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\author{Tomasz Natkaniec, W{\l}adys{\l}aw Wilczy{\'n}ski}


\title{SUMS OF PERIODIC DARBOUX FUNCTIONS AND MEASURABILITY}


\newcommand{\co}{{\mathfrak c}}
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\begin{document}
{
\renewcommand{\thefootnote}{}
 \footnotetext{AMS Subject Classification (1991): Primary {\bf
 26A15}; Secondary: 28A20}
}
{
{\renewcommand{\thefootnote}{}
\footnotetext{Key words: periodic functions; Lebesgue measurable functions; Baire property; $s$-measurable functions}
}

\date{}

\maketitle


\begin{quote}
{\bf Abstract. } We show that a function $f: \mR\to\mR$ can be represented
as a sum of $n$ Darboux periodic functions if and only if it can be
represented as a sum of $n$ periodic functions. Also it is shown that
even functions having very simple structure cannot be represented as a
sum of $n$ ($n\geq 2$) Lebesgue'e measurable periodic functions.
\end{quote}

\section{Introduction.}

S. Mortola and R. Peirone in [MP] have found a necessary and sufficient
condition under which a function $f: \mR\to\mR$ can be represented
as a sum of $n$ periodic functions.
(See Theorem~\ref{t1}.) In this note we shall improve their
result by showing that the summands can be also Darboux function.
However, in general it is not possible to require that the representation
consists of Lebesgue measurable (or having the Baire property) functions.

We shall say that a function $f: X\to R$ ($X\subset\mR$) is periodic of period $h\in\mR\setminus\{0\}$ ($h$-periodic, shortly) if for each $x\in X$ we have $x\pm h\in X$
and $f(x\pm h) = f(x)$. Recall also that a function $f: \mR\to\mR$
is a Darboux function if and only if for each interval $I\subset\mR$
(open, closed, semi-open, bounded or unbounded) the image $f(I)$ is a
connected set.

Throughout the paper $\mN$ will denote the set of natural numbers, $\mZ$ ---
the set of integers and $\mQ$ --- the set of rational numbers. The Lebesgue
measure on the real line will be denoted by $\lambda$. Following von Neumann, we will identify a cardinal $\kappa$ with the first ordinal of the size $\kappa$. $\co$ denotes the cardinality of $\mR$.

We will consider $\mR$ as a linear space over $\mQ$.
For $A\subset\mR$ the linear subspace of $\mR$ generated by $A$ will be denoted by ${\rm LIN}(A)$. We will need the fact that there are linearly independent subsets $H$ of $\mR$ that are perfect-dense in $\mR$, i.e., $H\cap (a,b)$ contains a perfect set for arbitrary interval $(a,b)$. Such a set can be easily constructed from a linearly independent perfect set, the description of which can be found in \cite[Theorem 2, p. 270]{MK}.
Notice that each such set $H$ is $\co$-dense in $\mR$, i.e. ${\rm card}(H\cap (a,b))=\co$ for each interval $(a,b)$, and it can be divided onto $\co$ many pairwise disjoint $\co$-dense linearly independent subsets of $\mR$.

Given $f\colon\mR\to\mR$ and $h_0,\ldots h_{n}>0$ define  $\Delta(h_0)f(x) = f(x+h_0) -f(x)$ and $\Delta(h_0, h_1,\ldots, h_n)f(x) =\Delta(h_0,\ldots,h_{n-1})(\Delta(h_n))f(x)$.

\begin{thm}~{\rm\cite[Theorem 3]{MP}\label{t1}}
A function $f\colon\mR\to\mR$ is the sum of finitely many periodic functions iff there exist positive numbers $h_0, \ldots, h_{n-1}$ such that $h_i/h_j\not\in\mQ$ if $i\not= j$ and $\Delta(h_0,\ldots, h_{n-1})f=0$.
\end{thm}

For example, for every positive numbers $h_0$, $h_1$ with $h_0/h_1\not\in\mQ$ the identity $f(x)=x$ can be written as the sum of two periodic functions with periods $h_0$ and $h_1$, respectively. On the other hand, it is clear that every continuous periodic function is bounded, so the identity cannot be written as the sum of finitely many {\em continuous} periodic functions. In this note we consider the question how ``good'' can be periodic summands of the identity in Mortola-Peirone Theorem.

