%December 7, 2000, version accepted 

\documentclass{rae}
\usepackage{amsmath}\usepackage{amssymb}

\newcommand{\ch}[1]{\marginpar{{\tiny #1}}} 

\markboth{
M. Balcerzak,
A. Bartoszewicz,
K. Ciesielski
}
{On Marczewski-Burstin representations of certain algebras}

\author{Marek Balcerzak%
\thanks{This work was partially supported by NSF Cooperative Research
Grant INT-9600548 and its Polish part financed by KBN.
\endgraf Papers authored or co-authored by a Contributing Editor are
managed by a Managing Editor or one of the other Contributing Editors.
}, Institute of Mathematics, {\L}\'od\'z Technical University,
al.~Politechniki 11, I-2, 90-924 {\L}\'od\'z, Poland, and
Faculty of Mathematics, University of {\L}\'od\'z, ul. Banacha~22,
90-238 {\L}\'od\'z, Poland (mbalce@krysia.uni.lodz.pl)
\\
Artur Bartoszewicz,
Institute of Mathematics, {\L}\'od\'z Technical University,
al.~Politechniki 11, I-2, 90-924 {\L}\'od\'z, Poland
(arturbar@ck-sg.p.lodz.pl)
\\
Krzysztof Ciesielski,
Department of Mathematics, West Virginia
University, Morgantown, WV 26506-6310, USA
(K\_Cies@math.wvu.edu)
{\tt http://www.math.wvu.edu/\~{}kcies}
}

\MathReviews{
Primary 03E35; Secondary 28A05}
\keywords{%Key words and phrases:
Borel sets, ultrafilters, Continuum Hypothesis, MB-representation.}

\title{On Marczewski-Burstin representations of certain algebras of sets}


\newtheorem{theorem}{Theorem}
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\newtheorem{definition}[theorem]{Definition}
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\newtheorem{Claim}[theorem]{Claim}

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\def\proof{\noindent {\sc Proof. }}

%\def\qed{\hfill$\Box$ \medskip}
\def\qed{\hfill$\square$ \medskip}


\newcommand{\la}{{\langle}}
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\newcommand{\continuum}{{\mathfrak c}}
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\newcommand{\C}{{\cal C}}
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\newcommand{\PR}{{\cal P}(\R)}
\newcommand{\F}{{\cal F}}
\newcommand{\G}{{\cal G}}
\newcommand{\I}{{\cal I}}
\newcommand{\K}{{\cal K}}
\newcommand{\Ll}{{\cal L}}
\newcommand{\Si}{\Sigma}
\newcommand{\SF}{{\cal S}({\cal F})}
\newcommand{\SO}{{\cal S}_0({\cal F})}
\newcommand{\U}{{\cal U}}
\newcommand{\V}{{\cal V}}
\newcommand{\dU}{\hat{\U}}
\newcommand{\dV}{\hat{\V}}
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\begin{document}
\maketitle
\begin{abstract}
We show that the Generalized Continuum Hypothesis GCH
(its appropriate part)
implies that many natural algebras on $\real$,
including the algebra $\B$ of Borel sets and the interval algebra $\Si$,
are outer Marczewski-Burstin representable by families of non-Borel sets.
Also we construct, assuming again an appropriate part of GCH,
that there are algebras on $\real$ which are not
MB-representable. We prove that some algebras
(including $\B$ and $\Si$) are not inner
MB-representable. We give examples of algebras
which are inner and outer MB-representable, or are inner but not outer
MB-representable.
\end{abstract}

\section{Introduction}

Our set theoretic notation is standard and follows that from~\cite{CiBook}.


For a fixed non-empty
set $X$ and a family $\F\subset\P(X)\setminus\{\emptyset\}$ define,
following the idea of Burstin and Marczewski,
\[
\SF=\{ A\subset X \colon
(\forall T\in\F )(\exists W\in\F )(W\subset T\cap A
\mbox{ or }W\subset T\cap A^c )\}
\]
and
\[
\SO=\{ A\subset X \colon
(\forall T\in\F )(\exists W\in\F )(W\subset T\cap A^c )\}.
\]
Then $\SF$ constitutes an algebra of sets and $\SO$ is
an ideal of subsets of~$X$.
(See \cite{BBRW}.) We will always assume that the whole space
$X$ is in an algebra; usually such a family is called a field of sets.
Burstin in \cite{B} proved that the $\sigma$-algebra
of Lebesgue measurable subsets of $\real$ is of the form $\SF$
for $\F$ being the family
of perfect subsets of $\real$ of positive measure. It can be also shown that,
for the same $\F$, the family $\SO$ consists of Lebesgue null sets.
(See \cite{R} or \cite{BET}.) On the other hand, if $\F$ is the family
of all perfect subsets of $\real$
then $\SF$ and $\SO$ constitute a quite different pair
of a $\sigma$-algebra and a $\sigma$-ideal on $\real$
which were introduced by Marczewski in
\cite{M}.

