\documentclass[12pt]{article}
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\newcommand{\UpdateDate}{July 27, 2003}


\title{Strong Fubini properties for measure and category}

\author{
Krzysztof Ciesielski%
\thanks{2000 {\it Mathematics Subject Classification}: 03E35, 26B20, 28A05 \endgraf
\ \ This project has been funded in part by the National Academy of Sciences
and under the Collaboration in Basic Science and Engineering Program
supported by Contract No. INT-0002341
from the National Science Foundation. 
In addition, the 
first author was also partially supported by 
2002/03 West Virginia University Senate Research Grant
while the second
author was partially supported by the Hungarian National
Foundation for Scientific Research  Grant No. T032042.
}\\
{\footnotesize Department of Mathematics,}
{\footnotesize West Virginia University,} \\
{\footnotesize Morgantown, WV 26506-6310, USA}\\
{\footnotesize e-mail: K\_Cies@math.wvu.edu}; %\\
{\footnotesize web page: {\tt http://www.math.wvu.edu/\~{}kcies}}
\and
Mikl\'os Laczkovich\\
{\footnotesize Department of Analysis,}
{\footnotesize E\"{o}tv\"{o}s Lor\'{a}nd University,} \\
{\footnotesize Budapest, P\'azm\'any P\'eter s\'et\'any 1/C, Hungary 1117}; %\\
{\footnotesize e-mail: laczko@renyi.hu}\\
{\footnotesize and}\\
{\footnotesize Department of Mathematics, University College London}\\
{\footnotesize WC1E 6BT London, England; e-mail: laczk@math.ucl.ac.uk}\\
}
\date{}

\pagestyle{myheadings}
\markboth{Strong Fubini properties for measure and category \ \ \ 
\UpdateDate }{Strong Fubini properties for measure and category \ \ 
\UpdateDate}

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\begin{document}

\maketitle


\begin{abstract}
Let (FP) abbreviate the statement that
$$\int_0^1 \left(\int_0^1 f\, dy\right) dx =
\int_0^1 \left(\int_0^1 f\, dx\right) dy$$
holds for every bounded function $f\colon[0,1]^2 \to {\mathbb R}$
whenever each of the integrals involved exists. We shall denote by (SFP)
the statement that the equality above holds for every
bounded function $f\colon[0,1]^2 \to {\mathbb R}$ having 
measurable vertical and horizontal sections.
It follows from well-known results that both of (FP) and (SFP) are independent
from the axioms of ZFC. We investigate the logical connections of these statements
with several other strong Fubini type properties of the ideal of null sets. In particular,
we establish the equivalence of (SFP) to the
nonexistence of certain sets with paradoxical properties, a phenomenon that
was already known for (FP).
We also give the category analogues of these statements and, whenever
possible, we try to put the statements in a setting of general ideals as
initiated by Rec\l aw and Zakrzewski.
\end{abstract}





\section{Introduction}\label{sec1}

In this paper we 
investigate the equality
\begin{equation}\label{eq1}
\int_0^1 \left(\int_0^1 f\, dy\right) dx =
\int_0^1 \left(\int_0^1 f\, dx\right) dy
\end{equation}
under different conditions on the function $f\colon\nl ^2\to\real$.
It was already known to Cauchy that (\ref{eq1}) holds for any continuous
function $f\colon[0,1]^2\to\real$.
Cauchy new also (see \cite{Cauchy}) that this is false
when we allow just one point of discontinuity for $f$:
if $f(x,y)=(x^2-y^2 )/ (x^2+y^2)^2$ then the two sides of (\ref{eq1})
equal $\pi /4$ and $-\pi /4,$ respectively. (Another simple example
is $f(x,y)=(x-y)/(x+y)^3 ,$ where the corresponding values are $-1/2$ and $1/2.$)
However, it was proved in 1883 by du Bois-Reymond \cite{duBois}
that if $f$ is Riemann integrable on $\nl ^2$ then both of the integrals
$\int_0^1 \left(\overline{\int_0}^1 f\, dy\right) dx$ and
$\int_0^1\left(\overline{\int_0}^1 f\, dx\right) dy$ are equal to the double
integral $\int \! \! \int _{\nl ^2} f(x,y) \, dx\, dy$, where $\oi$ denotes
the Darboux upper integral.
In particular, if $f$ is Riemann integrable on $\nl^2$
and the integrals $\int_0^1 f_x\, dy$ and $\int_0^1 f^y\ dx$ exist
for every $x,y\in[0,1]$ then (\ref{eq1}) holds.
Here $f_x,f^y\colon [0,1]\to\real$ are the sections of $f$;
that is, $f_x(y)=f(x,y)=f^y(x)$ for every
$x,y\in[0,1]$.

\msk
The most important theorem on (\ref{eq1}) is that 
of Fubini \cite{Fubini}\footnote{According 
to \cite[p.~161]{Hawkins} Fubini's proof was defective.
Later different proofs were provided by Hobson~\cite{Ho} and 
de la Vall\'ee-Poussin~\cite{delaV}.}
stating that (\ref{eq1}) holds provided that $f$ is summable on $\nl
\times \nl .$

\msk
The first theorem stating (\ref{eq1}) without assuming the measurability of
$f$ was proved in 1911 by L.~Lichtenstein~\cite{Li},\footnote{Note 
that Sierpi\'nski in~\cite{Si1} 
incorrectly refers this result to another paper of Lichtenstein of 1910. 
We like to thank Dr. J.~Trzeciak for helping us to spot this error. 
}
who showed 
that (\ref{eq1}) holds for every bounded function $f$ for which
$f^y$ is Riemann integrable for every
$y\in[0,1]$, and $f_x$ is measurable for every
$x\in[0,1]$.

\msk
Motivated by Lichtenstein's result Sierpi\'nski \cite{Si1} proved that, under the continuum
hypothesis (CH), there exists a 
function $f\colon[0,1]^2\to\{0,1\}$
for which $\int_0^1 f^y\;dx=0$ and $\int_0^1 f_x\;dy=1$
for every $x,y\in[0,1]$. In particular, the iterated integrals for this function
exist but are not equal, and so (\ref{eq1}) fails.
Actually, Sierpi\'nski proves the following statement under CH:
\begin{itemize}
\item[$\ S^2 :$] there exists a
set $H\su \nl ^2$ such that the horizontal sections of $H$ are of measure zero
and the vertical sections of $H$ are of full measure. That is,
$\la(H^y )=\la(\nl\se H_x)=0$ for every $x,y\in \nl$, where 
$\la$ is the Lebesgue measure, 
$H^y=\{x\colon\lan x,y\ra\in H\}$
and
$H_x=\{y\colon\lan x,y\ra\in H\}$.
\end{itemize}
Then Sierpi\'nski notes that the characteristic function of $H$ has the
properties described above. This example shows that in the theorems of
Fubini and Lichtenstein we cannot omit the conditions that
$f$ is summable or that
the sections $f^y$ are Riemann integrable.

\msk
Can Sierpi\'nski's example be constructed in ZFC?
The negative answer was given in
1980 by Friedman~\cite{Friedman}, who 
proved that there are models of set
theory ZFC in which the following statement,
which we call {\em Fubini property}, holds.
\begin{itemize}
\item[{~(FP)}]
If $f\colon[0,1]^2\to\real$ is a bounded function such that
$f_x$ and $f^y$ are measurable for every $x,y\in[0,1]$
and the mappings
$[0,1]\ni x\mapsto \int_0^1 f(x,y)\;dy$  and $[0,1]\ni y\mapsto \int_0^1
f(x,y)\;dx$
are measurable then (\ref{eq1}) holds.
\end{itemize}
Later it was shown independently by Laczkovich~\cite{LaPrep}
and Freiling~\cite{Fr} that (FP) is, in fact, {\it equivalent}\/ to the
negation of the statement $S^2 .$ Now it is easy to see that $S^2$
implies the inequality $\cov(\nnn)\le\non(\nnn)$.
Here $\cov(\nnn)$ denotes the minimum cardinality of a system of null sets that can
cover $\real$ and $\non(\nnn)$ is the minimum
cardinality of a set of positive outer measure. 
It was shown by Kunen \cite{Kunen} that the inequality
$\cov(\nnn)>\non(\nnn)$ is consistent with ZFC. Since this
inequality implies $\lnot S^2$ which, in turn, is equivalent to (FP), we find
that the consistency of (FP) also follows from Kunen's theorem.