\section{Darboux like summands.}
\begin{thm}\label{darboux}
Assume $h_0,\ldots, h_{n-1}>0$, $h_i/h_j\not\in\mQ$ for $i\not= j$. A function $f\colon\mR\to\mR$ is the sum of $n$ Darboux functions $f_0,\ldots, f_{n-1}$ such that $f_i$ is $h_i$-periodic for $i<n$, iff $\Delta(h_0,\ldots, h_{n-1})f=0$.
\end{thm}
\pf
The implication ``$\Rightarrow$" follows from Theorem~ref{t1}. To prove ``$\Leftarrow$" choose a sequence $H_0, \ldots, H_{n-1}$ of pairwise disjoint sets such that
\begin{description}
\item[(i)]
${\rm LIN}(H_0\cup\ldots\cup H_{n-1})\cap {\rm LIN}\{h_0,\ldots, h_{n-1}\}=\{ 0\}$.
\item[(ii)]
$H_0\cup\ldots\cup H_{n-1}$ is linearly independent over $\mQ$.
\item[(iii)]
For each $i<n$ the set $H_i$ is $\co$-dense in $\mR$.
\end{description}

For each $i<n$ let $\varphi^0_i\colon H_i\to\mR$ be a function with dense level sets, i.e., such that for any $y\in\mR$ the set $(\varphi^0_i)^{-1}(y)$ is dense in $\mR$. Observe that every extension $f\colon\mR\to\mR$ of $\varphi^0_i$ is a Darboux function. Let $H_n$ be a basis of the space ${\rm LIN}\{h_0,\ldots, h_{n-1}\}$ and $H$ be a Hamel basis containing all sets $H_i$, $i\leq n$.

Since $\Delta(h_0,\ldots, h_{n-1})f=0$, there are functions $g_0,\ldots, g_{n-1}\colon\mR\to\mR$ such that $g_i$ is $h_i$-periodic, $i<n$, and $\sum_{i<n}g_i=f$. (See~\cite[Theorem 3]{MP}.)
Define $\varphi_i\colon H\to\mR$, $i<n$, in the following way.
 $$
 \begin{array}{ll}
 \varphi_i(h)=\left\{\begin{array}{ll}
\varphi_i^0(h)-g_i(h) &\mbox{for $h\in H_i$,}\\
 g_{i+1}(h)-\varphi^0_{i+1}(h)& \mbox{for $h\in H_{i+1}$,}\\
 0 & \mbox{for $h\in H\setminus (H_i\cup H_{i+1}$),}
\end{array} \right.
&
\!\!\!\!\! i=0,\ldots,n-2\\
%\mbox{}\\
\mbox{}\\
 \varphi_{n-1}(h)=\left\{\begin{array}{ll}
 \varphi_{n-1}^0(h)-g_{n-1}(h) &\mbox{for $h\in H_{n-1}$,}\\
 g_{0}(h)-\varphi_{0}^0(h)& \mbox{for $h\in H_{0}$,}\\
 0 & \mbox{for $h\in H\setminus (H_0\cup H_{n-1}$),}
\end{array} \right.
\end{array}$$

For each $i<n$ let $\overline\varphi_i\colon\mR\to\mR$ be the linear extension of $\varphi_i$. Note that $\sum_{i<n}\varphi_i=0$, so $\sum_{i<n}\overline\varphi_i=0$. Moreover, observe that $\overline\varphi_i|H_n=0$, so $\overline\varphi_i(h_i)=0$ and therefore $\overline\varphi_i$ is $h_i$-periodic. Put $f_i=\overline\varphi_i+g_i$, $i<n$. Then every $f_i$ is $h_i$-periodic and $\sum_{i<n}f_i=\sum_{i<n}\overline\varphi_i+\sum_{i<n}g_i=f$. Moreover, $f_i|H_i=\varphi^0_i$, so $f_i$ is Darboux.
\qed

\medskip
\noindent
{\bf Remark. }
Darboux property in   Theorem~\ref{darboux} can
be replaced by extendability, so also by arbitrary another property of Darboux type from Gibson's diagram.  It is enough to assume in the proof that all sets $H_i$ are perfect-dense. (See e.g., \cite{GN} for definitions and  \cite{TN} for more details.)