If a given algebra $\A$ (an ideal ${\cal I}$, respectively)
on a set $X$
can be represented as $\SF$ (respectively, as $\SO$) for some family
$\F\subset\P(X)\setminus\{\emptyset\}$,
we say that it is {\em Marczewski-Burstin representable}\/ (or, briefly,
{\em MB-representable}) by $\F$. If additionally, $\F\subset\A$ (respectively,
$\F\cap\A=\emptyset$), we say that $\A$ is inner (outer) MB-representable
by $\F$.
Similarly, for $\I\subset\A$ we say that
the pair $\la\A,{\cal I}\ra$
is MB-representable if $\A =\SF$ and
$\I=\SO$ for the same family $\F$.
This notion is most often considered when $\I$ is the
ideal
\[
\He (\A )=\{ A\subset X\colon (\forall B\subset A)(B\in\A )\}
\]
of sets which hereditarily belong to $\A$.

Systematic studies of MB-representations of algebras and ideals were
initiated in \cite{R}, \cite{BET}, and \cite{BBRW}.
For instance, in \cite{BET} it is proved that the algebra of sets
in $\real$ with the Baire property is inner MB-representable by a
family of Borel sets, and in \cite{BBRW} it is shown that the
{\em interval algebra} $\Si$ generated by intervals $[a,b)$ with $a<b$,
is MB-representable by a family of Borel sets. Some necessary conditions
for MB-representability of a pair $\la\A ,\I\ra$ by a family of Borel sets
are given in \cite{BET} and \cite{ET}.
Until now, however, the following basic questions about MB-representability
(see \cite{BBRW}) were without an answer:
{\em Is every algebra of sets MB-representable?
What about some basic algebras,
like the algebra $\B$ of Borel subsets of $\real$?
Is it MB-representable, and if so, is it inner (outer) MB-representable?}

In this note we show that, assuming appropriate
set theoretical assumptions
(which follow from the Generalized Continuum Hypothesis GCH),
there are algebras (on $\real$ and other infinite sets)
which are not MB-representable. We also show, under
similar set theoretical assumptions, that many ``natural''
algebras, including the algebras $\B$ and $\Si$, are outer
MB-representable in some strong manner.
It has to be pointed out here
that our representation families $\F\subset\P(X)$, unlike those
studied in the earlier papers,
are not nice in a sense that they are not connected
with the Borel structure of a space.
The same is true for our example of algebras
which are not MB-representable.
Moreover, these facts are not proved in ZFC.
On the other hand
we prove that the algebras
$\B$ and $\Si$ are not inner MB-representable.
In Section 1 we show a simple criterion for an inner MB-representable
algebra to be outer MB-representable. We apply it to some classical
$\sigma$-algebras on $\real$.


\section{Algebras which are inner and outer MB-representable}
Recall that the algebras of Lebesgue measurable sets, of sets with
the Baire property, and of Marczewski $s$-sets are inner MB-representable.
We shall prove that they are also outer MB-representable.
To this end we propose some general scheme.

We need the following fact which easily results from the definition
of $\SF$ and $\SO$. (See \cite{BBRW}.)

\fact{f2}{For $\F_0,\F_1\subset\P (X)\setminus\{\emptyset\}$ assume that
$$(\forall i\in\{ 0,1\} )(\forall A\in\F_i)
(\exists B\in\F_{1-i})(B\subset A).$$
(We thus say that $\F_0,\F_1$ are mutually coinitial.) Then $S(\F_0)=S(\F_1)$
and $S_0(\F_0)=S_0(\F_1)$.}

