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\Volume{36} % publication information for your paper
\IssueNumber{2} % publication information for your paper
\Year{2010/2011} % publication information for your paper
\Editor{Paul D. Humke} % for your paper
\Received{April 8, 2010} % publication information for your paper
%\CoverAuthor{Jack Grahl and Togo Nishiura \\}
%\CoverTitle{A Factorization Problem}
\begin{Author}
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\FirstName{Jack}\LastName{Grahl}
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\PostalAddress{Department of Mathematics, University College London WC1E63T, London, U.K}
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\Email{jgrahl@math.ucl.ac.uk }
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%agency.}
\end{Author}
\begin{Author}
\FirstName{Togo}\LastName{Nishiura}
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\PostalAddress{Department of Mathematics and Computer Science, Dickinson College,
Carlisle, PA 17013, USA;
and
Department of Mathematics, Wayne State University,
Detroit, MI 48203, U.S.A}
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\Email{nishiura@dickinson.edu }
\end{Author}
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\keyword{measure preserving function}
\keyword{rearrangement of functions}
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\end{KeyWords}
\title{A Factorization Problem}
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\markboth{Jack Grahl and Togo Nishiura}{A Factorization Problem}
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\theoremstyle{plain}
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\newtheorem*{facprob}{Factorization Problem}
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\begin{document}
\maketitle
\begin{abstract}
A solution is presented of a problem proposed at the Summer
Symposium in Real Analysis XXXIII.
\end{abstract}
The factorization problem was proposed by the first author during the
Summer Symposium in Real Analysis XXXIII, which was held at
Southeastern Oklahoma State University.
The statement of the problem (given in Section~\ref{problem})
requires some definitions.
A \emph{partition} $\Pi$ of $[0,1)$ is a finite disjointed collection
$\{\,I_i\colon i=1,2,\dots,n\,\}$ whose union is $[0,1)$, where $I_i$
is a half-open interval of the form $[a,b)$.
The \emph{size} of a partition, $\delta(\Pi)$, is the maximum length
of the intervals of $\Pi$.
A function $\varphi\colon[0,1]\to[0,1]$ is called a
\emph{permutation} of a partition~$\Pi$ of $[0,1)$ if
$\Pi'=\{\,\varphi[I_i]\colon I_i\in\Pi\,\}$ is a partition of
$[0,1)$, $\varphi$ restricted to $I_i$ is a translation for each $i$,
and $\varphi(1)=1$.
Clearly a permutation is bijective and has the property that the
Lebesgue measures of $\varphi[E]$ and $\varphi^{-1}[E]$ are equal to
the Lebesgue measure of $E$ for every Lebesgue measurable set
$E\subset[0,1]$.
(Lebesgue measure will be denoted by $\lambda$, and the modifying
``Lebesgue'' will be dropped.)
The set of all permutations will be denoted by $\mathcal P$.
The set of all functions $h\colon [0,1]\to[0,1]$ that
are almost everywhere limits of sequences in $\mathcal P$ will be
denoted by~$\mathcal H$.
This is the same as the set of functions that are
limit-in-measure of sequences in $\mathcal P$.
(See \cite[Chapter 5]{G-Z}.)
The collection of partitions of $[0,1)$ can be used to define another
set $\mathcal H'$.
A function $h\colon[0,1]\to[0,1]$ is in $\mathcal H'$ if, for each
sequence $\Pi_n$ of partitions of $[0,1)$ such that $\delta(\Pi_n)\to
0$ as $n \to \infty$, there exists a permutation $\varphi_n$ of
$\Pi_n$ such that $\varphi_n$ converges almost everywhere to $h$.
Obviously, $\mathcal H'\subset\mathcal H$.
It will be shown in Section \ref{characterization} that $\mathcal
H=\mathcal H'$ and that $h\in\mathcal H$ if and only if $h$ is
$\lambda$-measurable and $\lambda=h_\#\lambda$.\footnote{For
any $\lambda$-measurable function $f\colon E\to\mathbb R$, where $E$
is a Borel set of $\mathbb R$, the measure $f_\#\lambda$ is defined on
the Borel sets $B$ of $\mathbb R$ by
$f_\#\lambda(B)=\lambda(f^{-1}[B])$.}
\section{Factorization problems}\label{problem}
The statement of the proposed factorization problem is the following:
Find a minimal\footnote{As usual, we mean minimal in the sense of the
partial order $\subset$. Of course, minimal sets need not be
unique.} set $\mathcal G$ of functions $g\colon[0,1]\to\mathbb R$
such that for each $\lambda$\nobreakdash-measurable function
$f\colon[0,1]\to\mathbb R$ there is an $h$ in~$\mathcal H$ and there
is a $g$ in $\mathcal G$ such that the composition\footnote{As
real-valued functions will not be multiplied, compositions will be
indicated in the multiplicative notation.}
$gh\colon[0,1]\to\mathbb R$ is $\lambda$-equivalent to $f$.
