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% last edit: jch 10/24/1998
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% set in production style: jch 10/24/1998
% Section Inroads editor Bullen
\documentclass{rae}
\usepackage{amsmath,amsthm,amssymb}
%\coverauthor{I. J. L. Garces, L. Peng-Yee and Z. Dongsheng}
%\covertitle{Moore-Smith Limits and the Henstock Integral}
\received{November 27, 1997}
\MathReviews{Primary: 26A39}
\firstpagenumber{1}
\markboth{I. J. L. Garces, L. Peng-Yee and Z. Dongsheng}
{Moore-Smith Limits and the Henstock Integral}
\author{Ian June L. Garces, Department of
Mathematics, Ateneo de Manila University,
P.O. Box 154, 1099 Manila, The Philippines
\and
Lee Peng-Yee and Zhao Dongsheng, Division of
Mathematics,
National Institute of Education, Singapore
259756, e-mail:
{\tt leepy@@am.nie.ac.sg}}
\title{MOORE-SMITH LIMITS AND THE HENSTOCK INTEGRAL}
%%%%%%Put Author's Definitions Below Here %%%%%%%%%%%
\newcommand{\consecutive}
{
\newtheorem {theorem} {Theorem}
\newtheorem {lemma}[theorem] {Lemma}
\newtheorem {cor}[theorem] {Corollary}
\newtheorem {example} [theorem]{Example}
}
\def\pf{\noindent{\sc Proof.} }
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\consecutive
\maketitle
\begin{abstract}
An integral is defined using the Moore-Smith limit and this
new integral is compared to the Henstock integral.
\end{abstract}
\bigskip
\noindent
It is well-known though not easily found in
the literature that the Riemann
integral can be defined by Moore-Smith limit
using divisions.
Then many properties of the Riemann integral
will have straightforward
proofs. In this paper, we shall investigate
whether the Henstock
integral can also be defined by means of
Moore-Smith limit involving
$\delta$-fine divisions. We assume that the
reader is familiar with
the definition of the Henstock integral~\cite3.
A division $D$ of $[a,b]$ is a finite set of
interval-point pairs $([u,v],\xi)$
such that the intervals $[u,v]$ from $D$ are
non-overlapping and their union is
$[a,b]$ and also $\xi \in [u,v]$ for each
$([u,v],\xi) \in D$.
Let $D_1 = \{ ([u,v],\xi) \}$ and $D_2 = \{
([s,t],\eta) \}$ be two divisions
of $[a,b]$. Then $D_2$ is
said to be finer than $D_1$ in the Riemann
sense, or in symbols,
$D_2 \sqsupseteq D_1$ if for each
$([s,t],\eta) \in D_2$ we have
$[s,t] \subset [u,v]$ for some $([u,v],\xi)
\in D_1$ and when $[s,t] = [u,v]$
we have $\eta = \xi$.
Then $({\cal D}, \sqsupseteq)$ is a directed
set of divisions $D$ of $[a,b]$ .
More precisely, the following conditions are
satisfied:
\begin{enumerate}
\item $D \sqsupseteq D$ for all $D \in {\cal
D}$;
\item if $D_1,D_2,D_3 \in {\cal D}$
with $D_1 \sqsupseteq D_2$ and $D_2
\sqsupseteq D_3$, then
$D_1 \sqsupseteq D_3$; \item if $D_1,D_2 \in
{\cal D}$ with
$D_1 \sqsupseteq D_2$ and $D_2 \sqsupseteq
D_1$, then $D_1 = D_2$; and
\item for every $D_1,D_2 \in {\cal D}$, there
exists $D_3 \in {\cal D}$
such that $D_3 \sqsupseteq D_1$ and $D_3
\sqsupseteq D_2$.
\end{enumerate}
Hence the Riemann
integral of $f$ on $[a,b]$ is the
Moore-Smith limit~\cite1
of Riemann sums using $({\cal D},
\sqsupseteq)$. In symbols,
$$ \int_a^b f = \lim_{D\in{\cal D}} \
(D)\sum f(\xi)(v - u).$$
We recall that a function $f$ is {\it
Henstock integrable} to $A$ on $[a,b]$ if
for every $\epsilon > 0$ there exists
$\delta(x) >0$ such that for
any $\delta$-fine division $D$ of $[a,b]$ we
have
$$ \left| (D)\sum f(\xi)(v - u) - A \right| <
\epsilon.$$
A division $D$ of $[a,b]$ is $\delta$-fine
if
$\xi \in [u,v] \subset (\xi - \delta(\xi),
\xi + \delta(\xi))$ for each $([u,v],\xi)
\in D$.
