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\newtheorem{uw}{Remark}
\newtheorem{lem}{Lemma}
\newtheorem{tw}{Theorem}
\newtheorem{prz}{Example}
\newtheorem{df}{Definition}
\newtheorem{wn}{Corollary}

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    \newcommand{\LIM}{{\rm LIM}_{\ooo_1}\,}
\newcommand{\pf}{{\noindent\sc Proof. }}



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 \begin{document}
{
\renewcommand{\thefootnote}{}
 \footnotetext{AMS Subject Classification (1991): Primary {\bf
 26A21}; Secondary: 03E50; 26A03; 03E75.}
}
{
{\renewcommand{\thefootnote}{}
\footnotetext{Key words: transfinite sequence of functions, transfinite
limit, Continuum Hypothesis, Martin Axiom, $\lambda$-set.}
}
\title{On the convergence of $\o_1$ sequences of real functions}
 \author{T.~Natkaniec\thanks{Partially supported  by NSF Cooperative
Research Grant INT-9600548 with its Polish part
financed by KBN. The paper was prepared during
a visit of the first author at West
Virginia University. We  would like to thank Professors Krzysztof
Ciesielski, Janusz Pawlikowski and Jerzy Wojciechowski for many valuable
remarks.} \hspace{3mm} and  \hspace{3mm} J.~Weso{\l}owska}
\date{}
\maketitle

\begin{abstract}
We study sets of points at which  $\o_1$ sequences of real functions from
a given class $\F$ converge. As $\F$ we consider continuous functions,
first class of Baire, Borel measurable functions, functions with Baire
property and Lebesgue measurable functions. Connections of those problem
with additional set-theoretic axioms are discussed.
\end{abstract}

\section{Introduction}
The notion of the convergence of transfinite sequences of
functions was introduced by W.~Sierpi{\'n}ski in \cite{WS}. So far,
some classes of functions were investigated on
closedness under taking transfinite limits of sequences of
functions. In particular, Sierpi\'{n}ski proved that the classes of all
continuous functions from $\R$ into $\R$ as well as the first
class of Baire are closed under this operation~\cite{WS}. He
also remarked that the second class of Baire has not this property.
Moreover, it is easy to observe that if the Continuum Hypothesis holds
then every function $f\colon\R\to\R$ is the transfinite limit of a
sequence of $B_2$ functions (similar problems were considered recently by
 P.~Komj{\'{a}}th in \cite{PK}). This problem was also considered by many
 mathematicians for other classes of functions, see e.g.,
 P.~Kostyrko~\cite{PKos}, T.~\v{S}alat~\cite{TS},
 A.~Neubrunnov\'{a}~\cite{AN}, J.~Lipi\'{n}ski~\cite{JL},
 Z.~Grande~\cite{ZG} and M.~Dindo\v{s}~\cite{MD}
 (from those papers the most interesting are papers of Lipi{\'{n}}ski
 and Dindo{\v{s}}, who showed that the problem whether the classes of all
 approximately continuous functions and of all derivatives are closed with
  respect to transfinite limits cannot be solved in ZFC).

Our study goes into another  direction, namely
we are interested in characterizing of sets of
convergence points of ${\o}_1$  sequences of functions from a given
class. Recall that if $A$ is a subset of  a Polish space $X$ then $A$ is
$\Pi^0_3$ iff there exists a sequence $(f_n)_n$ of continuous functions
from $X$ into $\R$ such that $A=\{ x\in X\colon (f_n(x))_n\;
\mbox{converges}\}$. (See~\cite[Theorem 23.18, p. 185]{AK}).
J.~Weso{\l}owska has proved recently the analogous result for sequences of
functions from the Baire class $\a$, for any $\a<\o_1$.
Note that all those results are obtained in ZFC. We will show that if we
consider sets of convergence of transfinite sequence of Baire one
functions then the results obtained under the Continuum Hypothesis CH and
under Martin's Axiom MA with the negation of CH do not coincide.

\section{Notations and definitions}

 We will use the standard terminology and notation as
in~\cite{Ci:book}.
In particular, ordinal numbers will be identified with the set of
their
predecessors and cardinal numbers with the initial ordinals.
Thus the first infinite cardinal $\o$ is identified
with the set of natural numbers.
The cardinality of the set $\R$ of real numbers is denoted by
$\co$.
 The letters $X$ and $Y$ will denote metric spaces. We shall consider
 mainly the subsets of the real line $\R$, so our spaces will be
 separable, but not necessary complete.

We will use the following abbreviations for some classes of
subsets of $X$:
\begin{description}
\item[${\P}(X)$] ---  all subsets of $X$;
\item[$ \Pi^0_1(X)\ ( \Sigma^0_1(X))$] ---  closed (open) subsets of $X$;
\item[$ \Pi^0_{ \a}(X)\ (\Sigma^0_{\a}(X))$]  ---
the multiplicative (additive) class $\a$ of
Borel sets, $0 < \a < \o_1$;
\item[$\Delta^0_{\a}(X) = \Pi^0_{ \a}(X) \cap \Sigma^0_{\a}(X)$]
--- ambiguous class $\a$, $0 < \a < \o_1$;
\item[${\B}orel(X)$] --- Borel sets;
\item[${\B}aire(X)$] --- sets with Baire property;
\item[${\M}(X),\ {\N}(X)$] --- $\sigma$-ideals of meager
(first-category) and Lebesgue measure-zero subsets of $X$;
\item[$ \L ebesgue$] --- Lebesgue measurable subsets of $\R$.
\end{description}
It will cause no confusion if we use the same abbreviations
for the classes of Borel, Baire measurable and Lebesgue
measurable functions, respectively.
Moreover, we will denote by ${\C}(X)$ the class of all continuous real
functions defined on $X$ and,
for $\xi<\o_1$, by  ${\B}_{\xi}(X)$ the class of Baire
class-$\xi$ functions.

For a function $f\colon X\to\R$ we denote by $C(f)$ the set of all
continuity points of $f$. We say that $f$ is {\em pointwise discontinuous}
if $C(f)$ is dense in $X$. We say that $f$ is {\em cliquish} iff for
each $x_0\in X$, $\varepsilon >0$ and a neighbourhood $W$ of $x_0$ there
is a non-empty open set $W_0\subset W$ such that $\osc f|_{W_0}
<\varepsilon$. Recall that every pointwise discontinuous function is
cliquish and  this two properties coincide if we assume that $X$ is a
Baire space. Moreover, each cliquish function has the Baire
property.(See e.g., \cite{TN}.)


The cardinality of a set $A$ is denoted by $|A|$. For any
infinite set $X$ and a cardinal $\kappa$ we use the
following symbols:
\begin{enumerate}
\item[ ]$[X]^{\kappa}=\{ Y\subset X: |Y| = \kappa\}$,
\item[ ]$[X]^{< \kappa}=\{ Y\subset X: |Y| < \kappa\}$,
\item[]$[X]^{\leq \kappa}=\{ Y\subset X: |Y| \leq \kappa\}$.
\end{enumerate}

\bigskip
For a family of sets ${\cal A}\subset {\P}(X)$ define

$$ ({\cal A})_{\o_1}=\left\{\bigcup_{\a < \o_1} A_{\a}\colon\;
\{A_{\a}: \a < \o_1\}
\subset {\cal A}\right\}.$$

Recall also some cardinal numbers connected with the
$\sigma$-ideal ${\cal I}$ of subsets of $X$.

\begin{enumerate}
\item[]
$\add({\cal I})=\min\{ |{\cal D}| : {\cal D} \subset {\cal I}\ \& \
\bigcup {\cal D} \not\in {\cal I}\}$,
\item[]
$\cov({\cal I})=\min\{ |{\cal D}| : {\cal D} \subset {\cal I}\  \& \
\bigcup {\cal D} = X \}$,
\item[]
$\non({\cal I})=\min\{ |X_0| : X_0 \subset X\ \& \ X_0
\not\in {\cal I}\}$.
\end{enumerate}

