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\documentclass{rae}
\usepackage{amsmath,amssymb,amsthm}
\received{December 3, 1996}
\author{Zbigniew Grande, Institute of Mathematics,
Pedagogical University, ul. Chodkiewicza 30, 85-064 Bydgoszcz,
Poland. e-mail: {\tt grande@wsp.bydgoszcz.pl}}
%\coverauthor{Zbigniew Grande}
\title{ON SOME SPECIAL NOTIONS OF APPROXIMATE QUASI-CONTINUITY}
%\covertitle{On Some Special Notions of Approximate Quasi-continuity}
\markboth{Zbigniew Grande}{Special Notions Approximate Quasi-continuity}
\MathReviews{26A15, 54C08, 54C30}
\keywords{continuity, strong quasicontinuity, density
topology, oscillation, sequences of functions, uniform convergence.}
\newcommand{\R}{{\mathbb R}}
\newcommand{\Qed}{\null\hfill $\Box\;\;$ \medskip}
\newtheorem {thm} {Theorem}
\newtheorem {lem} {Lemma}
\newtheorem {rem} {Remark}
\newtheorem {cor} {Corollary}
\def\pf{\noindent{\sc Proof. }}
\DeclareMathOperator{\osc}{osc}
\DeclareMathOperator{\aposc}{ap-osc}
\DeclareMathOperator{\Int}{int}
\DeclareMathOperator{\cl}{cl}
\firstpagenumber{1}
\begin{document}
\maketitle
\begin{abstract}
Some special notions of approximate quasi-continuity and cliquishness
are considered.
Moreover, uniform, pointwise and transfinite convergence of sequences
of such functions are investigated.
\end{abstract}
Let $\R$ be the set of all reals and let $\mu _e$ ($\mu $) denote
outer Lebesgue measure (Lebesgue measure) in $\R$. Let
$$ d_u(A,x) =
\limsup_{h \rightarrow 0^+} \mu _e(A \cap (x-h,x+h))/2h $$
$$(d_l(A,x) = \liminf_{h \rightarrow 0^+} \mu _e(A \cap (x-h,x+h))/2h) $$
denote the upper (lower) density of a set $A \subset \R$ at a point $x$.
A point $x \in \R$ is called a density point of a set $A \subset
\R$ if there exists a measurable (in the sense of Lebesgue) set $B
\subset A$ such that $d_l(B,x) = 1$.
The family
$${\cal T}_d = \{ A \subset \R; A
\text{ is measurable and each $x \in A$ is a density point
of $A$} \}$$
is a topology called the density topology \cite{1}.
Moreover, let ${\cal T}_e$ denote the Euclidean topology on
$\R$.
All considered functions will be real and defined on $\R$.
A point $x$ is a continuity point (an approximate continuity
point) of a function $f$ if it is a continuity point of $f$
considered as an application from $(\R,{\cal T}_e)$
($(\R,{\cal T}_d)$) to $(\R,{\cal T}_e)$. Denote by
$C(f)$ (resp. $A(f)$) the set of all continuity (approximate
continuity) points of $f$.
Define the following families of functions:
\begin{enumerate}
\item[--] $f \in {\cal A}_1$ ($f \in {\cal A}_2$) if for every
point $x$, for every positive real $\eta $, and for every set
$A \in {\cal T}_d$ containing $x$ there is a point
$u \in A \cap C(f)$ ($u \in A(f) \cap A$) such that
$|f(u) - f(x)| < \eta $;
\item[--] $f \in {\cal A}_3$ ($f \in {\cal A}_4$) if for every
point $x$, for every positive real $\eta$, and for every set
$A \in {\cal T}_d$ containing $x$ there is an open interval $I$
such that $I \cap A \neq \emptyset $, $I \cap A \subset C(f)$
($I \cap A \subset A(f)$) and $|f(u) - f(x)| < \eta $ for all points
$u \in I \cap A$;
\item[--] $f \in {\cal A}_5$ ($f \in {\cal A}_6$) if for every
nonempty set $A \in {\cal T}_d$ there is an open interval $I$ such that
$I \cap A \neq \emptyset $ and $I \cap A \subset C(f)$ ($I \cap
A \subset A(f)$).
\item[--] $f \in {\cal A}_7$ if for each positive real $\eta $ and for
each nonempty set $A \in {\cal T}_d$ there is an open interval $I$ such
that $I \cap A \neq \emptyset $, $I \cap A \subset A(f)$ and $\osc f \leq \eta
$ on $I \cap A$.
\end{enumerate}
Moreover, a function $f$ is said to be strongly quasi-continuous
[strongly cliquish] (abbreviated s.q.c. [s.c]) at a point $x$ if for
every set $A \in {\cal T}_d$ containing $x$ and for every positive
real $\eta$ there is an open interval $I$ such that
$I \cap A \neq \emptyset $ and $|f(t) - f(x)| < \eta $ for all
$t \in A \cap I$ [$\osc f < \eta $ on $I \cap A$] (\cite{4}).
Let $f:\R \longrightarrow \R$ be a function and let
$x \in \R$ be a point.
If there is an open set $U$ such that $d_u(U,x) > 0$ and the
restricted function $f|(U \cup \{ x \} )$ is continuous at $x$,
then $f$ is s.q.c. at $x$.
By elementary proofs, we obtain:
\begin{rem}
If all functions $f_n$, $n = 1,2, \ldots $, of some
uniformly converging sequence $(f_n)_n$ are s.q.c. at a point
$x$, then its limit $f$ is also s.q.c. at $x$.
