% LaTeX2e
% A. Miller   Oct 2007
\documentclass[12pt]{article}
\usepackage{amssymb}

\def\om{\omega}
\def\proof{\par\noindent Proof\par\noindent}
\def\qed{\par\noindent QED\par\bigskip}
\def\res{\upharpoonright}
\def\rmand{\mbox{ and }}
\def\su{\subseteq}
\def\s1{$S_1(\Gamma,{\mathcal O})$}
\def\sm{\setminus}
\def\uu{{\mathcal U}}
\def\ga{\gamma}

\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}

\begin{document}

\begin{flushright}
A.Miller \\
Oct 9, 2007
\end{flushright}

\begin{theorem}
Suppose $X$ is a Luzin set and $Y$ a Sierpinski set.  Then
$X\times Y$ is \s1.
\end{theorem}

\begin{lemma}
Assume $X\su 2^\om$ is a dense Luzin set and $Y\su 2^\om$ is measure dense
Sierpinski set (i.e. $Y$ meets every positive measure Borel set).  Suppose
$(\uu_n:n<\om)$ are families of open sets in $2^\om\times 2^\om$
which $\ga$-cover $X\times Y$.
Then there exists $(V_n\in\uu_n:n<\om)$ and countable $X_0\su X$ and
$Y_0\su Y$ such that:
$$(X\sm X_0)\times (Y\sm Y_0)\su \bigcup_n V_n.$$
\end{lemma}

\proof
Let $\{x_n\in X:n<\om\}$ be dense in $2^\om$.   For each $n$
let $\uu_n=\{U_{n,m}:m<\om\}$ and define:
$$U_{n,m}^{x_n}=\{y\in 2^\om: (x_n,y)\in U_{n,m}\}.$$
Since $$Y\su \bigcup_N\bigcap_{m>N}U_{n,m}^{x_n}$$ we can choose
$k_n$ so that:
 $$\mu(U_{n,k_n}^{x_n})> 1-{1\over 2^n}.$$
Let $C_n\su U_{n,k_n}^{x_n}$ be compact with 
$\mu(C_n)>1-{1\over 2^n}$. 
Since $Y$ is Sierpinski there exists a countable
$Y_0\su Y$ such that:
$$Y\sm Y_0\su \bigcup_N\bigcap_{n>N}C_n.$$

Choose $f:\om\to\om$ so that for every
$n$:
$$[x_n\res f(n)]\times C_n\su U_{n,k_n} .$$
Since $X$ is Luzin there is a countable $X_0\su X$ such that
$$\forall x\in X\sm X_0\;\;\; \exists^\infty n \;\; x\in [x_n\res f(n)].$$

It follows that 
$$(X\sm X_0)\times (Y\sm Y_0)\su \bigcup_n U_{n,k_n} .$$
\qed

If $X\su 2^\om$ is any Luzin set, then there exists a countable
$X_0\su X$ such that $X\sm X_0$ is homeomorphic to a dense Luzin set.
Given any $Y\su 2^\om$ Sierpinski we may find disjoint closed sets
$C_n$ such that 
$\bigcup_nC_n$ covers $Y$ and for each $n$ either $C_n\cap Y$ is
countable or $C_n$ has positive measure and $Y\cap C_n$ is
relatively measure dense in $C_n$.
Since the countable union of \s1 sets is \s1 it is enough
to prove the Theorem for dense Luzin sets and measure dense Sierpinski 
sets.

Given $\ga$-covers $\uu_n$ of $X\times Y$ first split $\om$ into
infinitely many infinite pieces $(P_n:n<\om)$.  Then apply the Lemma
to find $(V_n\in \uu_n:n\in P_0)$ and countable $X_0,Y_0$ such that:
$$(X\sm X_0)\times (Y\sm Y_0)\su \bigcup_{n\in P_0} V_n.$$
Using the remaining $P_k$ to cover each of the countably many sets:
$$\{x\}\times Y \rmand X\times\{y\}$$
for $x\in X_0$ and $y\in Y_0$.

\qed

\end{document}