\msk
We may also ask whether or not the following stronger version of (FP),
which we call {\em strong Fubini property},
is consistent with ZFC.
\begin{itemize}
\item[{(SFP)}]
If $f\colon[0,1]^2\to\real$ is a bounded function such that
$f_x$ and $f^y$ are measurable for every $x,y\in[0,1]$
then the mappings
$[0,1]\ni x\mapsto \int_0^1 f(x,y)\, dy$  and $[0,1]\ni y\mapsto \int_0^1
f(x,y)\, dx$ are measurable and (\ref{eq1}) holds.
\end{itemize}

Although we did not find in the literature (SFP) in its specific,
it follows easily from the following {\em strong approximation property}\/
(for measure).
\begin{itemize}
\item[{(SAP)}]
If $f\colon[0,1]^2\to\real$ is such that
$f_x$ and $f^y$ are measurable for every $x,y\in[0,1]$
then there is a measurable function $F\colon[0,1]^2\to\real$
such that $f_x=F_x$ a.e. and $f^y=F^y$ a.e. for every $x,y\in[0,1]$.
\end{itemize}
In \cite[p. 139]{RZ}, the theorem stating the consistency of (SAP) is
attributed to H. Woodin. The first published proof can be found in
\cite[Theorem~2.17(ii)]{RZ}. Therefore (SAP) and, consequently, (SFP)
are both consistent with ZFC.

\msk
As we shall see later, the statements (SAP) and (SFP) are, in fact,
equivalent.
One of the goals of this paper will be to
find several other statements equivalent to (SFP), including one that
states the nonexistence of certain sets with paradox properties, 
analogously to the equivalence (FP)$\iff \lnot S^2 .$
Since all these statements can be expressed in a setting of general ideals on Polish
spaces, we will do this in the next section.













\section{Kuratowski-Ulam type theorems for ideals}

We use standard set theoretic terminology as in \cite{CiBook}.

Let $X$ be a Polish space and let $\ii$ be an ideal in $X$.
We shall denote by $\B_\ii$ the $\si$-algebra generated by $\ii$ and by the
Borel subsets $\B$ of $X.$ 
The ideals that are the most interesting for us are
the $\sigma$-ideals $\N$ of Lebesgue measure zero subsets of $\real$ and
$\M$ of meager subsets of $\real$. 
Then $\B_\N$ and $\B_\M$ are the families of measurable sets, and 
the sets with the Baire property, respectively. 
The sections of a set $A\su X^2$ are defined by
$$A_x =\{y\in X\colon \lan x,y\ra\in A\} \ \  {\rm and} \ \
A^y =\{ x\in X\colon \lan x,y\ra\in A\}$$
for every $x,y\in X$.

Recall that the Kuratowski-Ulam theorem says that 
for every $A\subset\real^2$ with the Baire property,
if $A_x\in\M$  for every $x\in\real$ then $\{y\colon A^y\notin\M\}\in\M$.
Thus the following {\it Kuratowski-Ulam property}\/ of an arbitrary
ideal $\I$ is a stronger version of the theorem, when 
considered with $\I=\M$. 
\begin{itemize}
\item[(KU)] If $A\su X^2$ is such that $A_x\in\ii$ and $A^y \in\B_\ii$
for every $x,y\in X$ and $\{y\colon A^y \notin\ii\}\in\B_\ii$,
then $\{y\colon A^y\notin\ii\}\in\ii$.
\end{itemize}
We will also consider the following two 
{\it strong Kuratowski-Ulam properties}. 

\begin{itemize}
\item[(SKU)] If $A\su X^2$ is such that $A_x\in\ii$ and $A^y \in\B_\ii$
for every $x,y\in X$,
then $\{y\colon A^y\notin \ii\}\in\ii$.

\item[(SKU$^*$)] If $A\su X^2$ is such that $A_x\in\B_\ii$ and $A^y \in\B_\ii$
for every $x,y\in X$,
then $\{y\colon A^y\notin \ii\}\in\B_\ii$.
\end{itemize}
Note that Rec\l aw and Zakrzewski in~\cite{RZ}
refer to (SKU) as ``strong Fubini property'' and denote 
it as (SFP). However, we prefer to reserve the adjective ``Fubini'' to the properties
that refer explicitly to the properties of integrals. 

Our first aim is to establish the logical connections between these three 
statements and the nonexistence of sets having some paradoxical
properties with respect to the ideal $\ii$.
We shall say that a set $H\su A\times B$ is a {\it 0--1 set (in $A\times B$)}\/ provided 
$H^y\in\ii$ for every $y\in B$ and $B\se H_x\in\ii$
for every $x\in A$.
Note that 
\begin{equation}\label{eq3}
\mbox{if there is a 0--1 set in $A\times B$ then there is one also in $B\times A$.}
\end{equation}
Indeed, if
$H$ is a 0--1 set in $A\times B$ then 
$\{\lan x,y\ra\in B\times A\colon \lan y,x\ra\notin H\}$ 
is a 0--1 set in $B\times A$.

We shall use the notation $A^c =X\se A$ for every $A\su X$. 
We will also use the following propositions concerning 
the existence of various 0--1 sets. 

\begin{itemize}
\item[$S^2_\ii$:] There exists a 0--1 set in $X\times X.$

\item[$S^{2w}_\ii$:] There exists a set $A\subset X$ not in $\ii$ 
such that $A\times X$ contains a 0--1 set.

\item[$E_\ii$:] There exist sets $A,B\su X$  not in $\ii$ 
such that both $A\times B$ and $A^c \times B^c$ contain 0--1 sets. 
\end{itemize}

To state the next theorem, we need also the following classical 
cardinal numbers connected to $\I$.
\begin{align*}
\add(\I)& =\min\{|\F|\colon \F\subset\I\ \&\ \bigcup\F\notin\I\}\\
\cov(\I)& =\min\{|\F|\colon \F\subset\I\ \&\ \bigcup\F=X\}\\
\non(\I)& =\min\{|A|\colon A\subset X\ \&\ A\notin\I\}\\
\shr(\I)& =\min\{\kappa\colon \forall A\subset X (A\notin\I\Implies
[A]^\kappa\setminus\I\neq\emptyset)\}.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\end{align*}
In other words, $\shr(\I)$ is the smallest cardinal
$\kappa$ such that any subset $A$ of $X$ not belonging to $\I$ 
contains a subset $B$ such that $|B|\le \kappa$ and $B\notin\I$.
Clearly $\non(\I)\leq\shr(\I)$. 


\begin{theorem}\label{t1}
Let $\ii$ be an ideal in an uncountable Polish 
space $X$.  
Then we have the following implications:

\begin{picture}(300,100)
\thicklines

\put(179,80){\makebox(0,0){{\rm (SKU)}}}
\put(179,68){\vector(0,-1){40}}

\put(198,80){\vector(1,0){20}}
\put(233,80){\makebox(0,0){{\rm (KU)}}}
\put(233,68){\vector(0,-1){40}}


\put(30,15){\makebox(0,0){$[\shri<\covi]$}}

\put(78,15){\vector(1,0){20}}
\put(120,15){\makebox(0,0){$\lnot E_\ii$}}

\put(140,15){\vector(1,0){20}}
\put(179,15){\makebox(0,0){$\lnot S^{2w}_\ii$}}

\put(198,15){\vector(1,0){20}}
\put(233,15){\makebox(0,0){$\lnot S^2_\ii$}}

\put(247,15){\vector(1,0){20}}
\put(312,23){\makebox(0,0){$[\noni<2^\om\ \&$}}
\put(322,7){\makebox(0,0){$\ \addi <\covi]$}}
\end{picture}

\noindent Moreover, if $\ii$ satisfies the following condition:
\begin{itemize}
\item[{\rm ($*$)}] every set $B\in \B _\ii \setminus \ii$ contains a subset $S\notin\B_\ii$
\end{itemize}
then the chart can be expanded as

\begin{picture}(300,100)
\thicklines

\put(120,80){\makebox(0,0){{\rm (SKU$^*$)}}}
\put(120,68){\vector(0,-1){40}}

\put(140,80){\vector(1,0){20}}
\put(179,80){\makebox(0,0){{\rm (SKU)}}}
\put(179,68){\vector(0,-1){40}}