\end{rem}
\begin{rem}
If $f \in {\cal A}_i$, $i = 1,3,4$, then $f$ is s.q.c. at each
point.
\end{rem}
\begin{rem}
The inclusions
$${\cal A}_3 \subset {\cal A}_1 \subset {\cal A}_2;\;\; {\cal A}_3
\subset {\cal A}_4 \subset {\cal A}_2;\;\; {\cal A}_3 \subset
{\cal A}_5 \cup {\cal A}_4 \subset {\cal A}_7 \subset {\cal A}_6$$
are true.
\end{rem}
Since every function which is s.q.c. at each point is also
almost everywhere continuous \cite{4}, we can observe that a function $f
\in {\cal A}_1$ if and only if $f$ is s.q.c. at each point $x$ and that
the families ${\cal A}_i$, $i = 1, 3, 4$, contain only almost everywhere
continuous functions. It is obvious also that all functions belonging to
the family ${\cal A}_5$ are almost everywhere continuous.
Approximately continuous functions are in Baire class $1$, so they
belong to ${\cal A}_7 \cap{\cal A}_2$. Since there are approximately
continuous functions which are not almost everywhere continuous
(\cite{1}), in the families ${\cal A}_7$ and ${\cal A}_2$ there are
functions which are not almost everywhere continuous. However, if
$f \in {\cal A}_2 \cup {\cal A}_6$, then for every positive real $\eta $
and for every measurable set $A$ with $\mu (A) > 0$ there is a measurable
subset $B \subset A$ such that $\mu (B) > 0$ and $\osc f \leq
\eta $ on $B$. So, by Davies theorem (\cite{2,3}), every
function $f \in {\cal A}_2 \cup {\cal A}_6$ is measurable (in
the sense of Lebesgue). If $f \in {\cal A}_6$, then for every
open interval $I$ and for every positive real $\eta $ there is
an open interval $J \subset I$ such that $\osc f \leq \eta $ on
$J$. Thus, the set $C(f)$ of arbitrary function $f \in {\cal A}_6$
is dense, and consequently it is a residual $G_{\delta }$ set.
So, every function $f \in {\cal A}_6$ has the Baire property.
However, there are functions $f \in {\cal A}_2$ which do not have
the Baire property. Therefore, we adopt the following definition
(\cite{6}).
A function $f$ is called approximately quasi-continuous at a point
$x$ if for every positive real $\eta $ and for each set $A \in
{\cal T}_d$ containing $x$ there is a nonempty set $B \subset A$
belonging to ${\cal T}_d$ such that $|f(t) - f(x)| < \eta $ for
all $t \in B$.
Observe that a function $f$ is approximately quasi-continuous at every
point if and only if $f \in {\cal A}_2$. In \cite{6} it is proved that for
every measurable function $f$ there is a sequence of
approximately quasi-continuous functions $f_n$, $n = 1,2,\ldots
$ such that $f = \lim_{n\to \infty }f_n$. So, if $f$ is
measurable without the Baire property, then there is an index $k$
such that the function $f_k$ does not have the Baire property. Such
function $f_k$ is in the family ${\cal A}_2$ as an approximately
quasi-continuous function, but it doesn't have the Baire property.
From the above it follows that
$${\cal A}_2 \setminus {\cal A}_i \neq \emptyset ,\;\;\; i =
1,3,4,5,6,7.$$
Since there is an approximately continuous function $f$ with $\Int(C(f)) =
\emptyset $ (\cite{1}) we obtain
${\cal A}_7 \setminus {\cal A}_5 \neq \emptyset .$
The function $f(x) = 0$ for $x \neq 0$ and $f(0) = 1$ belongs to
${\cal A}_5 \subset {\cal A}_7$, but it is not in ${\cal A}_2
\supset {\cal A}_1 \supset {\cal A}_3$. Moreover, it is not in ${\cal
A}_4$. So,
$${\cal A}_5 \setminus {\cal A}_i \neq \emptyset ,\;\;\; i = 1,2,3,4.$$
If $g$ is an approximately continuous function such that $g(\R) = [0,1]$ and the set $C(f) = g^{-1}(0)$ is dense and its interior is
empty (\cite{1}), then the
function $f(x) = g(x)$ for $x \neq 0$ and $f(0) = 2$ belongs to
${\cal A}_7 \setminus {\cal A}_i$ for $i = 1,2,3,4,5$.
As an example of function $f \in {\cal A}_1 \setminus {\cal A}_6$
we can take any strictly monotone function which is
continuous from the right at every point and which is such that
the set $\R \setminus C(f)$ is dense.
So, we obtain the following assertion.
\begin{thm}
The inclusions
\begin{enumerate}
\item[--] ${\cal A}_3 \subset {\cal A}_1 \subset {\cal A}_2; \;\;\;
{\cal A}_3 \subset {\cal A}_4 \subset {\cal A}_2$;
\item[--] ${\cal A}_3 \subset {\cal A}_5 \subset {\cal A}_7; \;\;\;
{\cal A}_3 \subset {\cal A}_4 \subset {\cal A}_7$;
\end{enumerate}
are true.
Moreover, each of the above inclusions is strict and
$${\cal A}_1 \setminus {\cal A}_6 \neq \emptyset .$$
\end{thm}
The inclusion ${\cal A}_7 \subset {\cal A}_6$ is evident.
\begin{thm}
The relation $ {\cal A}_6 \setminus {\cal A}_7 \neq \emptyset$
is true.