\section{Measurable summands.}

In this section we will need the following generalization of Kronecker's Theorem (cf., \cite[Theorem 438, p. 375]{HW}).

\begin{lem}\label{l1}
For each positive reals $h_0,\ldots,h_k$ and for each $\delta>0$ there exist integers $m_0,\ldots, m_k$ such that $|m_0h_0-m_ih_i|<\delta$ for $i=1,\ldots,k$.
\end{lem}
\pf
For $x\in\mR$ let ${\rm E}(x)$ denote the integral part of $x$ and $r(x)=x-{\rm E}(x)$.
For a positive integer $k$ let $r_k\colon \mR^k\to [0,1]^k$ be the function defined by $r_k(x_1,\ldots,x_k)=(r(x_1),\ldots,r(x_k))$.
For $x=(x_1,\ldots,x_k)\in \mR^k$ set $\mN x=\{ (mx_1,\ldots,mx_k)\colon m\in{\mN}\}$. Observe that $r_k({\mN} x)$ is bounded, so it has a limit point. Consequently, for each $\varepsilon>0$ there are $m,n\in{\mN}$ with $|r_k(mx)-r_k(nx)|<\varepsilon$. Therefore for each $\varepsilon>0$ there are $m,n\in{\mN}$ such that $|r(mx_i)-r(nx_i)|<\varepsilon$ for $i=1,\ldots,k$.

Fix $\delta>0$ and put $x_1=h_0/h_1, \ldots, x_k=h_0/h_k$. Then for $\varepsilon=\min\{\delta/h_i\colon\; 1\le i\le k\}$ there are positive integers $m,n\in{\mN}$ such that $|mx_i-m'_i-nx_i+n'_i|<\delta/h_i$ for $i\le k$ and $m'_i={\rm E}(mx_i)$, $n'_i={\rm E}(nx_i)$. Thus $|(m-n)h_0-(m'_i-n'_i)h_i|<\delta$.
Finally, for $m_0=m-n$ and $m_i=m'_i-n'_i$ for $0<i<k$ we have the claim.
\qed

\noindent
{\bf Remark. } The number $m$ in the proof above can be chosen arbitrary big.



\begin{thm}
The identity $f(x)=x$ cannot be represented as the sum of finite number of  Lebesgue measurable periodic functions.
\end{thm}
\pf
Suppose $f= f_0+\ldots+f_{n-1}$ for some Lebesgue measurable periodic functions $f_0, \ldots, f_{n-1}$ of periods $h_0,\ldots, h_{n-1}$, respectively.

Then, according to the corollary of Lusin's theorem, (See, for example, \cite[Theorem~6.9, p. 278]{JF})
all $f_i$ are approximately continuous a.e.
Let $x_0$ be a common point of approximate continuity of
all $f_i$. Take $\varepsilon >0$. There exists $\delta >0$ such that for each  $i=0,\ldots n-1$ we have
\begin{eqnarray}\label{e1}
\lambda(\{x\in (x_0-\delta, x_0+ \delta): \; | f_i(x)-f_i(x_0)|< \frac{\varepsilon}{n}\})>2(1-\frac{1}{4n})\delta
\end{eqnarray}
 Obviously, we can assume that ${h_0}/{h_i}\notin\mQ$ for $i>0$. Fix $c\in\mR$.
 Lemma~\ref{l1} implies that there exist integers $m_0,\ldots, m_{n-1}$ such that $x_0+m_0h_0>c$ and $|m_0 h_0 -m_i h_i| <\delta/2$ for $i=1,\ldots, n-1$.
Set $J_i=(x_0+m_ih_i-\delta, x_0+m_ih_i+\delta)$ for $i=0,\ldots,n-1$ and $J=(x_0+m_0h_0-\delta/2, x_0+m_0h_0+\delta/2)$. Then $J\subset\bigcap_{i<n}J_i$ and, by virtue of (\ref{e1}) and periodicity of $f_i$ we have $\lambda(\{ x\in J_i\colon |f_i(x)-f_i(x_0)|\geq\varepsilon/n\})<\delta/2n$. Thus $\lambda(\{ x\in J\colon \; (\exists i<n)\; |f_i(x)-f_i(x_0)|\geq\varepsilon/n\})<\delta/2$ and consequently there is $x_c\in J$ such that $|f_i(x_c)-f_i(x_0)|<\varepsilon/n$ for all $i<n$.  Then we have $|f(x_c)-f(x_0)|\leq\sum_{i<n}|f_i(x_c)-f(x_0)|<\varepsilon$, so $f(x_c)\in (f(x_0)-\varepsilon,f(x_0)+\varepsilon)$, which contradicts the fact that $\lim_{x\to\infty} f(x)
=+\infty.$
\qed