\bigskip
Then for any $\sigma -$ideal ${\cal I}$ on $\R$ the
following diagram holds, in which the symbol $\rightarrow$ denotes the
inequality $\leq$ between cardinal numbers. (See, e.g., ~\cite{Fr}.)

\begin{picture}(300,100)
\put(1,50){${\o_1}$}
\put(30,50){\vector(1,0){30}}
\put(75,50){$\add({\cal I})$}
\put(130,55){\vector(4,1){30}}
\put(130,50){\vector(4,-1){30}}
\put(182,60){\makebox(30,10){$\cov({\cal I})$}}
\put(182,35){\makebox(30,10){$\non({\cal I})$}}
\put(230,63){\vector(4,-1){30}}
\put(230,42){\vector(4,1){30}}
\put(280,50){$2^{\co}$}

\end{picture}


The transfinite sequence $\{x_{\a}:\a < \o_1\}$
of real numbers is said to be convergent to $x\in \R$
if for every positive number $\eee$ there exists an
ordinal $\a_0 < \o_1$ such that $|x_{\a} - x| <
\eee$ for any $\a \geq \a_0$.
Then we say that $x$ is a limit of a sequence $\{x_{\a}:\a < \o_1\}$
and we denote this by $x=\lim_{\a} x_{\a}$.

%R1
\begin{uw}
A transfinite sequence $\{x_{\a}:\a < \o_1\}$
is convergent if and only if there exists an ordinal
$\a_0 < \o_1$ such that $x_{\a}=x_{\a_0}$
for each $\a \geq \a_0$.\Qed
\end{uw}

Let $\{f_{\a}: \a < \o_1\}$ be a transfinite
sequence of real functions. A function $f:\R \to \R$ is said to be the
$\o_1$-limit of this sequence if for every $x\in \R$
 $\{f_{\a}(x): \a < \o_1 \}$ is convergent
and $f(x)=\lim_{\a} f_{\a}(x)$. We write it
$f=\lim_{\a} f_{\a}$.

We will denote by $L(\{f_{\a}: \a <
\o_1\})$ the set of all convergence points of the
sequence $\{f_{\a}: \a < \o_1\}$, i.e.

$$L(\{f_{\a}: \a < \o_1\})=
\{x\in \R:\lim_{\a} f_{\a}(x)\ {\rm exists} \}.$$


For a family of functions ${\F}\subset {\R}^{X}$ define

$$
 \L ({\F})=\left\{ L(\{f_{\a}\colon \a < \o_1\})\colon \;
\{f_{\a}: \a < \o_1\} \subset {\F}\right\}.
$$


For two arbitrary functions $f,\ g\colon X \to \R$ let
$[f=g]=\{x\in X : f(x)=g(x)\}$. Then we have

\bigskip
\noindent
{\bf Basic Remark.}

 $$L(\{f_{\a}: \a < \o_1\})=
\bigcup_{\a < \o_1} \bigcap_{\b \geq \a}
[f_{\a}=f_{\b}].$$\Qed

Thus for any family of functions $\F$ we have
$$ \L (\F)\subset\left\{ \bigcup_{\a<\o_1}
\bigcap_{\b\geq\a} [f_{\b}=g_{\b}]\colon\;
f_{\b},g_{\b}\in\F\right\}.\leqno (*)$$


\begin{lem}\label{LEM GRUP}
If $\F$ is an additive subgroup of $\R^{\R}$, then
$$ \L (\F)=\left\{ \bigcup_{\a < \o_1}
\bigcap_{\b \geq \a}[f_{\b}=g_{\b}]\colon
f_{\b},g_{\b}\in {\F}\right\}.$$
\end{lem}
\pf
The implication $"\subset"$ follows from the Basic Remark.

$"\supset"$. Let $A=\bigcup_{\a < \o_1}
\bigcap_{\b \geq \a} A_{\b}$ and
$A_{\b}=[f_{\b}=g_{\b}]$, where $f_{\b}, g_{\b}\in {\cal
F}$. Then $h_{\b}=f_{\b}-g_{\b} \in {\F}$ and
$A=\bigcup_{\a < \o_1} \bigcap_{\b \geq \a} [h_{\b}=0]$.
Take a bijection $\varphi:\o_1\to 2\times \o_1$,
$\varphi=(\varphi_0,\varphi_1)$ and put
$p_{\a}=\varphi_0(\a)\cdot h_{\varphi_1(\a)}$ for $\a < \o_1$.
Then $p_{\a}$ are in $\F$ and $A=L(\{p_{\a}:\a <
\o_1 \})$.
\Qed

%TW 13
\begin{tw}\label{TW 13}
Let ${\cal A}$ be a $\sigma$-algebra of sets and ${\cal
M}_{\cal A}$ be the family of ${\cal A}$-measurable
functions. Then
$$ \L ({\M}_{\cal A})=\left\{\bigcup_{\a < \o_1}
\bigcap_{\b \geq \a}B_{\b}: B_{\b} \in {\cal A}\right\}.$$
\end{tw}
\pf
The inclusion $"\subset "$ follows from
Basic Remark.
By Lemma~\ref{LEM GRUP}, to prove the inclusion $"\supset "$, it is enough to remark, that every set $A\in {\cal A}$ is the zero-set of the characteristic function of the complement of $A$ and that this function is $\cal A$ measurable.
\Qed

\begin{wn}\label{CU}
$({\cal A})_{\o_1}\subset  \L ({\M}_{\cal A})$.
\end{wn}
\pf
Suppose that $A=\bigcup_{\gamma <\o_1}A_{\gamma}$, where all $A_{\gamma}$
belong to $\cal A$.
It is sufficient to use Theorem~\ref{TW 13} for
$B_{\b}=\bigcup_{\gamma \leq \b}A_{\gamma}$.\Qed

We will show in the end section that generally the inclusion
$(*)$ cannot be replaced by the equality.

\begin{uw}\label{moc}
For any family $\F\subset \R^{X}$ the class $ \L (\F)$ is of the size less or equal to $|\F|^{\o_1}$.
\Qed
\end{uw}

\section{Continuous functions}


%TW 1
\begin{lem}\label{TW 1}
Let $X$ be a metric space.
If $\{A_{\a}: \a < \o_1\} \subset \Pi^0_1(X)$, then
$$\bigcup_{\a < \o_1}
\bigcap_{\b \geq \a}A_{\b} \in \Pi^0_1(X).$$
\end{lem}
\pf
For each   $\a < \o_1$ put
$ F_{\a}=\bigcap_{\b \geq \a}A_{\b}.$
Then $\{ F_{\a}\colon \a < \o_1\}$ form an increasing sequence of closed
sets. Moreover, $F=\bigcup_{\a < \o_1}F_{\a}$ is a closed set. In
fact, if $\{x_n : n < \o \} \subset F$ and
$\lim_{n\to \infty} x_n = x$, then there exists
$\a < \o_1$ such that $x_n \in F_{\a}$ for each
$n < \o $, so $ x\in \cl(F_{\a})=F_{\a}\subset F$.
\Qed

%TW 2
\begin{tw}\label{TW 2}
Let $X$ be a metric space, $A\subset X$ and $f\colon A\to\R$. Then the
following conditions are equivalent.
\begin{description}
\item[(i)]
$A$ is closed and $f$ is continuous;
\item[(ii)]
there exists a sequence $\{ f_{\a}\colon \a<\o_1\}$ of continuous functions
such that $A=L(\{f_{\a}: \a < \o_1\})$ and $f=\lim_{\a}f_{\a}|A$.
\end{description}
\end{tw}
\pf
$(i)\Rightarrow (ii)$. Let $\varphi : \o_1 \to 2\times
\o_1$, $\varphi = ({\varphi}_0 ,{\varphi}_1)$ be an arbitrary bijection.
By Tietze Theorem, there exists  a continuous extension $F:X \to \R$ of
$f$.  For every $\a < \o_1$ define the function $g_{\a}\colon X\to\R$
by the formula $g_{\a}(x)= {\varphi}_0 (\a) \cdot \dist(x,A)$. Note
that $L(\{g_{\a}: \a < \o_1\})=\{x\in X\colon
\lim_{\a}g_{\a}(x)=0 \}=A$. Then for
$f_{\a}=g_{\a} + F$, $\a<\o_1$,  the assertion follows.