\prop{p0}{Assume that an algebra $\A$ on $X$ is inner MB-representable
by a family $\F\subset\A$ with the following properties:
\begin{itemize}
\item [{\rm (a)}] $(\forall F\in\F)(\exists F_1,F_2\in\F )(F_1\cup F_2\subset F\;
\&\; F_1\cap F_2=\emptyset)$;
\item [{\rm (b)}] $(\exists B\subset X)(\forall F\in\F) B\cap F\notin\A$.
\end{itemize}
Then $\A$ is outer MB-representable by the family
$$
\F_B=\{ F_1\cup (F_2\cap B)\colon F_1,F_2\in\F\ \&\ F_1\cap
F_2=\emptyset\},
$$
where $B$ is a set realizing (b).
}
\proof From (a) and the 
definition of $\F_B$ it follows that $\F$ and $\F_B$ are
mutually coinitial. So $\SF= S(\F_B)$ by Fact \ref{f2}. Condition (b)
implies that $\F_B\cap\A=\emptyset$.
\qed

%\newpage 

\cor{cor0}{The following algebras on $\real$
are outer MB-representable:
\begin{itemize}
\item the algebra of Lebesgue measurable sets,
\item the algebra of sets with the Baire property, and
\item the algebra  of Marczewski measurable sets.
\end{itemize}
}
\proof
We use Proposition \ref{p0}. For any algebra listed in the assertion, the
role of $\F$ is played by: the perfect sets of positive
measure, the comeager $G_{\delta}$ sets in nonempty open sets (see \cite{BET}),
and by all perfect sets, respectively. In every case, $\F$ satisfies
conditions (a) and (b) where $B$ in (b) stands for a Bernstein set.
\qed

\section{Algebras which are strongly outer MB-representable}

Let $\A$ be an algebra on $X$. If, for each family $\C\subset{\cal P}(X)$
with $\A\subset\C$ and $|\C |=|\A |$, there is an $\F\subset{\cal P}(X)
\setminus\C$ such that $\A=\SF$, we say that $\A$ is 
{\em strongly outer MB-representable}. 
If additionally, $\He (\A )=S_0 (\F )$, we say that the pair
$\langle\A ,\He (\A )\rangle$ is strongly outer MB-representable.

Let $X$ be an infinite set of cardinality $\kappa$.
The following is the main theorem of this section.

\thm{thm1}{Let $\A$ be an algebra of subsets of $X$ such that
$[X]^{<\kappa}\subset\A$. If $2^\kappa=\kappa^+$ and $|\A|<2^\kappa$
then the pair $\la\A,\He(\A)\ra$ is strongly
outer MB-re\-pre\-sen\-ta\-ble.}

 From this theorem we immediately obtain the following corollary.

\cor{cor2}{
If $2^\om =\om_1$ and $2^{\om_1}=\om_2$ then the pair $\la\B,\coun\ra$
is strongly outer MB-re\-pre\-sen\-ta\-ble.
}

In the sequel we will use the following fact
which is well known (see e.g. \cite[Lemma 2]{W}).
However we provide its easy proof for the reader's convenience.

\fact{fact3}{For every algebra $\A$ on $X$ and $Z\in\P(X)\setminus\A$
there exists an ultrafilter $\U_Z$ in $\A$ such that
$U\cap Z\notin\A$ for every $U\in\U_Z$.
}

\proof
Observe that the family 
${\cal G}=\{E\in\A\colon Z\setminus E\in\A\}$ 
is a filter in the
algebra $\A$. Consider an ultrafilter ${\cal U}_Z\supset{\cal G}$
in $\A$. Then ${\cal U}_Z$ is as desired.
\qed

\noindent{\sc Proof of Theorem~\ref{thm1}.}
To construct family $\F$
let $\{Z_\xi\colon\xi<\kappa^+\}$ be an enumeration
of $\P(X)\setminus\A$.
For each $\xi<\kappa^+$ use Fact~\ref{fact3} to choose
an ultrafilter $\U_\xi=\U_{Z_\xi}$ in $\A$ for which
\begin{equation}\label{star}
U\cap Z_\xi\notin\A\ \mbox{ for each }\ U\in\U_\xi.
\end{equation}
Fix a family $\C\subset{\cal P}(X)$ with $\A\subset\C$
and $|\C |=|\A |$.
By induction on $\xi<\kappa^+$ we construct a sequence
$\la D_\xi\subset X\colon \xi<\kappa^+\ra$
of ``very independent sets'' in a sense that
\begin{equation}\label{eq2}
|D_\xi\cap Y|=\kappa =|D^c_\xi\cap Y|
\end{equation}
for each set $Y\subset X$ of cardinality $\kappa$
which belongs to the algebra $\K_\xi$ of sets generated by the
family
\begin{equation}\label{eqq}
\{D_\zeta\colon\zeta<\xi\}\cup\{Z_\zeta\colon\zeta\le\xi\}\cup\C .
\end{equation}
Such $D_\xi$ can be chosen by an easy diagonal argument
(another transfinite induction) since $|\K_\xi|=\kappa$.
\[
\F=\bigcup_{\xi<\kappa^+}\{U\cap D_\xi\colon U\in\U_\xi\}.
\]
Note that by (\ref{eq2}) and (\ref{eqq})
we clearly have $\F\cap\C =\emptyset$.
The remaining properties of $\F$ will be
shown in the following three steps.

