% spellchecked:10-11-98pdh % last edit:10-11-98pdh % gallies sent: % gallies corrected: % set in production style: 10-11-98pdh % Section Inroads editor Thomson % Received 3/1/98 % AMS # 26A39, 26A42 \documentclass{rae} \usepackage{amsmath,amsthm,amssymb,latexsym,enumerate} %\coverauthor{F. S. Cater} %\covertitle{A Change of Variables Formula for Darboux Integrals} \received{March 1, 1998} \MathReviews{26A39, 26A42.} \keywords{Darboux integral, Riemann integral.} \firstpagenumber{1} \markboth{F. S. Cater}{A Change of Variables Formula for Darboux Integrals} \author{F. S. Cater, Department of Mathematics, Portland State University, Portland, Oregon 97207, USA} \title{A CHANGE OF VARIABLES FORMULA FOR DARBOUX INTEGRALS} %%%%%%Put Author's Definitions Below Here %%%%%%%%%%% \def\pf{\noindent{\sc Proof.} } \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{proposition}{Proposition} \newtheorem{corollary}{Corollary} \newtheorem{lemma}{Lemma} \theoremstyle{definition} \newtheorem{example}{Example} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \maketitle \begin{abstract} We offer an inequality involving upper and lower Darboux integrals of bounded real functions which implies a number of variations of the change of variables formula for Riemann integrals. \end{abstract} \bigskip Throughout this paper $Dg$ denotes one of the four Dini derivates of a continuous real valued function $g$ on a compact interval $[a,b]$ (the same Dini derivate at all $x$). Let $g(a) \le g(b)$ and let $Dg$ be bounded on $[a,b]$. Let $f$ be a bounded function on $g[a,b]$ such that for almost every $t \in [a,b]$, one or both of the functions $f(g(\cdot))$ or $Dg(\cdot)$ is continuous at $t$. \smallskip We offer the following change of variables formula for Darboux integrals. \begin{theorem}\label{T1} For upper and lower Darboux integrals we have the inequality: $$\label{Eq*} \begin{split} \overline{\int_a^b} f\bigl (g(t)\bigr ) \, Dg(t)\, dt & \ge \overline{\int_{g(a)}^{g(b)}} f(x)\, dx \\[5mm] & \ge \underline{\int_{g(a)}^{g(b)}} \, f(x)\, dx \ge \underline{\int_a^b} f\bigl (g(t)\bigr ) \, Dg(t)\, dt\,. \end{split} \tag*{(*)}$$ Moreover, if $f(g(t))\, Dg(t)$ is $R$-integrable on $[a,b]$, then $f(x)$ is $R$-integrable on $[g(a),g(b)]$ and $$\int_a^b f\bigl (g(t)\bigr )\, Dg(t)\, dt = \int_{g(a)}^{g(b)} \, f(x)\, dx\,.$$ \end{theorem} The inspiration for this work was references \cite{H15} and \cite{T12}, which provided essentially the following proposition. \begin{proposition}\label{P1} In Theorem \ref{T1}, let $Dg$ be continuous almost everywhere on $[a,b]$ and let $g$ be nondecreasing. Then we have $$\label{Eq**} \begin{split} \overline{\int_a^b} f\bigl (g(t)\bigr ) \, Dg(t)\, dt & = \overline{\int_{g(a)}^{g(b)}} f(x)\, dx \\[4mm] & \ge \underline{\int_{g(a)}^{g(b)}} \, f(x)\, dx = \underline{\int_a^b} f\bigl (g(t)\bigr ) \, Dg(t)\, dt\,. \end{split} \tag*{(**)}$$ \end{proposition} The continuity hypothesis in this paper is satisfied, for example, when $f$ is everywhere continuous; likewise it is satisfied when $Dg$ is continuous almost everywhere on $[a,b]$ as in Proposition \ref{P1}. The hypothesis of Theorem \ref{T1} subsumes both of these cases. \par Theorem \ref{T1} has immediate applications to Riemann integration. We offer some of these. \begin{corollary}\label{C1} Let $k$ be a real valued function on $[a,b]$ with a bounded difference quotient, and $k(a) \le k(b)$. Let $h$ be a bounded function on $k[a,b]$. If two of the three functions $h(k(t))\, Dk(t)$, $h(k(t))$, $Dk(t)$ are $R$-integrable on $[a,b]$, then $h$ is $R$-integrable on $[k(a),k(b)]$, $h(k(t))\, Dk(y)$ is $R$-integrable on $[a,b]$, and $$\int_a^b h \bigl ( k(t) \bigr ) \, Dk(t) \, dt = \int_{k(a)}^{k(b)} h(x)\, dx\,.$$ \end{corollary} To prove this from Theorem \ref{T1}, recall that a bounded function is $R$-inte-grable on $[a,b]$ if and only if it is continuous almost everywhere on $[a,b]$. Thus in particular, the product of two $R$-integrable functions is $R$-integrable on $[a,b]$. \begin{corollary}\label{C2} In Theorem \ref{T1} let $f(g(t))\, Dg(t)$ be $R$-integrable on $[a,b]$. Then $f$ is $R$-integrable on the interval $g[a,b]$. \end{corollary} To prove this apply Theorem \ref{T1} on the intervals $[a,s]$ and $[r,b]$ where $g$ takes its maximum at $s$ and its minimum at $r$. \begin{corollary}\label{C3} In Theorem \ref{T1} let $f$ be $R$-integrable on $g[a,b]$ and let $Dg$ be $R$-integrable on $[a,b]$. Then $f(g(t))\, Dg(t)$ is $R$-integrable on $[a,b]$ and $$\int_a^b f \bigl ( g(t) \bigr ) \, Dg(t) \, dt = \int_{g(a)}^{g(b)} f(x)\, dx\,.$$ (See also \cite{S36}). \end{corollary} \begin{proof} In view of Theorem \ref{T1}, it suffices to prove that $m(X) = 0$ where $$X = \bigl \{x \, : \, f\bigl (g(t)\bigr )\, Dg(t)~ \text{is continuous at x}\}\,.$$ Put $$X_1 = \bigl \{x\in X \, : \, Dg~ \text{is continuous at x}\}\,.$$ Then $m(X \setminus X_1) = 0$ because $Dg$ is $R$-integrable on $[a,b]$. Also $Dg(x) \neq 0$ for each $x\in X_1$ because $Dg$ is continuous at $x$ and $f$ is bounded. Moreover, $f$ is discontinuous at $g(x)$ for each $x\in X_1$. Now each $x\in X_1$ lies in an open interval $I_x$ with $Dg$ bounded away from $0$ on $I_x$. Thus $g$ has an absolutely continuous inverse $g^{-1}$ on $g(I_x)$. But $m(g(X_1)) = 0$ because $f$ is $R$-integrable. Hence $$m \bigl (I_x \cap X_1 \bigr ) \le m \Bigl ( g^{-1} \bigl (I_x \cap X_1 \bigr ) \Bigr ) = 0\,.