Observe that the set $\mathcal B_2$ of Baire class 2 functions
has the property that each $\lambda$-measurable function $f$ has a
factorization $f=gh$ almost everywhere with $g\in\mathcal B_2$ and $h$
being the identity function; but $\mathcal B_2$ is not minimal.
(The first author conjectured that a possible $\mathcal G$ is the
set of nondecreasing functions.)
See Theorems \ref{solution} and \ref{solution theorem} for our solution.
\begin{facprob}
Give a minimal set $\mathcal G$ of functions
$g\colon[0,1]\to[0,1]$ having the property that for each Borel measurable
function $f\colon [0,1]\to(0,1)$ there is an $h$ in $\mathcal H$ and
there is a $g$ in $\mathcal G$ such that $gh$ is $\lambda$-equivalent
to $f$.
\end{facprob}
Note that the $\lambda$-measurability of $f\colon[0,1]\to\mathbb R$ has been
replaced with Borel measurability of $f\colon[0,1]\to(0,1)$ and that $g\colon
[0,1]\to\mathbb R$ has been replaced by $g\colon [0,1]\to [0,1]$.
Clearly there is no loss in generality in making these replacements.
We shall show that the set $\mathcal G$ of continuous-from-above,
nondecreasing functions will solve the above factorization problem.
Also, we strengthen the solution to include connections between the
Baire classes of the functions $f$ and $h$.
Our solution of the factorization problem is achieved by
employing measures on certain linearly ordered spaces $(S,\prec)$
which are defined in Section~\ref{four}.
The order topology of $(S,\prec)$ will not be used, only its measure
theoretic properties induced by the order $\prec$ are exploited.
Before embarking on the construction of the function $h$ let us
characterize the sets $\mathcal H$ and $\mathcal H'$.
\section{Characterization of $\mathcal H$ and $\mathcal H'$}\label{characterization}
The 2 sets $\mathcal H$ and $\mathcal H'$ were introduced to
define the factorization problem.
$\mathcal H'\subset\mathcal H$ has been observed already.
We have the following characterisation.
\begin{theorem}
$\mathcal H'=\mathcal H$, and $h\in\mathcal H$ if and only if $h$ is a
$\lambda$-measurable function such that $\lambda=h_\#\lambda$.
\end{theorem}
\noindent
A $\lambda$-measurable function $h\colon[0,1]\to [0,1]$ is said to
be \textit{measure preserving}\footnote{Here, $h$ need not be
bijective; in ergodic theory, ``measure preserving'' requires that
$h$ be bijective and both $h$ and $h^{-1}$ be measurable and measure
preserving in our sense (see, for example,
\cite[page~7]{alpern-prasad}). Such $h$ is often called an
automorphism.}
if $\lambda=h_\#\lambda$.
Note that $\lambda(h^{-1}(1))=0$.
The next 2 propositions provide the proof of the
theorem.
\begin{proposition}\label{necprop}
If $h\in\mathcal H$, then $h$ is measure preserving.
\end{proposition}
\begin{proof}
Let $\varphi_n$ be a sequence of permutations
converging in measure to $h$, and let $\varepsilon_n$ be such that
$\lambda(E_n)<\varepsilon_n$, where $E_n=\{\,x\colon\lvert
\varphi_n(x)-h(x)\rvert\ge\varepsilon_n\,\}$, with $\varepsilon_n\to0$ as
$n\to\infty$.
For closed sets $K$ in $[0,1]$ let $K_n$ be the
$\varepsilon_n$-neighborhood of $K$.
As ${\varphi_n}^{-1}[K]\subset h^{-1}[K_n]\cup E_n$, it follows that $\lambda(K)\le
h_\#\lambda(K_n)+\varepsilon_n$, whence $\lambda(K)\le h_\#\lambda(K)$.
