%%%%%%%%%%%%%% DELETE %%%%%%%% % These items (%?%) are missing: % -- the address of the author % -- the first reference %%%%%%%%%%%%%% DELETE %%%%%%% % spellchecked:pdh 11/24/98 % last edit: jch 11/15/1998 % gallies sent: 11/24/98 % gallies corrected: % set in production style: jch 11/15/1998 % Section Inroads editor Thomson \documentclass{rae} \usepackage{amsmath,amsthm,amssymb} %\coverauthor{Zbigniew Grande} %\covertitle{On Sequences of Monotone Functions} \received{January 18, 1998} \MathReviews{Primary: 26A15, 26A21, 26A48} \keywords{continuity, monotone functions, convergence, monotone convergence, transfinite convergence, variation} \firstpagenumber{1} \markboth{Zbigniew Grande}{On Sequences of Monotone Functions} \author{Zbigniew Grande, Institute of Mathematics, Pedagogical University, plac Weyssenhoffa 11, 85-072, Bydgoszcz, Poland\\ e-mail: {\tt grande@wsp.bydgoszcz.pl}} \title{ON SEQUENCES OF MONOTONE FUNCTIONS} %%%%%%Put Author's Definitions Below Here %%%%%%%%%%% \newtheorem {thm} {Theorem} \newtheorem {lem} {Lemma} \newtheorem {rem} {Remark} \newtheorem {cor} {Corollary} \def\pf{\noindent{\sc Proof.} } \def\mpf#1{\noindent{\sc #1.} } \def\osc{\mbox{\rm osc\,}} \def\dist{\mbox{\rm dist\,}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \maketitle \begin{abstract} Several kinds of convergence (pointwise, monotone, a.c., uniform, \dots\!\!\!) in the family of monotone functions are investigated. \end{abstract} \bigskip Let ${\cal R}$ denote the set of all reals. Observe that the limit $f$ of a converging sequence of monotone functions $f_n:I \mapsto {\cal R}$, where $I$ is a nondegenerate interval, is a monotone function. Of course, there is a subsequence $(f_{n_k})_k$, where all functions $f_{n_k}$ are decreasing or increasing and consequently, the function $f$ is decreasing or respectively increasing as the limit of the subsequence $(f_{n_k})_k$. \begin{thm} If $f:[a,b] \mapsto {\cal R}$ is an increasing function (i.e. nondecreasing) then there are continuous increasing functions $f_n:[a,b] \mapsto {\cal R}$, $n = 1,2,\ldots$, such that $f_n(a) = f(a)$, $f_n(b) = f(b)$ for $n \geq 1$ and $\lim_{n \rightarrow \infty }f_n = f$. \end{thm} \pf Fix a positive integer $n$ and observe that the set $$A = \left\{ x \in [a,b]; \osc f(x) \geq \frac{1}{n}\right\}$$ is empty or finite. We can assume that $A$ is nonempty. Let $$A = \{ x_1,\ldots ,x_k\}, \;\;\; x_1 < \cdots < x_k.$$ There are closed intervals $I_i = [a_i,b_i]$, $i \leq k$, such that \begin{itemize} \item[] $b_{i-1} < a_i < x_i < b_i < a_{i+1}$ for $i = 2,\ldots ,k - 1$; \item[] if $a < x_1$ then $a < a_1 < x_1 < b_1$; \item[] if $a = x_1$ then $a = a_1 = x_1 < b_1$; \item[] if $x_k < b$ then $a_k < x_k < b_k < b$; \item[] if $x_k = b$ then $a_k < x_k = b_k = b$; \item[] $b_i - a_i < \frac{1}{n}$ for $i \leq k$; \item[] $f(b_i) - f(a_i) < \osc f(x_i) + \frac{1}{n}$ for $i \leq k$. \end{itemize} Since $\osc f(x) < \frac{1}{n}$ for each point $x \in [a,b] \setminus A$, there are points $c_{i,j} \in (b_{i-1},a_i)$, $i = 2,\ldots ,k$, $j \leq j(i)$, such that \begin{gather*} b_{i-1} = c_{i,1} < c_{i,2} < \cdots < c_{i,j(i)-1} < c_{i,j(i)} = a_i\mbox{ for }i = 2,\ldots ,k;\\ f(c_{i,j+1}) - f(c_{i,j}) < \frac{1}{n}\mbox{ for }i = 2,\ldots ,k. \end{gather*} Analogously, if $a < a_1$ ($b_k < b$) there are points $c_{1,j}$, $j \leq j(1)$, ($c_{k+1,j}$, $j \leq j(k+1)$,) such that \begin{align*} a = c_{1,1} < \cdots < c_{1,j(1)} & = a_1\\ (b_k = c_{k+1,1} < \cdots < c_{k+1,j(k+1)} & = b) \mbox{ for }j \leq j(1)\quad (j \leq j(k+1));\\ f(c_{1,j+1}) - f(c_{1,j}) & < \frac{1}{n}\\ \Bigl (f(c_{k+1,j+1}) - f(c_{k+1,j}) &< \frac{1}{n}\Bigr) \mbox{ for }j \leq j(1)\quad (j \leq j(k+1)). \end{align*} Define on the interval $[a,b]$ the following continuous increasing function \begin{displaymath} f_n(x)=\left\{ \begin{array}{cll} f(x) & \mbox{\rm for} & x \in \{ a,b,c_{i,j}\} ,\;\; i \leq k + 1,\;\;j \leq j(i)\\ f(x) & \mbox{\rm for} & x = x_i, \;\;i \leq k\\ \mbox{\rm linear} & \mbox{\rm otherwise\ on} & [a,b]. \end{array} \right. \end{displaymath} We will prove that $\lim_{n \rightarrow \infty }f_n = f$. If $x \in \{ a,b,x_i;i \leq k\}$ then $f_n(x) = f(x)$. Moreover, if $x \in [c_{i,j},c_{i,j+1}]$ then $|f_n(x) - f(x)| < \frac{1}{n}$. So, if $$x \in [a,b]\;\; \wedge \;\; \dist(x,A) = \inf \bigl\{ |u - x|;u \in A\bigr\} \geq \frac{1}{n}$$ then $$|f_n(x) - f(x)| < \frac{1}{n}.$$ For each point $x \in [a,b]$ which is a discontinuity point of the function $f$ there is a positive integer $n$ such that $\osc f(x) > \frac{1}{n}$. Consequently, for every $k > n$ we have $f_k(x) = f(x)$. Now we suppose that $x \in [a,b]$ is a continuity point of the function $f$. Fix a positive real $\eta$ and a positive integer $n$ with $\frac{2}{n} < \eta$. Let $$B = \left\{ x \in [a,b];\osc f(x) \geq \frac{1}{n}\right\} .$$ Since $x \in [a,b] \setminus B$ and $B$ is a closed set, there is a positive integer $k > n$ with $$\left(x - \frac{1}{k},x + \frac{1}{k}\right) \cap B = \emptyset .$$ Fix an integer $m > k$. If $$E = \left\{ y \in [a,b]; \osc f(y) \geq \frac{1}{m}\right\}$$ then $$|f_m(x) - f(x)| < \frac{1}{m} < \eta$$ if $\dist(x,E) \geq \frac{1}{m}$ and $$|f_m(x) - f(x)| < \osc f(y) + \frac{1}{m} < \frac{1}{n} + \frac{1}{m} < \frac{2}{n} < \eta$$ for some $y \in E \setminus B$, if $\dist (x,E) < \frac{1}{m}$. So, $\lim_{n \rightarrow \infty }f_n(x) = f(x)$ and the proof is completed.\qed\medskip It is well known that the limit of a decreasing (increasing) sequence of continuous functions is upper (lower) semicontinuous. \begin{thm} If $f:[a,b] \mapsto {\cal R}$ is an upper semicontinuous increasing function then there are continuous increasing functions $f_n:[a,b] \mapsto {\cal R}$, $n \geq 1$, such that $f_n \geq f_{n+1} > f$ for $n \geq 1$ and $f = \lim_{n \rightarrow \infty }f_n = f$. \end{thm} In the proof of the above theorem we apply the following sandwich lemma: \begin{lem} Let $f:[a,b] \mapsto {\cal R}$ be an upper semicontinuous increasing (decreasing) function and let $g:[a,b] \mapsto {\cal R}$ be a continuous function such that $f(x) < g(x)$ for each $x \in [a,b]$. Then there is a continuous increasing (decreasing) function $h:[a,b] \mapsto {\cal R}$ such that $f(x) < h(x) < g(x)$ for all $x \in [a,b]$. \end{lem} \mpf{Proof of Lemma~1} We suppose that the function $f$ is increasing. The proof for a decreasing function $f$ is analogous. From the upper semicontinuity of $f$ follows that $f$ is continuous from the right hand. Let $$r = \inf \bigl\{ g(x) - f(x);x \in [a,b]\bigr\} .