\begin{thm}\label{baire}
The identity is not a sum of finite number of periodic
functions having the Baire property.
\end{thm}
\pf
Suppose that there exist periodic functions $f_0,\ldots,  f_{n-1}$ of period $h_0,\ldots, h_{n-1}$, respectively, having the Baire property and such that $f = f_0+\ldots+ f_{n-1}$.
Then, according to the theorem of Kuratowski \cite[p. 400]{KK}, there exists a meager set $M\subset \mR$ such that $f_i|(\mR\setminus M)$ are continuous for all $i<n$.

Let $x_0\in \mR\setminus M$. Take $\varepsilon >0$. Then there exists
$\delta >0$ such that for each $i=0,\ldots,n-1$ the set
$\{x\in (x_0- \delta, x_0 +\delta)\colon \; |f_i(x)-f_i(x_0)|<\varepsilon\}$
is residual in $(x_0 -d, x_0 +d)$.

The rest of the proof is similar (even simpler) than in the measure case.
\qed


\section{Marczewski measurable summands.}
Recall that a function $f\colon\mR\to\mR$ is Marczewski measurable ($(s)$-measurable, shortly) if for each perfect set $P\subset\mR$ there exists a  perfect subset $Q$ of $P$ such that the restriction $f|Q$ is continuous. (See \cite{WS} and \cite{EM}.)
It is well-known that the class of $(s)$-measurable functions is closed with respect to the addition. Recall also that every Borel measurable function is Marczewski measurable but the three properties: Marczewski measurability, Lebesgue measurability and Baire property are pairwise independent~\cite{EM}.

In this section we will need some additional set theoretical assumptions.

\begin{lem}{\rm [A.W.~Miller, S.G.~Popvassilev]}\label{miller}
Let $L$ be an $F_{\sigma}$ linear subspace of $\mR$ and let $P$ be a perfect set such that $P\setminus {\rm LIN}(L\cup F)$ is dense in $P$ for any finite set $F\subset P\setminus L$. Then there exists a Cantor set $C\subset P\setminus L$ such that
\begin{description}
\item[(a)]
 $C$ is linearly independent and ${\rm LIN}(C)\cap L=\{ 0\}$;
\item[(b)]
if $K$ is linearly independent with ${\rm LIN}(K)=L$ then $K\cap C=\emptyset$ and $K\cup C$ is linearly independent.
\end{description}
\end{lem}
\pf
This  lemma was proved originally in an earlier version of \cite{MiP} but, unfortunately, it is omitted in a printed version
of the paper. The construction of $C$ is an easy modification of Jones construction of a perfect linearly independent set. (See \cite[p.~270]{MK}.) We will left the details to the readers.
\qed

\begin{lem}\label{miller2} {\rm [[A.W.~Miller, S.G.~Popvassilev]}
Assume CH. There exists a sequence $\{ C_{\alpha}\colon \alpha<\co\}$ of pairwise disjoint closed sets such that
\begin{itemize}
\item
$\bigcup_{\alpha<\co}C_{\alpha}$ is linearly independent;
\item
for every perfect set $P$ there exists $\alpha<\co$ such that $P\cap {\rm LIN}(\bigcup_{\beta<\alpha}C_{\beta})$ contains a perfect set.
\end{itemize}
\end{lem}
\pf
In fact, such a sequence was constructed in the proof of Theorem 10 in an earlier version of the paper \cite{MiP}. Since this proof also has been omitted in the final version of the paper, we decide to recall the construction.

Let $\{ P_{\alpha}\colon \alpha<\co\}$ be a list of all perfect subsets of $\mR$. Construct by induction sets $C_{\alpha}\subset P_{\alpha}$, $\alpha<\co$, such that:
\begin{description}
\item[(i)]
$C_{\alpha}=\emptyset$, or $C_{\alpha}$ is finite, or $C_{\alpha}$ is a Cantor set.
\item[(ii)]
$C_{\beta}\cap C_{\alpha}=\emptyset$ if $\beta<\alpha$.
\item[(iii)]
$\bigcup_{\beta\leq\alpha}C_{\beta}$ is linearly independent.
\item[(iv.a)]
If the set $(q_0C_{\beta_0}+\ldots+ q_nC_{\beta_n})\cap P_{\alpha}$ is uncountable for some $n\in\mN$, $q_0,\ldots, q_n\in\mQ\setminus\{ 0\}$ and $\beta_0<\ldots<\beta_n<\alpha$, then $C_{\alpha}=\emptyset$.

