On A. C. Limits of Decreasing Sequences of Continuous or Right Continuous Functions

Abstract

: 9/8/99 cew % last edit: 9/8/99 cew % gallies sent: 9/21/99 % gallies corrected: cew 10/6/99 % set in production style: 9/8/99 cew % Research Section from Editor Thomson \documentclass{rae} \usepackage{amsmath,amsthm,amssymb} %\coverauthor{Zbigniew Grande} %\covertitle{On A. C. Limits of Decreasing Sequences of Continuous or Right Continuous Functions} \received{August 3, 1998} \MathReviews{26A15, 26A21, 26A99} \keywords{upper semicontinuity, decreasing sequences of functions, $B_1^*$ class, right continuity.} \firstpagenumber{231} \markboth{Zbigniew Grande}{On A. C. Limits of Decreasing Sequences} \author{Zbigniew Grande\thanks {Supported by Bydgoszcz Pedagogical University grant 1998} , Institute of Mathematics, Pedagogical University, Plac Weyssenhoffa 11, 85-072 Bydgoszcz, Poland. e-mail: {\tt grande@wsp.bydgoszcz.pl}} \title{ON A. C. LIMITS OF DECREASING SEQUENCES OF CONTINUOUS OR RIGHT CONTINUOUS FUNCTIONS} %%%Put Autnor's Definitons Below Here%%% %\newcommand{\mathR}{\mathop{\rm I\kern-0.35ex R}\nolimits} \newtheorem {thm} {Theorem} \newtheorem {lem} {Lemma} \newtheorem {rem} {Remark} {\theoremstyle{definition} \newtheorem{example}{Example}} \DeclareMathOperator{\cl}{cl} \DeclareMathOperator{\a.c.}{a.c.} %%%%%% \newcommand{\rest}{\restriction} \DeclareMathOperator{\osc}{osc} \newcommand{\R}{\ensuremath{\mathbb R}} \newcommand{\pf}{\noindent{\sc Proof.~}} \begin{document} \maketitle \begin {abstract} The a.c. limits ({\bf i.e.} the discrete limits introduced by Cs{\'a}sz{\'a}r and Laczkovich) of decreasing sequences of continuous (resp. right continuous) functions are investigated. \end {abstract} Let ${\R}$ be the set of all reals. $(X,\tau )$ or $X$ in this paper always denotes a perfectly normal Hausdorff topological space. A function $f:X \to {\R}$ is a $B_1^*$ function (belongs to the class $B_1^*$) if there is a sequence of continuous functions $f_n:X \to {\R}$ with $f = \a.c.\lim_{n \rightarrow \infty }f_n$, {\bf i.e.} for each point $x \in X$ there is a positive integer $k$ such that $f_n(x) = f(x)$ for every $n > k$ (compare \cite{2, 3}). From the results obtained in \cite{2} it follows that the function $f:X \to {\R}$ belongs to ${\cal B}_1^*$ if and only if there are closed sets $A_n$, $n = 1,2,\ldots $, such that the restricted functions $f\restriction A_n$ are continuous and $X = \bigcup_{n=1}^{\infty }A_n.$ \section{The Discrete Limits of Decreasing Sequences of Continuous Functions.} In the first part of this article we will investigate $B_1^*$ functions which are upper semicontinuous. Recall that the function $f:X \to {\R}$ is upper semicontinuous if for every real $a$ the set $\{ x \in X;f(x) < a\} $ belongs to $\tau $. Evidently the pointwise limit of each decreasing sequence of upper semicontinuous functions $f_n:X \to {\R}$, $n = 1,2,\ldots $, is upper semicontinuous. The following theorem can be found on page 51 of \cite{R}. \begin{rem} If the function $f:X \to {\R}$ is upper semicontinuous, then there is a decreasing sequence of continuous functions $f_n:X \to {\R}$, $n = 1,2,\ldots $, such that $f = \lim_{n \rightarrow \infty }f_n$. \end{rem} %%% We will prove the following theorem. \begin{thm} Let $(X,\tau )$ be a perfectly normal $\sigma $-compact Hausdorff topological space. Then an upper semicontinuous function $f:X \to {\R}$ belongs to class $B_1^*$ if and only if there is a decreasing sequence of continuous functions $f_n:X \to {\R}$ such that $f = \a.c.