$$ Thus $X_1$ is the union of countably many subsets of measure $0$, so $m(X_1) = m(X) = 0$. \end{proof} \begin{corollary}\label{C4} Let \emph{\ref{Eq**}} hold in Theorem \ref{T1}. Then $f(g(t))\, Dg(t)$ is $R$-integrable on any interval $[u,v]$ for which $g(u) = g(v)$. \end{corollary} We defer the proof of Corollary \ref{C4} until we prove Theorem \ref{T1}. Finally we show by example that the equations in \ref{Eq**} need not hold in Theorem \ref{T1}. \par Our techniques will be quite different from those used for change of variables for the Lebesgue and other integrals. Compare with the work found in references \cite{Foran9}, \cite{G4}, \cite{G3}, \cite{K2}, \cite{S36}, for example. \par We turn now to some definitions we will use. We say that a finite increasing sequence $a = x_0 < x_1 < \ldots < x_n = b$ is a \emph{partition} of the interval $[a,b]$. The $x_i$ are called the \emph{dividing points} of this partition. By the \emph{norm} of the partition we mean $\max_{i=1}^n (x_i-x_{i-1})$. By the \emph{inherited partition} of the subinterval $[u,v]$ of $[a,b]$ we mean the partition whose dividing points are $u$, $v$ and all the $x_i$ between $u$ and $v$. We say that another partition of $[a,b]$, $a = y_0 < y_1 < \ldots < y_m = b$ is a \emph{refinement} of this one if for each $x_i$ there is a $y_j$ with $y_j = x_i$. \par We say that the partition $a = x_0 < x_1 < \ldots < x_n = b$ is a \emph{special partition} of $[a,b]$ if there is a finite increasing sequence of integers $0 = n_0 < n_1 < \ldots < n_k = n$ such that for each index $j = 0,1,\ldots,k-1$ either $g(x_{n_j}) = g(x_{n_{j+1}})$, or $n_{j+1} = n_j +1$ and $g(x_{n_j}) < g(x_{n_{j+1}})$ and $g(x_{n_j},x_{n_{j+1}}) = \bigl (g(x_{n_j}), g(x_{n_{j+1}})\bigr )$. \par By a \emph{Riemann sum} for $f(g(t)) \, Dg(t)$ on the partition $a = x_0 < x_1 < \ldots < x_n=b$ we mean a sum of the form $$\sum_{i=1}^n f \bigl ( g(t_i) \bigr ) \, Dg(t_i) (x_i-x_{i-1})$$ where $t_i$ is any point in the interval $[x_{i-1},x_i]$. If $f(g(t_i))\, Dg(t_i)$ is replaced by $$\sup \Bigl \{ f \bigl ( g(t) \bigr ) \, Dg(t) \, : \, t \in \bigl [ x_{i-1},x_i \bigr ] \Bigr \}$$ the sum is called an \emph{upper sum}. Likewise when $f(g(t_i))\, Dg(t_i)$ is replaced by $$\inf \Bigl \{ f \bigl ( g(t) \bigr ) \, Dg(t) \, : \, t \in \bigl [ x_{i-1},x_i \bigr ] \Bigr \}$$ the sum is called a \emph{lower sum}. Similar definitions are given for $f$ on partitions of the interval $[g(a),g(b)]$. \par By a \emph{mixed sum} on the partition $a = x_0 < x_1 < \ldots < x_n=b$ we mean a sum of the form $\sum_{i=1}^n f \bigl (g(s_i) \bigr ) \, Dg(t_i) (x_i-x_{i-1})$ where $s_i$ and $t_i$ are in $[x_{i-1},x_i]$. Thus a Riemann sum for $f(g(t))\, Dg(t)$ is a particular kind of mixed sum. \par Special partitions and mixed sums are ad hoc definitions in this paper. \begin{lemma}\label{L1} Let $a = x_0 < x_1 < \ldots < x_n = b$ be a partition of $[a,b]$. Then it has a refinement that is a special partition of $a,b]$. \end{lemma} \begin{proof} Assume $g(a) < g(b)$; otherwise $g(a) = g(b)$ and the given partition is a special partition. \par Let $y$ be the least number in the set $[a,b] \cap g^{-1}\bigl (g(b)\bigr )$, let $z$ be the greatest number in the set $[a,b] \cap g^{-1}\bigl (g(a)\bigr )$, and let $w$ be the greatest number in the set $[a,y] \cap g^{-1}\bigl (g(a)\bigr )$. \par We use induction on $n$. If $n=1$, the dividing points in the special partition are the distinct points among $a$, $b$, $y$ and $w$. Note that $g(w,y) = \bigl ( g(w),g(y) \bigr )$ if $w < y$, $g(a) = g(w)$ if $a < w$ and $g(y) = g(b)$ if $y < b$. \par Assume $n \ge 2$ and the conclusion holds for partitions of any subinterval of $[a,b]$ with fewer than $n+1$ dividing points. We have three cases. \par {\sc Case 1}. $g(x_{n-1}) \ge g(b)$. Then $y \le x_{n-1}$. Use $g(b) = g(y)$ and the induction hypothesis on the inherited partition of the interval $[a,y]$. \par {\sc Case 2}. $g(x_{n-1}) \le g(a)$. Then $x_{n-1} \le z$. Use $g(a) = g(z)$ and the induction hypothesis on the inherited partition of the interval $[z,b]$. \par {\sc Case 3}. $g(a) < g(x_{n-1}) < g(b)$. Use the induction hypothesis on the respective inherited partitions of the intervals $[a,x_{n-1}]$ and $[x_{n-1},b]$. \par This covers all cases, and the induction is proved. \end{proof} \begin{lemma}\label{L2} Let $a= x_0 < x_1< \ldots < x_n=b$ be a partition of $[a,b]$. Let $\delta > 0$. Then there is a refinement of this partition that is a special partition of $[a,b]$ of norm $< \delta$. \end{lemma} \begin{proof} Take a refinement of norm $<\delta$ and apply Lemma \ref{L1} to it. \end{proof} \begin{lemma}\label{L3} Let $a= x_0 < x_1< \ldots < x_n=b$ be any partition of $[a,b]$, let $g(a) = g(b)$, and $\epsilon > 0$. Then there is a refinement of this partition with a mixed sum on it $> -\epsilon$ ($< \epsilon$). \end{lemma} \begin{proof} The proof follows by induction on $n$. Let $n=1$. Choose any $x \in [a,b]$. We use $g(b)-g(a)=0$ and Dini's Theorem \cite{S3} to select $t\in [a,b]$ such that $$f \bigl ( g(x) \bigr ) \, Dg(t)(b-a) > -\epsilon$$ as follows. For $f(g(x)) > 0$ choose $t$ so that $$Dg(t) > \bigl (g(b)-g(a) \bigr ) (b-a)^{-1} - \epsilon f \bigl ( g(x) \bigr )^{-1} (b-a)^{-1}$$ and multiply by the positive number $f \bigl (g(x) \bigr ) (b-a)$; for $f \bigl (g(x) \bigr ) < 0$ choose $t$ so that $$Dg(t) < \bigl (g(b)-g(a) \bigr ) (b-a)^{-1} - \epsilon f \bigl ( g(x) \bigr )^{-1} (b-a)^{-1}$$ and multiply by the negative number $f \bigl (g(x) \bigr ) (b-a)$; for $f \bigl ( g(x) \bigr ) = 0$, any $t$ will do. \par Now assume that $n \ge 2$ and the conclusion holds when $[a,b]$ is replaced by any subinterval, $\epsilon$ is replaced by any positive number, and the partition has fewer than $n+1$ dividing points. \par {\sc Case 1}. $g(x_1) = g(a)$ or $g(x_{n-1}) = g(b)$. Use the induction hypothesis on the respective inherited partitions of the intervals $[a,x_1]$ and $[x_1,b]$, or $[a,x_{n-1}]$ and $[x_{n-1},b]$. \par {\sc Case 2}. $g(x_1) < g(a) < g(x_{n-1})$. Let $q \in (x_1,x_{n-1})$ such that $g(q) = g(a)$. Use the induction hypothesis on the respective inherited partitions of the intervals $[a,q]$ and $[q,b]$. \par {\sc Case 3}. $g(x_1) > g(a) > g(x_{n-1})$. Analogous to Case 2. \par {\sc Case 4}. $g(x_1) \ge g(x_{n-1}) > g(a)$. Let $u$ be the smallest number in the set $[a,b] \cap g^{-1}\bigl (g(x_{n-1}) \bigr )$. Then $u \le x_1$. use the Intermediate Value Theorem to select $v_1 \in [a,u]$ and $v_2 \in [x_{n-1},b]$ such that $g(v_1) = g(v_2)$. As in the proof for $n=1$, use Dini's Theorem \cite{S3} to choose $r\in [a,u]$ and $s \in [x_{n-1},b]$ such that $$\label{Eq1} f \bigl ( g(v_1) \bigr ) \, Dg(r)(u-a) > f \bigl ( g(v_1) \bigr ) \bigl ( g(u)-g(a) \bigr ) - \frac{1}{4} \epsilon\,,$$ $$\label{Eq2} f \bigl ( g(v_2) \bigr ) \, Dg(s)(b-x_{n-1}) > f \bigl ( g(v_2)\bigr ) \bigl ( g(b)-g(x_{n-1}) \bigr ) - \frac{1}{4} \epsilon\,.$$ Recall that $g(a) = g(b)$, $g(u) = g(x_{n-1})$, $g(v_1) = g(v_2)$, and add \eqref{Eq1} and \eqref{Eq2} to obtain $$\label{Eq3} f\bigl (g(v_1) \bigr ) \, Dg(r) (u-a) + f \bigl ( g(v_2) \bigr ) \, Dg(s) (b-x_{n-1}) > - \frac{1}{2} \epsilon\,.$$ Apply the induction hypothesis to the inherited partition of the subinterval $[u,x_{n-1}]$ (with $\frac{1}{2}\epsilon$), and add the inequality obtained to \eqref{Eq3}. \par {\sc Case 5}. $g(x_{n-1}) \ge g(x_1) > g(a)$. Let $u$ be the largest number in the set $[a,b] \cap g^{-1}\bigl (g(x_1) \bigr )$. Then $u \ge x_{n-1}$. Argue as in Case 4 and use the subintervals $[a,x_1]$, $[x_1,u]$ and $[u,b]$. \par {\sc Case 6}. $g(x_1) < g(a)$ and $g(x_{n-1}) < g(a)$. This is analogous to Cases 4 and 5, so we leave it. \par This covers all the cases to prove the induction for the first inequality (with $> -\epsilon$). The second inequality (with $< \epsilon$) is proved analogously. We leave the rest. \end{proof} \begin{lemma}\label{L4} Let $\epsilon > 0$ and $\delta > 0$. Then there is a special partition of $[a,b]$ with norm $< \delta$ and a mixed sum on it $$> \overline{\int_{g(a)}^{g(b)}} f(x)\, dx - \epsilon\,.