\end{thm}
\pf Let $A_{1} \subset I_1 = (0,1)$ be a Cantor set of positive
measure. In the second step in every component $I_{1,n}$, $n = 1,2,\ldots
$, of the set $I_1 \setminus A_{1}$ we find an open interval
$J_{1,n}$ having the same centers as $I_{1,n}$ and such that $|J_{1,n}| <
4^{-2}|I_{1,n}|$, where the symbol $|J_{1,n}|$ denotes the length of the
interval $J_{1,n}$. Next, in each open interval $J_{1,n}$ we find a
Cantor set $A_{1,n} \subset J_{1,n}$ of positive measure.
In general, in the $k^{th}$ step ($k > 2$) we consider all components
$I_{1,n_2,\ldots ,n_k}$ of the set $J_{1,n_2,\ldots ,n_{k-1}} \setminus
A_{1,n_2,\ldots ,n_{k-1}}$, $n_i \geq 1$ for $1 < i \leq k$, and we find open
intervals $J_{1,n_2,\ldots ,n_k} \subset I_{1,n_2,\ldots ,n_k}$ having
the same centers as $I_{1,n_2,\ldots ,n_k}$ and such that
$$|J_{1,n_2,\ldots ,n_k}| < 4^{-k}|I_{1,n_2,\ldots ,n_k}|. \leqno (1)$$
For $1 < i \leq k$ and $n_i = 1,2,\ldots $, let $A_{1,n_2,\ldots ,n_k}
\subset J_{1,n_2,\ldots ,n_k}$ be a Cantor set of positive measure and
let
$$B_{1,n_2,\ldots ,n_k} = \{ x \in A_{1,n_2,\ldots ,n_k};
d_l(A_{1,n_2,\ldots ,n_k},x) = 1\} .$$
Let $f(x) = 1$ for $x \in B_{1,n_2,\ldots ,n_k}$, whenever $k$ is even
and $n_i \geq 1$ for $1 < i \leq k$ and let $f(x) = 0$ otherwise on
$\R$. We shall prove that $f \in {\cal A}_6$. Let
$A \in {\cal T}_d$ be a nonempty set. Denote by $B$ the union of all sets
$B_{1,n_2,\ldots ,n_k}$, where $k = 2,3,\ldots $ and $n_i = 1,2,\ldots $
for $1 < i \leq k$. Observe that $B \subset A(f)$. So, if $A \subset B$,
then for every open interval $I$ such
that $I \cap A \neq \emptyset $ we have $I \cap A \subset A(f)$. If $A$
is not a subset $B$, then there is a point $x \in A \setminus B$. If $x$
is not in $\cl(B)$, where $\cl(B)$ denotes the closure of the set $B$, then
$f$ is continuous at $x$ and for every interval
$I \subset \R \setminus \cl(B)$ containing $x$ we obtain that $I \cap A
\neq \emptyset $ and $I \cap A \subset C(f) \subset A(f)$. So, we suppose
that $x \in cl(B)$. If $x = 0$ or $x = 1$, then $f$ is unilaterally
continuous at $x$ and there is an open interval $I \subset {\cal R }
\setminus [0,1]$ with $I \cap A \neq \emptyset $. Consequently, in this
case we have $f(t) = f(x) = 0$ for all $t \in I \cap A$.
If there are indexes $n_2,\ldots ,n_k$ such that $x
\in A_{1,n_2,\ldots ,n_k} \setminus B_{1,n_2,\ldots ,n_k}$, then, by $(1)$,
there is an open interval $$I \subset J_{1,n_2,\ldots ,n_k} \setminus
A_{1,n_2,\ldots ,n_k} \setminus \bigcup_{n_{k+1}=1}^{\infty
}J_{1,n_2,\ldots ,n_{k+1}}$$ such that $I \cap A \neq \emptyset $ and $I \cap A
\subset C(f) \subset A(f)$. If not, there is a sequence of
indexes $n_2,\ldots ,n_k,\ldots $, such that
$$x \in \bigcap_{k=1}^{\infty }I_{1,n_2,\ldots ,n_k} = \bigcap
_{k=1}^{\infty }J_{1,n_2,\ldots ,n_k}.$$
Since $d_l(A,x) = 1$, there is an index $m$ such that for each $k \geq m$
we have
$$\mu (A \cap I_{1,n_2,\ldots ,n_k}) > |I_{1,n_2,\ldots ,n_k}|/2. \leqno
(2)$$
By $(1)$, there is an index $j > m$ such that for each $k \geq j$ we
obtain
$$4|J_{1,n_2,\ldots ,n_k}| < |I_{1,n_2,\ldots ,n_k}|. \leqno(3)$$
Fix $k > j$ and observe that from $(2)$ and $(3)$ it follows that
$$A \cap (I_{1,n_2,\ldots ,n_k} \setminus cl(J_{1,n_2,\ldots ,n_k})) \neq
\emptyset .$$
Consequently, there is an open interval $I$ such that $A \cap I \neq
\emptyset $ and $I \cap A \subset C(f) \subset A(f)$. So, $f \in {\cal
A}_6$.
For the proof that $f$ is not in ${\cal A}_7$ it suffices to observe that
$B \in {\cal T}_d$ and for every open interval $I$ with $I \cap B \neq
\emptyset $ the oscillation $\osc f$ on $I \cap B$ is greater than $a$ for
every positive real $a < 1$. \qed
\begin{rem}
Observe that if $f \in {\cal A}_6$ is of the first Baire class, then $f
\in {\cal A}_7$. Consequently, every approximately continuous function
belongs to ${\cal A}_7$.