\lim_{n \rightarrow \infty }f_n$. \end{thm} We start from the following lemma. \begin{lem} Let $f:X \to {\R}$ be a function. If there are sets $A_n$ and continuous functions $f_n:X \to {\R}$ such that $A_1 \subset A_2 \subset \cdots$, $X = \bigcup_nA_n,$ $f_n \geq f$ and $f_n\rest A_n = f\rest A_n$ for $n = 1,2,\ldots,$ then there is a decreasing sequence of continuous functions $g_n:X \to {\R}$ with $f = \a.c.\lim_{n \rightarrow \infty }g_n$. \end{lem} \pf Of course, the functions $g_n = \min_{k \leq n}f_n$ satisfy all required conditions.\qed \medskip {\noindent{\sc Proof of Theorem 1.~} If $f$ is the discrete limit of a decreasing sequence of continuous functions $f_n:X \to {\R}$, then evidently $f \in {\cal B}_1^{*}$. So, we assume that $f \in {\cal B}_1^*$. Since $f$ is upper semicontinuous and $X$ is perfectly normal, by Remark 1 there is a decreasing sequence of continuous functions $f_n:X \to {\R}$ which converges to $f$ at each point $x \in X$. On the other hand $f$ is the discrete limit of continuous functions; so there are closed sets $A_n$, $n = 1,2,\ldots $, such that every restricted function $f\rest A_n$ is continuous and $X = \bigcup_{n=1}^{\infty }A_n.$ We can assume that $A_n$ is compact for each $n = 1,2,\ldots $. Fix a positive integer $k$. On $A_k$ the sequence $(f_n)$ tends uniformly to $f$ due to Dini's lemma. So we can also assume that $$\max\{ (f_n(x) - f_{n+1}(x));x \in A_k\} \leq 2^{-n}.$$ By Tietze's theorem for $n = 1,2,\ldots $ there is a continuous extension $g_n:X \to [0,2^{-n}]$ of the restricted function $(f_n - f_{n+1})\rest A_k$. Let $$h_n = \min(g_n,f_n - f_{n+1})\text{ for }n = 1,2,\ldots ,$$ and let $l_k = f_1 - \sum_{n=1}^{\infty }h_n.$ Since the series $\sum_{n=1}^{\infty }h_n$ converges uniformly, the function $l_k$ is continuous. Moreover, for $k = 1,2,\ldots $ we have $l_k \geq f$ and $f\rest A_k = l_k\rest A_k.$ So, by Lemma 1 we obtain our theorem. \qed \bigskip Theorem $1$ in the presented form and its proof was proposed by the referee. My formulation concerned the function $f:[a,b] \to {\R}$ and the Euclidean topology and its proof was more complicated. \section{Decreasing Sequences of Right Continuous Functions} In this part we assume that $X = [a,b)$ and $\tau $ is the topology of right continuity. This topology $\tau $ is perfectly normal and Hausdorff but is not $\sigma $-compact. So, the limit $f$ of a decreasing sequence of right upper semicontinuous functions $f_n$, $n = 1,2,\ldots $, is a right upper semicontinuous function and Remark 1 is valid for $(X,\tau )$. Thus we have the following assertion. \begin{rem} For every right upper semicontinuous function $f$ there is a decreasing sequence of right continuous functions $f_n$, $n = 1,2,\ldots $, such that $f = \lim_{n \rightarrow \infty }f_n$. \end{rem} From the last remark by an elementary proof we obtain the next assertion. \begin{rem} If a function $f:[a,b) \to {\R}$ is right upper semicontinuous, then there is a decreasing sequence of functions $f_n:[a,b) \to {\R}$ such that \begin{itemize} \item[] the functions $f_n$ are right continuous ; \item[] $f = \lim_{n \rightarrow \infty }f_n$; \item[] all functions $f_n$, $n = 1,2,\ldots $, are locally constant from the right, {\bf i.e.