Now let ${\cal D}$ be the family of
$\delta$-fine divisions of $[a,b]$
for some given $\delta (x) > 0$. For
$D_1,D_2 \in {\cal D}$,
we write $D_2 \geq D_1$ and say that $D_2$
is finer than $D_1$ in the Henstock
sense using $\delta$ if for every
$([s,t],\eta) \in D_2$ we have
$[s,t] \subset [u,v]$ for some $([u,v],\xi)
\in D_1$, and
$\{ \xi : ([u,v],\xi) \in D_1 \} \subset \{
\eta : ([s,t],\eta) \in D_2 \}.$
Then $({\cal D}, \geq)$ is a directed set.
A function $f$ is said to be {\it
$H_1$-integrable} to $A$ on $[a,b]$ if
$A$ is the Moore-Smith limit of the Riemann
sums using the directed set
$({\cal D}, \geq)$. More precisely,
there exists $\delta(x) > 0$ such that for
every $\epsilon > 0$ there exists
a $\delta$-fine division $D_0$ such that for
every $\delta$-fine
division $D \geq D_0$ we have
$$ | (D)\sum f(\xi)(v - u) - A | <
\epsilon.$$
We say that $A$ is the $H_1$-integral of $f$
on $[a,b]$ and
that $f$ is $H_1$-integrable on $[a,b]$
using $\delta$.
Note the difference that here we choose
$\delta(x)$ first then
$\epsilon$, whereas in the definition of the
Henstock integral
$\delta(x)$ comes after $\epsilon$.
\begin{example}\rm A Riemann integrable
function on $[a,b]$ is
$H_1$-integrable there using an arbitrary
$\delta(x) > 0$,
and the two integrals are equal.
\end{example}
\begin{example}\rm
The Dirichlet function, given by $f(x) = 1$
when $x$ is rational and 0 when $x$ is
irrational, is $H_1$-integrable on $[0,1]$,
using $\delta$ where $\delta(r_i) = 2^{-i}$
for $i=1,2,\ldots$ and
$\{ r_1, r_2, \ldots \}$ are the rational
numbers in $[0,1]$.
\end{example}
\begin{example}\rm Let $f(x) = F^{\prime}(x)$
where $F(0)=0$ and
$F(x)= x^2 \sin x^{-2}$ where $x \ne 0$.
Then $f$ is $H_1$-integrable
on $[0,1]$, using $\delta$ where $\delta(x)
= \delta_n$ when $x \in (1/(n+1),1/n]$
for some suitable $\delta_n$ and
$n=1,2,\ldots$, and arbitrary when $x=0$.
\end{example}
It is easy to see that every
$H_1$-integrable function on $[a,b]$ is also
Henstock integrable there. We will be using
this fact frequently in the
succeeding discussion.
For convenience, we say that $f$ is
$H_1$-integrable on a set $X \subset [a,b]$
if $f {\cal X}_X$ is $H_1$-integrable on
$[a,b]$ where ${\cal X}_X$ denotes the
characteristic function of $X$.
We may define the primitive $F$ of $f$ on
$[a,b]$ with $f(x)=0$ for
$x \in [a,b]-X$.
It is easy to see that if $f$ is
$H_1$-integrable
on $X_1$ using $\delta_1$ and on $X_2$ using
$\delta_2$ then $f$ is
$H_1$-integrable on the union $X_1 \cup X_2$
using $\delta = \min \{ \delta_1,
\delta_2 \}$. However, we have the
following.
\begin{lemma} Let $f$ be
$H_1$-integrable on a closed set
$X_1 \subset [a,b]$ using $\delta_1$, and
on another closed set $X_2 \subset [a,b]$,
with
$f(x) = 0$ for $x \not\in X_1 \cup X_2$.
If the primitive $F$ of $f$ on $[a,b]$
is absolutely continuous there, then $f$ is
$H_1$-integrable on $X_1 \cup X_2$
using $\delta$, where
$\delta(x) = \delta_1(x)$ when $x \in
X_1$.\end{lemma}
\pf We may assume $X_1 \subset
X_2$.
Suppose $f$ is $H_1$-integrable on $X_2$
using~$\delta_2$.