Consequently, $\lambda(U)\le h_\#\lambda(U)$ for every open set $U$ in
$[0,1]$.
It now follows that $\lambda(K)= h_\#\lambda(K)$ since
$h_\#\lambda([0,1])=1$.
\end{proof}
It remains to prove that if $h$ is measure preserving then
$h\in\mathcal H'$.
The proof is a ``pigeonhole'' process.
That is, a partition $\Pi$ of pigeons are to be assigned to a partition
$\Pi'$ of pigeonholes under certain rules $\varphi$.
Unfortunately the pigeons may be too fat to achieve this assignment.
The final step of the proof of the characterization will depend on the
following modified pigeonhole lemma.
\begin{lemma}
Let $\Pi'=\{\,I'_i\colon i=1,2,\dots,n\,\}$ be a
partition of~$[0,1)$ and let $\{\,K_i\colon i=1,2,\dots,n\,\}$ be a
disjointed collection of compact sets of $[0,1)$ such that
$\lambda(I'_i)>\lambda(K_i)$ for each $i$.
If $\varepsilon>0$, then there is a $\delta>0$ such that for each
partition $\Pi=\{\,I_j\colon j=1,2,\dots,m\,\}$
with $\delta(\Pi)<\delta$ there is a permutation
$\varphi\colon [0,1]\to[0,1]$ of $\Pi$ such that, for each $i$,
\[
\lambda(I'_i)-2\delta(\Pi)>\lambda(\varphi[H_i])>
\lambda(\varphi[K_i])-\varepsilon/n,
\]
where $H_i=\bigcup_j\{\,I_j\colon\varphi[I_j]\subset I'_i\textup{ and }
I_j\cap K_i\ne\emptyset\,\}$.
\end{lemma}
\begin{proof}
Let $\gamma$ be such that $0<3\gamma<\lambda(I'_i)-\lambda(K_i)$ and
$\lambda(U_i\setminus K_i)<\varepsilon/n$ for each~$i$, where
$U_i$ is the $\gamma$-neighborhood of $K_i$, and such that
$3\gamma$ is less than the distances between distinct $K_i$'s.
With $W(i,\Pi)=\bigcup\{\,I_j\in\Pi\colon I_j\cap K_i\ne\emptyset\,\}$,
observe that $\lambda(I'_i)-\lambda(W(i,\Pi))+\lambda\big(W(i,\Pi)\setminus
K_i\big)=\lambda(I'_i)-\lambda(K_i)>3\gamma$.
As $\lambda\big(W(i,\Pi)\setminus K_i\big)\to 0$ as $\delta(\Pi)\to 0$, there
is a $\delta$ such that
$0<\delta<\gamma$ and such that
$\lambda\big(W(i,\Pi)\setminus K_i\big)<\gamma$ whenever $\delta(\Pi)<\delta$.
So, if $\delta(\Pi)<\delta$, then
\[
\lambda(I'_i)-2\delta(\Pi)>\textstyle{\sum}\{\,\lambda(I_j)\colon
I_j\in\Pi,\ I_j\cap K_i\ne\emptyset\,\}.
\]
Let us construct the required $\varphi$.
Let $\delta(\Pi)<\delta$.
Note that no interval $I_j$ intersects more than one of the
$K_i$.
We separate the intervals $I_j$ into classes $B_i$, which
consists of those intervals which intersect $K_i$, $1 \leq i\leq n$,
and $C$, those intervals which do not intersect any $K_i$.
Denote $\sum\,\{\,\lambda(I_j)\colon I_j\in B_i\,\}$ by $l_i$
and index the collection $C$ as $J_1, J_2,\dots, J_m$.
Then $\sum_{i=1}^nl_i+\sum_{k=1}^m\lambda(J_k)=1$.
Call $a_i$ the right endpoint of $I'_i$.
Let us describe the first step of the construction of $\varphi$.
As $l_12\,\delta_{m+1}$.
For each $m$ let
\[
D_m=h^{-1}(1)\cup\textstyle\bigcup_{m\le m'}\bigcup\{\,h^{-1}[I'(m',i)]\setminus K(m',i)\colon
I'(m',i)\in\Pi'_{m'}\,\}.
\]
Note $D_{m'}\subset D_m$ whenever $m'\ge m$, and $\lambda(D_m)\le
2^{-(m-1)}$.