If (iv.a) is not the case, let $L_{\alpha}={\rm LIN}(\bigcup_{\beta<\alpha}C_{\beta})$ and note that
$$L_{\alpha}=\bigcup\{ q_0C_{\beta_0}+\ldots+ q_nC_{\beta_n}\colon n\in\mN,\; q_i\in\mQ\setminus\{ 0\}; \beta_0<\ldots<\beta_n<\alpha\}$$
is an $F_{\sigma}$ set.
\item[(iv.b)]
If not (iv.a), and there is a finite set $F\subset P_{\alpha}\setminus L_{\alpha}$ such that ${\rm LIN}(L_{\alpha}\cup F)\cap P_{\alpha}$ contains a perfect set, then $C_{\alpha}\subset P_{\alpha}\setminus L_{\alpha}$ is such a finite set with $\bigcup_{\beta\leq \alpha}C_{\beta}$ linearly independent.
\item[(iv.c)]
If neither (iv.a) nor (iv.b), then $C_{\alpha}\subset P_{\alpha}\setminus L_{\alpha}$ is a Cantor set such that
$\bigcup_{\beta\leq \alpha}C_{\beta}$ is linearly independent. Such a set exists by Lemma~\ref{miller}.
\end{description}
Finally, it is easy to verify that the sequence $\{ C_{\alpha}\colon \alpha<\co\}$ satisfies our assertion.
\qed

\begin{lem}\label{l-sm}
Assume CH. If $h_0,h_1>0$, $h_0/h_1\not\in\mQ$, $g\colon\mR\to\mR$ is $h_0$-periodic and $(s)$-measurable, then there exists $f\colon\mR\to\mR$ such that $f$ is $h_0$-periodic and $(s)$-measurable, and $\Delta(h_1)f=g$.
\end{lem}
\pf
Let $\{ C_{\alpha}\colon \alpha<\co\}$ be a sequence of pairwise disjoint closed sets as in Lemma~\ref{miller2}.
Obviously, we can assume that $C_0=\{h_0,h_1\}$.
Let $H$ be a Hamel basis containing all sets $C_{\alpha}$, $\alpha<\co$. Put $H_0=H\setminus C_0$.

 We will verify that the function $f$ defined in the proof of \cite[Lemma]{MP} has all properties we need.
Recall the definition of $f$ for convenience of readers. Every $x\in\mR$ has unique representation in the form $x=p(x)+n(x)h_1$, where $p(x)=b+qh_1$, $b\in {\rm LIN}(H_0\cup\{h_0\})$, $q\in [0,1)\cap\mQ$ and $n(x)\in\mZ$. Then

$$ \begin{array}{l}
 f(x)=\left\{\begin{array}{ll}
 \sum_{i=0}^{n(x)}g(p(x)+ih_1) &\mbox{if $n(x)>0$,}\\
 0 & \mbox{if $n(x)=0$,}\\
 -\sum_{i=n(x)}^{-1}g(p(x)+ih_1) &\mbox{if $n(x)>0$.}
\end{array} \right.
\end{array} $$
The function $f$ is $h_0$-periodic and $\Delta(h_1)f=g$~\cite{MP}. Thus we have to prove that $f$ is $(s)$-measurable. Fix a perfect set $P\subset\mR$.


The choice of a sequence $C_{\alpha}$, $\alpha<\co$, guarantees  that there exist $n\in\mZ$ and $r\in[0,1)\cap\mQ$ such that $P\cap((n+r)h_1+{\rm LIN}(H_0\cup\{h_0\}))$ contains a perfect set $P_0$.
%In fact, since $\mR=\bigcup_{q\in\mQ}(qh_1+{\rm %LIN}(H_0\cup\{h_0\}))$, there are $n\in\mZ$ and $r\in %[0,1)\cap\mQ$ such that the intersection %$P\cap((n+r)h_1+{\rm LIN}(H_0\cup\{h_0\}))$ is of size $\co$.
%Now, the set ${\rm LIN}(H_0\cup\{h_0\}))$ is the union of %$\omega_1$ closed sets $E_{\alpha}$, $\alpha<\omega_1$. %Since $\co=\omega_2$, there is $\alpha<\omega_2$ such %that $P\cap E_{\alpha}$ has cardinality $\co$. Since %$P\cap  E_{\alpha}$ is closed, it contains a perfect set.