\medskip

\noindent{\bf Step 1.} If $Z=Z_\xi\in\P(X)\setminus\A$ then $Z\notin\SF$.
\medskip

To see this let $T=D_\xi$. Then $T=X\cap D_\xi\in\F$. We shall prove that
$W\not\subset T\cap Z$ and $W\not\subset T\cap Z^c$ for all $W\in\F$.
Thus let $W\in\F$, say $W=U\cap D_\eta$ for some $U\in\U_\eta$,
$\eta <\kappa^+$. Consider three cases:
\begin{itemize}
\item If $\eta <\xi$ then $W\not \subset T$ since, by (\ref{eq2}),
$W\cap T^c=(U\cap D_\eta )\cap D^c_\xi\neq\emptyset$.
\item If $\eta >\xi$ then once again we have $W\not\subset T$ since
condition (\ref{eq2}) implies that
$W\cap T^c=(U\cap D_\eta )\cap D^c_\xi =
D_\eta\cap (U\cap D_\xi^c)\neq\emptyset$.
\item If $\eta =\xi$ then by (\ref{star}) we have $U\cap Z\notin\A$.
So, 
$|U\cap Z|=|U\cap Z^c|=\kappa$. Consequently, by (\ref{eq2}),
$|D_\xi\cap (U\cap Z)|=|D_\xi\cap (U\cap Z^c)|=\kappa$.
Thus we have $W\not\subset T\cap Z$ since
$$W\not\subset T\cap Z\iff U\cap D_\xi\not\subset D_\xi\cap Z\iff
U\cap D_\xi\cap Z^c\neq\emptyset ,$$
and also $W\not\subset T\cap Z^c$ since
$$W\not\subset T\cap Z^c\iff U\cap D_\xi\not\subset D_\xi\cap Z^c\iff
U\cap D_\xi\cap Z\neq\emptyset .$$
\end{itemize}
This completes Step~1.
\medskip

\noindent{\bf Step 2.} If $V\in\A$ then $V\in\SF$. \medskip

Let $T\in\F$, say $T=U\cap D_\xi$ where $\xi <\kappa^+$ and $U\in\U_\xi$. Since
$\U_\xi$ is an ultrafilter in $\A$, we have
either $V\in\U_\xi$ or $V\notin\U_\xi$.
If $V\in\U_\xi$ then $U\cap V\in\U_\xi$ and thus for $W=(U\cap V)\cap D_\xi
=T\cap V$ we have $W\in\F$. If $V\notin\U_\xi$ then $V^c\in\U_\xi$ and thus
for $W=(U\cap V^c)\cap D_\xi =T\cap V^c$ we have $W\in\F$. Hence in the
both cases $V\in\SF$. Step~2 has been completed.
\medskip

\noindent{\bf Step 3.} $\SO =\He(\A)$. \medskip

We clearly have
\[
\SO\subset\He(\SF)=\He(\A).
\]
To show that $\He(\A)\subset\SO$ consider  $V\in\He(\A)$ and let
$T=U\cap D_\xi\in\F$ where $\xi <\kappa^+$ and $U\in\U_\xi$.
Since $V\in\He(\A)$, by (\ref{star}) we have
$V\notin\U_\xi$. So $V^c\in\U_\xi$ and
$W=(U\cap V^c)\cap D_\xi =T\cap V^c$ belongs to $\F$.
Hence $V\in\SO$.
This finishes the proof of Theorem~\ref{thm1}.
\qed

\noindent
{\bf Remark.}
It is worth to point out here that we do not need a full strength of
the assumption $2^\kappa=\kappa^+$ to prove Theorem~\ref{thm1}.
In fact there are models of ZFC in which
$2^\kappa>\kappa^+$ and we can find sets $D_\xi$ satisfying
(\ref{eq2}) for any family of less than $2^\kappa$-many sets
of cardinality $\kappa$. In such models the proof presented
above remains valid.

\medskip

The structural assumptions on $\A$ in Theorem~\ref{thm1}
were that $|\A|<2^\kappa$ and $[X]^{<\kappa}\subset\A$.
Although the example presented in the next section
clearly violates both of these assumptions, it is
worth to mention that the second assumption
can be modified with resulting statement still being true.
This is stated in the next theorem.