$$ \end{lemma} \begin{proof} We can dismiss the case in which $g(a) = g(b)$; for here we just apply Lemma \ref{L3} to any partition of $[a,b]$ of norm $< \delta$. So assume $g(a) < g(b)$. \par Let $a = x_0 < x_1 < \ldots < x_n = b$ be a special partition of $[a,b]$ of norm $< \delta$ (Lemma \ref{L2}). Say $0 = n_0 < n_1 < \ldots < n_k =n$ such that for any $j = 0,1,\ldots,k-1$, either $g(x_{n_j}) = g(x_{n_{j+1}})$ or $g(x_{n_j}) < g(x_{n_{j+1}})$ and $g\bigl (x_{n_j},x_{n_{j+1}}\bigr ) = \bigl (g(x_{n_j}), g(x_{n_{j+1}})\bigr )$ and $n_{j+1} = n_j +1$. For any such $j$ we consider the two possibilities. \par {\sc Case 1}. $g(x_{n_j}) = g(x_{n_{j+1}})$. Use Lemma \ref{L3} to obtain a refinement of the inherited partition of the interval $[x_{n_j},x_{n_{j+1}}]$ and a mixed sum on it $$> \overline{\int_{g(x_{n_j})}^{g(x_{n_{j+1}})}} f(x) \, dx - \epsilon (2k)^{-1}\,.$$ \par {\sc Case 2}. $g\bigl (x_{n_j},x_{n_{j+1}}\bigr ) = \bigl (g(x_{n_j}), g(x_{n_{j+1}})\bigr )$. Let $M$ denote the $\sup$ of $f$ on the interval $\bigl [ g(x_{n_j}),g(x_{n_{j+1}}) \bigr ]$. We use Dini's Theorem \cite{S3} to find a $t\in [x_{n_j},x_{n_{j+1}}]$ such that $$M\,Dg(t)\bigl (x_{n_{j+1}} - x_{n_j} \bigr ) > M\Bigl (g\bigl (x_{n_{j+1}}\bigr ) - g\bigl (x_{n_j}\bigr ) \Bigr ) - \epsilon (2k)^{-1}$$ as follows. Choose $t$ so that $$Dg(t) > \Bigl ( g \bigl (x_{n_{j+1}} \bigr ) - g \bigl ( x_{n_j} \bigr ) \Bigr ) \bigl (x_{n_{j+1}} - x_{n_j} \bigr )^{-1} - \epsilon (2k)^{-1} M^{-1} \bigl (x_{n_{j+1}} - x_{n_j} \bigr )^{-1}$$ an multiply by $M(x_{n_{j+1}} - x_{n_j})$ if $M$ is positive; choose $t$ so that $$Dg(t) < \Bigl ( g \bigl (x_{n_{j+1}} \bigr ) - g \bigl ( x_{n_j} \bigr ) \Bigr ) \bigl (x_{n_{j+1}} - x_{n_j} \bigr )^{-1} - \epsilon (2k)^{-1} M^{-1} \bigl (x_{n_{j+1}} - x_{n_j} \bigr )^{-1}$$ an multiply by $M(x_{n_{j+1}} - x_{n_j})$ if $M$ is negative; choose any $t$ if $M=0$. Then we choose $s \in [x_{n_j},x_{n_{j+1}} ]$ so that $$f\bigl (g(s)\bigr ) \, Dg(t) \bigl (x_{n_{j+1}}-x_{n_j}\bigr ) > M\Bigl (g(x_{n_{j+1}}) - g(x_{n_j})\Bigr ) - \epsilon (2k)^{-1}$$ $$\ge \overline{\int_{g(x_{n_j})}^{g(x_{n_{j+1}})}} f(x)\, dx - \epsilon (2k)^{-1}\,.$$ \par >From Cases 1 and 2 we easily see that there is a refinement of $a = x_0 < x_1 < \ldots < x_n = b$ and a mixed sum on it $$> \sum_{j=0}^{k-1} \overline{\int_{g(x_{n_j})}^{g(x_{n_{j+1}})}} f(x) \, dx - \epsilon = \overline{\int_{g(a)}^{g(b)}} f(x)\, dx - \epsilon\,.$$ \end{proof} So far we have not used the continuity hypothesis on $f$ and $Dg$. Now we use this hypothesis to change from mixed sums to Riemann sums and complete the proof of Theorem \ref{T1}. \begin{lemma}\label{L5} Let $\epsilon > 0$. Then there is a $\delta > 0$ such that for any partition $a = x_0 < x_1 < \ldots < x_n = b$ of $[a,b]$ with norm $< \delta$ and for any mixed sum $W$ on this partition, there is a Riemann sum $W_0$ on the same partition such that $|W-W_0| < \epsilon$. \end{lemma} \begin{proof} Use the Vitali Covering Theorem to find finitely many intervals $I_1,I_2,$ $\ldots,I_k$ such that $$m \Bigl ([a,b] \setminus \bigl (\cup_{j=1}^k I_j \bigr ) \Bigr ) < \epsilon$$ where $m$ is Lebesgue measure, and such that either $$\bigl |f\bigl (g(u)\bigr ) - f\bigl (g(v)\bigr ) \bigr | < \epsilon ~~~\text{or}~~~ \bigl | Dg(u) - Dg(v) \bigr | < \epsilon$$ when $u$ and $v$ lie in an interval concentric with any $I_j$ and having twice the length of $I_j$ ($1 \le j \le k$). \par Now $f$ and $Dg$ are bounded. Choose $M$ with $M > |f|$ and $M > |Dg|$. Let $$\delta = \frac{1}{2} \min_{j=1}^k \,(\text{length}\, I_j)$$ and let the partition $a = x_0 < x_1 < \ldots < x_n = b$ have norm $< \delta$. Let $$W = \sum_{i=1}^n f \bigl (g(s_i) \bigr ) \, Dg(t_i) (x_i-x_{i-1}) ~~~~\Bigl (s_i,t_i \in [x_{i-1},x_i] \Bigr ).$$ Now \setcounter{equation}{0} $$\label{Eq4} \begin{split} \Bigl | \sum\nolimits^* f \bigl (g(s_i) \bigr ) \, Dg(t_i) \bigl (x_i-x_{i-1}\bigr ) \Bigr | \le M^2 \sum\nolimits^* (x_i-x_{i-1} ) < M^* \epsilon\,,\\[2mm] \Bigl | \sum\nolimits^* f \bigl (g(t_i) \bigr ) \, Dg(t_i) \bigl (x_i-x_{i-1}\bigr ) \Bigr | \le M^2 \sum\nolimits^* (x_i-x_{i-1} ) < M^* \epsilon\,\,, \end{split}$$ where $\sum^*$ means sum on those indices $i$ for which $[x_{i-1},x_i]$ does not meet any $I_j$. But if $[x_{i-1},x_i]$ does meet some $I_j$, either {\small $$\label{Eq5} \begin{split} \Bigl |f \bigl (g(s_i) \bigr ) \, Dg(t_i) \bigl (x_i-x_{i-1}\bigr ) \!-\! f \bigl (g(t_i) \bigr ) \, Dg(t_i) \bigl (x_i-x_{i-1}\bigr ) \Bigr | \le M (x_i-x_{i-1} )\epsilon,~\text{or}\\[2mm] \Bigl |f \bigl (g(s_i) \bigr ) \, Dg(t_i) \bigl (x_i-x_{i-1}\bigr ) - f \bigl (g(s_i) \bigr ) \, Dg(s_i) \bigl (x_i-x_{i-1}\bigr ) \Bigr | \le M (x_i-x_{i-1} )\epsilon\:.~~\\[2mm] \end{split}$$} We use \eqref{Eq4} and \eqref{Eq5} to select $r_i = s_i$ or $t_i$ in such a way that if $$W_0 = \sum_{j=1}^n f \bigl ( g(r_i) \bigr ) \, Dg(r_i)(x_i-x_{i-1})\,,$$ then $$\bigl |W-W_0 \bigr | < 2M^2 \epsilon + M \epsilon \sum_i \bigl (x_i-x_{i-1}\bigr ) < 2M^2 \epsilon + M\epsilon (b-a)\,.$$ We leave the rest. \end{proof} \begin{proof}[Proof of Theorem \ref{T1}] We deduce the first inequality from Lemmas \ref{L4} and \ref{L5}. The last inequality is proved similarly by reversing appropriate inequalities in Lemma \ref{L4} and its proof. We leave the rest. \end{proof} \begin{proof}[Proof of Corollary \ref{C4}] Let $a \le u < v \le b$ and let \ref{Eq**} hold. Let $g(u) = g(v)$. We define a function $h$ on $[a+v-u,b]$ as follows: put $h(x) = g(x-v+u)$ for $a+v-u \le x \le v$ and $h(x) = g(x)$ for $v < x \le b$. Note that $h(b) = g(b)$, $h(v) = g(u) = g(v)$, $h(a+v-u) = g(a)$, $$\overline{\int_{a+v-u}^v} f\bigl (h(t)\bigr )\, Dh(t) \, dt = \overline{\int_a^u} f\bigl (g(t)\bigr )\, Dg(t)\, dt\,,$$ $$\overline{\int_v^b} f\bigl (h(t)\bigr )\, Dh(t) \, dt = \overline{\int_v^b} f\bigl (g(t)\bigr )\, Dg(t)\, dt\,.