\put(198,80){\vector(1,0){20}}
\put(233,80){\makebox(0,0){{\rm (KU)}}}
\put(233,68){\vector(0,-1){40}}


\put(30,15){\makebox(0,0){$[\shri<\covi]$}}

\put(78,15){\vector(1,0){20}}
\put(120,15){\makebox(0,0){$\lnot E_\ii$}}

\put(140,15){\vector(1,0){20}}
\put(179,15){\makebox(0,0){$\lnot S^{2w}_\ii$}}

\put(198,15){\vector(1,0){20}}
\put(233,15){\makebox(0,0){$\lnot S^2_\ii$}}

\put(247,15){\vector(1,0){20}}
\put(312,23){\makebox(0,0){$[\noni<2^\om\ \&$}}
\put(322,7){\makebox(0,0){$\ \addi <\covi]$}}
\end{picture}






\end{theorem}

\pf (SKU$^*$) $\akkor$ (SKU):
The argument for this implication is closely related to that for~\cite[Theorem 2.11]{RZ}.
Let $A\su X^2$ and suppose that $A_x \in \ii$ and $A^y \in\B_\ii$ for every $x,y\in X$.
By (SKU$^*$) we have $C:=
\{y\colon A^y \notin\ii\}\in\B_\ii$. 
If $C\notin \ii$ then, by ($*$), there is a subset $D\su C$ such that $D\notin\B_\ii$. 
Put
$$
H=A\cap (X\times D).
$$
Then $H^y\in\B_\ii$ for every $y\in X$, since $y\in D$ implies
$H^y=A^y$ and $y\notin D$ implies $H^y =\emp$. 
Since $H\su A$, we also have $H_x \in\ii$ for every $x\in X$. 
Applying (SKU$^*$) again we find that
$$
\{y\colon H^y \notin\ii\}=D\in\B_\ii,
$$
which is impossible. 


\msk \noi
(SKU$^*$) $\akkor \lnot E_\ii$: Suppose that $E_\ii$ holds. 
Then there are sets $A,B \su X$, $H\su A\times B$, and $K\su A^c \times B^c$ such that 
$A\notin\ii$, $B\notin\ii$, $H$ is a 0--1 set in $A\times B$, and $K$ is a 0--1 set in
$A^c\times B^c$. 

First assume that $A,B\in\B_\ii$.
Then $H_x,H^y\in\B_\ii$ for every $x,y\in X$. 
By ($*$), there is a set $D\su B$ such that $D\notin\B_\ii$.
Putting $E=H\cup (A\times D)$ we have $E_x,E^y\in\B_\ii$ for every $x,y\in X$,
and $\{y\colon E^y\notin\ii\}=D\notin\B_\ii$. 
This, however, contradicts (SKU$^*$).

Therefore we may assume that at least one of $A$ and $B$ does not belong to $\B_\ii$. 
By symmetry we may suppose $B\notin\B_\ii$.
Put
$$
M=H\cup((A^c\times B^c)\se K)\cup (A\times B^c)
$$
and notice that $M$ refutes (SKU$^*$). 
Indeed, to see that the sections of $M$ belong to $\B_\ii$ note that
if $y\in B$ then $M^y=H^y\in\I$ and for $y\in B^c$ we have $X\se M^y=K^y\in \I$.
Similarly, if $x\in A$ then $X\se M_x=B\se H_x\in \I$
and for $x\in A^c$ we have $M_x=A^c\se K_x\in \I$. 
However, $\{y\colon M^y\notin\ii\}=B\notin\B_\ii$ contradicts (SKU$^*$).



\msk \noi
(SKU) $\akkor$ (KU): This implication is immediate from the definitions.

\msk

All the remaining implications will be proved  by contraposition. 


\msk \noi
(SKU) $\akkor \lnot S^{2w}_\ii$: Suppose that $S^{2w}_\ii$ holds. 
Then, by (\ref{eq3}), there are sets $A\su X$ and $H\su X\times A$ such that
$A\notin\ii$ and $H$ is a 0--1 set in $X\times A$.
Let $K=(X\times A) \se H$. 
Then $K_x\in\ii$ and $K^y \in\B_\ii$ for every $x,y\in X$
while $\{y\colon K^y \notin\ii\}=A\notin\ii$, which contradicts (SKU).


\msk \noi
(KU) $\akkor \lnot S^2_\ii$: If $S^2_\ii$ holds then there is a 0--1 set
in $X^2$ which clearly contradicts~(KU).


\msk \noi
$\lnot S^2_\ii \akkor [\noni<2^\om\ \&\ \addi<\covi]$: 
A proof that there is a 0--1 set if either 
$\noni=2^\om$ or $\addi = \covi$ can be found, for example, in~\cite[Theorem 2]{La}.
(Note that the proof given in~\cite{La} is valid for every ideal.)

\msk \noi
$\lnot S^{2w}_\ii \akkor \lnot S^w_\ii$: 
The implication $S^2_\ii \akkor S^{2w}_\ii$ is obvious. 


\msk \noi $\lnot E_\ii\akkor\lnot S^{2w}_\ii$: 
Assume that $S^{2w}_\ii$ holds. 
Then there are sets $A\su X$ and $H\su A\times X$ such that
$A\notin\ii$ and $H$ is a 0--1 set in $A\times X$.
If $A^c \in \ii$ then $E_\ii$ is satisfied with $B=A$. 
If $A^c \notin \ii$ then $E_\ii$ is satisfied with $B=A^c$,
since $H\cap(A\times A^c)$ is a 0--1 set in $A\times A^c$,
and the existence of a 0--1 set in $A^c\times A$
follows from (\ref{eq3}). 
 
\msk \noi $[\shri<\covi]\akkor\lnot E_\ii$: 
For the proof of $E_\ii \akkor [\shri \ge \covi ]$ we shall need
the following lemma, which is implicit 
in~\cite[p. 193]{Fr} and~\cite[Lemma 2.7]{RZ}. 
For the sake of completeness we give the simple proof.
In the lemma we use the notation $\I|A$ to denote the ideal
$\{B\in\I\colon B\subset A\}$. 

\begin{lemma}\label{l4}
If $A,B\su X$, $A,B\notin\ii$, and there is a 0--1 set in $A\times B$ then
$\cov(\ii|A) \le \non(\ii |B).$
\end{lemma}

\pf  Let $H$ be a 0--1 set in $A\times B$. 
Then $H^y\in\ii$ for every $y\in B$ and $B\se H_x \in \ii$ for every $x\in A$.
If $D\su B$, $|D|=\non(\ii |B)$, and $D\notin\ii$ then $A=\bigcup_{y\in D} H^y$, 
since $x\in A\setminus\bigcup_{y\in D}H^y$ would imply
$H_x \cap D =\emp$ contradicting $B\se H_x\in\ii$.
Thus $\cov(\ii|A)\le|D|=\non(\ii |B)$. \qed

Now we turn to the proof of $E_\ii\akkor[\shri\ge\covi]$.
By $E_\ii$, there are sets $A,B\notin\ii$ such that 
both of $A\times B$ and $A^c\times B^c$ contain 0--1 sets. 
If $A^c\notin\ii$ and $B^c\notin\ii$ then Lemma~\ref{l4} gives
$\cov(\ii|A)\le\non(\ii|B)\le\shri$ and
$\cov(\ii|A^c)\le\non(\ii|B^c)\le\shri$, and hence
$$
\covi\le\max\{\cov(\ii|A),\cov(\ii|A^c)\}\le\shri.
$$
On the other hand, if $A^c\in\ii$ then there is a 0--1 set in $X\times B$
and thus $\covi\le\non(\ii|B)\le\shri$.
We have the same conclusion if $B^c\in\ii$. \qed 

For a semigroup $G$ of Borel functions from $X$ to $X$
and an ideal $\I$ on $X$ we say that $\I$ is
{\em $G$-invariant}\/ provided $g^{-1}(A)\in\I$ for every $g\in G$ and $A\in \I$;
and $\I$ is {\em $G$-ergodic}\/ when $X\setminus\bigcup_{g\in G} g^{-1}(A)\in\I$
for every $A\in\B_\I\setminus\I$. 
Clearly if $G$ is the group of rational translations in $\real$ then the 
ideals $\N$ and $\M$ are $G$-invariant and $G$-ergodic.