\thm{thm2}{Let $\A$ be an algebra of subsets of $X$ such that
$\A\cap[X]^{<\kappa}=\{\emptyset\}$. If $2^\kappa=\kappa^+$ and $|\A|<2^\kappa$
then the pair $\la\A,\He(\A)\ra=\la\A,\{\emptyset\}\ra$
is strongly outer MB-re\-pre\-sen\-ta\-ble.
}

\noindent{\sc Sketch of proof.}
Put
$\Aa =\{A\cup M\colon A\in\A\ \&\ M\in [X]^{<\kappa}\}$,
where $\A$ is as above,
and notice that the following version of Fact~\ref{fact3} remains true:
\begin{quotation}
\noindent
For every $Z\in\P(X)\setminus\Aa$
there exists an ultrafilter $\U_Z$ in $\A$ such that
$U\cap Z\notin\Aa$ for every $U\in\U_Z$.
\end{quotation}
Indeed, similarly as in Fact~\ref{fact3},
it is enough to show that
if $\V$ is a maximal filter in $\A$
such that
$V\cap Z\notin\Aa$ for each $V\in\V$
then $\V$ is an ultrafilter in $\A$.
But if $\V$ is not an ultrafilter in $\A$
then there are
$V_0,V_1\in\V$ such that $V_0\cap A\cap Z\in\Aa$
and $V_1\cap A^c\cap Z\in\Aa$.
Hence there are $A_0,A_1\in\A$ and $M_0,M_1\in[X]^{<\kappa}$ such that
\[
V_0\cap V_1\cap A\cap Z=A_0\cup M_0 \in\A
\mbox{ and }V_0\cap V_1\cap A^c\cap Z=A_1\cup M_1\in\A
\]
which implies that $V_0\cap V_1\cap Z=(A_0\cup A_1)\cup(M_0\cup M_1)\in\Aa$
where $V_0\cap V_1\in\V$, a contradiction.

Then proceed as in the proof above listing as sets $Z_\xi$
only the sets from $\P(X)\setminus\Aa$.
This will clearly results with $\F\cap\C =\emptyset$,
$\A\subset\SF$, and with $(\P(X)\setminus\Aa )\cap\SF=\emptyset$.
To finish the proof it is enough to notice that
if $Z\in\Aa\setminus\A$ then $Z^c\notin\Aa$,
as $\A\cap[X]^{<\kappa}=\{\emptyset\}$, and
so $Z^c\notin\SF$. Thus we have also $Z\notin\SF$. \qed

It was proved in \cite{BBRW} that the interval algebra $\Si$ on $\real$
is outer representable by a family of Borel sets.
If we use Theorem \ref{thm2} with $\A=\Si$,
we obtain the following

\cor{cor29}{If $2^\continuum=\continuum^+$ then the interval
algebra $\Si$ on $\real$ is strongly outer MB-representable.
In particular, it is outer MB-representable by a family of
non-Borel sets.
}

\section{Algebras which are not MB-representable}

The key step towards constructing an algebra
which is not MB-representable is the following fact.

\prop{prop31}{Let $X$ be an infinite set of cardinality $\kappa$
and let $\A$ be an algebra on $X$ having the following properties:
\begin{itemize}
\item[{\rm (i)}] $\A\cap[X]^{<\kappa}=\{\emptyset\}$;
\item[{\rm (ii)}] for every $E\in [X]^{<\kappa}$ and $x\notin E$
there exists an $A\in\A$ such that $x\in A\subset X\setminus E$;
\item[{\rm (iii)}] for every $Z\in [X]^{\kappa}$ and $x\notin Z$
there exists an $A\in\A\setminus\{\emptyset\}$ such that
either $|A\cap Z^c|<\kappa$ or $x\in A$ and  $|A\cap Z|<\kappa$.
\end{itemize}
If $\F\subset\P(X)\setminus\{\emptyset\}$ is such that
$\A\subset\SF$ then $\SF$ contains a singleton.
In particular algebra $\A$ is not MB-representable.
}