Also, if $x\notin D_m$ and $m'>m$, then $x$ is in some $K(m',i)$,
whence $h(x)\in I'(m',i)\in\Pi'_{m'}$.
Suppose that $\Pi_k$ is a sequence of partitions such that
$\delta(\Pi_k)\to0$ as $k\to0$.
Let $k_m$ be the least $k$ such that $\delta(\Pi_{k'})<\delta_m$
whenever $k'0$ and let $m$ be such that
$2^{-(m-1)}<\varepsilon$.
Suppose $x\notin D_m$ and $m'\ge m$.
Let $m''$ and $k$ be such that $k_{m'}=k_{m''}k_{m''}$.
There is an $I(k,j)$ in $\Pi_k$ such that $x\in I(k,j)\cap
K(m'',i)$ for some $i$.
By the modified pigeonhole lemma, $\varphi_k[I(k,j)]\subset
I'(m'',i)\in\Pi'_{m''}$.
Hence $h(x)$ and $\varphi_k(x)$ are in the same $I'(m'',i)$.
As $\delta(\Pi'_{m''})<2^{-m''}$,
$\left|\varphi_k(x)-h(x)\right|<2^{-m''}$ whenever $k_{m''+1}\ge
k>k_{m''}$.
We infer from this that $\varphi_k$ converges to $h$ except on a
subset of $D_m$.
Hence the set in which $\varphi_k$ does not converge to $h$ has
measure less than $\varepsilon$.
\end{proof}
\section{Measure induced by $f\colon [0,1]\to[0,1]$}\label{section f}
For a Borel measurable function $f\colon[0,1]\to[0,1]$ let
$g\colon[0,1]\to[0,1]$ be its nondecreasing, continuous-from-above
distribution function (see Remark~\ref{H-L} below) that satisfies
\[
g_\#\lambda([0,y])=f_\#\lambda([0,y]),\quad y\in [0,1].
\]
For each $y$ in $[0,1]$, the level sets of $g$ and $f$ are given by
$g^{-1}(y)$ and $f^{-1}(y)$, respectively.
Moreover, $f_\#\lambda(\{y\})=\lambda(g^{-1}(y))$.
As $g$ is nondecreasing and continuous-from-above, each nonempty level
set $g^{-1}(y)$ is a connected set such that
\[
a(y)=\min g^{-1}(y)\text{\qquad and\qquad} b(y)=\sup g^{-1}(y)
\]
\noindent satisfy
$\lambda\big([0,a(y))\big)=\lambda\big(f^{-1}\big[[0,y)\big ]\big)$
and $\lambda(g^{-1}(y))= b(y)-a(y)=\lambda(f^{-1}(y))$.
It will be shown in the following section that the set
$\mathcal G$ of all nondecreasing, continuous-from-above functions
$g\colon[0,1]\to[0,1]$ fulfills the requirements of the factorization.
Indeed, for each $f$, a measure-preserving $h$ is constructed so that
$f = gh$ almost everywhere.
Moreover, if $f\in\mathcal G$ then the $g$ in $\mathcal G$ and the
constructed $h$ yielding $f=gh$ almost everywhere are $f$ and the
identity function, respectively.
To see that $\mathcal G$ is a minimal set, it is enough to observe
that if two nondecreasing, continuous-from-above functions $g_1$ and
$g_2$ are different, then they differ on a set of positive measure.
Hence there exists some $r$ such that
\[
\lambda(\{\,x: g_1(x) > r\,\}) \neq \lambda(\{\,x: g_2(x) > r\,\}).
\]
So there can be no measure preserving function $h$ such that $g_1h = g_2$
almost everywhere.
Consequently, upon the successful construction of the measure preserving
$h$ in the next section, the following theorem yields a solution of the
factorization problem.
\begin{theorem}\label{solution theorem}
The set $\mathcal G$ of all functions $g\colon[0,1]\to[0,1]$ that
are nondecreasing and continuous-from-above is a minimal set having
the property that for each Borel measurable function $f\colon
[0,1]\to(0,1)$ there is an $h$ in $\mathcal H$ and there is a $g$ in
$\mathcal G$ such that $gh$ is $\lambda$-equivalent to $f$.
\end{theorem}
\noindent (Observe: If each $g$ in $\mathcal G$ of the theorem is
reassigned the value $y_0$ at $0$ and at~$1$, then the resulting set is
also minimal.
