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\begin{document}


\setbeamercovered{dynamic}

\title{Disintegration of $\sigma$-finite measures}



\author{S.\ Jackson}



\institute{
(Joint with {\color{red}K.\ Backs} and {\color{red}D.\ Mauldin}) \\[10pt]
Department of Mathematics\\
University of North Texas\\
%jackson@unt.edu
}






\date{
September 2012\\
Luminy}


\begin{frame}
\titlepage
\end{frame}





\section{Introduction}
\begin{frame}
In 1950 {\color{red}Maharam} asked whether every disintegration of a 
$\sigma$-finite measure is necessarily uniformly $\sigma$-finite.
\mys

The answer has been shown to be yes under special conditions on the 
disintegration. 
\mys

We show that it is consistent that the answer is {\color{blue}no}. Specifically, 
we show the answer is no assuming $V=L$. 
\mys

We do this by constructing,
assuming $V=L$, a $\bp^1_1$ set in the plane with certain special properties.
\end{frame}




\begin{frame}
Let $(X, \sB(X))$, $(Y,\sB(Y))$ be standard Borel spaces with their $\sigma$-algebras
of Borel sets. Let $\varphi\colon Y\to X$ be  measurable and $\mu$, $\nu$ 
be measures on $\sB(Y)$, $\sB(X)$ respectively. 




\begin{defn}
A {\color{blue}disintegration} of $\mu$ with respect to $(\nu,\phi)$ is a
family, $\{\mu_x:x \in X\}$, of measures on $(Y,\mathcal{B}(Y))$
satisfying:
\begin{enumerate}
    \item
        $\forall B \in \mathcal{B}(Y),\,\, x \mapsto \mu_x(B) \text{
      is } \mathcal{B}(X) \text{-measurable}$
    \item
        $\forall x \in X,\,\, \mu_x(Y \setminus \phi^{-1}(x)) = 0$ and
    \item
        $\forall B \in \mathcal{B}(Y),\,\, \mu(B) = \int \mu_x (B)
      d\nu(x)$.
\end{enumerate}
\end{defn}

\begin{rem}
The collection $\{ \mu_x \colon x \in X\}$ is sometimes called a 
{\color{blue}transition kernel}.
\end{rem}
\end{frame}

\begin{frame}
Note that if we have a disintegration $\{ \mu_x\colon x \in X\}$ of $\mu$ 
with respect to $(\nu,\varphi)$, then $\mu \circ \varphi^{-1} \ll \nu$. 
\mys 

The converse is also true according to the following theorem.
\mys

\begin{thm}
Suppose $(X,\mathcal{B}(X))$ and $(Y,\mathcal{B}(Y))$ are standard
Borel spaces, $\mu$ is a $\sigma$-finite measure on $\mathcal{B}(Y)$,
$\nu$ is a $\sigma$-finite measure on $\mathcal{B}(X)$, and $\phi\colon Y
\rightarrow X$ is a Borel measurable function. If $\mu \circ \phi^{-1}
\ll \nu$ then there exists a $\sigma$-finite disintegration $\{\mu_x \colon
x \in X\}$ of $\mu$ with respect to $(\nu,\phi)$. Moreover this
disintegration is unique in the sense that if $\{\hat{\mu}_x : x \in
X\}$ is any $\sigma$-finite disintegration of $\mu$ with respect to
$(\nu,\phi)$, then there exists $N \subseteq X$ such that $\nu(N) = 0$
and $\forall x \not \in N \,\, \mu_x = \hat{\mu}_x$.
\end{thm}
\end{frame}

\begin{frame}
We say the disintegration $\{ \mu_x \colon x \in X\}$ is {\color{blue}$\sigma$-finite} if 
each $\mu_x$ is $\sigma$-finite. 
\mys


We say the disintegration  $\{ \mu_x \colon x \in X\}$ is 
{\color{blue} uniformly $\sigma$-finite} if 
there exists a sequence, $(B_n)$, from $\mathcal{B}(Y)$ such that

\begin{enumerate}
    \item
    $\forall n \in \mathbb{N} \,\, \forall x \in X, \,\, \mu_x(B_n) < \infty$ and
    \item
    $\forall x \in X, \,\, \mu_x(Y \setminus \bigcup_n B_n) = 0$.
\end{enumerate}
\mys


{\color{blue}Problem}({\color{red}Maharam})