\end{rem}
For a family $ {\Phi }$ let ${\cal B}({\Phi })$ (${\cal B}_u({\Phi })$)
denote the family of all limits of converging (of uniformly converging)
sequences of functions from $ \Phi $. Moreover, let $Q_s$ denote the
family of all functions which
are s.q.c. at every point. By Remark 1 the family $Q_s = {\cal A}_1$ is
uniformly closed.
\begin{thm}
The equality ${\cal B}_u({\cal A}_3) = Q_s$ is true.
\end{thm}
\pf Let $f \in Q_s$ be a function. It suffices to prove
that for every positive real $\eta $ there is a function $g \in
{\cal A}_3$ such that $|f - g| \leq \eta $. Fix a real $\eta > 0$.
Let
$$E = \{ y; \mu (cl(f^{-1}(y))) > 0\} .$$
Since $f$ is almost everywhere continuous, the set $E$ is
countable. There is a sequence $(c_k)_{k=-\infty }^{\infty }$ of
reals $c_k \in \R \setminus E$ such that $0 < c_{k+1} -
c_k < \eta /2$ for all integers $k$ and $$\R =
\bigcup_{k=-\infty }^{\infty }[c_k,c_{k+1}).$$
If $x$ is such that $c_k
\leq f(x) < c_{k+1}$, then let $h(x) = c_k$. If $h$ is not s.q.c.
at $x$ and $h(x) = c_k$, then we put $g(x) = c_{k-1}$. Otherwise let $g(x) =
h(x)$. Since $f$ is s.q.c. at each
point $x \in \R$, the function $g$ is the same on $\R$. But the image $g(\R)$ is a discrete set, so we obtain
$g \in {\cal A}_3$. Evidently, $|f - g| \leq \eta $.\qed
\medskip
Since ${\cal A}_3 \subset {\cal A}_4 \subset Q_s,$
we obtain the following.
\begin{cor}
The equality ${\cal B}_u({\cal A}_4) = Q_s$ is true.
\end{cor}
\begin{rem}
The equality ${\cal B}_u({\cal A}_2) = {\cal A}_2$ is true.
\end{rem}
\pf Fix $f \in {\cal B}_u({\cal A}_2)$. There is a
sequence of functions $f_n \in {\cal A}_2$, $n = 1,2,\ldots $,
which converges uniformly to $f$. Fix a point $x$, a positive
real $\eta $ and a set $A \in {\cal T}_d$ containing $x$. Let
$k$ be an index for which $|f_k - f| < \eta /4$. Since $f_k \in
{\cal A}_2$, there is a point $u \in A \cap A(f_k)$ such that
$|f_k(u) - f_k(x)| < \eta /4$. But $u \in A(f_k)$, so there is a
measurable subset $B \subset A$ such that $\mu (B) > 0$ and
$|f_k(t) - f_k(u)| < \eta /4$ for each point $t \in B$. The
function $f$ is measurable as the limit of the sequence of
measurable functions $f_n$. Consequently, there is a point $w
\in B \cap A(f) \subset A \cap A(f)$. Moreover, we have
\begin{align*}
|f(w) - f(x)| \leq& |f(w) - f_k(w)| + |f_k(w) - f_k(u)| + |f_k(u)
- f_k(x)|\\ &+ |f_k(x) - f(x)|
< \eta /4 + \eta /4 + \eta /4 + \eta/4 = \eta.\tag*{\qed}
\end{align*}
Let ${\cal C}_{ae}$ denote the family of all functions $f$ for
which $\mu (\R \setminus C(f)) = 0$. We have the following.
\begin{rem}
The equality ${\cal B}_u({\cal A}_5) = {\cal C}_{ae}$ is true.
\end{rem}
\pf It suffices to prove that for every $f \in {\cal
C}_{ae}$ and for every positive real $\eta $ there is a function
$h \in {\cal A}_5$ with $|f - h| \leq \eta $. Fix $f \in {\cal
C}_{ae}$ and a positive real $\eta $. Define the function $h$
the same as that in the proof of Theorem 4 and observe that $h
\in {\cal A}_5$, because it is almost everywhere continuous and
its image $h(\R)$ is a discrete set. \qed
\begin{thm}
A function $f \in {\cal B}_u({\cal A}_7)$ if and only if it is strongly
cliquish at each point $x \in \R$.
\end{thm}
\pf Assume, to the contrary, that there is a function $f
\in {\cal B}_u({\cal A}_7)$ such that there are a nonempty set $A \in
{\cal T}_d$ and a positive real $\eta $ with $\osc f > \eta $ on
$A \cap I$ for every open interval $I$ such that $I \cap A \neq
\emptyset $. Since $f \in {\cal B}_u({\cal A}_7)$, there is a function
$g \in {\cal A}_7$ such that $|f - g| < \eta /4$. For every open
interval $I$ such that $I \cap A \neq \emptyset $ we obtain that
$\osc g > \eta /2$ on the set $I \cap A$. But $g \in {\cal
A}_7$, so there is an open interval $I$ such that $I \cap A \neq
\emptyset $ and $I \cap A \subset A(g)$ and $\osc g < \eta /2$ on $ I
\cap A$. This contradiction shows that if $f \in {\cal B}
_u({\cal A}_7)$, then $f$ is s.c. at each point.
Now, suppose the function $f$ is s.c. at each point $x \in \R$. Fix
a positive real $\eta $. We shall prove that there is a function
$g \in {\cal A}_7$ such that $|f - g| < \eta $. Let $I_1$ be an open
interval with rational endpoints such that $\osc f < \eta $ on $I_1$.