If we strengthen a bit the assumptions of the main 
part of Theorem~\ref{t1}
then we can get few more implications. 

\begin{corollary}\label{cor4}
Let $\ii$ be a $\sigma$-ideal
in the Polish space $X$ and suppose that 
there is a countable semigroup $G$ of Borel functions from $X$ to $X$
such that $\ii$ is $G$-invariant and $G$-ergodic. 
Then we have the following implications:

\begin{picture}(300,100)
\thicklines

\put(179,80){\makebox(0,0){{\rm (SKU)}}}
\put(179,68){\vector(0,-1){40}}
\put(179,28){\vector(0,1){40}}

\put(198,80){\vector(1,0){20}}
\put(233,80){\makebox(0,0){{\rm (KU)}}}
\put(233,68){\vector(0,-1){40}}
\put(233,28){\vector(0,1){40}}


\put(30,15){\makebox(0,0){$[\shri<\covi]$}}

\put(78,15){\vector(1,0){20}}
\put(120,15){\makebox(0,0){$\lnot E_\ii$}}

\put(140,15){\vector(1,0){20}}
\put(179,15){\makebox(0,0){$\lnot S^{2w}_\ii$}}

\put(198,15){\vector(1,0){20}}
\put(233,15){\makebox(0,0){$\lnot S^2_\ii$}}

\put(247,15){\vector(1,0){20}}
\put(312,23){\makebox(0,0){$[\noni<2^\om\ \&$}}
\put(322,7){\makebox(0,0){$\ \addi <\covi]$}}
\end{picture}
\end{corollary}

\pf The equivalence (SKU) $\iff \lnot S^{2w}_\ii$ is proved in~\cite[Lemma 2.8]{RZ}. 
The same argument gives $\lnot S^{2}_\ii \iff$ (KU).
The other implications follow from Theorem~\ref{t1}. \qed





We do not know if the hypothesis above also implies (SKU$^*$) $\iff\lnot E_\ii$
even in the presence of ($*$). 
However, as we shall see later, this equivalence holds for $\ii =\nnn$ and $\ii=\mmm$.

The condition $(*)$ was only used in the proof of Theorem~\ref{t1} in the implications
(SKU$^*$) $\akkor \lnot E_\ii$ and (SKU$^*$) $\akkor$ (SKU). 
The following example shows that it is really needed in these arguments. 

\begin{example}
Let $X=\real$ and let $\ii$ be a prime ideal containing all sets
of cardinality less than $2^\om$. 
For this ideal {\rm (SKU$^*$)} holds while all other statements of the diagram are false.
\end{example}

\pf Since $\ii$ is prime, we have $\B_\ii={\cal P}(\real)$. Thus (SKU$^*$) holds. 
On the other hand, clearly $\noni=2^\om$, so 
the statements of the diagram are false. \qed


We shall also investigate the following {\em Borel approximation property}:
\begin{itemize}
\item[{(BAP)}] 
if $A\su X^2$ is such that $A_x\in\B_\ii$ and $A^y\in\B_\ii$
for every $x,y\in X,$ then there is a Borel set $B\su X^2$ such that
$(A\De B)_x \in \ii$ for $\ii $-a.e. $x\in X$ and
$(A\De B)^y \in \ii$ for $\ii $-a.e. $y\in X.$
\end{itemize}
We will leave the following simple fact without a proof. 

\begin{fact}\label{fact6}
If $\I$ is a $\sigma$-ideal on $X$ then 
{\rm (BAP)} is equivalent to the following statement.
\begin{itemize}
\item
If $f\colon X^2\to\real$ is such that
$f_x$ and $f^y$ are $\B_\I$-measurable for every $x,y\in X$  
then there is a Borel function $g\colon X^2\to\real$
such that $f_x=g_x$ $\I$-a.e. and $f^y=g^y$ $\I$-a.e. for $\I$-almost every $x,y\in X$.
\end{itemize}
Also, {\rm (BAP)} is equivalent to {\rm (SAP)} for $\I=\N$ and for $\I=\M$. 
\end{fact}

We state the equivalence to (SAP) only for the ideals $\N$ and $\M$
to avoid a problem of defining the meaning of $\B_\I$-measurability  
of a function from $X^2$ into $\real$. 
(A generalization of (SAP) to the category case is natural.)

\msk 
Notice that it would not be reasonable to require in (BAP) the
existence of a Borel set $B$ such that
$(A\De B)_x \in \ii$ and $ (A\De B)^y \in \ii$
for {\it every} $x$ and $y.$ In fact, this stronger form of
(BAP) is satisfied neither by $\nnn$ nor by $\mmm $ as the following
simple example shows.

Let $H$ be a non-Borel subset of the Cantor set, and put $A=H\times\nl$.
Then $A_x\in\B_\ii$ and $A^y\in\B_\ii$ hold
for every $x,y\in\nl$ for both $\ii=\nnn$ and $\I=\mmm$. 
Suppose there is a Borel set $B\su\nl^2$ such that $(A\De B)_x \in \nnn$ for every $x\in \nl$.
Then $\{x\in\nl\colon B_x\notin\nnn\} =H$ 
contradicting the fact that this set must be Borel
whenever $B$ is Borel. (See e.g. \cite[(16.1) Theorem, p.~94]{Ke}.)
The same argument works for $\I=\mmm$. 

In what follows we will prove that (SAP) is equivalent to (SKU$^*$)
for $\ii=\nnn$ and $\ii=\mmm$. 
We start in this direction noticing the following simple fact,
which will be left without a proof. 


\begin{proposition}\label{p1}
Suppose that $\{y\colon A^y \notin \ii \} \in\B_\ii$ for every Borel
set $A\su X^2 .$ Then {\rm (BAP) $\akkor$ (SKU$^*$).}
\end{proposition}

Note that the assumption of 
Proposition~\ref{p1} is not satisfied by every ideal. For example,
if $\ii$ equals the ideal of countable sets or that of the
first category null sets than there is a Borel set 
$A\su \real^2$
such that $\{y\colon A^y \notin\ii\}\notin\B_\ii$.

We do not know whether or not (BAP) implies (SKU$^*$)
for every ideal. 

\section{Strong Fubini properties for measure}

It was proved by C. Freiling in \cite{Fr} (see also Laczkovich~\cite{LaPrep})
that (FP) is equivalent to $\lnot S^2_\nnn$. 
Our aim here is to give a similar characterization of (SFP),
by showing that it is equivalent to $\lnot E_\nnn$ 
as well as to several other strong Fubini type properties, including (BAP) for $\N$. 
Recall that $\B_\nnn$ coincides with the $\si $-algebra of measurable
subsets of $\rr$.

Notice also that the chart implies that all the properties considered
there are independent of the axioms of set theory ZFC. 
This is the case, as since the relations $\add(\N)=\cov(\N)=\non(\N)=2^\omega$
and $\shr(\N)<\cov(\N)$ are both consistent with ZFC:
the first is a consequence of the Martin's Axiom, while the second
holds in the random reals model. 
(For the proof of $\shr(\N)<\cov(\N)$ see \cite{LM} or \cite{KY}.
This was certainly already known to Kunen~\cite{Kunen}.)