$(ii)\Rightarrow (i)$. By Basic Remark and Lemma~\ref{TW 1}, $A$ is closed in $X$.
Suppose that $f$ is discontinuous at $x\in A$. Then there exists a
sequence $\{x_n\colon n<\o\}$ in $A$ such that $\lim_{n\to \infty}
x_n=x$ and $\lim_{n\to \infty} f(x_n)\neq f(x)$. Fix $\a<\o_1$
such that $f(t)=f_{\a}(t)$ for each $t\in\{ x\}\cup\{ x_n\colon n<\o\}$.
Then $f_{\a}(x)\neq \lim_{n\to \infty}f_{\a}(x_n)$, contrary with
continuity of $f_{\a}$.
\Qed

% WN 1
\begin{wn}\label{WN 1}
$ \L ({\C}(X))= \Pi^0_1(X)$.\Qed
\end{wn}
Note that the results above are obtained in ZFC.
The situation will be different in next sections.


\section{First class of Baire}

%TW 3
\begin{tw}\label{TW 3}
For every $\xi < \o_1$
$$ \L ({\B}_{\xi})=\left\{\bigcup \limits_{\a < \o_1}
\bigcap \limits_{\b \geq \a}A_{\b}: A_{\b} \in \Pi^0_{\xi +1}\right\}.$$
\end{tw}
\pf
By Lemma~\ref{LEM GRUP}, it is enough to notice that
for each $B \in \Pi^0_{\xi +1}$ there exists $h\in {\cal B}_{\xi}$ such that $B=[h=0].$ In
fact, $\R \setminus B \in \Sigma^0_{\xi +1}$, so $\R \setminus
B= \bigcup_{n < \o}B_n$ for some sequence $\{B_n:
n < \o \}\subset \Delta^0_{\xi +1}.$ We can assume that
$B_n \cap B_m =\emptyset$ for $n\ne m.$ Then
$$
 h(x)=\cases {{1\over{n+1}} & for $x\in B_n,\ n<\o$,\cr
 0 & for $x\in B$. \cr}
$$
\mbox{}\Qed

%WN 2
\begin{wn}\label{WN 2}
$ \L ({\B}_1)=\left\{\bigcup_{\a < \o_1}
\bigcap_{\b \geq \a}A_{\b}: A_{\b} \in \Pi^0_2\right\}.$\Qed
\end{wn}

\begin{tw}\label{th-bor}
For every $\xi<\o_1$, if $A\in (\Pi^0_{\xi})_{\o_1}$, then $\R \setminus A \in \L ({\B}_{\xi})$.
\end{tw}
\pf
Let $\R \setminus
A=\bigcap_{\a <\o_1}H_{\a}$, where
$H_{\a}\in\Sigma^0_{\xi}$. Put $A_{\a}=\bigcap_{\b \leq \a }H_{\b}$.
Then $A_{\a}\in \Pi^0_{\xi+1}$ and $\R \setminus A = \bigcup_{\a < \o_1}\bigcap_{\b \geq \a}A_{\b}\in  \L ({\B}_{\xi})$.
\Qed

\begin{wn}\label{WN 3}
If $A\in (\Pi_1^0)_{\o_1}$, then $\R\setminus A\in \L(\B_1)$.
\Qed
\end{wn}

Since each set $A\in\Sigma^0_2$ is the union of countable many closed sets, we have the following.
%WN 4
\begin{wn}\label{WN 4}
$\Pi^0_2 \subset  \L ({\B}_1).$\Qed
\end{wn}

%WN 5
\begin{wn}\label{WN 5}
If $A\in [\R]^{\leq \o_1}$, then $\R \setminus A \in  \L ({\B}_1).$\Qed
\end{wn}

Since there are  non-Borel subsets of $\R$ of the size $\o_1$, we have the following.
%WN 6
\begin{wn}\label{WN 6}
$ \L ({\B}_1) \not\subset {\B}orel.$\Qed
\end{wn}

%WN 7
\begin{wn}\label{WN 7}
If CH holds, then $ \L ({\B}_1)={\P}(\R)$.\Qed
\end{wn}

%LEM 1
\begin{lem}\label{LEM 1}
Assume that $X$ is a separable space with
$\cov({\M}(X))> \o_1$,
$\{ f_{\a}\colon X \to \R\colon\; \a < \o_1\}$ is a sequence of pointwise
discontinuous function
and $A=L(\{ f_{\a}: \a < \o_1\}).$ If $A$ is
dense in $X$, then $X\setminus A \in ({\M}(X))_{\o_1}.$
\end{lem}
\pf
Let $f=\lim_{\a}{f_{\a}}|_{A}.$
For each $\a < \o_1$ let $M_{\a}= X\setminus C(f_{\a})
\in {\M}(X).$ Then  $M=\bigcup_{\a < \o_1}M_{\a}\in
({\M}(X))_{\o_1}.$ Let $D$ be a countable dense
subset of $A$. There exists $\a_0 < \o_1$ such that
${f_{\a}}|_{D} = f|_{D}$ for each $\a \geq
\a_0.$ Fix $x\in X\setminus M$ and $\a \geq
\a_0.$ Then $x\in C(f_{\a})$, so $ f_{\a}(x)=
\lim_{t\in D, t\to x}f_{\a}(t)=
\lim_{t\in D, t\to x}f_{\a_0}(t)=
f_{\a_0}(x).$
Thus $x\in A$ and consequently,
$X\setminus A\subset M \in ({\M}(X))_{\o_1}.$
\Qed

%WN 8
\begin{tw}\label{WN 8}
Let $X$ be a separable Baire space with $\cov({\M}(X))> \o_1$. If $A\in
{\M}(X)$ is dense in $X$ then $A\not\in  \L ({\B}_1(X)).$
\end{tw}
\pf
This is a consequence of Lemma~\ref{LEM 1}, because each $\B_1$ function
defined on a Baire space is pointwise discontinuous.\Qed

%TW 4
\begin{wn}
If $\cov({\M}) > \o_1$, then $\Q\not\in   \L ({\B}_1)$.
Thus $\Sigma^0_2  \not\subset  \L ({\B}_1)$. \Qed
\end{wn}

%TW 5
\begin{tw}\label{TW 5}
If $\cov({\M})> \o_1 = \add({\M})$, then
$ \L ({\B}_1)\not\subset {\B}aire.$
\end{tw}
\pf
Since $\add(\M)=\o_1$, there exists a sequence $\{E_{\a}:
\a < \o_1\}$ of closed, nowhere dense sets such that
$E=\bigcup_{\a < \o_1}E_{\a}\not\in {\M}.$ Since
$\cov({\M}) > \o_1$, $E\not\in {\B}aire$, so
$\R \setminus E \not\in {\B}aire.$ By Corollary~\ref{WN 3},
$\R \setminus E \in  \L ({\B}_1).$
\Qed

\begin{tw}\label{cov=o}
If $\cov({\M}) = \o_1$, then $ \L ({\B}_1)\setminus
{\B}aire \ne \emptyset .$
\end{tw}
\pf
Since $\cov({\M}) = \o_1$, $\R=\bigcup_{\a<\o_1}M_{\a}$,
where all $M_{\a}$ are closed and nowhere dense. We will show that there
is a set $\Omega\subset\o_1$ such that $\bigcup_{\a\in\Omega}M_{\a}
\not\in {\B}aire$.
\footnote{This fact has been proved  by Janusz Pawlikowski.}
Consider the following ideal $\cal J$ defined on $\o_1$:
$A\in {\cal J}$ iff $\bigcup_{\a\in A}M_{\a}\in {\B}aire$.
Then $\cal J$ satisfies all assumptions of Ulam's Theorem, so there exists
$\Omega\not\in {\cal J}$. (See, e.g., \cite[Theorem 6.11, p.79]{KKun}.)