} for each point $x \in [a,b)$ there is a positive real $r_{x,n}$ such that $$I_{x,n} = [x,x+r_{x,n}] \subset [a,b)\text{ and }f\rest I_{x,n} \text{ is constant };$$ \item[] if $\limsup_{t \rightarrow x+}f(t) < f(x)$, then for $n$ sufficiently large $f_n(x) = f(x)$; \item[] for every integer $n$ the inclusion $f_n([a,b)) \subset \cl(f[a,b))), $ (where $\cl$ denotes the closure operation) holds. \end{itemize} \end{rem} \pf The set $A$ of all points $x$ at which $\limsup_{t \rightarrow x+}f(t) < f(x)$ is countable, i.e. if $A \neq \emptyset,$ then $A = \{ x_1,x_2,\ldots \} $. By Remark 2 there is a decreasing sequence of right continuous functions $g_n$ such that $f = \lim_{n \rightarrow \infty }g_n$. Fix a positive integer $n$ and observe that there is a sequence of intervals $I_{i,n} = [u_{i,n},v_{i,n})$, $i = 1,2,\ldots $, such that: \begin{itemize} \item[] $[a,b) = \bigcup_iI_{i,n}$; \item[] $I_{i,n} \cap I_{j,n} = \emptyset $ for $i \neq j$; \item[] $u_{i,n} = x_i$ for $i \leq n$; \item[] $\osc g_n < \frac{1}{n}$ on each interval $I_{i,n}$; \item[] $g_n(x) > f(x)$ if $x \in I_{i,n}$ and $i \leq n$. \end{itemize} Let $$h_n(x) =\begin{cases} f(x_i)&\text{for }x \in I_{i,n},\;\;i \leq n\\ \sup_{I_{i,n}}g_n&\text{for } x \in I_{i,n}\;\;i > n.\end{cases}$$ Then the functions $f_n = \min(h_1,h_2,\ldots ,h_n), $ for $n = 1,2,\ldots ,$ satisfy all required conditions.\qed \begin{thm} If $f = \a.c.\lim_{n \rightarrow \infty }f_n$, where all functions $f_n$, $n = 1,2,\ldots $, are right continuous, then $f$ satisfies the following condition. \begin{itemize} \item[(1)] For each nonempty perfect set $A \subset [a,b)$ there is an open interval $I$ such that $A \cap I \neq \emptyset $ and the restricted function $f\rest(I \cap B)$, where $$B = \{ x \in A;x\text{ is a right limit point of } A\} ,$$ is right continuous at each point of the intersection $B \cap I$. \end{itemize} \end{thm} \pf Let $A \subset [a,b)$ be a nonempty perfect set and let $B$ denote the set of all right limit points of $A$. For each point $x \in [a,b)$ there is a positive integer $n(x)$ such that $f_n(x) = f(x)$ for $ n \geq n(x).$ For $n = 1,2,\ldots $ put $A_n = \{ x \in [a,b);n(x) = n\} $ and observe that $[a,b) = \bigcup_{n=1}^{\infty }A_n.$ So there are an open interval $I$ and a positive integer $k$ such that $I \cap B \neq \emptyset$ and $A_k \cap I \cap B$ is dense in $B \cap I.$ Thus $f(x) = f_k(x)$ for each point $x \in I \cap A$ which is a right limit point of $A$ and consequently $f\rest(B \cap I)$ is right continuous at each point of $I \cap B$. \qed \medskip The above proof of Theorem $2$ is short. However the referee related this statement to the result of Cs{\'a}sz{\'a}r and Laczkovich (Theorem $13$ of \cite{2}, pp. 469) which says that if $X$ is a Baire space, the functions $f_n:X \to {\R}$, $n = 1,2,\ldots $, are continuous and $f = \a.c.