Let $\{\mu_x\colon x \in X\}$ be a $\sigma$-finite disintegration
of $\mu$ with respect to $(\nu,\phi)$. Is this disintegration
uniformly $\sigma$-finite?
\end{frame}


\begin{frame}
Every $\sigma$-finite disintegration $\{ \mu_x \colon x \in X\}$ of 
the $\sigma$-finite measure $\mu$ on $Y$ with respect to $(\nu,\varphi)$ 
is ``almost-everywhere'' uniformly $\sigma$-finite in the following sense.
\mys



\begin{fact}
Suppose $\{\mu_x: x \in X\}$ is a $\sigma$-finite disintegration of the
$\sigma$-finite measure $\mu$ with respect to $(\nu,\phi)$. There
exists a uniformly $\sigma$-finite disintegration $\{\hat{\mu}_x \colon  x
\in X\}$ of $\mu$ with respect to $(\nu,\phi)$ such that $\mu_x =
\hat{\mu}_x$ for $\nu$-almost every $x \in X$.
\end{fact}
\mys

{\color{blue}Proof.} Let $B_n \in \sB(Y)$ with $Y=\bigcup_n B_n$ and 
$\mu(B_n)<\infty$ for all $n$. 
\mys


Let $E= \bigcap_n \{ x \in X \colon \mu_x(B_n)<\infty\}$. So,
$\nu(X-E)=0$. 
\mys

Let $B'_n=B_n \cap \varphi^{-1}(E)$, so $\mu(B_n-B'_n)=0$. 
\end{frame}

\begin{frame}
There is $E' \subseteq E$ with $\nu(Y-E')=0$ such that for $x \in E'$, 
$\mu_x(Y-\bigcup_n B_n)=0$. 
\mys

Let 

\[
\hat{\mu}_x= \begin{cases} \mu_x & \text{ if } x \in E' \\
0 & \text{ if } x \notin E' \end{cases}
\]
\mys

So, Maharam's problem concerns the behavior on measure zero sets.
\end{frame}



\section{Special sets}


\begin{frame}
\frametitle{Proof assuming special set}
We give the proof of the theorem assuming the existence of a {\color{blue}special}
$\bp^1_1$ set in the plane. 
We assume henceforth that $X=Y=\omega^\omega$. 
\mys

Let $P$ be a closed subset of $X \times Y$ such that $\forall x \in X$,\,
$P_x$ is nonempty and perfect and if $x \neq x',\, P_x \cap P_{x'} =
\emptyset.$ We say $G$ is a special coanalytic set for $P$ provided
$G \subseteq P$ is a $\mathbf{\Pi}^1_1$
set with the following properties:

\begin{enumerate}
\item
$\forall x \in X \,\, |G_x| = \omega_0$,

\item
$G$ is not the union of countably many $\mathbf{\Pi}^1_1$ graphs over
$X$,
\item
For every $n \in \omega$ and for every $B \in \mathcal{B}(Y),\ \{x \in X:
|B \cap G_x| = n\} \in \mathcal{B}(X)$.
\item
There exists a Borel set $H \subseteq X$ with such that $G \cap (H
\times Y)$ is the union of countably many pairwise disjoint Borel
graphs over $H$.
\end{enumerate}
\end{frame}

\begin{frame}
\frametitle{Proof assuming special set}
\begin{rem}
Properties (1)-(3) are the main properties, (4) is for convenience. 
A set satisfying (1) and (2) can be constructed in $\zfc$.
\end{rem}
\mys

Let $P=\{ (x,y) \colon \ (y)_0=x\}$.
\mys

Assume there is a special set $G\subseteq \ww \times \ww$ for $P$.
\mys

Let $\varphi \colon Y \to X$ be defined by $\varphi(y)=(y)_0$. 
\mys


For $B \in \sB(Y)$ and $x \in X$, define $\mu_x(B)=|B \cap G_x|$,
that is, counting measure on $G_x$. From (1) we have that each 
$\mu_x$ is $\sigma$-finite.
\end{frame}

\begin{frame}
Since $G \subseteq P$ we have that $\mu_x(Y-\varphi^{-1}(\{ x\}))=0$.
\mys

Let $B \in \sB(Y)$. Then 

\[
\{ x \in X \colon \mu_x(B)\geq n\}
=\{ x \in X \colon |B \cap G_x|\geq n\} \in \sB(X)
\]
as $G$ is special. 
\mys