$$ By hypothesis, $$\overline{\int_a^u} f\bigl (g(t)\bigr )\, Dg(t)\, dt + \overline{\int_u^v} f\bigl (g(t)\bigr )\, Dg(t)\, dt + \overline{\int_v^b} f\bigl (g(t)\bigr )\, Dg(t)\, dt$$ $$= \overline{\int_{g(a)}^{g(b)}} f(x)\, dx\,.$$ By Theorem \ref{T1}, $$\overline{\int_{a+v-u}^v} f\bigl (h(t)\bigr )\, Dh(t) \, dt + \overline{\int_v^b} f\bigl (h(t)\bigr )\, Dh(t) \, dt \ge \overline{\int_{h(a+v-u)}^{h(b)}} f(x)\, dx\,.$$ Take the difference of the last two (in)equalities and obtain $$\overline{\int_u^v} f\bigl (g(t)\bigr )\, Dg(t)\, dt \le 0\,.$$ Similar arguments on the lower integrals give $$\underline{\int_u^v} f\bigl (g(t)\bigr )\, Dg(t)\, dt \ge 0\,.$$ It follows that $f(g(t))\,Dg(t)$ is Riemann integrable on $[u,v]$. \end{proof} We conclude this paper with some examples. \begin{example}\label{Ex1} Let $g(x) = |x|$ for $-1 \le x \le 1$, let $f(x) = 1$ if $x$ is rational, and $f(x) = 0$ if $x$ is irrational. Then $g(-1) = g(1) = 1$, and $$\overline{\int_{g(-1)}^{g(1)}} f(x)\, dx = 0 < 1 = \overline{\int_{-1}^1} f\bigl (g(t)\bigr )\, Dg(t)\, dt\,,$$ So the equations in \ref{Eq**} need not hold in Theorem \ref{T1} when $g$ is not monotone. \end{example} \begin{example}\label{Ex2} Let $E$ be a measurable subset of $[0,1]$ such that any subinterval of $[0,1]$ meets $E$ in a set of positive measure, and meets its complement $\complement E$ in a set of positive measure. let $g$ be the indefinite integral of the characteristic function of $E$: $$g(x) = \int_0^x \chi_{{}_E} (t)\, dt\,.$$ Then $g$ is strictly increasing, and the sets $A = \{t \, : \, g^\prime (t) = 1\}$ and $B = \{t \, : \, g^\prime (t) = 0\}$ are dense in $[0,1]$; hence $g(A)$ and $g(B)$ are dense in $g[0,1]$. Now let $f$ be identically $1$. It follows that $$\overline{\int_0^1} f\bigl (g(t)\bigr ) \,Dg(t)\, dt = 1 > m(E) = g(1) - g(0) = \overline{\int_{g(0)}^{g(1)}} f(x)\, dx\,.$$ So the equations in \ref{Eq**} need not hold in Theorem \ref{T1} when $Dg$ is not Riemann integrable. \end{example} \begin{example}\label{Ex3} Let $E$ and $g$ be as in Example \ref{Ex2}, and put $k = g$. On the range of $k$ put $h(x) = 1$ if $x \in k(B)$ and $h(x) = 0$ if $x \notin k(B)$. Then $h\bigl (k(t)\bigr )\, Dk(t) = 0$ for all $t$, and $$\overline{\int_0^1} h\bigl (k(t)\bigr )\, Dk(t)\, dt = 0 < m(E) = k(1) - k(0) = \overline{\int_{k(0)}^{k(1)}} h(x)\, dx\,.$$ Of course the continuity hypothesis of the paper is not satisfied by $h$ and $Dk$. \end{example} \begin{thebibliography}{999} \bibitem[F]{Foran9} J.~Foran, \emph{A chain rule for the approximate derivatives and change of variables for the {(D)}-integral}, Real Analysis Exchange \textbf{8} (1982-1983), 443--454. \bibitem[G1]{G4} G.~S. 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