Then $E=\bigcup_{\a\in\Omega}M_{\a}\not\in {\B}aire$ and by Corollary~\ref{WN 3}, $\R\setminus E\in  \L ({\cal B}_1)\setminus
{\B}aire$. \Qed

Note that the assumption $\cov(\M)=\o_1$ does not imply the equality
$ \L ({\B}_1)={\P}(\R)$.
\footnote{Such a problem has been formulated by Jerzy Wojciechowski.}
This is a consequence of the next theorem and the fact that there are
models of ZFC in which $\cov(\M)=\o_1$ and $2^{\o_1}<2^{\co}$.
(In fact, this holds if we add $\o_2$ random reals to a model of
ZFC with the Generalized Continuum Hypothesis GCH.
\footnote{This fact has been communicated to us by Krzysztof Ciesielski.}
(See, e.g., \cite{BJ}.))

\begin{tw}\label{moc-b-1}
If $2^{\o_1}<2^{\co}$, then $ \L ({\B}_1)\neq {\P}(\R)$.
\end{tw}
\pf
By Remark~\ref{moc}, $| \L ({\B}_1)|\leq \co^{\o_1}=
(2^{\o})^{\o_1}=2^{\o_1}$, while $|{\P}(\R)|=2^{\co}$. \Qed

\begin{pr}\label{pro-B-1}
Does the statement $ \L ({\B}_1)={\P}(\R)$ imply CH?
\end{pr}


%TW 8
\begin{tw}\label{TW 8}
If $\add({\M})> \o_1$, then  $  \L ({\B}aire)\subset
{\B}aire.$
\end{tw}
\pf
It is well known that the  assumption $\add({\M})> \o_1$ implies that
the ${\B}aire$ class is closed under $\o_1$ sums and
intersections. (See, e.g., \cite{JSh}.) Thus
the inclusion follows from Basic Remark.
\Qed

%WN 10
\begin{wn}\label{WN 10}
If $\add({\M})> \o_1$, then  $  \L ({\B}_1)\subset
{\B}aire.$\Qed
\end{wn}

By Lemma~\ref{LEM 1} we also obtain.
%TW 7
\begin{tw}\label{TW 7}
If $\add({\M})> \o_1$, then every set $A\in  \L ({\B}_1)$
satisfies the following condition

\begin{description}
\item[$(*)$] for each non-empty perfect set $P\subset \R$,
if $P\subset \cl(A\cap P)$, then $P\setminus A\in {\M}(P)$.
\end{description}
\mbox{}\Qed
\end{tw}

\bigskip
For a family ${\F}\subset \R^X$ define $${\rm LIM}_{\o_1}({\F})=
\left\{ \lim_{\a}f_{\a}\colon\;
\{f_{\a}: \a < \o_1\}\subset {\F}\right\}.$$

Let $A$ be a subset of $\R$. The following statements follow easily
from Sierpi\'{n}ski's paper~\cite{WS}.
\begin{tw}\label{sierp}
\mbox{}
\begin{itemize}
\item
${\rm LIM}_{\o_1}(\C(A))=\C(A)$;
\item
If $A\in \Pi^0_2$, then ${\rm
LIM}_{\o_1}({\B}_1(A))=\B_1(A)$.\Qed
\end{itemize}
\end{tw}

In particular, if $A=\R$ then $\LIM(\B_1)\neq\R^{\R}$. On the other
hand, if $|A|\leq\o$ then $\B_1(A)=\R^A$, so
$\LIM({\B}_1(A))=\B_1(A)=\R^A$.

%TW 10
\begin{tw}\label{TW 10}
Assume CH. There exists a subset $A$ of $\R$ such that ${\B}_1(A)\ne
{\R}^A$ and
${\rm LIM}_{\o_1}({\B}_1(A)) = {\R}^A.$
\end{tw}
\pf
Let $A\subset \R$ be an uncountable $\lambda$ - set, i.e. such that
every countable subset of $A$ is a ${\cal G}_{\delta}$ set relative
to $A$. (See, e.g., \cite{KKur}.) Let $A=\{x_{\a}:\a < \o_1\}$.
Take an arbitrary function $f\colon A\to \R$ and  for $\a < \o_1 $ put
$$
 f_{\a}(x)=\cases {f(x) & for $x=x_{\b},\ \b \leq \a $,\cr
 0 & otherwise. \cr}
$$
Then $f_{\a}\in {\B}_1(A)$ and
$f=\lim_{\a}f_{\a}$, so $f\in {\rm
LIM}_{\o_1}({\B}_1(A))$. Of course, $ {\B}_1(A)\ne \R^A $,
because $|{\B}_1(A)|=\o_1< 2^{\o_1}=|\R^A| $.
\Qed

Note that Theorem~\ref{TW 10} cannot be proved in ZFC.

\begin{tw}\label{ma}
Assume MA$+\neg$CH. There exists no $A\subset\R$ such that
${\B}_1(A)\ne {\R}^A$ and ${\rm LIM}_{\o_1}({\B}_1(A)) = {\R}^A.$
\end{tw}
\pf
Let us consider two cases.

Case 1: $ A\in[\R ]^{< \co} .$ Then $A$ is a $Q-set$, i.e. every
subset of $A$ is a relative $\Sigma_0^2$.
(See, e.g., \cite{MS}.)
Consequently, if $f:A\to \R$, then $f\in {\B}_1(A)$, so $\B_1(A)=\R^A$.

Case 2:  $A\in [\R ]^{\co}$. Then ${\rm LIM}_{\o_1}({\cal
B}_1(A)) \neq {\R}^A$, because
$ |{\rm LIM}_{\o_1}({\B}_1(A))| = {\c}^{\o_1} = \c <
2^{\c} = |{\R}^A|$.
\Qed

The situation is different if we consider a $\o_1$-sequence of
$\B_1$ functions defined on whole real line which converge on a set
$A\subset\R$.

%TW 11
\begin{tw}\label{TW 11}
For every transfinite sequence $\{f_{\a}: \a < \o_1\} \subset
{\B}_1(\R)$ and $A\subset L(\{f_{\a}: \a <
\o_1\})$ the function $f=\lim_{\a}{f_{\a}}|_{A}$ is cliquish.
\end{tw}
\pf
Suppose, to obtain a contradiction, that $f$ is not cliquish.
Then there exists $n<\o$ such that $A_n = \{x\in A:
\osc_f(x)\geq {1\over n}\}$ is a closed non-nowhere dense subset of $A$.
Thus there is an interval $(a,b)$ such that $\emptyset \neq
(a,b)\cap A \subset A_n.$ Let $B=(a,b)\cap A$ and $C=\{c_k : k < \o \}$
be a dense subset of $B.$ Then for any $k,\ i <\o$ there are
$d'_{k,i},\ d''_{k,i}\in (c_k - {1\over i},\ c_k + {1\over
i}) \cap B$ such that $|f(d'_{k,i}) - f(d''_{k,i})| \geq
{1\over{2n}}.$ Thus $D=\{d'_{k,i},\ d''_{k,i} : k,\ i < \o
\}$ is  dense in $B$ and for every $x\in \cl(B)$
$\osc_f(x)\geq {1\over{2n}}.$ Moreover, since $D$ is a
countable subset of $A$, there is $\a_0 < \o_1$ such that
$f|_D=f_{\a_0}|_D.$ Consequently, $f_{\a_0}|_{\cl(B)}$ has no
continuity points, which contradicts the assumption that
$f_{\a_0}\in {\B}_1(\R).$
\Qed

The next theorem is a slight modification of  Lebesgue's well-known
characterization of $\B_1$ functions. (See, e.g., \cite{RSik}.)