Thus, $x \mapsto \mu_x(B)$ is $\sB(X)$ measurable, so $\{ \mu_x \colon x \in X\}$
is a transition kernel. 
\end{frame}


\begin{frame}
From (4) of special we have a Borel $H \subseteq X$ and Borel functions 
$f_n \colon X \to Y$ such that for $x \in H$ we have $G_x=\bigcup_n \{ f_n(x)\}$.
Each $f_n$ is one-to-one over $H$.
\mys


Let $\nu$ be a Borel probability measure on $X$ with $\nu(H)=1$. 
\mys



Define $\mu$ on $\sB(Y)$ by:

\[
\mu(B)=\int \mu_x(B)\, d\nu(x)
\]
\mys


So, $\{ \mu_x \colon x \in X\}$ is a $\sigma$-finite disintegration of $\mu$
with respect to $(\nu,\varphi)$. 
\end{frame}


\begin{frame}
We show that $\mu$ is $\sigma$-finite. Let $B_n=f_n(H) \subseteq Y$. Each $B_n$
is Borel as $f_n$ is one-to-one on $H$. 
\mys

For $x \in H$, $G_x =\bigcup_n (B_n\cap G_x)$. Since $\nu$ concentrates on $H$ 
it follows that 
\begin{equation*}
\begin{split}
\mu(Y-\bigcup_n B_n)&= \int \mu_x(Y-\bigcup_n B_n) \, d\nu(x) \\
&= \int \mu_x(G_x-\bigcup_n B_n) \, d\nu(x)=0.
\end{split}
\end{equation*}
\mys


So, $\{ \mu_x\colon x \in X\}$ is a $\sigma$-finite disintegration of the 
$\sigma$-finite measure $\mu$ with respect to $(\nu,\varphi)$.
\end{frame}


\begin{frame}
However, this disintegration is not uniformly $\sigma$-finite.
\mys

For suppose $\{ B_n\}$ witnessed $\{ \mu_x\colon x \in X\}$ being uniformly 
$\sigma$-finite. So, for all $x \in X$ we have $\mu_x(B_n)<\infty$ 
and $\mu_x(Y-\bigcup_n B_n)=0$. 
\mys


For each $n$, $G \cap (X \times B_n)$ is $\bp^1_1$ with
finite sections and is thus a countable union of $\bp^1_1$
graphs  implying that $G = \bigcup_n G \cap B_n$ is a
countable union of $\bp^1_1$ graphs, a contradiction.
\mys


This constructs a counterexample assuming the existence of a special 
$\bp^1_1$ set in the plane.
\end{frame}

\section{Construction}

\begin{frame}
\frametitle{Construction of a special set}
Assuming $V=L$ we now construct a special $\bp^1_1$ set in $\ww \times \ww$.
\mys


Let $\varphi_n$ enumerate the formulas in the language of set theory. Let 
$\varphi_{m}$ define the canaonical wellordering of $L$. For $x \in 2^\omega$
let $\thr_x=\{ \varphi_n \colon x(n)=1\}$. 

\mys

The set of {\color{blue}codes} $C$ is defined by
$x \in C$ iff $x \in 2^\omega$, $\thr_x$ is a consistent, complete extension of
$\zf_N +V=L$, and $x(m)=1$. 
\mys


For each $\varphi_n=\exists w\ \varphi(w, x_1,\dots,x_k)$ let $\tau_n(z)$
be the coresponding formula defining the skolem term, that is, 

\[
\tau_n(z)=(\exists w\ \varphi \wedge z \text{ is the } <_L \text{ such } w)
\vee (\neg \exists w\ \varphi_n \wedge z=0)
\]
\end{frame}



\begin{frame}
For $x \in C$, let $M_x$ be the model whose elements are equivalence classes of terms 
$[\tau_n]$ where $\tau_n \equiv \tau_m$ iff $\thr_x \proves 
\exists z\ [\tau_n(z)\wedge \tau_m(z)]$. 
\mys

Let $E_x$ be the natuaral binary relation on $M_x$: $[\tau_n] E_x [\tau_m]$
iff $\thr_x \proves  \exists z\ \exists w\ (\tau_n(z)\wedge \tau_m(w) 
\wedge z \in w)$.
\mys

So $M_x$ is countable and $(M_x,E_x) \models \zf_N+V=L$. 
\mys


Also, $M_x=\hull^{M_x}(\emptyset)$.
\mys


If $(M_x,E_x)$ is wellfounded then there is an $\alpha <\omega_1$
such that $(M_x,E_x) \cong (L_\alpha,\epsilon)$.
\mys