$$ Since the function $g - f$ is positive and lower semicontinuous, the real $r$ is positive. Define the set $$A = \left\{ x \in [a,b]; \osc f(x) \geq \frac{r}{5}\right\}$$ and we observe that it is empty or finite. We can assume that $A$ is nonempty. Let $$A = \{ x_{1},\ldots ,x_{k}\}, \;\;\; x_{1} < \cdots < x_{k}.$$ There are closed intervals $I_{i} = [a_{i},x_{i}]$, $i \leq k$, such that \begin{gather*} a < a_{i} < x_{i} < a_{i+1} < x_{i+1} \leq b \mbox{ for } i = 1,\ldots ,k - 1;\\ f(x_{i}) - f(a_{i}) < \osc f(x_{i}) + \frac{r}{5}\mbox{ for } i \leq k;\\ |g(x) - g(x_i)| < \frac{r}{5}\mbox{ for }x \in I_i, i \leq k. \end{gather*} Let $x_{0} = a$. Since $\osc f(x) < \frac{r}{5}$ for each point $x \in [a,b] \setminus A$, there are points $c_{i,j} \in [x_{i-1},a_{i}]$, $i = 1,\ldots ,k$, $j \leq j(i)$, such that \begin{gather*} x_{i-1} = c_{i,1} < c_{i,2} < \cdots < c_{i,j(i)-1} < c_{i,j(i)} = a_{i} \mbox{ for }i = 1,\ldots ,k;\\ f(c_{i,j+1}) - f(c_{i,j}) < \frac{r}{5}\mbox{ for }i = 1,\ldots ,k \mbox{ and }j \leq j(i) - 1. \end{gather*} Analogously, if $x_{k} < b$ there are points $c_{k+1,j}$, $j \leq j(k+1)$, such that \begin{gather*} x_{k} = c_{k+1,1} < \cdots < c_{k+1,j(k+1)} = b;\\ f(c_{k+1,j+1}) - f(c_{k+1,j}) < \frac{r}{5}\mbox{ for } j < j(k+1). \end{gather*} Define on the interval $[a,b]$ a continuous increasing function in the following way: \begin{displaymath} g_1(x)=\left\{ \begin{array}{cll} f(x) & \mbox{for}& x \in \{ a,b,c_{i,j}\} ,\;\; i \leq k + 1,\;\;j \leq j(i)\\ f(x) & \mbox{for} & x = x_{i}, \;\;i \leq k\\ \mbox{linear} & \mbox{otherwise on} & [a,b]. \end{array} \right. \end{displaymath} If $x \in \{ a,b,x_{i};i \leq k\}$ then $g_1(x) = f(x)$. Moreover, if $x \in [c_{i,j},c_{i,j+1}]$ then $|g_1(x) - f(x)| < \frac{r}{5}$. Let $$h(x) = g_1(x) + \frac{r}{4}, \;\;\; x \in [a,b].$$ Then $h$ is a continuous increasing function and for $$x \in [a,b] \setminus \bigcup_{i \leq k}I_i$$ the inequalities $$f(x) < g_1(x) + \frac{r}{5} < h(x) < f(x) + r \leq g(x)$$ are true. If $x \in I_i$ for some $i \leq k$ then \begin{align*} f(x) \leq f(x_i-) &= f(a_i) + \frac{r}{5} < g_1(a_i) + \frac{r}{4} = h(a_i) \leq h(x) \\ &= f(x_i) + \frac{r}{4} \leq g(x_i) - r + \frac{r}{4} < g(x_i) - \frac{r}{5} < g(x).\end{align*} So, the function $h$ satisfies to all requirements.\qed \medskip \mpf{Proof of Theorem 2} Since the function $f$ is upper semicontinuous, there are continuous functions $g_n:[a,b] \mapsto {\cal R}$ such that $$f(x) < g_{n+1}(x) < g_n(x), \;\;\; x \in [a,b], \;\;\; n \geq 1,$$ and $f = \lim_{n \rightarrow \infty }g_n$ (\cite{1}). By Lemma 1 there is a continuous increasing function $f_1:[a,b] \mapsto {\cal R}$ with $f < f_1 < g_1$. Let $h_2 = \min(f_1,g_2)$. By Lemma 2 there is a continuous increasing function $f_2:[a,b] \mapsto {\cal R}$ with $f < f_2 < h_2 = \min(f_1,g_2)$. Next by induction, for each positive integer $n > 2$ there is a continuous increasing function $f_n:[a,b] \mapsto {\cal R}$ with $f < f_n < \min(f_{n-1},g_n)$. Consequently, the sequence $(f_n)_n$ satisfies all requirements and the proof is completed.