%TW 12
\begin{tw}\label{TW 12}
Let $X$ be a second countable perfect topological space.
If for every closed subset
$F$ of $X$ the restriction $f|_{F}$ of a function $f:X\to\R$
is cliquish, then $f\in {\B}_1(X).$
\end{tw}
\pf
Fix a non-empty open set $G\subset\R$. For $n<\o$ define
$F_n=\{y\in\R: \dist(y, \R\setminus G) \geq {1\over n}\}.$ Then $F_n$ are
closed, $G=\bigcup_{n<\o}F_n$ and $\dist(F_n,\R\setminus G)\geq {1\over n}$.
We can assume that $G$ is bounded and have to show that

\begin{description}
\item[$(*)$] for every $i<\o$ there is
$H_i\in \Sigma^0_2(X)$ such that $f^{-1}(F_i)\subset H_i\subset f^{-1}(G)$,
\end{description}
which gives $f^{-1}(G)=
\bigcup_{i<\o} H_i \in \Sigma^0_2(X)$.

Let $\{B_k: k<\o\}$ be a base of $X$. Fix $i<\o$ and let $K=F_i$. Put
$$M=\{n<\o: \exists H_n \in \Sigma^0_2(X)\ (f^{-1}(K)\cap B_n
\subset H_n\subset f^{-1}(G)\cap B_n)\}.$$
Let $U=\bigcup_{n\in M} B_n$, $H=\bigcup_{n\in
M} H_n$ and $F=X\setminus U$. Then $F$ is closed, $H\in
\Sigma^0_2(X)$ and $f^{-1}(K)\cap U \subset H \subset
f^{-1}(G)\cap U$.
We will prove that $F=\emptyset$. To obtain a
contradiction,
suppose that $F\neq \emptyset$. By assumption, $f|_F$ is
cliquish. We have two cases to consider.

Case 1:\quad There exists $m<\o$ such that $F\cap B_m \neq \emptyset$
and $f(F\cap B_m)\subset G.$  Then $H'=(H\cap B_m)\cup (F\cap B_m)\in
\Sigma^0_2(X)$ and $f^{-1}(K)\cap B_m \subset H'\subset
f^{-1}(G)\cap B_m.$ Thus $m\in M$ and consequently $B_m
\subset U$, so $F\cap B_m=\emptyset$, a contradiction.

Case 2:\quad For every $m<\o$ such that $F\cap B_m \neq \emptyset$
we have $f(F\cap B_m)\not\subset G$.  Then there is $m_0<\o$ such that
$F\cap B_{m_0} \neq \emptyset$ and $f(F\cap B_{m_0})\cap K = \emptyset.$
Indeed, if we assume otherwise, then both $F\cap
f^{-1}(K)$ and $F\cap f^{-1}(\R\setminus G)$ are dense in
$F.$ Since $K$ is compact, $\eee=\dist(K, \R\setminus G)> 0$, so
$\osc_f(x)\geq \varepsilon$ for every $x\in F$, which contradicts the
cliquishness of $f|_F$. Therefore, $H'=H\cap B_{m_0}\in
\Sigma^0_2(X)$ and $f^{-1}(K)\cap B_{m_0} \subset H'\subset
f^{-1}(G)\cap B_{m_0}$, so $m_0 \in M$,  $B_{m_0}
\subset U$ and $F\cap B_{m_0}=\emptyset$, a contradiction.

Finally,  we have $F=\emptyset$, so $U=X$ and $H$ satisfies
$(*)$ for fixed $K=F_i$.
\Qed

%WN 11
\begin{wn}
For every transfinite sequence $\{f_{\a}: \a < \o_1\} \subset
{\B}_1(\R)$ and $A\subset L(\{f_{\a}: \a < \o_1\})$ we have  $f=
\lim_{\a}{f_{\a}}|_{A}\in {\B}_1(A)$.
\end{wn}
\pf
This is a consequence of Theorems~\ref{TW 11} and \ref{TW 12}.\Qed


\section{Baire measurable functions}

Theorem~\ref{TW 13} and Corollary~\ref{CU} imply the following two results on sequences of functions with Baire property.

\begin{tw}\label{baire}
$\L(\B aire) =\left\{\bigcup_{\a<\o_1}\bigcap_{\b\geq\a}B_{\b}\colon\;
B_{\b}\in\B aire\right\}$\Qed
\end{tw}

%WN 12
\begin{wn}\label{WN 12}
$({\B}aire)_{\o_1} \subset  \L ({\B}aire)$.\Qed
\end{wn}

%WN 13
\begin{wn}\label{WN 13}
If $\cov({\M})=\o_1$, then $ \L ({\B}aire)={\P}(\R)$.
\end{wn}
\pf
Let $\R = \bigcup_{\a < \o_1}
M_{\a}$, where $M_{\a}\in {\M}$.
Fix $A\subset \R$. Then  we have $A=\bigcup_{\a < \o_1}
\bigcap_{\b \geq \a}B_{\b}$ for $B_{\b}=A\cap
\bigcup_{\xi \leq \b}M_{\xi} \in {\B}aire$. By
Theorem~\ref{baire}, $A\in  \L ({\B}aire)$.
\Qed

A set $A\subset\R$ is called a {\em Bernstein set} if $B$ and
$\R\setminus B$ meet every non-empty perfect set $P\subset\R$.

\begin{lem}\label{bernstein}
Assume $\cov(\M) > \o_1$. There is no Bernstein set $B\in (\M)_{\o_1}$.
\footnote{This fact has been proved  by Janusz Pawlikowski.}
\end{lem}
\pf
We consider the space $K(\R)$ of all compact subsets of the real line with
the Hausdorff metric. (See, e.g., \cite[p. 25]{AK}.)
Recall that this is a Polish space (\cite[Theorem 4.25, p.~26]{AK}), so
$\cov(\M(K(\R)))=\cov(\M)$.

Observe that if $M\subset\R$ is a nowhere dense closed set then
$\R\setminus M$ is open and dense. Consequently, $\{ P\in K(\R)\colon
P\subset \R\setminus M\}$ is open and dense in $K(\R)$, so $\P_M=
\{ P\in K(\R)\colon P\cap M\neq\emptyset\}$ is nowhere dense in $K(\R)$.
Suppose that $B\in (\M)_{\o_1}$, $B=\bigcup_{\a<\o_1}M_{\a}$, where all
sets $M_{\a}$ are nowhere dense in $\R$. Then
$$\P_B=\{ P\in K(\R)\colon P\cap B\neq\emptyset\}=\bigcup_{\a<o_1}
\P_{M_{\a}}\in (\M(K(\R)))_{\o_1}.$$
Since $\cov(\M(K(\R)))>\o_1$, $K(\R)\setminus\P_B\neq\emptyset$.
Thus there exists a non-empty perfect set $P$ such that
$P\cap B=\emptyset$, so $B$ is not a Bernstein set.\Qed

%TW 14
\begin{tw}\label{TW 14}
If $\cov({\M}) > \o_1 $, then $ \L ({\B}aire)\ne {\P}(\R)$.
\end{tw}
\pf
We will prove that no Bernstein set belongs to $ \L ({\cal B}aire)$.
Assume, to obtain a contradiction, that $B\in  \L ({\cal
B}aire)$. Then, by Theorem~\ref{baire},  $B= \bigcup_{\a < \o_1}
\bigcap_{\b \geq \a}B_{\b}$, where $ B_{\b} \in
{\B}aire$. For each $\a < \o_1$, $B_{\a}=G_{\a} \cup
M_{\a}$, where $G_{\a}\in \Pi^0_2$ and $M_{\a}\in {\M}$.
By Lemma~\ref{bernstein}, $B\setminus \bigcup_{\a < \o_1}
M_{\a}$ is dense in $\R$. Let $D\subset B\setminus \bigcup_{\a < \o_1}
M_{\a}$ be a countable dense subset of $\R$ and let $\a <
\o_1$ be such that $D\subset \bigcap_{\b \geq \a
}G_{\b} \subset B $. Then each $G_{\b}$ is residual, so $\R \setminus B
\subset \bigcup_{\b \geq \a}(\R \setminus G_{\b}) \in
({\M})_{\o_1}$, contrary to Lemma~\ref{bernstein}.
\Qed

By Theorem~\ref{TW 8} and Corollary~\ref{WN 12} we obtain the following.