\newpage

\begin{theorem}\label{t2}
The following statements are equivalent to each other.
\begin{itemize}
\item[{\rm (i)}] If $f$ is a bounded
function on $\nl ^2$ such that the sections $f_x$ and $f^y$
are measurable for every $x,y\in \nl$ then the mapping
$x\mapsto \int_0^1 f(x,y)\, dy$ is measurable on $\nl$.
\item[{\rm (ii)}] The strong Fubini property {\rm (SFP)} holds.
\item[{\rm (iii)}] $\nnn$ has property {\rm (SAP).}
\item[{\rm (iv)}] $\nnn$ has property {\rm (BAP).}
\item[{\rm (v)}] $\nnn$ has property {\rm (SKU$^*$).}
\item[{\rm (vi)}] $E_\nnn$ is false.
\end{itemize}
In particular, for $\I=\N$ we have the following relations. 

\begin{picture}(300,170)
\thicklines

\put(5,140){\makebox(0,0){{\rm (BAP)}}}
\put(24,140){\vector(1,0){20}}
\put(44,140){\vector(-1,0){20}}

\put(62,140){\makebox(0,0){{\rm (SAP)}}}
\put(81,140){\vector(1,0){20}}
\put(101,140){\vector(-1,0){20}}

\put(120,140){\makebox(0,0){{\rm (SFP)}}}
\put(120,132){\vector(0,-1){40}}
\put(120,92){\vector(0,1){40}}

\put(120,80){\makebox(0,0){{\rm (SKU$^*$)}}}
\put(120,68){\vector(0,-1){40}}
\put(120,28){\vector(0,1){40}}

\put(140,80){\vector(1,0){20}}
\put(179,80){\makebox(0,0){{\rm (SKU)}}}
\put(179,68){\vector(0,-1){40}}
\put(179,28){\vector(0,1){40}}

\put(198,80){\vector(1,0){20}}
\put(233,80){\makebox(0,0){{\rm (KU)}}}
\put(233,68){\vector(0,-1){40}}
\put(233,28){\vector(0,1){40}}

\put(233,140){\makebox(0,0){{\rm (FP)}}}
\put(233,132){\vector(0,-1){40}}
\put(233,92){\vector(0,1){40}}

\put(30,15){\makebox(0,0){$[\shr(\N)<\cov(\N)]$}}

\put(78,15){\vector(1,0){20}}
\put(120,15){\makebox(0,0){$\lnot E_\N$}}

\put(140,15){\vector(1,0){20}}
\put(179,15){\makebox(0,0){$\lnot S^{2w}_\N$}}

\put(198,15){\vector(1,0){20}}
\put(233,15){\makebox(0,0){$\lnot S^2_\N$}}

\put(247,15){\vector(1,0){20}}
\put(312,23){\makebox(0,0){$[\non(\N)<2^\om\ \&$}}
\put(322,7){\makebox(0,0){$\ \add(\N) <\cov(\N)]$}}
\end{picture}

\end{theorem}

\pf The implications in the chart follow from the main part of the theorem,
from Freiling's result quoted above, as well as 
from Theorem~\ref{t1} and Corollary~\ref{cor4}.

In the course of the proof the sign $\int$ will
abbreviate $\int_0^1$ unless another domain is indicated.

\msk\noi
(i)$\akkor$(ii): Let $f$ be a bounded
function on $\nl ^2$ such that the sections $f_x$ and $f^y$ are measurable for every $x,y\in\nl$. 
By (i), for every $x,y\in\nl$ the mappings $x\mapsto \int f(x,y)\, dy$ and
$y\mapsto \int f(x,y)\, dx$ are measurable. 
We need to prove that the integrals $I_1 =\int(\int f(x,y)\, dy )\, dx$ and
$I_2=\int (\int f(x,y) \, dx )\, dy$ are equal.

By way of contradiction assume that $I_1 \ne I_2$. Then, by
Freiling's theorem \cite{Fr}, there exists a 0--1 set $A\su\nl^2$.
After removing from $A$ some vertical
sections we obtain a set $B$ such that $B^y\in\nnn$
for every $y,$ and $\{x\in[0,1]\colon [0,1]\setminus B_x\in\nnn\}$ 
is not measurable. But then the characteristic
function of $B$ refutes (i).

\msk \noi
(ii)$\akkor$(iii): Suppose (SFP).
Let $f\colon \nl^2\to\real$ be a function with measurable sections. 
We have to show that there is a measurable function
$F\colon \nl ^2 \to\real$ such that $F_x =f_x$ a.e. for every $x\in \nl$ and
$F^y =f^y$ a.e. for every $y\in\nl$. 
In the proof we may assume that $f$ is bounded. 
For $x,y\in \nl$ put $G(x,y)=\int_0^y f(x,t)\, dt$.
Then $G$ is well defined and measurable. 
Indeed, by (SFP), $G^y$ is measurable for every $y\in\nl$. 
Also, $G_x$ is continuous for every $x\in\nl$.
Thus, by Lebesgue's theorem \cite{Le}, $G$ is measurable. 
Let $H=\frac{\partial}{\partial y}G$ where $\frac{\partial}{\partial y} G$ exists, and $H=0$ elsewhere.
Then $H$ is measurable, and for every $x\in\nl$ we have 
$H_x(y)=f_x(y)$ for almost all $y\in\nl$. Then, by (SFP) and the Fubini theorem,
\begin{equation*}
\begin{split}
\int_0^b \left( \int_0^a f^y (x)\, dx \right) \, dy & =
\int_0^a \left( \int_0^b f_x (y)\, dy \right) \, dx \\
&= \int_0^a \left( \int_0^b H_x (y)\, dy \right) \, dx \\
&= \int_0^b \left( \int_0^a H^y (x)\, dx \right) \, dy
\end{split}
\end{equation*}
for every $a,\, b\in \nl$. Thus, for every $a\in\nl$ and almost
every $y\in \nl$ we have $\int_0^a f^y (x) \, dx=\int_0^a H^y (x) \, dx.$
So, for almost every $y\in \nl$ we have
$\int_0^a f^y (x) \, dx=\int_0^a H^y (x) \, dx$ for every rational $a\in \nl$. 
Thus, by the continuity of the integral, for almost every $y\in \nl$ we have
$\int_0^a f^y (x) \, dx=\int_0^a H^y (x) \, dx$ for every $a\in \nl$. 
Then, for almost every $y\in \nl$, that is,
for $y'$s outside some null set $T\su \nl$, we have
$H^y (x)=f^y (x)$ for almost every $x\in \nl$.
To get $F$ as desired define it as equal to $f$ on $\nl \times T$ and equal to $H$ otherwise.


\msk \noi
(iii)$\akkor$(iv): This is obvious, and follows from Fact~\ref{fact6}.


\msk \noi
(iv)$\akkor$(v): If $A\su \nl ^2$ is Borel then the set
$\{y\in\nl\colon A^y\notin\nnn\}$ is measurable by Fubini's theorem.
Thus we may apply Proposition~\ref{p1}.

\msk \noi
(v)$\akkor$(vi): This was proved in Theorem~\ref{t1}.


\msk \noi
(vi)$\akkor$(i): Suppose that (i) is false. 
Then there exists a bounded function $f\colon \nl ^2 \to \real$
such that the sections of $f$ are measurable, but the function
$h(x)=\int f_x \, dy$, defined for $x\in \nl$, is not. 
Fix a real number $b$ such that
the set $H=\{x\in\nl\colon h(x) <b\}$ is not measurable, 
and define \linebreak $K=\{x\in \nl\colon h(x)\ge b\}$. 
Then $\la(H)+\la(K)>1$, where $\la$ stands for Lebesgue outer measure. 
Thus, there exists a real number $a<b$ such that $\la(H_a)+\la(K)>1$,
where $H_a =\{x\in\nl\colon h(x) \le a\}$.

Select measurable sets $M$ and $N$ such that 
$H_a\su M$, $\la(H_a )=\la (M)$, $K\su N$, and $\la (K)=\la (N)$. 
Then $\la(M)+\la(N) =\la(H_a )+\la (K)>1$, and thus $P=M\cap N$ is of positive measure. 
Clearly, $\la(P)=\la(H_a\cap P)= \la(K\cap P)$.
The measure space $\lan P,\frac{1}{\la(P)} (\la \restriction P)\ra$
is isomorphic to $\lan \nl ,\la \ra$. 
Let $\phi$ be an isomorphism from $\nl$
onto $P$, and put $f_1(x,y)=f(\phi(x),y)$ for $x,y\in\nl$.
Then the sections of $f_1$ are measurable and both of the sets
$U=\{x\in \nl\colon \int f_1 (x,y) \, dy \le a\}$ and
$V=\{x\in \nl\colon \int f_1 (x,y) \, dy \ge b\}$ have outer measure~1.