Then for every $\epsilon >0$
there exists a $\delta_i$-fine division
$D_i$ on $[a,b]$, $i=1,2$, such that
for any $\delta_i$-fine division $D \geq
D_i$ we have
$$\Bigl|(D)\sum_{\xi \in X_i} f(\xi)(v-u) - A_i\Bigr|
< \epsilon,$$
where $A_i$ denotes the $H_1$-integral of
$f$ on $X_i$.
We may assume $\delta_2(x) \leq \delta_1(x)$
for all $x \in [a,b]$.
Put $\delta(x) = \delta_1(x)$ when $x \in
X_1$ and
$\delta_2(x)$ when $x \in [a,b] - X_1$.
We may modify $\delta_2(x)$, if necessary,
so that
$(x-\delta_2(x),x+\delta_2(x)) \cap X_1 =
\emptyset$ when $x \notin X_1$.
Since $F$ is absolutely continuous on
$[a,b]$,
there exists $\eta > 0$ such that
for any partial division $D$ of $[a,b]$
we have
$$\Bigl|(D) \sum F(u,v)\Bigr| < \epsilon \quad\hbox{
whenever }\quad (D)\sum |v-u| < \eta,$$
where $F(u,v) = F(v) - F(u)$.
Note that the Saks-Henstock Lemma~\cite[p.~11]{3},
for the $H_1$-integral holds. If, in
addition,
$D$ is $\delta_2$-fine partial division of
$[a,b]$ with $\xi \in X_2$
such that
$$\Bigl|(D)\sum \{ f(\xi)(v-u) - F(u,v) \} \Bigr| <
\epsilon,$$
then $(D)\sum |v-u| < \eta$
implies $|(D)\sum f(\xi)(v-u)| < 2
\epsilon$.
Now, take a $\delta$-fine division $D_0$ of
$[a,b]$ such that
$D_0 \geq D_1,D_2$ using $\delta_1$
and a subset $E$ of $D_0$ covers $X_1$ with
$|E - X_1| < \eta$.
For any $\delta$-fine division $D \geq D_0$,
take a $\delta_2$-fine
division $D_3$ of those intervals in $D$
which are $\delta_1$-fine, namely, those
with $\xi \in X_1$. Note that
$(D_3)\sum f(\xi)(v-u)$ and $(D)\sum_{\xi
\in X_1} f(\xi)(v-u)$
are sums over the same intervals, in which
$D_3$ may contain $([u,v],\xi)$
with $\xi \in X_1$ and $\xi\in X_2-X_1$.
Then we have
\begin{align*}
\Bigl|(D)\sum_{\xi \in X_2}& f(\xi)(v - u) - A_2\Bigr|
\leq \Bigl|(D)\sum_{\xi \in X_1}
f(\xi)(v - u) - A_1\Bigr|\\
& \quad + \Bigl|A_1 - (D_3)\sum_{\xi \in X_1}
f(\xi)(v - u)\Bigr|\\
& \quad + \Bigl|(D_3)\sum_{\xi \in X_2} f(\xi)(v-u)
+
(D)\sum_{\xi\in X_2 - X_1} f(\xi)(v-u) -
A_2\Bigr|\\
& \quad + \Bigl|(D_3)\sum_{\xi \in X_2 - X_1}
f(\xi)(v-u)\Bigr|\\
&< 5 \epsilon.
\end{align*}
The proof is complete. \qed
\begin{theorem} Let $f$ be
$H_1$-integrable on
a closed set $X_n$ with primitive $F_n$ for
$n=1,2,\ldots$,
and $X=\cup_{n=1}^{\infty} X_n$.
If $f$ is non-negative on $[a,b]$ and
$F_n(b)-F_n(a) \rightarrow A$ as
$n \rightarrow \infty$,
then $f$ is $H_1$-integrable on~$X$.\end{theorem}
\pf We may assume that $X_n \subset
X_{n+1}$ for each $n=1,2,\ldots$.
Since $f$ is $H_1$-integrable on $X_n$,
there exists $\delta_n(x) > 0$
such that for any $\delta_n$-fine division
$D$ of $[a,b]$ we have
$$(D)\sum_{\xi \in X_n} |f(\xi)(v - u) -
F_n(u,v)| < \frac{1}{2^n},$$
where $F_n(u,v) = F_n(v) - F_n(u)$ and $F_n$
is the primitive of
$f{\cal X}_{X_n}$ on $[a,b]$.