\proof Let $\A$ and $\F$ be as in the assumptions.
If $\{x\}\in\SF$ for every $x\in X$ then there is nothing to prove.
So assume that there exists an $x\in X$ for which $\{x\}\notin\SF$.
This means that there exists a $Z\in\F$ for which
neither $W\subset Z\cap\{x\}$ nor $W\subset Z\setminus\{x\}$
for every $W\in\F$. Thus $x\in Z$ and
\begin{equation}\label{eq31}
x\in W\ \mbox{ for every $W\in\F$ with $W\subset Z$.}
\end{equation}
Next note that
\begin{equation}\label{eq32}
\mbox{there is no $A\in\A$ containing $x$ with $|A\cap Z|<\kappa$.}
\end{equation}

Indeed, if there is such an $A$ then, by (ii) used with $E=A\cap
Z\setminus\{x\}$
we can find an $A_1\in\A$ with $A\cap A_1\cap Z=\{x\}$.
Since $A\cap A_1\in\A\subset\SF$, there exists a $W\in \F$
such that
either $W\subset Z\cap (A\cap A_1)=\{x\}$ or
$W\subset Z\setminus(A\cap A_1)=Z\setminus\{x\}$.
But, by (\ref{eq31}), the second case is impossible.
Thus, $\{x\}=W\in\F$ and so $\{x\}\in\SF$,
contradicting our choice of~$x$.

Note that the condition (\ref{eq32}) works also for $Z$ replaced 
with $Z'=Z\setminus\{x\}$, which implies that $|Z'|=\kappa$. 
Thus, applying (iii) to $Z'$ and our $x$, we conclude that 
there is an $A\in\A\setminus\{\emptyset\}$ 
such that $|A\cap (Z')^c|<\kappa$. 
For this $A$ we have also $|A\cap Z^c|<\kappa$. 
Now, using (ii) if necessary to decrease $A$, we can
additionally assume that $A\cap Z^c=\emptyset$ and $x\notin A$.
(Indeed, pick an $x_0\in A\cap Z$, $x_0\neq x$, and let
$E=(A\cap Z^c)\cup\{ x\}$. Then $x_0\notin E$ and, by (ii), there
exists an $A^*\in\A$ such that $x_0\in A^*\subset X\setminus E$.
Then $\hat{A}=A\cap A^*\ni x_0$ is as required.)
So $A\subset Z\setminus\{x\}$. Thus, by (\ref{eq31}),
$A$ contains no $W\in\F$. Since $A\in\SF$, this implies that
$A\in\SO$. So $\{a\}\in\SF$ for every $a\in A$. \qed

\prop{prop32}{If $2^\kappa=\kappa^+$ and $|X|=\kappa$ then there exists
an algebra $\A$ on $X$ satisfying conditions (i)--(iii)
from Proposition~\ref{prop31}.
}

\proof Let $\{\la Z_\xi,x_\xi\ra\colon \xi<\kappa^+\}$
be an enumeration of all pairs $\la Z,x\ra\in\P(X)\times X$ with $x\notin Z$.
Similarly as in the proof of Theorem ~\ref{thm1},
by induction on $\xi<\kappa^+$, we construct a sequence
$\la D_\xi\subset X\colon \xi<\kappa^+\ra$
such that
\begin{equation}\label{eq33}
|D_\xi\cap A|=\kappa =|D^c_\xi\cap A|
\end{equation}
for every non-empty set $A$ which belongs
to the algebra $\Ll_\xi$ of sets generated by the
family $\{D_\zeta\colon\zeta<\xi\}$.
In addition, if there is no $A\in \Ll_\xi\setminus\{\emptyset\}$
with $|A\setminus Z_\xi|<\kappa$ then
we will additionally require that
$z_\xi\in D_\xi\subset X\setminus Z_\xi$.

Such $D_\xi$ can be chosen by an easy diagonal argument
since $|\Ll_\xi|\le\kappa$ and sets $D_\xi$ and $D^c_\xi$
need to intersect all sets $A\in \Ll_\xi\setminus\{\emptyset\}$
and, if additional requirement is
claimed, they need also to intersect all
sets $A\setminus Z_\xi\in[X\setminus Z_\xi]^\kappa$
for $A\in \Ll_\xi\setminus\{\emptyset\}$.
Let $\A$ denote the algebra generated by all sets
$\{D_\xi\colon\xi<\kappa^+\}$. Then $\A$ has all the desired properties.

Indeed, (i) is obvious.

To see (ii) let $E\in [X]^{<\kappa},x\notin E$, and take an $\xi<\kappa^+$
with $\la Z_\xi,x_\xi\ra=\la E,x\ra$.
Then from $|Z_\xi|<\kappa$ and (6) it follows that
$|A\setminus Z_\xi|=\kappa$ for all
$A\in \Ll_\xi\setminus\{\emptyset\}$. 
So $x=z_\xi\in D_\xi\subset X\setminus Z_\xi=X\setminus E$
and $A=D_\xi\in\A$ is as desired.