\lim_{n \rightarrow \infty }f_n$, then the points of discontinuity of $f$ constitute a nowhere dense set in $X$. \medskip The connection between these two results is the following assertion. \medskip Let $(X,{\cal T},\tau )$ be a bitopological space such that $\tau $ is finer than ${\cal T}$ and $(X,{\cal T})$ is a Baire space. Assume that for every nonempty set $A \in \tau $ there is a nonempty set $B \in {\cal T}$ such that $B \subset A$. Then every $\tau $-nowhere dense set is ${\cal T}$-nowhere dense and $(X,{\cal T})$ is a Baire space. \medskip We arrive at Theorem $2$ at once if we observe that the sets $X \subset [a,b)$ having no right isolated points satisfy the conditions of the previous statement. So the quoted theorem of Cs{\'a}sz{\'a}r and Laczkovich can be applied. \medskip \begin{example}\label{ex1} Let $C$ be the Cantor ternary set and let $I_n = (a_n,b_n)$, $n = 1,2,\ldots $, be an enumeration of all components of the set $[0,1) \setminus C$ such that $I_n \cap I_m = \emptyset$ for $n \neq m,\;n,\,m = 1,2,\ldots .$ Put \begin{displaymath} f(x) = \left\{ \begin{array}{ccl} 1 & \text{for}& x \in B = C \setminus \{ a_n;n \geq 1\} \\ 0 &\text{for} & x \in [0,1) \setminus B. \end{array} \right. \end{displaymath} Observe that the function $f$ is not of Baire class one. For $n \geq 1$ let \begin{displaymath} f_n(x) = \left\{ \begin{array}{ccl} 0 &\text{for} & x \in B_n = \bigcup_{i \leq n}[a_i,b_i) \\ 1 &\text{for}& x \in [0,1) \setminus B_n. \end{array} \right. \end{displaymath} Then all functions $f_n$, $n = 1,2,\ldots $, are right continuous, $f_n \geq f_{n+1}$ for $ n = 1,2,\ldots $ and $\a.c.\lim_{n \rightarrow \infty }f_n = f.$ \end{example} Now we introduce the following condition $(1')$. \begin{itemize} \item[(1')] A function $f$ satisfies condition $(1')$ if for every nonempty closed set $A \subset [0,1)$ there is an open interval $I$ such that $I \cap A \neq \emptyset $ and the restricted function $f\rest(A \cap I)$ is right continuous. (If $x \in A$ is right isolated in $A$, then $f\rest A$ is right continuous at $x$ by default.) \end{itemize} Observe that the implication $(1') \Longrightarrow (1)$ is true. The function $f$ from Example $1$ satisfies condition $(1)$ but it does not satisfy condition $(1')$. Observe also that, by Baire's theorem on Baire $1$ functions, every function $f$ satisfying condition $(1')$ is of Baire $1$ class. \begin{thm} A function $f$ satisfies condition $(1')$ if and only if it satisfies the following condition. \begin{itemize} \item[(2)] There is a sequence of nonempty closed sets $A_n \subset [a,b)$ such that all restricted functions $f\rest A_n$, $n = 1,2,\ldots $, are right continuous and $[a,b) = \bigcup_{n=1}^{\infty }A_n.$ \end{itemize} \end{thm} \pf $(1') \Longrightarrow (2)$. We will apply transfinite induction. Let $I_0$ be an open interval with rational endpoints such that the restricted function $f\rest I_0$ is right continuous. Fix an ordinal number $\alpha > 0$ and suppose that for every ordinal number $\beta < \alpha $ there is an open interval with rational endpoints $I_{\beta }$ such that $H_{\beta } = I_{\beta } \setminus \bigcup_{\gamma < \beta }I_{\gamma } \neq \emptyset$ and the restricted function $f\rest H_{\beta }$ is right continuous. If $G_{\alpha } = [a,b) \setminus \bigcup_{\beta < \alpha }I_{\beta } \neq \emptyset,$ then by $(1')$ there is an open interval $I_{\alpha }$ with rational endpoints such that $I_{\alpha } \cap G_{\alpha } \neq \emptyset$ and the restricted function $f\rest (I_{\alpha } \cap G_{\alpha })$ is right continuous. Let $\xi $ be the first ordinal number $\alpha $ such that $[a,b) \setminus \bigcup_{\beta < \xi }I_{\beta } = \emptyset .