Let $\pi_x$ denote the transitive collapse on $\wf(M_x)$. 
\end{frame}


\begin{frame}
\begin{fact}
We have the following properties of the coding.
\begin{enumerate}
\item
If $x,x' \in C$ and $M_x \cong L_\alpha \cong M_{x'}$, then $x=x'$.
\item
If $L_{\alpha+1} \cap \ww \neq L_\alpha \cap \ww$, then there is an $x \in C$
such that $M_x \cong L_\alpha$. 
\end{enumerate}
\end{fact}
\mys


(1) follows from the fact that $M_x$ determines its theory. (2) 
follows from the fact that such an $\alpha$ projects to $\omega$.
\mys

Define 

\begin{equation*}
\begin{split}
V &= \{ x \in C \colon \omega \in \wf(M_x) \}\\
U&= \{ x \in C \colon (M_x,E_x) \text{ is wellfounded } \}
\end{split}
\end{equation*}
\mys

So, $U \subseteq V \subseteq C$, $V,C \in \bd^1_1$, $U\in \bp^1_1$.
\end{frame}


\begin{frame}
For $x \in \ww$, say $\beta$ is {\color{blue}good} with respect to $x$ 
if $L_\beta(x) \models \zf_N+V=L$.
\mys

Recall $P=\{ (x,y) \in \ww \times \ww \colon (y)_0=x\}$.
\mys

Fix Borel functions $f_n \colon \ww \to \ww$ such that for all $x$
and $n \neq m$ we have $f_n(x)\neq f_m(x)$, and $f_n(x) \in P_x$.
\end{frame}


\begin{frame}
Define the set $G' \subseteq \ww \times \ww$ by $(x,y) \in G' \iff$
\begin{equation*}
\begin{split}
[x &\notin V \wedge \exists n (y = f_n(x))] \vee [x \in V \wedge
  (x,y) \in P \wedge [``y \in M_x \text{"} \vee \\ &\exists \text{ a
      well-founded extension } M \text{ of } M_x \,\, \exists \alpha <
    \omega_1 \\ &\quad (M_x \subseteq M \cong L_{\alpha} \wedge y \in
    L_{\alpha}(x) \wedge \\ &\quad [\forall \gamma < \alpha (\neg
      (\gamma \text{ is good and a limit of good ordinals})
      \vee\\ &\qquad \exists \phi \in \Sigma^1_2 \,\, \exists \tau >
      \gamma \, (L_{\gamma}(x) \models \neg \varphi \wedge L_{\tau}(x) \models
       \varphi))])]].
\end{split}
\end{equation*}
\end{frame}

\begin{frame}

Easily $G'$ is $\boldsymbol{\Sigma}^1_2$ and all sections of $G'$ are countable. 
\mys

Let $G'(x,y) \leftrightarrow \exists z \ G(x,\langle y,z\rangle)
\leftrightarrow \exists ! z\ G(x,\langle y,z\rangle)$.
\mys

We may assume $N$ is large enough so that if $\Omega'$ and $\Omega$ are the 
$\Sigma^1_2$, $\Pi^1_1$ statements defining $G'$, $G$, then the above equivalences
are theorems of $\zf_N$. 
\mys


Clearly the sections of $G$ are countable.
\end{frame}


\begin{frame}
We verify property (3) of $G$. 
\mys


Fix $B\subseteq Y=\ww$. Let $K_n=\{ x \in X\colon |B\cap G_x|\geq n\}$.
Let $b \in \ww$ be a Borel code for $B$. By $V=L$, let $\tau <\omega_1$
be such that $b \in L_\tau$. 
\mys

It suffices to show that $K_n \cap V \in \boldsymbol{\Delta}^1_1$. 
Let 

\[ \psi(x)=\exists \text{ distinct } a_1,\dots, a_n\in M_x\ 
(a_1,\dots, a_n \in B).
\]
\mys

$\psi$ correctly computes $K_n$ on $V-U$ by definition of $G$.
\end{frame}

\begin{frame}


Let $V=D \cup E$ where $D=\{ x \in V \colon b \in M_x\}$ and $E=V-D$.
\mys

If $x \in U\cap D$, then by 
$\boldsymbol{\Sigma}^1_1$ absoluteness we have that $\psi$ still 
correctly computes $K_n$.
\mys


However, $U \cap E$ is countable. So, we have a Borel definition of 
$K_n\cap V$ which works except on a countable set. So, $K_n \cap V$ is Borel.











\end{frame}















\end{document}