%TW 15
\begin{tw}\label{TW 15}
If $\add({\M}) > \o_1 $, then $ \L ({\B}aire)= {\B}aire$.
\Qed
\end{tw}

%TW 16
\begin{tw}\label{TW 16}
If $\non({\M}) = \o_1 $, then $ \L ({\B}aire)\setminus
{\B}aire \ne \emptyset$.
\end{tw}
\pf
By Corollary~\ref{WN 13} it is enough to
consider the case that $\cov({\M}) > \o_1$. Since
$\non({\M}) = \o_1 $, there
exists $A\subset \R$ such that $ |A|=\o_1$ and $A\not\in
{\M}$. Consequently,  $A\not\in {\B}aire$ but, by
Corollary~\ref{WN 5}, $\R\setminus A \in  \L ({\B}_1) \subset
 \L ({\B}aire)$.\Qed


\section{Borel functions}

By Theorem~\ref{TW 13} we obtain the following characterization of the
class $ \L ({\B}orel)$.

%TW 17
\begin{tw}\label{TW 17}
$ \L ({\B}orel)=\left\{\bigcup_{\a < \o_1}
\bigcap_{\b \geq \a}B_{\b}: B_{\b} \in {\B}orel \right\}$.
\Qed
\end{tw}

Corollary~\ref{CU} implies the following fact.
%WN 15
\begin{wn}\label{WN 15}
$(\B orel)_{\o_1}\subset  \L ({\B}orel)$.\Qed
\end{wn}

Denote by $\Sigma^1_1$ and $\Pi^1_1$ the sets of all
analytic and co-analytic subsets of $\R$, respectively.

%WN 16
\begin{wn}\label{WN 16}
$\Sigma^1_1\cup \Pi^1_1 \subset  \L ({\B}orel).$
\end{wn}
\pf
It follows immediately from Corollary~\ref{WN 15},
because by Lusin-Sierpi\'{n}ski Theorem every analytic (co-analytic) set
is the union of $\o_1$ many of Borel sets. (See, e.g.,
\cite[Theorem 25.16, p.~201]{AK}.)\Qed

%WN 17
\begin{wn}
$\B orel\not\subset \L ({\B}orel)\not\subset {\B}orel$.
\end{wn}
\pf
The first implication follows from Corollary~\ref{WN 7}, the second
one -- from Corollary~\ref{WN 16}.\Qed

Corollary~\ref{WN 7} implies the following.

%TW 18
\begin{tw}\label{TW 18}
If CH holds,
then $ \L ({\B}orel)={\P}(\R)$.\Qed
\end{tw}

\begin{tw}
If $\cov({\M}) > \o_1 $, then ${\B}aire \not\subset  \L
({\B}orel).$
\end{tw}
\pf
Let $C \subset \R$ be the ternary Cantor set and
let $B\subset C$ be a Bernstein set in $C$.
 Suppose that $B=L(\{f_{\a}:\a <
\o_1\})$ for some sequence $\{f_{\a}:\a < \o_1\}$  of Borel
functions from $\R$ into $\R$. In particular, each $f_{\a}|_C$ has the
Baire property. However, as in the proof of Theorem~\ref{TW 14} we can
prove that $B$ is not the set of convergence of a $\o_1$ sequence of
Baire functions defined on $C$, a contradiction. \Qed

%WN 18
\begin{tw}\label{WN 18}
The following conditions are equivalent:
\begin{description}
\item[(i)]
$\add({\M}) > \o_1$;
\item[(ii)]
 $ \L (\B_1)\subset {\B}aire$;
\item[(iii)]
 $ \L ({\B}orel)\subset {\B}aire$;
 \item[(iv)]
 $ \L ({\B}aire) = {\B}aire$.
\end{description}
\end{tw}
\pf
The implication $(i)\Rightarrow (iv)$ it is exactly Theorem~\ref{TW 15}.

Since $\B_1\subset\B orel\subset \B aire$ and the operation $ \L (.)$
is monotone, we have the inclusions
$$ \L (\B_1)\subset \L (\B orel)\subset \L (\B aire).$$
Thus we have the implications $(iv)\Rightarrow (iii) \Rightarrow (ii)$.

To see the implication $(ii)\Rightarrow (i)$
suppose that $\add({\M}) = \o_1$. Then two cases are
possible.

Case 1: $\cov({\M}) = \o_1 $. Then by Theorem~\ref{cov=o},
$ \L ({\B}_1)\setminus {\B}aire \ne \emptyset$.


Case 2:  $\cov({\M}) > \o_1 $. Then by Theorem~\ref{TW  5},
$ \L ({\B}_1)\not\subset {\B}aire$.

\noindent Both of them contradict our assumption.
\Qed


Recall that a set $A\subset\R$ has {\em the Baire property in the
restricted sense} if for each perfect set $P\subset \R$, $A\cap
P\in\B aire(P)$. (See \cite[p. 61]{KKur}.)

\begin{tw}\label{Marcz}
If $\add({\M})> \o_1$, then every set $A\in  \L ({\cal B}_1)$ has
the Baire property in the restricted sense.
\end{tw}
\pf
This is a consequence of Theorems~\ref{TW 13}, \ref{WN 18} and the fact
that for any perfect set $P\subset \R$ cardinals $\add(\M)$ and $\add(\M(P))$ coincide. \Qed

In particular, every set $A\in  \L ({\B}orel)$ has the
{\em property $(s)$ of Marczewski}, i.e. every perfect set $P$
contains a perfect subset $Q$ such that either  $Q\subset A$
or $A\cap Q = \emptyset.$ (See \cite{BC}.) It implies that the inclusions (ii) and (iii) in Theorem~\ref{WN 18} cannot be replaced by the equalities.


\section{Lebesgue measurable functions}

By Theorem~\ref{TW 13} and Corollary~\ref{CU} we obtain the following two results.

%TW 22
\begin{tw}\label{TW 22}
$ \L ( \L ebesgue)=\left\{\bigcup_{\a < \o_1}
\bigcap_{\b \geq \a}B_{\b}: B_{\b} \in  \L ebesgue\right\}.$\Qed
\end{tw}


%WN  19
\begin{wn}\label{WN 19}
$ (\L ebesgue)_{\o_1} \subset  \L ( \L ebesgue)$.\Qed
\end{wn}

In the same way as Corollary~\ref{WN 13} we can prove the following
%WN  20
\begin{wn}\label{WN 20}
If $\cov({\N})=\o_1$, then $ \L ({\cal
L}ebesgue)={\P}(\R)$.\Qed
\end{wn}


%TW  23
\begin{tw}\label{TW 23}
If $\add({\N})> \o_1$, then  $  \L ( \L ebesgue)=
 \L ebesgue$.
\end{tw}
\pf
The implication ``$\subset$'' follows from Basic Remark and the well-known
fact that under the assumption $\add({\N})> \o_1$ the algebra
$ \L ebesgue$ of Lebesgue measurable sets is closed under $\o_1$ sums and
intersections. (See \cite{JSh}.) The other implication follows from
Corollary~\ref{WN 19}.\Qed

Note that the measure analogue of Lemma~\ref{bernstein} is false. In fact,
let $V$ be a model of ZFC+GCH. In the extension of $V$ on $\o_2$ random
reals there exists a Bernstein set $B$ that is the union of $\o_1$ sets
of measure zero.\footnote{This fact has been communicated to us by Janusz
Pawlikowski.} (See \cite{BJ}.) We can construct a Bernstein set
$B\not\in (\N)_{\o_1}$, however we are unable to prove the analogue of
Theorem~\ref{TW 14}.