Put $\delta(x) = \delta_n(x)$ when $x \in
X_n - X_{n-1}$ with $X_0=\emptyset$,
otherwise arbitrary.
We may modify $\delta_n(x)$, if necessary,
as in Lemma~4.
Since $f$ is non-negative on $[a,b]$,
$f{\cal X}_{X_n}$ is absolutely
$H_1$-integrable on $[a,b]$ and so $f{\cal
X}_{X_n}$
is absolutely Henstock integrable on
$[a,b]$. Thus, $F_n$ is absolutely
continuous on $[a,b]$, and the result of
Lemma~4 applies.
Given $\epsilon > 0$, there exists an
integer $N>0$ such that
$$|A - F_N(a,b)| < \epsilon
\quad \hbox{ and } \quad
\sum_{n=N+1}^{\infty}\frac {1}{{2^n}} <
\epsilon.$$
Further, there exists a $\delta$-fine
division $D_N$ of $[a,b]$
such that for any $\delta$-fine $D \geq D_N$
of $[a,b]$ we have
$$\Bigl|(D)\sum_{\xi \in X_N} \{ f(\xi)(v - u) -
F_N(u,v) \}\Bigr| < \epsilon.$$
Here we have used Lemma~4 to obtain $D_N$
and the last inequality.
Note that $F_n(u,v) \rightarrow F(u,v)$ as
$n \rightarrow \infty$ and
$0 \leq F(u,v) - F_n(u,v) \leq F(u,v) -
F_N(u,v)$ for $n \geq N$.
Here $F(a,b)=A$.
Then for any $\delta$-fine division $D \geq
D_N$ of $[a,b]$ we obtain
\begin{align*}
\Bigl|(D)\sum_{\xi\in X} f(\xi)(v - u) - A\Bigr|
&\leq
\Bigl|(D)\sum_{\xi \in X_N} \{ f(\xi)(v - u) -
F_N(u,v) \}\Bigr|\\
& \quad + \sum_{n=N+1}^{\infty} \Bigl|(D)\sum_{\xi
\in X_n - X_{n-1}} f(\xi)(v-u)
- F_n(u,v)\Bigr|\\
& \quad + |A - F_N(a,b)|\\
&\leq 3 \epsilon.
\end{align*}
Hence $f$ is $H_1$-integrable on~$X$.
\medskip
With the idea presented in the proofs of
Lemma~4 and Theorem~5, we can now look
at the $H_1$-integrability of a Henstock
integrable function.
Note that the Cauchy Criterion~\cite[p.~10]{3},
also holds for $H_1$-integral.
We give first the following lemmas.
\begin{lemma} Let $X$ be a closed
subset of $[a,b]$. If $f$ is
$H_1$-integrable and bounded on $[a,b]$,
then $f$ is $H_1$-integrable on $X$.\end{lemma}
\pf Let $|f(x)| \leq M$ for all $x
\in [a,b]$. By the Cauchy
Criterion, there exists $\delta(x) > 0$ such
that for each $\epsilon>0$ there
is a $\delta$-fine division $D_0$ of $[a,b]$
such that for any $\delta$-fine
divisions $D,D' \geq D_0$ of $[a,b]$ we have
$$\Bigl|(D)\sum f(\xi)(v-u)-(D')\sum f(\xi)(v-u)\Bigr|
< \epsilon.$$
Further, there exists a finite union $E$ of
closed intervals such that
$E \supset X$ and $|E-X| < \frac{\epsilon}{
M}$. We can assume that a subset
of $D_0$ forms a division of $E$. For every
$\delta$-fine divisions $D_1,D_2 \geq D_0$
of $[a,b]$, let $D_1^*$ and
$D_2^*$ be the respective subsets of $D_1$
and $D_2$ which form divisions of~$E$. Note that
$(D_i)\sum_{\xi\in
X}f(\xi)(v-u)=(D_i^*)\sum_{\xi\in
X}f(\xi)(v-u)$ for
$i=1,2$.