We shall denote by $\Om$ the product space $\nl^\om$
equipped with the product measure $\mu$ and define 
$\ol\mu(H)=\inf\{\mu(M)\colon H\su M \mbox{ and $M$ is $\mu$-measurable}\}$ 
for every $H\su\Om$.
For $x\in\nl$ and $v=\lan v_1,v_2,\ldots\ra\in \Om$ let
$$
F(x,v)=\limsup_{n\to\infty}\frac{f_1(x,v_1)+\cdots+f_1(x,v_n)}{n}.
$$
By the strong law of large numbers for every $x\in[0,1]$ the equation
$F(x,v)=\int f_1(x,y) \, dy$ holds for $\mu$-a.e. $v\in \Om .$
Thus, $F_x$ is $\mu$-a.e. constant for every $x\in[0,1]$ since
$F_x(v)=\int f_1(x,y) \, dy=F_x(v')$ holds for $\mu$-a.e. $v,v'\in\Om$. 
Moreover, for $x\in U$ we have $F(x,v)\le a$ for $\mu$-a.e. $v\in \Om$,
and for $x\in V$ we have $F(x,v)\ge b$ for $\mu$-a.e. $v\in \Om$.
Also, $F^v(x)$ is measurable for every $v\in \Om$,
being  a $\limsup$ of measurable functions 
$\frac{1}{n}[(f_1)^{v_1}(x)+\cdots+(f_1)^{v_n}(x)]$.
Now put
$$
G(u,v)=\limsup_{n\to\infty} \frac{F(u_1,v)+\cdots+F(u_n,v)}{n}
$$
for every $u,v\in \Om$. 
Then, by the strong law of large numbers, for every $v\in \Om$ we have 
$G(u,v)=\int F^v dx$ for $\mu$-a.e. $u\in \Om$.
In particular, $G^v$ is $\mu$-a.e. constant for every $v\in\Om$ since
$G^v(u)=\int F^v dx=G^v(u')$ holds for $\mu$-a.e. $u,u'\in\Om$. 
Also, $G_u(v)$ is $\mu$-a.e. constant for every $u\in\Om$ since
it is a $\limsup$ of $\mu$-a.e. constant functions 
$\frac{1}{n}[F_{u_1}(v)+\cdots+F_{u_n}(v)]$.

Put $E=\{\lan u,v\ra\in\Om^2\colon G(u,v)\le a\}$.
Since every vertical and horizontal section of $G$ is $\mu$-a.e. constant, it
follows that every vertical and horizontal
section of $E$ is either null or of full measure.
Let us define 
$A_1=\linebreak \{u\in\Om\colon\mu(E_u)=1\}$, 
$A_2=\{u\in\Om\colon\mu(E_u)=0\}$, 
$B_1=\{v\in\Om\colon\mu(E^v)=1\}$, and 
$B_2=\{v\in\Om\colon\mu(E^v)=0\}$.
By \cite[254~L~Theorem, p.~249]{Fremlin}, we have $\ol\mu(U^\om)=\ol\mu(V^\om)=1$. 
If $u\in U^\om$, then $G(u,v)\le a$ for $\mu $-a.e. $v$; that is, $\mu (E_u )=1$. 
Similarly, if $u\in V^\om$ then $\mu (E_u )=0$. 
Therefore $U^\om\su A_1$ and $V^\om\su A_2$ and thus $\ol\mu(A_1)=\ol\mu(A_2)=1$.

Suppose $\ol\mu(B_2)>0$. 
Then $E\cap(A_1\times B_2)$ is a 0--1 set in $A_1\times B_2$
and  $(A_2\times B_1)\se E$ is a 0--1 set in $A_2\times B_1$. 
On the other hand, if $\ol\mu(B_1)>0$, then
$(A_2\times B_1)\se E$ is a 0--1 set in $A_2\times B_1$
and $E\cap (A_1 \times B_2 )$ is a 0--1 set in $A_1 \times B_2$.
We obtain that the statement of $E_\nnn$ is true apart from the fact that the
sets $A_i$ and $B_i$ are in $\Om$ instead of $\nl$.
However, the measure spaces $\Om$ and $\nl$ are isomorphic, therefore we can find
sets in $\nl$ with the same properties. 
Thus $E_\nnn$ holds, which completes the proof. \qed




\section{The category case} 
In this section we will prove that the category analogue of 
Theorem~\ref{t2} is true with one obvious modification:
the integral conditions (FP) and (SFP) have no meaning in this case. 

Recall that $\B_\mmm$ coincides with the $\si$-algebra of sets $H\su\rr$
having the Baire property,
and that (SAP) is equivalent to the following statement:
if $A\su\nl^2$ is such that $A_x$ and $A^y$ have the Baire property
for every $x,y\in\nl$ then there is a set $B\su\nl^2$
having the Baire property such that $(A\De B)_x\in\mmm$ and
$(A\De B)^y\in\mmm$ for every $x,y\in\nl$.

Notice also that the chart below implies that all the properties considered
there are independent of the axioms of set theory ZFC. 
This is the case, as since the relations $\add(\M)=\cov(\M)=\non(\M)=2^\omega$
and $\shr(\M)<\cov(\M)$ are both consistent with ZFC:
the first is a consequence of the Martin's Axiom, while the second
holds in the Cohen model. 
(For the proof of $\shr(\M)<\cov(\M)$ see \cite{kom} or \cite{KY}.)



\begin{theorem}\label{t3}
The following statements are equivalent to each other.
\begin{itemize}
\item[{\rm (i)}] $\M$ has property {\rm (SAP).}
\item[{\rm (ii)}] $\M$ has property {\rm (BAP).}
\item[{\rm (iii)}] $\M$ has property {\rm (SKU$^*$).}
\item[{\rm (iv)}] $E_\M$ is false.
\end{itemize}
In particular, for $\I=\M$ we have the following relations. 

\begin{picture}(300,170)
\thicklines

\put(61,140){\makebox(0,0){{\rm (BAP)}}}
\put(81,140){\vector(1,0){20}}
\put(101,140){\vector(-1,0){20}}

\put(120,140){\makebox(0,0){{\rm (SAP)}}}
\put(120,132){\vector(0,-1){40}}
\put(120,92){\vector(0,1){40}}

\put(120,80){\makebox(0,0){{\rm (SKU$^*$)}}}
\put(120,68){\vector(0,-1){40}}
\put(120,28){\vector(0,1){40}}

\put(140,80){\vector(1,0){20}}
\put(179,80){\makebox(0,0){{\rm (SKU)}}}
\put(179,68){\vector(0,-1){40}}
\put(179,28){\vector(0,1){40}}

\put(198,80){\vector(1,0){20}}
\put(233,80){\makebox(0,0){{\rm (KU)}}}
\put(233,68){\vector(0,-1){40}}
\put(233,28){\vector(0,1){40}}

\put(30,15){\makebox(0,0){$[\shr(\M)<\cov(\M)]$}}

\put(83,15){\vector(1,0){20}}
\put(120,15){\makebox(0,0){$\lnot E_\M$}}

\put(140,15){\vector(1,0){20}}
\put(179,15){\makebox(0,0){$\lnot S^{2w}_\M$}}

\put(198,15){\vector(1,0){20}}
\put(233,15){\makebox(0,0){$\lnot S^2_\M$}}

\put(247,15){\vector(1,0){20}}
\put(312,23){\makebox(0,0){$[\non(\M)<2^\om\ \&$}}
\put(322,7){\makebox(0,0){$\ \add(\M) <\cov(\M)]$}}
\end{picture}
\end{theorem}

In the proof of the theorem we will need the following lemma. 

\begin{lemma}\label{l2}
If $A\su\nl^2$ is Borel then the set $\{y\in\nl\colon A^y\notin\mmm\}$ has the Baire property.
\end{lemma}

\pf  There is an open set $G\su\nl^2$ such that $M=G\De A$ is meager. 
Then the set $\{y\in\nl\colon G^y\notin\mmm\}=\{y\in\nl\colon G^y \ne\emp\}$ 
is open, and the set $\{y\in\nl\colon M^y\notin\mmm\}$ is meager
by the Kuratowski-Ulam theorem. Since
\begin{equation*}
\{y\in\nl\colon A^y \notin\mmm\} \De
\{y\in\nl\colon G^y \notin\mmm\} \su
\{y\in\nl\colon M^y \notin\mmm\},
\end{equation*}
it follows that $\{y\in\nl\colon A^y\notin\mmm\}$ has the Baire property. \qed

\bsk \noi
{\sc Proof of Theorem~\ref{t3}.}
The implications in the chart follow from the main part of the theorem,
Theorem~\ref{t1}, and Corollary~\ref{cor4}.


\msk \noi
(i)$\Longleftrightarrow$(ii): This follows from Fact~\ref{fact6}. 

\msk \noi
(ii)$\akkor$(iii): This follows from Lemma~\ref{l2} and Proposition~\ref{p1}.