\begin{pr}\label{pro-m}
Assume $\cov(\N)>\o_1$. Does there exist $A\in\P(\R)\setminus  \L
( \L  ebesgue)$?
\end{pr}

%TW  24
\begin{tw}\label{TW 24}
If $\non({\N}) = \o_1 $, then $ \L ( \L ebesgue)\setminus
 \L ebesgue \ne \emptyset$.
\end{tw}
\pf The proof is similar to that of
Theorem~\ref{TW 16}.\Qed

\begin{tw}\label{miara}
If $\cov(\N)>\o_1=\add(\N)$, then $ \L ( \L ebesgue)\not\subset
 \L ebesgue$.
\end{tw}
\pf
Since $\add(\N)=\o_1$, there exist $\{ N_{\a}\colon \a<\o_1\}$ with
$N=\bigcup_{\a<\o_1}N_{\a}\not\in\N$.
Since $\cov(\N)>\o_1$, $N\not\in  \L ebesgue$. By Corollary~\ref{WN 19},
$N\in  \L ( \L ebesgue)$.\Qed

%WN 22
\begin{wn}\label{WN 22}
The following conditions are equivalent:
\begin{description}
\item[(i)]
$\add({\N}) > \o_1$;
\item[(ii)]
$ \L ( \L ebesgue)= \L ebesgue$.
\end{description}
\end{wn}
\pf
The implication $(i)\Rightarrow (ii)$
it is exactly Theorem~\ref{TW 23}. To see the other implication
suppose that $\add({\N}) = \o_1$. Then we have  two
possibilities.

Case 1: $\cov({\N}) = \o_1 $. Then by Corollary~\ref{WN 20},
$ \L ( \L ebesgue)={\P}(\R)$.

Case 2:  $\cov({\N}) > \o_1 $. Then by Theorem~\ref{miara},
$ \L ( \L ebesgue)\not\subset  \L ebesgue$.

This contradicts our assumption.\Qed


\section{Quasi-continuous functions}

A function $f\colon X\to\R$ is {\em quasi-continuous} if for each
$x_0\in X$, $\varepsilon >0$ and a neighbourhood $W$ of $x_0$ there
is a non-empty open set $W_0\subset W$ such that $|f(x_0)-f(x)|
< \varepsilon$ for $x\in W_0$.

Recall that each quasi-continuous function is cliquish. (See, e.g.,
\cite{TN}.) Moreover, if $X$ is a complete space (e.g., for $X=\R$),
then $f$ is quasi-continuous iff $f|_{C(f)}$ is dense in $f$.
(No distinction is made between a function and its graph.)

The class of all quasi-continuous functions defined on $X$
will be denoted by $\QC(X)$ (or $\QC$ when $X=\R$).
The family of all cliquish functions is denoted by $\C liquish(X)$
(respectively, $\C liquish$ if $X=\R$).

\begin{lem}\label{LEM QC1}
Let $J$ be an open interval, $J=(a,b).$ Then for each $\varepsilon > 0$
and every countable set $Q$ dense in
$[0,\varepsilon]$ there exists a quasi-continuous function $g:J\to Q$ such
that the cluster sets of $g$ at the end-points of
$J$ are $[0,\varepsilon].$
\end{lem}
\pf
Choose sequences $\{a_n : n<\o\}$,
$\{b_n : n<\o\}$ such that $a_n \searrow a$, $b_n \nearrow
b$ and $a_0=b_0$. Let $Q=\{q_n: n<\o\}$ and $\varphi : \o
\to \o \times \o$, $\varphi = (\varphi_0, \varphi_1)$ be a
bijection. Define  $g(x) = q_{\varphi_1(n)}$ for $x \in (a_{n+1},a_n]\cup
[b_n,b_{n+1})$. Then $g$ satisfies required conditions.
\Qed

\begin{tw}\label{TW QC}
The class $\L ({\QC})$ is equal to the family of all sets of
the form $(G\setminus M)\cup N$, where $G$ is open, $M\in \M$
and $N$ is nowhere dense. Therefore, $\L ({\QC})\subset
{\B}aire$.
\end{tw}
\pf
"$\subset$". Let $A=L(\{f_{\a}: \a <
\o_1 \})$ for some sequence $\{f_{\a}: \a < \o_1 \} \subset {\QC}$ and
$f=\lim_{\a}f_{\a}|_A$. Consider a sequence  $\{I_n:
n < \o \}$ of components of $\int(\cl(A))$. Then $A=(\bigcup_{n<\o}I_n\cap A)
\cup (A\setminus \int(\cl(A)))$ and $A\setminus \int(\cl(A))$ is nowhere dense.
Fix $n< \o$. Let $A_0=I_n\cap A$.
Choose a countable set $C_0\subset A_0,$ dense in $A_0$.
Then there is $\a_0 < \o_1$ such that
\begin{description}
\item[($*$)] $f_{\a}|_{C_0}=f|_{C_0}$ for every $\a \geq \a_0$.
\end{description}
Notice that $C(f_{\a_0})$ is residual, by quasi-continuity of
$f_{\a_0}$. Now, fix $x\in I_n\cap C(f_{\a_0})$ and
$\a \geq \a_0$. We will show that
$f_{\a}(x)=f_{\a_0}(x).$ By quasi-continuity of $f_{\a}$ at
$x$, for every $k < \o$ there is a non-empty open set $G\subset I_n \cap
(x-{1\over k},x+ {1\over k})$ such that $f_{\a}(G)\subset
(f_{\a}(x)-{1\over k},f_{\a}(x)+ {1\over k})$. Since $C_0$
is dense in $I_n$, there is $x_k\in C_0\cap G$ such that
$f_{\a}(x_k)\in (f_{\a}(x)-{1\over k},f_{\a}(x)+ {1\over
k})$. Then we have a sequence $\{x_k: k < \o \} \subset C_0$
such that
$\lim_{k\to \infty}x_k=x$ and
$\lim_{k\to \infty}f_{\a}(x_k)=f_{\a}(x)$.
Consequently, by $(*)$ and continuity of $f_{\a_0}$ at $x$
we have
$$f_{\a}(x)=\lim_{k\to \infty}f_{\a}(x_k)=
\lim_{k\to \infty}f_{\a_0}(x_k)=f_{\a_0}(x)$$
 for every $\a \geq \a_0$, so $x\in A$.

Thus for every $n<\o$,
$I_n \cap C(f_{\a_0}) \subset A\cap I_n$. Since $C(f_{\a_0})$ is
residual in $I_n$, $ I_n \setminus A \in \M$. Putting
$G=\bigcup_{n<\o}I_n$, $M=\bigcup_{n<\o}(I_n\setminus A)$,
$N=A\setminus \int(\cl(A))$ we have the sets we claimed.

"$\supset$". Fix $A=(G\setminus M)\cup N$, where $G$ is open,
$M\in \M$ and $N$ is nowhere dense.  Let $\{J_k: k<\o\}$ be a
sequence of all components of $G$. Fix $k<\o$ and $J=J_k$.
Then $B=J\setminus A \in {\M}$, so $B\subset\bigcup_{n<\o}F_n$,
where $F_n$ are closed and nowhere dense subsets of $J$. Fix $n<\o$. Then,
by Lemma~\ref{LEM QC1} there is $g_n\colon J\to [0,{1\over{2^n}}]$ such that
\begin{enumerate}
\item
$g_n \in {\QC}(J)$;
\item
  $g_n(x)=0$ for $x\in F_n$;
\item
 if $(a,b)$ is a component of $J\setminus F_n$,
then $C^+(g_n,a)=C^-(g_n,b)=[0,{1\over{2^n}}]$.
\end{enumerate}
(here $C^+(g,x)$ and $C^-(g,x)$ denote the right and left hand cluster set
of $g$ at $x$, respectively).