$ Since the family of all intervals with rational endpoints is countable, $\xi $ is a countable ordinal number. Every set $H_{\alpha }$, $\alpha < \xi $, is an $F_{\sigma }$ set; so there are closed sets $H_{k,\alpha }$, $k = 1,2,\ldots $, such that $H_{\alpha } = \bigcup_{k=1}^{\infty }H_{k,\alpha } .$ Evidently, all restricted functions $f\rest H_{k,\alpha },\;k = 1,2,\ldots \text{ and } \alpha < \xi ,$ are right continuous. Now enumerate in a sequence $(A_n)$ all sets $$ H_{k,\alpha },\;\;k = 1,2,\ldots \;\; {\rm and} \;\; \alpha < \xi ,$$ and observe that this sequence satisfies all requirements. $(2) \Longrightarrow (1')$ Fix a nonempty closed set $A \subset [a,b)$. If $A$ contains isolated points, then condition $(1')$ is satisfied. So we assume that $A$ is a perfect set. By $(2)$ there is a sequence of closed sets $A_n$, $n = 1,2,\ldots $, such that $[a,b) = \bigcup_nA_n$ and all restricted functions $f\rest A_n$, $n = 1,2,\ldots $, are right continuous. Since $A = \bigcup_{n=1}^{\infty }(A \cap A_n),$ there are a positive integer $k$ and an open interval $I$ such that $I \cap A = I \cap A_k \neq \emptyset .$ But the restricted function $f\rest(A \cap I)$ is right continuous; so condition $(1')$ is satisfied.\qed \begin{thm} If $f$ satisfies condition $(1')$ (or equivalently $(2)$) from the last theorem, then there is a sequence of functions $f_n$, $n = 1,2,\ldots $, which are right continuous and for which $\a.c.\lim_{n \rightarrow \infty }f_n = f$. \end{thm} \pf There is a sequence of nonempty closed sets $A_n$, $n = 1,2,\ldots $, such that $[a,b) = \bigcup_{n=1}^{\infty }A_n,\; A_1 \subset A_2 \subset \cdots$ and all restricted functions $f\rest A_n$, $n = 1,2,\ldots $, are right continuous. By Tietze's theorem for $n = 1,2,\ldots $ there is a right continuous function $f_n:[a,b) \to {\R}$ which is equal to $f$ on the set $A_n$. Then $\a.c.\lim_{n \rightarrow \infty }f_n = f.$\qed \begin{thm} If a function $f$ is upper semicontinuous from the right and satisfies condition $(1')$ (or equivalently $(2)$), then there is a decreasing sequence of right continuous functions $f_n$, $n = 1,2,\ldots $, such that $\a.c.\lim_{n \rightarrow \infty }f_n = f.$ \end{thm} \pf Let a function $f$ satisfies the hypothesis of our theorem. There is a sequence of nonempty closed sets $A_n$, $n = 1,2,\ldots $, such that $[a,b) = \bigcup_{n=1}^{\infty }A_n$ and all restricted functions $f\rest A_n$, $n = 1,2,\ldots $, are right continuous. Without loss of the generality we can suppose that $a \in A_1 \subset A_2 \subset \ldots \subset A_n \subset \ldots .$ Fix a positive integer $n$ and enumerate in a sequence $(I_{n,k})_k$ all components of the set $[a,b) \setminus A_n$. If $I_{n,k} = (a_{n,k},b_{n,k})$, $ a_{n,k} \in A_n$ and $\limsup_{t \rightarrow a_{n,k}+}f(t) = f(a_{n,k})$, then we find points $c_{n,k,i}$, $i = 1,2,\ldots $, such that $$b_{n,k} = c_{n,k,1} > \ldots > c_{n,k,i} > \ldots \searrow a_{n,k}.