To see (iii) let $E\in[X]^{\kappa},x\notin E$, and take a $\xi<\kappa^+$
such that  $\la Z_\xi,x_\xi\ra=\la Z,x\ra$.
If there exists an  $A\in \Ll_\xi\setminus\{\emptyset\}$
such that $|A\cap Z^c|=|A\setminus Z_\xi|<\kappa$
then (iii) holds. Otherwise
$x=z_\xi\in D_\xi\subset X\setminus Z_\xi=X\setminus Z$
and $A=D_\xi\in\A$ is as desired since $A\cap Z=\emptyset$.
\qed

\cor{cor33}{If $2^\kappa=\kappa^+$ and $|X|=\kappa$ then there exists
an algebra $\A$ on $X$ which is not MB-representable.
}

It is also worth to notice that, as in the case of Theorem~\ref{thm1},
Corollary~\ref{cor33} remains valid also in some models with
$2^\kappa>\kappa^+$.
However, it is also worth to note that if $\A$
is an example as in Proposition~\ref{prop32}
and $X\subset Y$, then the algebra $\A_Y$ on $Y$ generated by $\A$
still is not MB-representable.
Thus if $X$ is such that there exists an infinite $\kappa\leq|X|$
with $2^\kappa=\kappa^+$ then there exists
an algebra $\A$ on $X$ which is not MB-representable.

\section{Algebras which are not inner MB-representable}

Now, we shall prove that the algebras
$\Si$ and $\B$ are not inner MB-representable.


\prop{propWr}{{\rm (S. Wro\'nski~\cite{Wr2})}
The interval algebra $\Si$ is not inner MB-re\-pre\-sen\-tab\-le.}



\proof Suppose that $\Si=S(\F )$ for some $\F\subset \Si$. Let $\G$
stand for the family of all intervals $[a,b)$ with $a<b$. Evidently,
$\Si$ and $\G$ are mutually coinitial, so $S(\Si )=S(\G )$ by Fact \ref{f2}.
Since $S(\G)$ contains singletons, we have $S(\G )\setminus\Sigma\neq
\emptyset$. Thus, by Fact \ref{f2}, $\G$ and $\F$ cannot be mutually coinitial,
and since $\F\subset\Sigma$, it follows that there is a $A\in\G$ such
that $P\setminus A\neq\emptyset$ for each $P\in\F$.
Let $A=[a,b)$. To obtain
a contradiction we shall show that $\{ a\}\in S(\F )$. Let $P\in\F$.
If $a\notin P$ then obviously $P\subset P\cap\{ a\}^c$. Let $a\in P$. 
Since $A\in S(\F )$ 
and since we cannot find a $Q\in\F$ such that $Q\subset P\cap A$, 
there is a $Q\subset P\cap A^c$, so $Q\subset P\cap\{ a\}^c$.
Consequently $\{a\}\in \SF\subset\SO$.
\qed


\thm{thm4}{
Let $X$ be an infinite set of cardinality $\kappa$. Let $\A$ be
an algebra on $X$ such that:
\begin{itemize}
\item[{\rm (I)}] $\He (\A )\subset [X]^{<\kappa}$;
\item[{\rm (II)}] $\A\cap [X]^{<\kappa}\subset\He (\A )$;
\item[{\rm (III)}] for $\A^*=\A\setminus [X]^{<\kappa}$ we have
$S(\A^*)\setminus\A\neq\emptyset$.
\end{itemize}
Then $\A$ is not inner MB-representable.}

\proof Suppose that $\A=S(\F )$ for some $\F\subset \A$. Put
$\F^*=\F\setminus [X]^{<\kappa}$.
First we shall prove that
\begin{equation}\label{e6}
(\forall B\in\A^*)(\exists F\in\F^*)\ F\subset B.
\end{equation}
Suppose it is not the case and let $B\in\A^*$ witness that (\ref{e6}) is
false. We have $|B|=\kappa$ and so, $B\notin\SO$ since, by (I),
\begin{equation}
\SO\subset\He (\SF )=\He (\A )\subset [X]^{<\kappa }.
\end{equation}
 From $B\notin\SO$ it follows that there are sets $Q\in\F$, $Q\subset B$.
Since (\ref{e6}) is false, we have $|Q|<\kappa$ for each set $Q\in\F$
contained in $B$. We shall show that $B\in\He (\SF )$ which yields a
contradiction since $|B|=\kappa$ and $\He (\SF )\subset [X]^{<\kappa}$
by (I). Let $Z\subset B$ and $P\in\F$.
We have to find $T\in\F$ such that  either $T\subset P\cap Z^c$ or
$T\subset P\cap Z$.