\msk \noi
(iii)$\akkor$(ii): Suppose that the sections of a set $A\su\nl^2$ have the Baire property. 
It is clear from (SKU$^*$) that for every interval $J$ the set
$H_J=\linebreak\{y\in\nl\colon J\se A^y\in\mmm\}$ has the Baire
property. 
Let
$$
E=\bigcup\{J\times H_J\colon J=(p,q)\su\nl\mbox{ for some rational numbers }  p<q \}.
$$
Then $E$ has the Baire property, and it is easy to see that
for $F=E\De A$ we have $F^y=E^y\De A^y\in\mmm$ for every $y\in\nl$.
It is also clear that $F_x =E_x \De A_x$ has the Baire property for every $x\in\nl$.
Since $\mmm$ has property (SKU$^*$), by Theorem \ref{t1} it also
has property (KU) and so $U:=
\{x\colon F_x\notin\mmm\}\in\mmm$.
Let $H=E\se(U\times\nl)$. 
Then $H$ has the Baire property and we have $(A\De H)_x\in\mmm$ and
$(A\De H)^y\in\mmm$ for every $x,y\in\nl$. 

\msk \noi
(iii)$\akkor$(iv): This was proved in Theorem~\ref{t1}.

\msk \noi
(iv)$\akkor$(iii): We prove that if (SKU$^*$) is false then $E_\mmm$
is true. We will work in $\rr$ instead of $\nl$.

Suppose that there exists a set $A\su\rr^2$ 
such that $A_x$ and $A^y$ have the Baire property
for every $x,y\in\rr$, and $H=\{y\in\rr\colon A^y\notin \mmm\}\notin\B_\mmm$. 
By Banach's theorem, there exists a nonempty open set
$G\su \rr$ such that $H\se G\in \mmm$ and $H$ is of second category
in every nonempty open subset of $G$. (See \cite[\S 10, V. pp. 83--85]{Ku}.)
Since $H\notin\B_\mmm$, the set $G\se H$ is not meager, and thus we can select an
open interval $I\su G$ such that $G\se H$ is of second category
in every nonempty open subinterval of $I$. 
We conclude that both $H$ and $\rr\se H$ are of second category
in every nonempty open subinterval of $I$.
Let $\phi$ be a homeomorphism from $I$ onto $\rr$, and put
$$
C=\{\lan x,\phi(y)\ra\colon\lan x,y\ra\in A\cap(\rr\times I)\}.
$$
Then $C_x$ and $C^y$ have the Baire property for every $x,y\in\rr$, and the set
$K=\{y\in\rr\colon C^y \notin\mmm\}$ has the property that
both $K$ and $\rr\se K$ are of second category in every subinterval of $\rr$. 
Let
$$
D=\{\lan x+r,y\ra\colon\lan x,y\ra\in C \ \& \ r\in\rational\}.
$$
Then $D^y\in\mmm$ if $y\notin K$ and $\rr\se D^y\in\mmm$ if $y\in K$. 
Also, the set \linebreak
$D_x=\bigcup_{r\in\rational}C_{x -r}$ 
has the Baire property for every $x\in\rr$.
Let us put $U=\{x\in\rr\colon \rr\se D_x\in\mmm\}$.
We shall distinguish between two cases.

\msk \noi
Case I: $\rr\se U\in\mmm$. Then the set $D\cap[U\times(\rr\se K)]$
is a 0--1 set in $U\times(\rr\se K)$, and $(\rr\se U)\times K$
is a 0--1 set in $(\rr\se U)\times K$. That is, $E_\mmm$ holds in this case.

\msk \noi
Case II: $\rr\se U\notin\mmm$. 
If $x\in\rr\se U$ then $\rr\se D_x\notin\mmm$ and, as $D_x$ has the Baire property, 
there exists an open interval $J$ with rational endpoints such that $D_x\cap J\in\mmm$.
Since $\rr\se U\notin\mmm$, it follows that we can fix an
open interval $J$ with rational endpoints such that the set
$V=\{x\in\rr\colon D_x\cap J\in\mmm\}$ is of second category.
Let $\psi$ be a homeomorphism from $J$ onto $\rr$, and put
$$
E=\{\lan x,\psi(y)\ra\colon \lan x,y\ra\in D\cap(\rr\times J)\}.
$$
Then $E^y\in\mmm$ if $y\notin\psi[K\cap J]$ and
$\rr\se E^y\in\mmm$ if $y\in\psi[K\cap J]$.
Also, the set $W=\{x\in\rr\colon E_x\in\mmm\}$ is of second category.
Now we define
$$
F=\{\lan x,y+r\ra\colon\lan x,y\ra\in E\ \&\ r\in\rational\}.
$$
Then either $F^y\in\mmm$ or $\rr\se F^y\in\mmm$ for every $y$. 
If $Y=\{y\colon\rr\se F^y\in\mmm\}$ then $Y\notin\mmm$, as $\psi[K\cap J]\su Y$. 
Also, we have $F_x\in\mmm$ if $x\in W$ and $\rr\se F_x\in\mmm$ if $x\notin W$.
Thus $(W\times Y)\se F$ is a 0--1 set in $W\times Y$ and
$[(\rr\se W)\times(\rr\se Y)] \cap F$ is a 0--1 set in $(\rr\se W)\times(\rr\se Y)$. 
Therefore, $E_\mmm$ holds in this case as well. \qed



\section{More on condition $E_\ii$} 

The definition of $E_\ii$ does not involve the topology of the space $X$; 
it is meaningful for any ideal of subsets of a set $X.$ We begin with an equivalent
form of $E_\ii$. We shall use the notation $A^c =X\se A$.

\begin{proposition}\label{p2}
For every ideal $\ii$ the statement $E_\ii$ is equivalent to the following:
\begin{itemize}
\item[$E_\ii^*:$] Either there exists a set $A\su X$ such that $A,A^c\notin\ii$
and there is a 0--1 set in $A\times A^c$, or there exists a set $A\su X$
such that there are 0--1 sets both in $A\times A$ and in $A^c\times A^c$.
\end{itemize}
\end{proposition}

\pf Suppose $E^*_\ii$. If $H$ is a 0--1 set in $A\times A^c$ where $A,A^c \notin \ii ,$
then $E_\ii$ is satisfied with $B=A^c .$
If there are 0--1 sets both in $A\times A$ and in $A^c \times A^c$ 
then we may assume $A\notin \ii$, since otherwise we replace $A$ by $A^c$. 
Then $E_\ii$ is satisfied with $B=A$.

Next suppose $E_\ii$. Suppose that $A,B\notin \ii$,
$H$ is a 0--1 set in $A\times B$ and $K$ is a 0--1 set in $A^c \times B^c .$
If $A\se B\in \ii$ and $B\se A\in\ii$ then there are 0--1 sets both in $A\times A$ and in
$A^c \times A^c$, and then $E^*_\ii$ holds.

Therefore we may assume that at least one of the sets $A\se B$ and
$B\se A$ does not belong to $\ii$. 
By symmetry, we may assume that $D:=
A\se B\notin\ii$. 
We also have $D^c \notin \ii$, since $D^c \supset B.$
We prove that there is a 0--1 set in $D\times D^c$.
Indeed, we have $D^c =A^c \cup B$, and thus
$$
D\times D^c=(D\times A^c)\cup(D\times B).
$$
Since $(D\times A^c)\su(B^c\times A^c)$ and $(D\times B) \su (A\times B)$, 
it follows that there are 0--1 sets both in $D\times A^c$ and in $D\times B$, 
and then there is also one in $D\times D^c$. \qed

In the sequel we shall assume that $X$ is an Abelian group and $\ii$
is a translation invariant ideal in $X$.
The paper \cite{La} investigated the logical connections between several statements including
$S^2_\ii$ and $S^{2w}_\ii$ and the following three conditions.
\begin{itemize}

\item[$S^1_\ii :$]
There exists a set $A\su X$ such that $A\notin\ii$, $X \se A\notin \ii$, and
$(A+h)\se A \in\ii$ for every $h\in X$.

\item[$S^{1s}_\ii :$]
There exists an $f\colon X \to X$ such that
$\{x\in X\colon f(x+h)-f(x)\ne 0\}\in\ii$ for every $h\in X$, and
$f\am(\{ y\})\in\ii$ for every $y\in X$.