\qed \begin{rem} If the function $f$ is upper semicontinuous and increasing (decreasing) then there are continuous increasing (decreasing) functions $g_n:[a,b] \mapsto {\cal R}$ such that $g_n(a) = f(a)$, $g_n(b) = f(b)$, $g_n \geq g_{n+1}$ for $n \geq 1$ and $\lim_{n \rightarrow \infty }g_n = f$. \end{rem} Without loss of the generality we can suppose that $$b = u = \inf \bigl\{ x \in [a,b];f(x) = f(b)\bigr\}$$ and $$a = v = \sup\bigl \{ x \in [a,b];f(x) = f(a)\bigr\} ,$$ since in the contrary case we can consider the reduced function $f/[u,v]$. We will prove the remark for the case of an increasing function $f$, because the case of a decreasing $f$ is analogous. Let $(a_n)_n$ and $(b_n)_n$ be sequences such that $$a < a_{n+1} < a_n < \cdots < a_1 < b_1 < \cdots < b_n < b_{n+1} < b,$$ and $$a = \lim_{n \rightarrow \infty }a_n, \\;\; b = \lim_{n \rightarrow \infty }b_n.$$ By Theorem 2, there is a decreasing sequence of continuous increasing functions $f_n:[a,b] \mapsto {\cal R}$ with $f = \lim_{n \rightarrow \infty }f_n$ and $f_n > f$ for $n = 1,2,\ldots$. Find a strictly increasing sequence $(n_k)_k$ of positive integers such that \begin{align*} &\quad \lim_{k \rightarrow \infty }n_k = \infty ;\\ \frac{f_{n_k}(a_k) - f(a_k)}{a_k - a} &< \min_{i 1;\\ \frac{f_{n_k}(b_k) - f(b_k)}{b - b_k} & < \min_{i 1. \end{align*} For $k \geq 1$ let \begin{align*} h_k(x) &= \frac{f_{n_k}(a_k) - f(a_k)}{a_k - a}(x - a) \;\; {\rm for}\;\; x \in [a,a_k],\\ h_k(x) &= \frac{f_{n_k}(b_k) - f(b_k)}{b - b_k}(b - x) \;\; {\rm for}\;\; x \in [b_k,b]\end{align*} and \begin{displaymath} g_k(x)=\left\{ \begin{array}{lcl} f(x) + h_k(x) & \mbox{for} & x \in [a,a_k]\\ f_{n_k}(x) & \mbox{for} & x \in [a_k,b_k]\\ f(x) + h_k(x) & \mbox{for} & x \in [b_k,b]. \end{array} \right. \end{displaymath} The sequence $(g_k)_k$ satisfies all requirements and the proof is completed.\qed \begin{rem} If a function $f:[a,b] \mapsto {\cal R}$ is increasing (decreasing) and lower semicontinuous then there is a increasing sequence of continuous increasing (decreasing) functions $f_n:[a,b] \mapsto {\cal R}$ such that $f_n(a) = f(a)$, $f_n(b) = f(b)$ for $n \geq 1$ and $\lim_{n \rightarrow \infty }f_n = f$. \end{rem} \pf It suffices to apply Remark 1 to the function $(-f)$.\qed \medskip We will write that $\mbox{a.c.}\lim_{n \rightarrow \infty }f_n = f$ (\cite{2,3}) if for each point $x$ there is a positive integer $n(x)$ such that for $n > n(x)$ the equality $f_n(x) = f(x)$ is true. Since monotone functions have only countable sets of discontinuity points, we prove the following theorem: \begin{thm} Suppose that functions $f,f_n:[a,b] \mapsto {\cal R}$ satisfy the following conditions: \begin{itemize} \item $f = \mbox{\rm a.c.}\lim_{n \rightarrow \infty }f_n$; \item for each integer $n \geq 1$ the set $D(f_n)$ of all discontinuity points of the function $f_n$ is countable. \end{itemize} Then for each nonempty closed set $F \subset [a,b]$ there are an open interval $I$ and a positive integer $k$ such that $I \cap F \neq \emptyset$ and for each point $x \in (F \cap I) \setminus \bigcup_nD(f_n)$ and for each integer $n > k$ the equality $f(x) = f_n(x)$ is true. \end{thm} \pf Since $\mbox{a.c.}\lim_{n \rightarrow \infty }f_n = f$, for each point $x \in [a,b]$ there is a positive integer $n(x)$ such that $f(x) = f_n(x)$ for all integers $n > n(x)$. For each integer $m \geq 1$ let $$A_m = \bigl\{ x \in [a,b];n(x) = m\bigr\} .