Since $B\in\A =\SF$, there
is a $Q\in\F$ such that either $Q\subset P\cap B^c$ or $Q\subset P\cap B$.
If $Q\subset P\cap B^c$ then $T=Q$ is as desired.
So, assume that $Q\subset P\cap B$. Then $|Q|<\kappa$ by our supposition.
Thus, by
(II), we have $Q\cap Z\in\A =\SF$ and so, there is
a $T\in\F$ such that either $T\subset Q\cap(Q\cap Z)\subset P\cap Z$ or
$T\subset Q\cap (Q\cap Z)^c\subset P\cap Z^c$. Consequently, $Z\in\SF$
and thus $B\in\He (\SF )$ as desired.

By Fact \ref{f2}, condition (\ref{e6}) together with an obvious inclusion
$\F^*\subset\A^*$ imply that $S(\F^*)=S(\A^*)$.

Next, we shall show that $S(\F^*)\subset S(\F )$. Assume that
$B\in S(\F^*)$. Let $P\in\F$. Consider two cases:
\begin{itemize}
\item if $P\in\F^*$ then $B\in S(\F^*)$ implies that we can find a
$Q\in\F^*$ (hence $Q\in\F$) with $Q\subset P\cap B$ or $Q\subset P\cap B^c$;
\item if $|P|<\kappa$ then, by (II),
$P\cap B\in\A =\SF$, so there is a $Q\in\F$ with
$Q\subset P\cap (P\cap B)$ or $Q\subset P\cap (P\cap B)^c=P\cap B^c$.
\end{itemize}
Hence $B\in\SF$.

Finally, we have $S(\A^*)=S(\F^*)\subset \SF =\A$ which contradicts (III).
\qed

 From the result by Elaloui-Talibi \cite[Thm. 1.1]{ET}
it follows that there is no $\F\subset\B$ with $\B=\SF$
and $[\real]^{\le\omega}=\SO$. We can derive a bit more

\cor{cor41}{The algebra $\B$ is not inner MB-representable.}

\proof It suffices
to check (III). Here $\B^*$ is the family of uncountable Borel sets.
Thus $\B^*$ and the family of perfect sets are mutually coinitial. So, by
Fact~\ref{f2}, $S(\B^*)$ is exactly the algebra of classical Marczewski
$(s)$-sets. Since there is a non-Borel $(s_0)$-set \cite{Mi}, therefore
it belongs to $S(\B^*)\setminus\B$.
\qed

\noindent
{\bf Example.} Let us observe that 
the condition (III) in Theorem \ref{thm4}
is essential. It was shown in \cite{BBRW} that if $\I$ is an ideal in
${\cal P}(X)$ then, for the dual filter $\F_{\I}=\{ E^c\colon E\in\I\}$
and the algebra $\A_{\I}=\I\cup\F_{\I}$, we have $\A_{\I}=S(\F_{\I})$
and $\I=S_0(\F_{\I})$. Hence $\A_{\I}$ is inner MB-representable.
Let us consider a special case. Assume that $\lambda$ is an infinite
cardinal with $\lambda\le\kappa =|X|$, and put $\I =[X]^{<\lambda}$,
$\A=\A_{\I}$. Then conditions (I) and (II), stated in Theorem \ref{thm4},
are obviously satisfied. However, $\A$ is inner MB-representable, and
(III) is false since $\A^*=\F_{\I}$ and so $\A=S(\F_{\I})=S(\A^*)$.
Finally, note that $\A$ is not outer MB-representable. Indeed, suppose
that $\A =S(\G )$ and $\G \cap\A =\emptyset$. Thus $|E|\ge\lambda$ and
$|E^c|\ge\lambda$ for all $E\in\G$. Since for each $A\in\F_{\I}$ there is
an $E\in\G$ such that $E\subset A$, therefore, by Fact \ref{f2},
from $S(\F_{\I})=\A =S(\G )$ it follows that for each $E\in\G$
there is an $A\in\F_{\I}$ such that $A\subset E$.
This however is impossible.

\noindent
{\bf Acknowledgements.} We would like to thank S. Wro\'nski for a fruitful
discussion. He has allowed us to include Proposition \ref{propWr} 
in our paper.

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\end{document}