Fix an ordinal number $\alpha > 1$ and suppose that for every ordinal
number $\beta < \alpha $ there is an open interval $I_{\beta }$ with
rational endpoints such that $\mu (I_{\beta } \setminus G_{\beta }) > 0$,
where
$G_{\beta } = \cup_{\gamma < \beta }I_{\gamma },$
and $\osc f < \eta $ on the set
$$H_{\beta } = \{ x \in I_{\beta } \setminus G_{\beta };d_l(I_{\beta }
\setminus G_{\beta },x) = 1\} .$$
Since the function $f$ is s.c. at every point, there is an open
interval $I_{\alpha }$ with rational endpoints such that $\mu (I_{\alpha }
\setminus G_{\alpha }) > 0$ and $\osc f < \eta $ on the set $H_{\alpha
}$. By transfinite induction we find a transfinite sequence of such open
intervals $(I_{\alpha })_{\alpha < \alpha _0}$ with rational endpoints,
where $\alpha _0$ is the first ordinal number for which $\mu (\R
\setminus G_{\alpha _0}) = 0$.
Since the family of all open intervals with rational endpoints is
countable, the ordinal number $\alpha _0$ is also countable. For every
$\alpha < \alpha _0$ we find a point $x_{\alpha } \in H_{\alpha }$ and
let $g(x) = f(x_{\alpha })$ for $x \in H_{\alpha }$ and $g(x) = f(x)$
otherwise on $\R$. If $x \in H_{\alpha }$, then
$$|f(x) - g(x)| = |f(x) - f(x_{\alpha })| < \eta .$$
In the remaining case $f(x) = g(x)$, so $|f - g| < \eta $.
For the completeness of the proof it suffices to show that the
function $g \in {\cal A}_7$.
For this, fix a nonempty set $A \in {\cal T}_d$ and a positive real
$\varepsilon $. There is an ordinal number $\beta < \alpha _0$ such that
$I_{\beta } \cap A \neq \emptyset $ and $I_{\alpha } \cap A = \emptyset $
for $\alpha < \beta $. Then $\emptyset \neq A \cap I_{\beta } \subset
H_{\beta }$, since otherwise we have $G_{\beta } \cap A \neq
\emptyset$, a contradiction. Consequently, $g(x) = f(x_{\beta })$ for each
point $x \in I_{\beta } \cap A$. So, $\osc f = 0 < \varepsilon $ on
$I_{\beta } \cap A$. \qed
\begin{rem}
The function $f$ from the proof of Theorem 2 is such that $f \in {\cal
A}_6 \setminus {\cal B}_u({\cal A}_7)$, since $\osc f = 1$ on the sets
$I \cap B$ for every open interval $I$ with $I \cap B \neq \emptyset $.
\end{rem}
Observe that if $f \in {\cal B}_u({\cal A}_6)$, then $f$ is measurable and the
set $\R \setminus C(f)$ is of the first category. Moreover, if
$$\aposc f(x) = \inf \{ \osc_Af;\emptyset \neq A \in {\cal T}_d, \;\; x
\in A\}, $$
then we have the following.
\begin{thm}
If a function $f$ belongs to ${\cal B}_u({\cal A}_6)$, then for every positive
real $\eta $ and for every nonempty set $A \in {\cal T}_d$ the set $\{ x
\in A; \aposc f(x) \geq \eta \} $ is nowhere dense in $A$.
\end{thm}
\pf Assume, to the contrary, that there are a function $f \in
{\cal B}_u({\cal A}_6)$, a nonempty set $A \in {\cal T}_d$,
and a positive real $\eta $ such that for
every open interval $J $ with $J \cap A \neq \emptyset $ there
is a point $x \in J \cap A$ at which $\aposc f(x) \geq \eta $. Since $f
\in B_u({\cal A}_6)$, there is a function $g \in {\cal A}_6$ such that
$|f - g| < \eta /3$. Observe that if $\aposc f(x) \geq \eta $, then
$\aposc g(x) \geq \eta /4$. So, for every open interval $J $ with $J
\cap A \neq \emptyset $ there is a point $x \in J \cap A$ at which
$\aposc g(x) \geq \eta /4$. But $g \in {\cal A}_6$, so there is an open
interval $J $ with $J \cap A \neq \emptyset $ and $J \cap A
\subset A(f)$. Consequently, for each point $x \in J \cap A$ we obtain
$\aposc g(x) = 0 < \eta /4$, a contradiction. \qed
\medskip
\noindent{\bf Problem.} Characterize the class ${\cal B}_u({\cal A}_6)$.
\begin{rem}
Since every function $f \in {\cal A}_2$ is measurable and every
measurable function is the limit of a sequence of approximately
quasi-continuous functions (which belong to ${\cal A}_2$) (\cite{6}), we
obtain that ${\cal B}({\cal A}_2)$ is the family of all measurable
functions.
\end{rem}
In \cite{8} Mauldin shows that $f \in {\cal B}({\cal C}_{ae})$ if and only
if there are a function $g$ of Baire class 1 and an
$F_{\sigma }$-set $A$ of measure zero such that $\{ x; f(x) \neq g(x)\}
\subset A$.
\begin{thm}
The equality
${\cal B}({\cal A}_5) = {\cal B}({\cal C}_{ae})$
is true.