This will be useful for Theorem~\ref{solution}.)
\section{Linearly ordered spaces}\label{four}
To prove Theorem \ref{solution theorem} let $g\in\mathcal G$
correspond to the function $f$.
Our task is to construct a measure preserving function $h$ such that
$gh=f$ almost everywhere.
It will be convenient to designate by $Z$ the space $[0,1]$ through
which the factorization is accomplished.
That is, $h\colon [0,1]\to Z$ and $g\colon Z\to[0,1]$.
Turning to the construction, we will need a Baire class $1$ function
$\eta$ that retracts $[0,1]$ onto $g[Z]$.
To this end, define
\[
\eta(y)=
\begin{cases}
\inf\,[y,1]\cap g[Z], &\text{if $[y,1]\cap g[Z]\ne\emptyset$;}\\
\sup\,g[Z], &\text{if $[y,1]\cap g[Z]=\emptyset$.}
\end{cases}
\]
Clearly $\eta$ is nondecreasing (hence is Baire class $1$); and, as
$g_\#\lambda([0,y])$ is continuous-from-above,
$g[Z]=\eta\big[g[Z]\big]$ and $\eta\eta=\eta$.
Moreover, $\eta f$ is equal to $f$ almost everywhere and $\eta
f\big[[0,1]\big]\subset g[Z]$.
The tool used in the construction of the function $h\colon
[0,1]\to Z$ is the following linear order\footnote{This is a
dictionary order on $[0,1]^2$, though not the usual one. If
$f\colon[0,1]\to[0,1]$ is nondecreasing, then $\psi\colon
x\mapsto(x,f(x))$ is an increasing map of $[0,1]$ into the linearly
ordered space $([0,1]^2,\prec)$. As the factorization problem is a
measure theoretic one, the emphasis will be on the $\sigma$-algebra
$\overline{\mathcal B}$ of $([0,1]^2,\prec)$ generated by the open
half-rays $H(\bar x,\bar y)=\{\,(x,y)\colon(x,y)\prec(\bar x,\bar
y)\,\}$. $\overline{\mathcal B}$ is contained in the collection of
the usual Borel sets of $[0,1]^2$. Moreover, the sets of
$\overline{\mathcal B}$ that are contained in horizontal lines of
$[0,1]^2$ are exactly the usual Borel sets of the lines.}
$\prec$ of $[0,1]^2$.
\[
(s,t)\prec(s',t')\quad\text{ if $t0$ the definition of $h_0(x,y)$ is a slight
modification of the above.
That is, if $x\in (\eta f)^{-1}(y)$, define $h_0(x,y)$ as before
whenever $b(y)-a(y)>\int^x_0I_y\,dt$ and define $h_0(x,y)$ to be
$(a(y),y)$ whenever $b(y)-a(y)\le\int^x_0I_y\,dt$.
As $\lambda(\{\,x\in(\eta f)^{-1}(y)\colon
b(y)-a(y)\le\int^x_0I_y\,dt\,\})=0$, the above defined $h_0$ maps into
$\textup{Graph}(g)$.
The $\nu$-measurability of $h_0$ is easily shown by the statements 1
and 2 above.
Moreover, $h_0[(\eta f)^{-1}(y)\times y]\subset g^{-1}(y)\times y$.
Observe that if $f=g$, where $g\in\mathcal G$, then the function that
corresponds to $f$ in $\mathcal G$ is $g$ itself and the constructed
$h_0$ is the identity map.
Let us return to the linearly ordered space
$(\textup{Graph}(g),\prec_g)$.
As this space is linearly isomorphic to $(Z,<)$ there is a natural
``Lebesgue'' measure $\lambda_0$ on it generated by
$\lambda_0(\{\,(z',g(z'))\colon(z',g(z'))\prec_g(z,g(z))\,\})=\lambda([0,z))=z$.
Let us verify $\lambda_0=(h_0)_\#\nu$.
To this end, for $y_0=g(z_0)$, we exhibit the mutually exclusive and
exhaustive cases of $M(z_0)={h_0}^{-1}\big[\{\,(z,g(z))\colon
(z,g(z)) \prec_g(z_0,y_0)\,\}\big]$.
Denoting the set $\{\,(x,\eta f(x))\colon \eta f(x)