$$ By Theorem 2 and Remark 1 for every $i \geq 1$ there is right constant function $h_{n,k,i}:[c_{n,k,i+1},c_{n,k,i}) \to {\R}$ such that $h_{n,k,i} \geq f/[c_{n,k,i+1},c_{n,k,i})$ and $h_{n,k,i}(c_{n,k,i}) < f(c_{n,k,i}) + \frac{1}{ik}.$ Let $g_{n,k,i}(x) = \max(h_{n,k,i}(x),f(a_{n,k}))$ for $x \in [c_{n,k,i+1},c_{n,k,i}).$ Observe that $g_{n,k,i}([c_{n,k,i+1},c_{n,k,i})) \subset \cl(f([a_{n,k},b_{n,k}))), \; i = 1,2,\ldots$ Next in every such interval $I_{n,k}$ we define the function $g_{n,k}$ by $$g_{n,k}(x) = g_{n,k,i}(x)\;\; {\rm for}\;\; x \in [c_{n,k,i+1},c_{n,k,i}), \;\;i = 1,2,\ldots .$$ If $I_{n,k} = [a_{n,k},b_{n,k}),\; a_{n,k} \in A_n,$ and $\limsup_{x \rightarrow a_{n,k}+}f(x) < f(a_{n,k}),$ then by Remarks $2$ and $3$ there is a right constant function $h_{n,k}:[a_{n,k},b_{n,k}) \to {\R}$ such that $h_{n,k} \geq f\rest [a_{n,k},b_{n,k})$, $ h_{n,k}(a_{n,k}) = f(a_{n,k})$ and $h_{n,k}([a_{n,k},b_{n,k})) \subset \cl(f([a_{n,k},b_{n,k})))).$ Let $g_{n,k}(x) = \max(h_{n,k}(x),f(x))$ for $ x \in [a_{n,k}, b_{n,k}).$ Put \begin{displaymath} g_n(x) = \left\{ \begin{array}{ccl} f(x) &\text{for} & x \in A_n\\ g_{n,k}(x) &\text{for} & x \in I_{n,k}\;\;i,k = 1,2,\ldots . \end{array} \right. \end{displaymath} Then the function $g_n$ is right continuous, $g_n \geq f$ and $g_n\rest A_n = f\rest A_n.$ So, by Lemma 1 we obtain our theorem.\qed \medskip Observe that the last theorem is not a corollary of Theorem $1$, since the topology $\tau $ is not $\sigma $-compact. \medskip \noindent{\bf Acknowledgment} I would like to thank to the referee for several suggestions, the reformulation of Theorem 1 and its proof. \noindent{\bf Problem.} Is Theorem 5 true if we replace condition $(1')$ by $(1)$? \begin{thebibliography}{9} \bibitem{1} A.~M.~Bruckner, J.~B.~Bruckner and B.~S.~Thomson, {\it Real Analysis}, Prentice--Hall International, INC, Upper Saddle River, New Jersey, 1996. \bibitem{2} A. Cs{\'a}sz{\'a}r and M. Laczkovich, {\it Discrete and equal convergence}, Studia Sci. Math. Hungar. 10 (1975), 463--472. \bibitem{3} R. J. O'Malley, {\it Approximately differentiable functions. The r topology}, Pacific J. Math. 72 (1977), 207--222. \bibitem{R}H. L. Royden, {\it Real Analysis\/}, Third Edition, MacMillan, 1988. %\bibitem{4} Sikorski R.;{\it Real Functions} Vol.I., (in Polish), Warsaw %PWN 1958. \bibitem{5} B. S. Thomson, {\it Real Functions}, Lectures Notes in Math. 1170 (1985), Springer--Verlag. \end{thebibliography} \par\newpage$\quad$\end{document} This remark is probably known, but I can not give any reference. So, I give an elementary proof. \bigskip By Th. $3.6$ from Sikorski's book \cite{4} (pp. 178) it suffices to prove that the characteristic function $\kappa _A$ of arbitrary set $A \in \tau $ is the limit of an increasing sequence of continuous functions. Fix a nonempty set $A \in \tau $. Since $X$ is perfectly normal, there are $\tau $-closed sets $A_1 \subset \ldots \subset A_n \subset \ldots $ such that $$A = \bigcup_{n=1}^{\infty }A_n .$$ By Tietze's theorem there are continuous functions $$g_n:X \to [0,1], \;\; n = 1,2,\ldots ,$$ such that $$g_n(A_n) = \{ 1\} \;\;\;{\rm and}\;\;\;g_n(X \setminus A) = \{ 0\} \;\;{\rm for}\;\;n = 1,2,\ldots .$$ Then the sequence of continuous functons $$f_n = \max (g_1,\ldots ,g_n) \;\;\; n = 1,2,\ldots ,$$ is creasing and converges to $\kappa _