We may assume further that
$$(D_1)\sum_{\xi \in \overline{[a,b]-E}}
f(\xi)(v-u) =
(D_2) \sum_{\xi \in \overline{[a,b]-E}}
f(\xi)(v-u).$$
Then
\begin{align*}
\Bigl|(D_1)\sum_{\xi \in X} f(\xi)(v-u)&-
(D_2)\sum_{\xi\in X} f(\xi)(v-u)\Bigr| \\
&\leq \Bigl|(D_1)\sum f(\xi)(v-u)-(D_2)\sum
f(\xi)(v-u)\Bigr| \\
&\quad + (D_1^*)\sum_{\xi \not\in
X}|f(\xi)|(v-u)
+ (D_2^*)\sum_{\xi\not\in X}|f(\xi)|(v-u) \\
&< 3\epsilon.
\end{align*}
By the Cauchy Criterion again, the above
inequalities imply that $f$ is\break
$H_1$-integrable on~$X$.\qed
\begin{lemma} Let $f$ be a measurable
function on $[a,b]$. Then there exists
a sequence $\{ X_i \}$ of closed subsets of
$[a,b]$ such that
$f$ is $H_1$-integrable on each $X_i$ and
$|[a,b] - \cup_{i=1}^{\infty} X_i|=0$.\end{lemma}
\pf It is well-known
\cite[p.~192]{2} that there exists a sequence
$\{ \varphi_n \}$ of continuous functions on
$[a,b]$ such that
$\varphi_n(x) \rightarrow f(x)$ almost
everywhere in $[a,b]$.
By Egoroff's Theorem and Lemma~6,
for each $i=1,2,\ldots$, there is a closed
set $X_i \subset [a,b]$ with
$|[a,b] - X_i| < \frac{1}{i}$ such that $f$
is $H_1$-integrable
on $X_i$. Obviously,
$$|[a,b] - \cup_{i=1}^{\infty} X_i| \leq
|[a,b] - \cup_{i=1}^N X_i| \leq |[a,b] -
X_N| <
\frac{1}{ N} \rightarrow 0$$
as $N \rightarrow \infty$, that completes
the proof of the lemma.\qed
\medskip
Let a function $F$ be defined on $[a,b]$ and
$X \subset [a,b]$.
The function $F$ is said to be $AC^*(X)$ if
for every $\epsilon > 0$
there exists $\eta >0$ such that for any
partial division $D=\{([u,v],\xi)\}$
of $[a,b]$ with $u$ or $v$ in $X$
$$(D)\sum|v-u|<\eta \qquad \hbox{implies}
\qquad (D)\sum|F(u,v)| < \epsilon.$$
Further, $F$ is said to be $ACG^*$ on
$[a,b]$ if $[a,b]$ is the union of
$X_1,X_2,\ldots$ such that $F$ is
$AC^*(X_i)$ for each $i$. This definition
is equivalent to the definition in
\cite[p.~29]{3} with $F$ being
continuous on $[a,b]$ and to the classical
definition in the book by Saks \cite4,
that is, if $F$ is $ACG^*$ on $[a,b]$, then
$F$ is continuous there. It was shown in \cite[p.~34]{3} that is $f$ is
a Henstock integrable function on $[a,b]$,
then its primitive $F$ is
$ACG^*$ on $[a,b]$ and we can assume that
$[a,b]=\cup X_i$ such that $F$
is $AC^*(X_i)$ and $X_i$ is closed for each~$i$.\qed
\medskip
On the other hand, a sequence $\{F_n\}$ of
functions defined on $[a,b]$ is
said to be $UAC^*(X)$ where $X \subset
[a,b]$ if, in the definition of
$AC^*(X)$, $\eta >0$ is independent of~$n$.
Further, $\{F_n\}$ is
$UACG^*$ on $[a,b]$ if $[a,b]=\cup X_i$ such
that $\{F_n\}$ is
$UAC^*(X_i)$ for each $i$ and we can assume
that $X_i$ is closed for each $i$.
Futhermore, the sequence $\{F_n\}$ is said
to be
{\it oscillation-convergent} to some
function $F$ defined on $[a,b]$ if
$[a,b]=\cup X_i$ with $X_i$ being closed and
for every $i$ and $\epsilon >0$ there
is an integer $N>0$ such that for any
partial division
$D=\{([u,v],\xi)\}$ of $[a,b]$ with $\xi$ in
$X_i$ we have
$$\sum|F_n(u,v)-F(u,v)|<\epsilon$$
whenever $n \geq N$. The following lemma was
proved in \cite[p.~56]{3}\qed
\begin{lemma} Let $\{f_n\}$ be a
sequence of Henstock integrable functions
on $[a,b]$ and is {\it control-convergent}
to some function $f$ on $[a,b]$;
that is, the following conditions are
satisfied:
\begin{enumerate}
\item $f_n(x) \rightarrow f(x)$ almost
everywhere in $[a,b]$ as
$n \rightarrow \infty$;
\item the sequence $\{F_n\}$ of primitives
of $\{f_n\}$ is $UACG^*$ on
$[a,b]$; and
\item $\{F_n\}$ converges uniformly on
$[a,b]$.