\end{thm}
\pf Since every $f \in {\cal A}_5$ belongs to ${\cal C}_{ae}$,
by Mauldin's theorem we obtain the inclusion
${\cal B}({\cal A}_5) \subset {\cal B}({\cal C}_{ae}).$
Let $f \in {\cal B}({\cal C}_{ae})$ be a function. By Mauldin's theorem
there are a function $g$ of the first class of Baire and an
$F_{\sigma }$-set $A$ of measure zero such that $\{ x;f(x) \neq g(x)\}
\subset A$. Let $h = f - g$. Then $h(x) = 0$ for each point $x$ which is
not in $A$. There are closed sets $A_n$, $n = 1,2,\ldots $, such that
$A_1 \subset \ldots \subset A_n \subset \ldots $ and $A = \bigcup_nA_n$.
For $n = 1,2,\ldots $ let $h_n(x) = h(x)$ for $x \in A_n$ and let $h(x) =
0$ otherwise on $\R$. Since every set $A_n$, $n = 1,2,\ldots $, is
closed and of measure zero, it is nowhere dense and consequently every
function $h_n \in {\cal A}_5$, $n = 1,2,\ldots $. For the function $g$
there is a sequence of continuous functions $g_n$, $n = 1,2,\ldots $,
such that $g = \lim_{n \rightarrow \infty }g_n$. Observe that every function
$f_n = g_n + h_n$, $n = 1,2,\ldots $, belongs to ${\cal A}_5$ as the sum
of the continuous function $g_n$ and the function $h_n$ belonging to
${\cal A}_5$. Since
$$f = g + h = \lim_ng_n + \lim_nh_n = \lim_nf_n,$$
the proof is complete.\qed
\medskip
In \cite{5} the following theorem is proved.
\begin{thm}
Let $f$ be a function such that there is a Baire 1 function $g$ such that
for every positive real $\eta $ and for each point $x$ such that $|f(x) -
g(x)| \geq \eta $ there is a closed interval $I(x)$ containing $x$ and
such that $\mu (I(x) \setminus cl(\{ t;|f(t) - g(t)| \geq \eta \} )) =
0$. Then there is a sequence of functions $f_n \in Q_s$ such that
$\lim_nf_n = f$.
\end{thm}
Since every function $g$ from $Q_s$ is the limit of a sequence of functions
from ${\cal A}_3$ which uniformly converges to $g$, every function
satisfying the hypothesis of the above theorem belongs to ${\cal B}({\cal
A}_3)$.
We can prove the following.
\begin{thm}
Let $f$ be a function such that there are a Baire 1 function $g$ and an
$F_{\sigma }$ set $B$ of measure zero such that $\{ t;f(t) \neq g(t)\}
\subset B$ and for every positive real $\eta $ and for each point $x$
such that $|fx) - g(x)| \geq \eta $ the upper density
$d_u(cl(\{ t;|f(t) - g(t)| \geq \eta \} ),x) = 0$.
Then there is a sequence of functions $f_n \in {\cal A}_3$, $n =
1,2,\ldots $, which converges to $f$.
\end{thm}
\pf Let $h = f - g$ and let $B = \cup_nB_n,$ where
every set $B_i$ is closed and $B_i
\subset B_{i+1}$ for $i = 1,2,\ldots $. For $n = 1,2,\ldots $ let
$A_n = \{ x;|h(x)| \geq 1/n\} .$
Fix a positive integer $n$. By our hypothesis there is a family of
disjoint closed intervals $I_{k,l,i}$, $k \leq n$, $l,i =
1,2,\ldots $ such that
\begin{enumerate}
\item[--]$I_{1,l,i} \subset \R \setminus (\cl(A_n) \cap B)$ for
$l,i \geq 1$;
\item[--] $I_{k,l,i} \subset \R \setminus \cl(A_n) $
for $1 < k \leq n$, $l,i = 1,2,\ldots $;
\item[--] for each $k \leq n$ the inclusion $$I_{k,l,i} \subset
A(\cl(A_k) \cap B_k,1/k) (= \{ t;\inf\{ |u - t|;u \in \cl(A_k) \cap B_k\} \leq
1/n\} $$ is true for
$l,i \geq 1$;
\item[--] if $k \leq n$, $l \geq 1$ and $x \in \cl(A_k) \cap B_k$, then
$d_u\bigl(\bigcup_{i=1}^{\infty }I_{k,l,i},x\bigr) > 0;$
\item[--] for each $k \leq n$ and for each
$x \in \R \setminus (A_k \cap B_k)$ there is an open set $U$
containing $x$ such that the set
$\{ (k,l,i);U \cap I_{k,l,i} \neq \emptyset \} $ is finite.
\end{enumerate}
Next, in every interval $\Int(I_{k,l,i})$, $k \leq n$,
$l,i = 1,2,\ldots $,
we find a closed interval $J_{k,l,i}$ such that if $k \leq n$, $l \geq 1$
and $x \in \cl(A_k) \cap B_k$, then
$d_u\bigl(\bigcup_{i=1}^{\infty }J_{k,l,i},x)\bigr) > 0.$
Let $(w_{1,l})_{l=1}^{\infty }$ be a sequence of all rationals
with $w_{1,1} = 0$ and for $k
> 1$ let $(w_{k,l})_{l=1}^{\infty }$ be a sequence of all rationals
belonging to the interval $[-1/(k-1),1/(k-1)]$ with $w_{k,1} = 0$.
Put
\begin{displaymath}
h_n(x) = \begin{cases}
w_{k,l}& x \in J_{k,l,i}, k \leq n, l,i \geq 1\\
h(x)& x \in A_n \cap B_n\\
\text{linear on the components}&\\
\text{of the sets } I_{k,l,i} \setminus J_{k,l,i},& l,i \geq 1, \; k \leq n\\
0&\text{ otherwise on } \R.