Without loss of generality we can assume that $n>0$.
Since $g$ is $(s)$-measurable, there exists a perfect set $Q_0\subset P_0-nh_1$ such that $g|Q_0$ is continuous. For $i=1,\ldots, n$ choose a perfect set $Q_{i}\subset Q_{i-1}$ such that $g|(Q_{i}+ih_1)$ is continuous. Observe that
for $g_i|Q_{i}$ is continuous for $g_i=g(p(x)+ih_1)$. Finally, put $Q=Q_n+nh_1$. Then $Q\subset P$,  all $g_i|Q$ are continuous, $n(x)=n$ for $x\in Q$ and consequently, $f|Q=\sum_{i\leq n}g_i|Q$ is continuous.
\qed

\begin{thm}\label{t-sm}
Assume CH. If $h_0,\ldots, h_{n-1}$ is a sequence of positive numbers such that $h_i/h_j\not\in\mQ$ if $i\not= j$. Then a function $f\colon\mR\to\mR$ is the sum of $(s)$-measurable periodic functions of period $h_0,\ldots, h_{n-1}$, respectively, iff $f$ is $(s)$-measurable and $\Delta(h_0,\ldots, h_{n-1})f=0$.
\end{thm}
\pf
The implication $``\Rightarrow''$ follows from Theorem~\ref{t1}. To prove $``\Leftarrow''$ we will proceed by induction, analogously to the proof of \cite[Theorem 2]{MP}.

Obviously, our claim is true for $n=1$. Suppose $\Delta(h_0,\ldots, h_n)f=0$. Then $\Delta(h_0,\ldots, h_{n-1})(\Delta(h_n)f)=0$. Note that $\Delta(h_n)f$ is $(s)$-measurable whenever $f$ is so. Thus by the inductive hypothesis, there exist $(s)$-measurable functions $g_i\colon\mR\to\mR$, $i<n$, such that each $g_i$ is $h_i$-periodic and $\Delta(h_n)f=\sum_{i<n}g_i$. By Lemma~\ref{l-sm}, for each $i<n$ there exists an $(s)$-measurable, $h_i$-periodic function $f_i$ such that $\Delta(h_n)f_i=g_i$. Put $g=\sum_{i<n}f_i$ and $f_n=f-g$. Then $g$ is $(s)$-measurable, so $f_n$ is $(s)$-measurable, $\Delta(h_n)f_n=0$, so $f_n$ is $h_n$-periodic, and $f=\sum_{i\le n}f_i$.
\qed

\begin{cor}\label{cor-sm}
Assume CH. Then the identity can be written as the sum of two periodic, $(s)$-measurable functions.
\end{cor}

\noindent
{\bf Remark. }The Miller-Popvassilev construction of the sequence $\{ C_{\alpha}\colon \alpha<\co\}$ from Lemma~\ref{miller2}  works also under weaker assumption that the union of less than continuum many meager sets is meager, so e.g. under MA. (In fact, it is enough to notice that then the subspace $L_{\alpha}$ is meager, thus it is included in some $F_{\sigma}$ linear subspace of $\mR$.) A sequence $\{ C_{\alpha}\colon \alpha<\co\}$ satisfying the assertion of Lemma~\ref{miller2} can be easily constructed also if we assume that $\co=\omega_2$ and there exists a Hamel basis being the union of $\omega_1$ closed sets. These two facts are a consequence of the CPA axiom introduced and studied recently by Ciesielski and Pawlikowski and occur e.g. in the iterated perfect set (Sacks) model~\cite{CP}. (Recall that in this model Martin Axiom fails.)  However,
we are unable to determine whether Theorem~\ref{t-sm} or even Corollary~\ref{cor-sm}
can be proved in ZFC, without extra set-theoretic assumptions.



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\end{thebibliography}

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

\bigskip
\noindent
\sc W{\l}adys{\l}aw Wilczy\'{n}ski,\\
 Faculty of Mathematics, Chair of Real Functions, \\
{\L}\'{o}d\'{z} University, Stefana Banacha 22, 90-238
{\L}\'{o}d\'{z}, Poland.\\
E-mail: wwil@krysia.uni.lodz.pl
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