$$ Let $F \subset [a,b]$ be a nonempty closed set. If the set $F$ has an isolated point then the condition of our theorem is satisfied. So, we can assume that $F$ is a perfect set. Since $$F = \bigcup_m(A_m \cap F),$$ by the Baire category theorem there is an integer $k \geq 1$ such that the set $A_k \cap F$ is of the second category in $F$. Consequently, there is an open interval $I$ such that $I \cap F \neq \emptyset$ and for every open interval $J \subset I$ with $J \cap F \neq \emptyset$ the set $J \cap F \cap A_k$ is of the second category in $F$. Since the set $$E = \bigcup_nD(f_n)$$ is countable, the set $$B = (I \cap F \cap A_k) \setminus E$$ is dense in $I \cap F$. The restricted functions $f_n/([a,b] \setminus E)$, $n \geq 1$, are continuous and for $m,n > k$ and $x \in B$ the equalities $$f_n(x) = f_m(x) = f(x)$$ are true. So, for $m,n > k$ and for $x \in (I \cap F) \setminus E$ we obtain $f_m(x) = f_n(x) = f(x)$ and the proof is finished.\qed \begin{cor} If functions $f_n:[a,b] \mapsto {\cal R}$ are continuous and increasing (decreasing) and $a.c.\lim_{n \rightarrow \infty }f_n = f$ then the function $f$ is increasing (decreasing) and in the class $B^*_1$ (i.e.\ for every nonempty closed set $F \subset [a,b]$ there is an open interval $I$ such that $I \cap F \neq \emptyset$ and the restricted function $f/(F \cap I)$ is continuous~{\rm\cite{2,3}}). \end{cor} \pf This corollary is an evident consequence of the last theorem.\qed \begin{thm} Suppose that the function $f:[a,b] \mapsto {\cal R}$ is increasing (decreasing) and in the class $B^*_1$. Then there is a sequence of continuous increasing (decreasing) functions $f_n:[a,b] \mapsto {\cal R}$ with $f = \mbox{\rm a.c.}\lim_{n \rightarrow \infty }f_n$. \end{thm} \pf Observe that there are nonempty closed sets $F_n$, $n \geq 1$, such that $$[a,b] = \bigcup_nF_n,$$ $$F_n \subset F_{n+1}, \;\;\; n \geq 1,$$ and the restricted functions $f/F_n$ are continuous (\cite{2}). For each integer $n \geq 1$ the functions $f/F_n$ can be extended to a continuous increasing (decreasing) function $f_n:[a,b] \mapsto {\cal R}$ such that $f_n(a) = f(a)$ and $f_n(b) = f(b)$. Evidently, $$f = \mbox{a.c.}\lim_{n \rightarrow \infty }f_n$$ and the proof is completed.\qed \begin{thm} Let $f:[a,b] \mapsto {\cal R}$ be a function. The following conditions are equivalent: \begin{itemize} \item[(a)] $f$ is increasing (decreasing); \item[(b)] There are increasing (decreasing) functions $f_n:[a,b] \mapsto {\cal R}$ such that $f_n(a) = f(a)$, $f_n(b) = f(b)$ and the sets $D(f_n)$ of all discontinuity points of $f_n$, $n \geq 1$, are finite and $\lim_{n \rightarrow \infty }V(f_n - f,a,b) = 0$, where $V(f_n - f,a,b)$ denotes the total variation of $f_n - f$ on $[a,b]$; \item[(c)] There is a sequence of increasing (decreasing) functions $f_n:[a,b] \mapsto {\cal R}$ which uniformly converges to $f$ on $[a,b]$ and for which $f_n(a) = f(a)$, $f_n(b) = f(b)$ and the sets $D(f_n)$, $n \geq 1$, are finite. \end{itemize} \end{thm} \pf The implication $(c) \Rightarrow (a)$ is evident. Since for each point $x \in [a,b]$ we have $$|f_n(x) - f(x)| \leq V(f_n - f,a,b),$$ we obtain the implication $(b) \Rightarrow (c)$. So, it suffices to prove the implication $(a) \Rightarrow (b)$. Fix an increasing function $f$ and a positive real $\eta$. Observe that the set $D(f)$ is countable. We may assume that $D(f)$ is nonempty. Let $$D(f) = \{ a_1,\ldots ,a_k,\ldots \} .