\end{cases}
\end{displaymath}
Evidently, the function $h_n$ is continuous at each point $x$ which is
not in $\cl(A_n) \cap B_n$. Fix a positive real $\eta $, a point
$x \in \cl(A_n) \cap B_n$ and a set $A \in {\cal T}_d$ containing $x$.
If there is an integer $k \leq n$ such that $x \in A_k \cap B_k$,
then there is a rational $w_{k,l}$ such that $|h(x) - w_{k,l}| < \eta $. Since
$d_u\bigl(\bigcup_iJ_{k,l,i},x\bigr) > 0,$
there is an interval $J_{k,l,i}$ such that $J_{k,l,i} \cap A \neq
\emptyset $. Every point $t \in A \cap J_{k,l,i}$ is a continuity
point of $h_n$ and
$$|h_n(t) - h_n(x)| = |w_{k,l} - h(x)| < \eta .$$
In the remaining case we obtain that $h_n(x) = 0$ and $x \in
\cl(A_k) \cap B_k$ for some positive integer $k$. Since $w_{k,1}
= 0$ and since
$d_u\bigl(\bigcup_iJ_{k,1,i},x\bigr) > 0,$
there is an interval $J_{k,1,i}$ such that $J_{k,1,i} \cap A \neq
\emptyset $. For each point $t \in J_{k,1,i} \cap A$ the function $h_n$
is continuous at $t$ and $|h_n(t) - h_n(x)| = 0 < \eta .$
So, $h_n \in {\cal A}_3$.
Now we will prove that $\lim_{n \rightarrow \infty }h_n = h$. If there is
a positive integer $n$ with $x \in A_n \cap B_n$, then $h_k(x) = h(x)$
for $k \geq n$. If not, we have $h(x) = 0$. Fix a positive
real $\eta $ and a positive integer $n$ with $1/n < \eta $.
If $x \in \cl(A_m) \cap B_m$ for some positive integer $m$, then $h_k(x) =
0$ for all $k \geq m$. So, we suppose that $x$ is not in
$\cl(A_m) \cap B_m$ for $m \geq 1$.
Since $x$ is
not in the set $cl(A_n) \cap B_n$, there is a positive integer
$m > n$ such that $|x - y| > 1/m$ for every point $y \in
\cl(A_n)$. Consequently, $$h_k(x) \leq 1/(m-1) \leq 1/n < \eta $$
for every $k \geq m$. This completes the proof that
$h=\lim_{n\to\infty}h_n.$
Since $g$ is a Baire 1 function, there is a sequence of continuous
functions $g_n$ such that $g = \lim_{n \rightarrow \infty }g_n$.
Evidently, the functions $f_n = h_n + g_n$, $n = 1,2,\ldots $, belong to
the family ${\cal A}_3$ and
\begin{equation}
\lim_{n \rightarrow \infty }f_n = \lim_{n \rightarrow \infty }h_n +
\lim_{n \rightarrow \infty }g_n = h + g = f.\tag*{\qed}
\end{equation}
\begin{cor}
If the function $f:\R \longrightarrow \R$ is almost everywhere
continuous, then $f \in {\cal B}({\cal A}_3)$.
\end{cor}
Denote by ${\cal P}_s$ the family of all functions which are s.c. at each
point.
\noindent{\bf Problem.} Is it true that
${\cal P}_s \cap {\cal B}({\cal C}_{ae}) = {\cal B}({\cal A}_3)?$
Now, we will investigate the transfinite convergence of sequences.
Let $\omega _1$ denote the first uncountable ordinal number. A
transfinite sequence of functions $f_{\alpha }$, $\alpha < \omega _1$,
converges to a function $f$ ($\lim_{\alpha }f_{\alpha } = f$) if for
each point $x$ there is an ordinal number $\beta < \omega _1$ such that
$f_{\alpha }(x) = f(x)$ for each countable ordinal $\alpha > \beta $.
\begin{thm}
Let ${\cal K}$ be a family of functions such that if a function $f$ is
not in ${\cal K}$, then there is a countable set $A$ such that for every
function $g \in {\cal K}$ there is a point $x \in A$ with $g(x) \neq
f(x)$.
Then the limits of all converging transfinite sequences of functions from
the family ${\cal K}$ belong to ${\cal K}$.
\end{thm}
\pf Let $(f_{\alpha }) \in {\cal K}$, $\alpha < \omega _1$, and let
$\lim_{\alpha }f_{\alpha } = f$. Suppose, to the contrary, that $f$ is not
in ${\cal K}$. Then there is a countable set $A = \{ x_1,x_2,\ldots \} $
such that for each function $g \in {\cal K}$ there is a point $x \in A$
with $g(x) \neq f(x)$. For each positive integer $n$ there is a countable
ordinal number $\beta _n$ such that $f_{\alpha }(x_n) = f(x_n)$ for
$\beta _n < \alpha < \omega _1$. There is a countable ordinal number
$\beta $ such that $\beta _n < \beta $ for all positive integers $n$. So,
$f_{\beta }(x_n) = f(x_n)$ for $n = 1,2,\ldots $. Since
$f_{\beta } \in {\cal K}$,
we obtain a contradiction.\qed
\begin{rem}
Observe that the families $P_s$ and ${\cal A}_i$, $i = 1,3,5$, satisfy the
hypothesis of the above Theorem 9.