\item[$D_\ii :$] There exists a partition $\{A,B\}$ of $X$ such that
$\cov(\ii|A)\le\non(\ii|B)$  and $\cov(\ii |B)\le\non(\ii|A)$.
\end{itemize}

In \cite[Theorem 2]{La} it was shown that if $\ii$ is a $\si $-ideal
and $|X|$ is less than the first (2-valued) measurable cardinal
then the following implications hold.

\begin{picture}(300,140)
\thicklines


\put(15,120){\makebox(0,0){$S^{1s}_\ii$}}
        \put(25,120){\vector(1,0){120}}
               \put(160,120){\makebox(0,0){$S^1_\ii$}}
                      \put(175,120){\vector(1,0){120}}
                          \put(310,120){\makebox(0,0){$D_\ii$}}



\put(15,105){\vector(0,-1){80}}

\put(310,105){\vector(0,-1){80}}



                           \put(310,10){\makebox(0,0){$[\cov(\ii)\le\shri]$}}
                      \put(177,10){\vector(1,0){80}}
                \put(160,10){\makebox(0,0){$S^{2w}_\ii$}}
         \put(25,10){\vector(1,0){120}}
\put(15,10){\makebox(0,0){$S^2_\ii$}}


                                            \put(215,70){\vector(2,1){80}}

          \put(25,115){\vector(2,-1){80}}
                         \put(160,65){\makebox(0,0){$[\cov(\ii)\le\non(\ii)]$}}
          \put(25,20){\vector(2,1){80}}
                                            \put(215,60){\vector(2,-1){80}}
\end{picture}
\begin{center}
Diagram~\Dcounter.\label{Diag:A}
\end{center}
\bigskip 

In the next theorem we try to locate the position of $E_\ii$
in this diagram. As we shall see, it must be somewhere in the
middle column.

\begin{theorem}\label{t4}
Let $\ii$ be an invariant ideal in the Abelian group $X$.
Then we have the following  implications:
\begin{center}
$\left[ S^1_\ii \ \ {\rm or} \ \  S^{2w}_\ii \right] \ \ \akkor \ \ 
E_\ii \ \ \akkor \ \ 
\left[ S^1_\ii \ \ {\rm or} \  \ (\covi \le \noni ) \ \ {\rm or} \ \    
S^{2w}_\ii \right].$
\end{center}
\end{theorem}


In the prove of the theorem we will use the following lemma.

\begin{lemma}\label{l3}
Let $\ii$ be an invariant ideal in an Abelian group $X$.
If there is a partition $\{A,B\}$ of $X$  such that
$\cov(\ii |A) \le\non(\ii |A)$ and
$\cov(\ii |B) \le\non(\ii |B)$ then $\covi\le\noni$.
\end{lemma}

\pf  If $\non(\ii|A) = \non(\ii|B)=\kappa$ then $\noni =\kappa$ and
$$
\covi =\max\{\cov(\ii |A),\cov(\ii |B)\}\le \kappa .
$$
Therefore we may assume that $\non(\ii|A) \ne \non(\ii |B)$.
By symmetry we may suppose $\non(\ii |A) < \non(\ii |B)$. 
Let $\non(\ii |A)=\la$ and let $H\su A$ be a set with $|H|=\la$ and $H\notin \ii$. 
Then, for every $x\in X$ we have $H+x\notin\ii$ and $|H+x|=\la<\non(\ii |B)$. 
Thus $H+x$ cannot be a subset of $B$.
Therefore $(H+x)\cap A\ne \emp$ for every $x$. 
This implies that for every $x$ there is an $h\in H$ such that $h+x\in A;$
that is, $x\in A-h.$ Thus $X=\bigcup_{h\in H} (A-h)$.

Since $A-h$ can be covered by $\cov(\ii |A)\le\la$ elements of $\ii$ and $|H|=\la$,
it follows that $X$ also can be covered by $\la$ elements of $\ii$. 
In other words, $\covi \le \la =\noni$. \qed



\noi
{\sc Proof of Theorem~\ref{t4}.} $S^{2w}_\ii  \akkor E_\ii$ was proved in Theorem~\ref{t1}.

\msk \noi
$S^1_\ii \akkor E_\ii:$
Suppose $S^1_\ii$, and let $A\su X$ be a set such that 
$A\notin \ii$, $X\se A \notin \ii$, and $(A+h)\se A\in\ii$ for every $h\in X$.
We claim that there is a 0--1 set in $A\times A^c$.
Indeed, let $H=\{\lan x,y\ra\in A\times A^c\colon y\notin(A-x)\se A\}$.
For every $x\in A$ we have $A^c \se H_x =(A-x)\se A\in\ii$. 
If $y\in A^c$ then we have
\begin{equation*}
\begin{split}
\lan x,y\ra\in H 
&\akkor y\notin A-x \akkor x\notin A-y \\
&\akkor x\in A\se (A-y) =[(A+y )\se A]-y\in \ii,
\end{split}
\end{equation*}
and thus $H^y \in \ii$ for every $y\in A^c$. 
Therefore, $H$ is a 0--1 set in $A\times A^c$.

\msk
Now we turn to the proof of the second implication of Theorem~\ref{t4}. 
Suppose that $E_\ii$ holds.
By Proposition \ref{p2}, one of the following statements is true:
(i) there exists a set $A\su X$ such that $A,A^c \notin \ii$
and there is a 0--1 set in $A\times A^c$, 
or 
(ii) there exists a set $A\su X$ such that there are 0--1 sets both in
$A\times A$ and in $A^c \times A^c$.

\msk
Suppose (ii). If $A^c\in\ii$ then there is a 0--1 set in $X\times X$, 
and thus $S^{2w}_\ii$ (even $S^2_\ii$) holds. 
We have the same conclusion if $A\in \ii$. If $A,X\se A\notin \ii$ then, by
Lemma~\ref{l4}, we obtain $\cov(\ii |A) \le\non(\ii |A)$ and
$\cov(\ii |A^c) \le \non(\ii |A^c )$.
Therefore, by Lemma \ref{l3}, we get $\covi \le \noni .$

\msk
Next suppose (i). 
Then there exists a set $A\su X$ such that $A,A^c \notin \ii$,
and there is a 0--1 set in $A\times A^c$. 
If $(A+h)\se A\in \ii$ for every $h\in X$ then we have $S^1_\ii$. 
Therefore we may assume that there is an $h\in X$ such that 
$B:=
(A+h)\se A=(A+h) \cap A^c \notin \ii$. 
Since
$$
B\times X\su  [(A+h)\times A^c ]\cup [A^c \times A]
$$
and there are 0--1 sets both in $(A+h)\times A^c$ and $A^c \times A$,
it follows that there is a 0--1 set in $B\times X$, and thus $S^{2w}_\ii$ holds. \qed

We conclude with another simple observation.

\begin{proposition}
For every invariant ideal $\ii$ on an Abelian group $X$ 
we have $S^{2w}_\ii \akkor D_\ii$ and $E_\ii \akkor D_\ii$.
\end{proposition}

\pf Suppose that $S^{2w}_\ii$ holds. 
Then there exists a set $A\su X$ such that $A\notin \ii$, and there is a 0--1 set in $A\times X$.
If $A^c\in\ii$ then $S^2_\ii$ holds, which implies $D_\ii$ by Diagram~\ref{Diag:A}. 
Therefore we may assume $A^c \notin\ii$.

Since there is a 0--1 set in $A\times X$, there is one also
in $A\times A^c$ and hence in $A^c\times A$ as well. 
Then, by Lemma \ref{l4}, it follows that $\cov(\ii |A)\le\non(\ii|A^c)$ and
$\cov(\ii |A^c )\le\non(\ii | A)$; that is, $D_\ii$ holds.

Now each of the statements $S^1_\ii$, $\covi \le \noni$, and $S^{2w}_\ii$ implies $D_\ii$. 
(For the first two of these implications we again refer to Diagram~\ref{Diag:A}.) 
Therefore, by Theorem \ref{t4}, we have
$E_\ii \akkor D_\ii$. \qed

We remark that the implication $S^{2w}_\ii \akkor D_\ii$ was not noted in \cite{La}. 
It would be interesting to decide
whether or not there are other implications that can be inserted into Diagram~\ref{Diag:A}.

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\end{document}                                        *
