R(x))$. Similarly, if $x\in[m,1]$ then $R(x) \leq R(m) = m$ so $R(x)\in[0,m]$ and $T(x) = F^{-1}(2(1-F(x))) = F^{-1}(2F(R(x))) = T(R(x))$.\qed \begin{lemma}\label{prop9.5} $T$ preserves $\Pr$. \end{lemma} \pf It will be sufficient to prove that for any $b\in[0,1]$ that $\Pr([b,1]) = \Pr(T^{-1}([b,1]))$. Fix such a $b$ and put $a_b = \inf(\{x:T(x)\geq b\})$ and $c_b = \sup(\{x:T(x)\geq b\})$. Observe that $T$ is right continuous on $[0,m]$, left continuous on $[m,1]$, and $T(m) = 1$. Therefore $a_b \leq m\leq c_b$ and $T^{-1}([b,1]) = [a_b,c_b]$. Once we show that $F(b) = 2F(a_b)$ and $F(b) = 2(1-F(c_b))$ we will be done, since averaging these equations gives $F(b) = F(a_b) + 1 - F(c_b)$, which in turn shows \begin{align*} \Pr([b,1]) & = 1 - F(b) = 1 - [F(a_b) + 1 - F(c_b)]\\ & = F(c_b) - F(a_b) = \Pr([a_b,c_b]). \end{align*} (Note the use of our assumption that $\Pr$ is non-atomic.) To see that $F(b) = 2(F(a_b))$ apply Lemma~\ref{lemma9.1} with $g(x) = 2F(x)$ on $[0,m]$ and $z = a_b$, and to see that $F(b) = 2(1-F(c_b))$ apply Lemma~\ref{lemma9.1} with $g(x) = 2(1-F(x))$ on $[m,1]$ with $z=c_b$.\qed \begin{thebibliography}{9} \bibitem{1} P. Collet and J. P. Eckmann, {\em Iterated Maps on the Interval as Dynamical Systems}, Birkh\"{a}user, Boston, 1980. \bibitem{2} W. 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