\end{rem}
\pf If $f$ is not in ${\cal A}_i$, $i = 1$ or $3$ or $5$, then
every countable set $A$ such that the set $\{ (x,f(x));x \in A\} $ is
dense in the graph of the function $f$ satisfies all requirements. \qed
\begin{thm}
Assume the Continuum Hypothesis HC. For every function $f$ there is a
transfinite sequence of functions $f_{\alpha } \in {\cal A}_2$,
$\alpha < \omega _1$, such that $f = \lim_{\alpha }f_{\alpha }$.
\end{thm}
\pf Let $(x_{\alpha })_{\alpha < \omega _1}$ be a transfinite
sequence of all reals numbers. Fix an ordinal number $\alpha < \omega _1$
and let $(t_n)_n$ be a sequence of all numbers $x_{\beta }$ with $\beta \leq
\alpha $ such that $t_i \neq t_j$ for $i \neq j$, $i,j = 1,2,\ldots $.
For every positive integer $n$ there are closed intervals $I_n$, $J_n$ and
a closed set $A_n$ such that
\begin{enumerate}
\item[--] $t_n$ is an endpoint of $I_n$ and $J_n $;
\item[--] $J_n \subset I_n$ and $|J_n| = |I_n|/2$;
\item[--] $A_n \subset J_n \setminus \{ t_k;k \neq n$ and $k = 1,2,\ldots \} $;
\item[--] $I_n \cap A_k = \emptyset $ for $k < n$;
\item[--] $d_u(A_n,t_n) > 0$;
\item[--] $\mu (A_n) < |J_n|/4^n$.
\end{enumerate}
For a construction of such $I_n$, $J_n$ and $A_n$ it suffices to find a
nowhere dense closed set
$$A_n \subset \R \setminus \bigcup_{k \neq n}\{t_k\} \setminus
\bigcup_{k 0$
and next fix some closed intervals $I_n$ and $J_n$ satisfying all
requirements.
Denote by $B_n$ the set of all points $t \in A_n$ at which
$d_u(A_n,t) > 0$. Let
\begin{displaymath}
f_{\alpha }(x) =
\begin{cases}
f(t_n) &\text{if } x \in B_n, n = 1,2,\ldots \\
0 &\text{otherwise on } \R
\end{cases}
\end{displaymath}
Then $f_{\alpha } \in {\cal A}_2$ and $f = \lim_{\alpha }f_{\alpha }$.
\begin{thm}
If functions $f_{\alpha } \in {\cal A}_4$ for $\alpha < \omega _1$ and $f
= \lim_{\alpha }f_{\alpha }$, then $f \in {\cal A}_4$.
\end{thm}
\pf Assume, by a contrary, that $f$ is not in ${\cal A}_4$. Since
by Theorems 1, 9 and Remark 9 the function $f \in {\cal A}_1$, there are a
positive
real $\eta $, a point $x$ and a set $A \in {\cal T}_d$ such that $x \in
A$ and for every open interval $I$ with $I \cap A \neq \emptyset $ and
$f(I \cap A) \subset (f(x) - \eta ,f(x) + \eta )$ there is a point $t \in
I \cap A$ at which the function $f$ is not approximately continuous. Let
$(x_n)_n$ be a sequence of points such that the set $\{ (x_n,f(x_n);n =
1,2,\ldots \} $ is dense in the graph of the function $f$ and for each
open interval $I$ with $I \cap A \neq \emptyset $ and $f(I \cap A)
\subset (f(x) - \eta ,f(x) + \eta )$ there is a point
$x_{n(I)} \in I \cap A$ at which
$f$ is not approximately continuous. There is a countable ordinal number
$\beta $ such that $f_{\alpha }(x_n) = f(x_n)$ for all countable ordinal
numbers $\alpha \geq \beta $ and $n = 1,2,\ldots $. Consequently, the
functions $f$ and $f_{\beta }$ are almost everywhere equal. Since $f \in
{\cal A}_1$, there is an open interval $I$ such that $I \cap A \neq
\emptyset $ and $f(A \cap I) \subset (f(x) - \eta ,f(x) + \eta )$. From
the relation $f_{\beta } \in {\cal A}_4$ we obtain that there is an open
interval $J \subset I$ such that $J \cap A \neq \emptyset $,
$f_{\beta }(J \cap A) \subset (f(x) - \eta ,f(x) + \eta )$ and the
function $f_{\beta }$ is approximately continuous at every point $t \in J
\cap A$. Let a point $u = x_k \in A \cap J$ be such that the function $f$ is
not approximately continuous at $u$. Since $f_{\beta }(u) = f(u)$ and the
functions $f$ and $f_{\beta }$ are equal at almost all points, the
function $f_{\beta }$ must be approximately continuous at $u$, a
contradiction. \qed
\medskip
Observe that there are Baire 1 functions which are not in ${\cal A}_6$.
Since every Baire 1 function is the limit of a transfinite sequence of
approximately continuous functions (see \cite{7}), then the classes ${\cal
A}_i$, $i = 6,7$, are not closed under the transfinite convergence. But
if a function $f$ is the limit of a transfinite sequence of functions
$f_{\alpha } \in {\cal A}_6$ ($f_{\alpha } \in {\cal A}_7$), then $f$ is
pointwise discontinuous (pointwise discontinuous on each nonempty set
belonging to ${\cal T}_d$).
\medskip
\noindent{\bf Problem.} Characterize the functions being the limits of
transfinite sequences of functions belonging to ${\cal A}_i$, $i = 6,7$.
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\end{document}
\end{document}