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% 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
probablity 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-\hskip0cmpreserving
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-\hskip0cmpreserving
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
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\bibitem{1} P. Collet and J. P. Eckmann, {\em Iterated Maps on the
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\bibitem{2}
W. Rudin, {\em Real and Complex Analysis, Third Edition},
McGraw-Hill Book Company, New York, 1987.
\bibitem{3}
P. Walters,
{\em An Introduction to Ergodic Theory},
Springer-Verlag, New York, 1982.
\end{thebibliography}
\end{document}