Put also

$$
h_n(x)=\cases {g_n(x) & if $x\not\in F_n$,\cr 0 & if $x\in F_n\cap A$, \cr
{1\over{2^n}} & if $x\in F_n\setminus A$. \cr}
$$
Then the functions $g,\ h :J\to \R$ given by formulas
$g(x)=\sum_{n<\o}g_n(x)$, $h(x)=\sum_{n<\o}h_n(x)$ are
quasi-continuous. For  $\a < \o_1$ put

$$
f_{\a}|_J=\cases {g & if $\a$ is even, \cr
h & if $\a$ is odd. \cr}
$$
Then $L(\{f_{\a}|_J: \a < \o_1\})=J\cap A$. Moreover, cluster sets of
$f_{\a}|J$ at end-points of $J$ are equal to the unit interval.

Now, let $\{I_m: m<\o\}$ be a sequence of components of the set
$\int(\R\setminus G)$. Fix $m<\o$ and put $I=I_m$.
Then $I\cap A \subset N\cap I\subset F$ and
 $F$ is a  closed, nowhere dense subset of $\cl(I)$. Fix $n<\o$.
Let $P=\sqrt{2}\Q$. Then,
by Lemma 4 there are $t, k:I\to [0,1]$ such that:
\begin{enumerate}
\item
$t, k \in {\QC}(I)$;
\item
$k(x)=0$ for $x\in F$;
\item
 if $(a,b)$ is a component of $I\setminus F$,
then $C^+(t,a)=C^-(t,b)=C^+(k,a)=C^-(k,b)=[0,1]$;
\item
$t(I)\subset P\cap (0,1]$ and $k(I)\subset \Q\cap [0,1]$.
\end{enumerate}
Define also $p \in \QC(I)$ by the formula
$$
p(x)=\cases {t(x) & if $x\not\in F$,\cr 0 & if $x\in F\cap A$, \cr
1 & if $x\in F\setminus A$. \cr}
$$
  For  $\a < \o_1$ put
$$
f_{\a}|_I=\cases {k & if $\a$ is even, \cr
p & if $\a$ is odd. \cr}
$$
Then $L(\{f_{\a}|_I: \a < \o_1\})=I\cap A.$

Notice that $\R = \bigcup_{k<\o}J_k\cup
\bigcup_{m<\o}I_m \cup M$, where $M$ is nowhere dense.
Moreover, the cluster sets of each $f_{\a}|_{\R\setminus M}$ at every point
of $M$ equal $[0,1]$. For $\a<\o_1$ define $f_{\a}|_M$
such that $f_{\a}|_M(M\cap A)=\{0\}$ and
$$
f_{\a}|_M(M\setminus A)=\cases {0 & if $\a$ is even, \cr
1 & if $\a$ is odd. \cr}
$$
Putting $f_{\a}=\bigcup_{k<\o}f_{\a}|_{J_k}\cup
\bigcup_{m<\o}f_{\a}|_{I_m} \cup f_{\a}|_M$ we get a required sequence.
\Qed


\begin{lem}\label{LEM CL}
$ \L ({\C}liquish)=\left\{ \bigcup_{\a < \o_1}
\bigcap_{\b \geq \a}[f=g]\colon\;
f,g\in {\QC}\right\}$.
\end{lem}
 \pf
 It is well-known (and easy to verify) that the class $\C liquish$ is
closed with respect to addition. (See, e.g., \cite{ZG1}.)
Thus Lemma~\ref{LEM GRUP} implies that
$ \L ({\C}liquish)=\left\{ \bigcup_{\a < \o_1}
\bigcap_{\b \geq \a} A_{\b}\colon\; A_{\b}=[f=0],
f\in {\C}liquish\right\}$.

Fix $f\in\C liquish$. Then $f$ is the sum sum of two quasi-continuous
functions $g$ and $h$~\cite{AMal}.
Hence $[f=0]=[g=-h]$ and consequently,
$$\left\{ [f=g]\colon f,g\in \C liquish\right\}=\left\{ [f=0]\colon
f\in\C liquish\}=\{ [g=h]\colon g,h\in \QC\right\}.$$
\Qed


Now we will show that in the case $\F=\QC$ the inclusion
$$ \L (\F)
\subset\left\{ \bigcup_{\a<\o_1}\bigcap_{\b\geq\a}[f=g]\colon\;
f,g\in\F\right\}$$
is not reversible. In fact, if we assume CH, then $ \L ({\cal C}liquish)=
{\P}(\R)$ (this is a consequence of  Corollary~\ref{WN 7}, because
$\B_1\subset \C liquish$). So Lemma~\ref{LEM CL} implies that under CH,
$\left\{ \bigcup_{\a<\o_1}\bigcap_{\b\geq\a}[f=g]\colon\;
f,g\in\QC\right\}=\P(\R)$. On the other hand, $ \L (\QC)\subset \B aire$.\Qed

Finally note that this fact cannot be proved in ZFC. Namely,
we have the following.
\begin{tw}\label{tw-cliquish}
If $\add(\M)>\o_1$ then $\L (\QC) = \L (\C liquish)$.
\end{tw}
\pf Since $\QC\subset\C liquish$,  by Theorem~\ref{TW QC}
it is enough to prove that every set $A\in \L (\C liquish)$ is of the form
$(G\setminus M)\cup N$, where $G$ is open, $M\in \M$ and $N$ is nowhere dense.
Let $A=L(\{f_{\a}: \a < \o_1 \})$ for some sequence $\{f_{\a}: \a < \o_1 \}
\subset \C liquish$ and
$f=\lim_{\a}f_{\a}|_A$. Consider a sequence  $\{I_n:
n < \o \}$ of components of $\int(\cl(A))$. Then $A=(\bigcup_{n<\o}I_n\cap A)
\cup (A\setminus \int(\cl(A)))$ and $A\setminus \int(\cl(A))$ is nowhere dense.
Fix $n< \o$. Let $A_0=I_n\cap A$.
Choose a countable set $C_0\subset A_0,$ dense in $A_0$.
Then there is $\a_0 < \o_1$ such that
\begin{description}
\item[($*$)] $f_{\a}|_{C_0}=f|_{C_0}$ for every $\a \geq \a_0$.
\end{description}
Notice that $C(f_{\a})$ is a dense $G_{\delta}$ set for every $\a\geq \a_0$,
by cliquishness of $f_{\a}$. Since $\add (\M)>\o_1$, $\bigcap_{\a\geq\a_0}C(f_{\a})$ is residual in $I_n$.
 Now, take $x\in I_n\cap \bigcap_{\a\geq\a_0}C(f_{\a})$. Then for every
$\a \geq \a_0$ by $(*)$ and continuity of $f_{\a}$ at $x$
we have
$$f_{\a}(x)=\lim_{t\to x}f_{\a}(t)=
\lim_{t\to x, t\in C_0}f_{\a}(t)=
\lim_{t\to x, t\in C_0}f_{\a_0}(t)=
f_{\a_0}(x).$$
Thus
$I_n \cap \bigcap_{\a\geq\a_0}C(f_{\a}) \subset A\cap I_n$ for every $n<\o$, so  $ I_n \cap A $ is residual in $I_n$. For
$G=\bigcup_{n<\o}I_n$, $M=\bigcup_{n<\o}(I_n\setminus A)$,
$N=A\setminus \int(\cl(A))$ we have $A=(G\setminus M)\cup N$, which finishes the proof.
\Qed
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\vskip5mm

\noindent
\sc Tomasz Natkaniec, Department of Mathematics, Gda\'{n}sk
University, Wita Stwosza 57, 80--952 Gda\'{n}sk, Poland. E-mail: mattn@ksinet.univ.gda.pl

\bigskip
\noindent
\sc Jolanta Weso{\l}owska, Department of Mathematics, Gda\'{n}sk
University, Wita Stwosza 57, 80--952 Gda\'{n}sk,  Poland. E-mail: jwes@ksinet.univ.gda.pl



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