Real Analysis Exchange

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

Zbigniew Grande

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 _

Article information

Source
Real Anal. Exchange, Volume 25, Number 1 (1999), 231-238.

Dates
First available in Project Euclid: 5 January 2009

Permanent link to this document
https://projecteuclid.org/euclid.rae/1231187602

Mathematical Reviews number (MathSciNet)
MR1758004

Zentralblatt MATH identifier
1015.26006

Citation

Grande, Zbigniew. On A. C. Limits of Decreasing Sequences of Continuous or Right Continuous Functions. Real Anal. Exchange 25 (1999), no. 1, 231--238. https://projecteuclid.org/euclid.rae/1231187602