$$ Define $g(a) = 0$ and for $x \in (a,b]$ let $$g(x) = \sum_{a_i < x}\osc f(a_i) + (f(x) - f(x-)).$$ Put $$h(x) = f(x) - g(x)\;\; {\rm for} \;\; x \in [a,b].$$ Then the function $h$ is increasing and continuous and $f = h + g$. Since $$\sum_{i}\osc f(a_i) \leq f(b) - f(a) < \infty ,$$ there is a positive integer $k$ with $$\sum_{i > k}\osc f(a_i) < \frac{\eta }{2}.$$ Put $g_1(a) = 0$, $g_1(b) = g(b)$ and for $x \in (a,b)$ let $$g_1(x) = \sum_{a_i < x;i \leq k}\osc f(a_i) + (f(x) - f(x-)).$$ If $$f_1(x) = h(x) + g_1(x)\;\;{\rm for}\;\; x \in [a,b],$$ then the function $f_1$ is increasing and $$f_1(a) = f(a), \;\;\;f_1(b) = f(b),$$ the set $D(f_1) \subset \{ a_1,\ldots ,a_k,b\}$ is finite, and $$V(f_1 - f,a,b) = 2\sum_{i > k}\osc f(a_i) < 2\frac{\eta }{2} = \eta.$$ This completes the proof for the increasing functions. If $f$ is a decreasing function on $[a,b]$ then we can use the proved part to the function $(-f)$. So, the proof is completed.\qed \medskip Now, denote by $\omega _1$ the first uncountable ordinal number and consider a transfinite sequence of monotone functions $f_{\alpha }:[a,b] \mapsto {\cal R}$, $\alpha < \omega _1$. We will say that the sequence $(f_{\alpha })_{\alpha < \omega _1}$ converges to a function $f$ (then we write $\lim_{\alpha }f_{\alpha } = f$) if for each point $x \in [a,b]$ there is a countable ordinal $\alpha (x)$ such that $f(x) = f_{\alpha }(x)$ for $\alpha > \alpha (x)$ (\cite{4}). \begin{thm} If a function $f:[a,b] \mapsto {\cal R}$ is the limit of a transfinite sequence of monotone functions $f_{\alpha }$, $\alpha < \omega _1$, then there is a countable ordinal $\beta$ such that $f = f_{\alpha }$ for $\alpha > \beta$. \end{thm} \pf The assumptions imply the monotonicity of the function $f$. Let $A \subset [a,b]$ be a countable set containing $D(f) \cup \{ a,b\}$ which is dense in $[a,b]$. There is a countable ordinal $\beta$ such that $$f_{\alpha }(x) = f(x),\;\;\;x \in A,\;\;\;\alpha > \beta .$$ If $\alpha > \beta$ is a countable ordinal then $f_{\alpha } = f$. Of course, if there is a point $x \in [a,b]$ with $f_{\alpha }(x) \neq f(x)$ then $x \in [a,b] \setminus A$. Consequently, $f$ is continuous at $x$ and there is a positive real $r$ such that $f_{\alpha }(x)$ is not in the interval $(f(x) - r,f(x) + r)$. Since the graph of the restricted function $f/A$ is dense in the graph of $f$, there are points $u,v \in A$ with $$f(x) - r < f(u) < f(x) < f(v) < f(x) + r.$$ But $$f_{\alpha }(u) = f(u), \;\;\;f_{\alpha }(v) = f(v)$$ and $f_{\alpha }$ is monotone, so $$f_{\alpha }(x) \in (f(x) - r,f(x) + r),$$ a contrary. This completes the proof.\qed \medskip Since each nondegenerate interval $I$ is the union of closed intervals $I_n$, $n \geq 1$, such that $\mbox{int\,}(I_n) \cap \mbox{int\,}(I_m) = \emptyset$ for $n \neq m$, we obtain that \begin{rem} Theorems~$1$, $4$, $5$ and $6$ and Remarks~$1$ and~$2$ are true for monotone functions $f:I \mapsto {\cal R}$ with $f_n(a+) = f(a+)$ and $f_n(b-) = f(b-)$. \end{rem} \begin {thebibliography} {9} \bibitem{1} A. M. Bruckner, J. B. Bruckner and B. S. 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