\end{enumerate}
Then $\{F_n\}$ is oscillation-convergent to
the primitive $F$ of $f$ on
$[a,b]$.\end{lemma}
\begin{lemma} Let $f$ be Henstock
integrable on $[a,b]$ with
primitive $F$. Then there exists a sequence
$\{ F_n \}$ of absolutely
continuous functions that is
oscillation-convergent to $F$ on $[a,b]$.\end{lemma}
\pf Since $f$ is Henstock
integrable on $[a,b]$, $F$ is $ACG^*$ on
$[a,b]$; that is, there exists a sequence
$\{ X_n \}$ of closed subsets of
$[a,b]$ such that $F$ is $AC^*(X_n)$ for
each $n$. We may assume that
$X_n \subset X_{n+1}$ and $a,b \in X_n$ for
each $n$. Since $X_n$ is closed,
we can write $(a,b) - X_n =
\cup_{k=1}^{\infty}(a_k,b_k)$ and put
\[
F_n(x) = \left\{
\begin{array}{ll}
F(x) & $$\hbox{ when } x \in X_n$$; \\
F(a_k)+\frac{F(b_k)-F(a_k)}{b_k-a_k}(x-a_k)
& $$\hbox{ when } x \in (a_k,b_k)
\hbox{ for all } k.$$
\end{array}\right.
\]
Since $f$ is Henstock integrable on $[a,b]$,
its primitive $F$ and thus
$F_n$ are continuous on $[a,b]$.
Further, we can assume that $F_n \rightarrow
F$ uniformly as
$n \rightarrow \infty$.
Since $F_n$ is $AC^*(X_n)$ for each $n$,
given an $\epsilon >0$ there exists
$\eta_n >0$ such that for any partial
division $\pi_n$ of $[a,b]$ with
$u$ or $v$ in $X_n$
$$(\pi_n)\sum |v-u| < \eta_n \quad \hbox{
implies } \quad
(\pi_n)\sum |F_n(u,v)| < \epsilon.$$
Let $N$ be fixed. For $n \geq N$,
$F_n(x)=F_N(x)$ for all $x \in X_N$.
Thus we can choose $\eta=\min_{1 \leq i \leq
N}\{\eta_i\}$ for all $F_n$
so that $\{F_n\}$ is $UAC^*(X_N)$. Hence
$\{F_n\}$ is $UACG^*$ on $[a,b]$.
Now define
\[
f_n(x) = \left\{
\begin{array}{ll}
f(x) & $$\hbox{ when } x \in X_n$$; \\
\frac{F(b_k)-F(a_k)}{b_k-a_k} & $$\hbox{
when } x \in (a_k,b_k)
\hbox{ for all } k.$$
\end{array}\right.
\]
It is easy to see that $f_n$ is Henstock
integrable on $[a,b]$ and $F_n$
is the primitive of $f_n$ for each $n$.
Since $X_n \uparrow [a,b]$,
$f_n(x) \rightarrow f(x)$ almost everywhere
in $[a,b]$.
From the above discussion, $\{f_n\}$ is
control-convergent to $f$ on $[a,b]$.
Thus, by Lemma~8, $\{F_n\}$ is
oscillation-convergent to $F$ on $[a,b]$.
The proof is complete.\qed
\medskip
We now give the main result of the paper.
\begin{theorem} Let $f$ be Henstock
integrable on $[a,b]$.
Then there is an\break $H_1$-integrable function
$g$ such that $f(x)=g(x)$
almost everywhere in $[a,b]$.\end{theorem}
\pf By Lemma 7, there exists a
sequence $\{X_i\}$ of closed
subsets of $[a,b]$ such that $f$ is
$H_1$-integrable on each $X_i$ and
$|[a,b]-\cup_{i=1}^{\infty} X_i|=0$. Let
$X=\cup_{i=1}^{\infty} X_i$ and
$X_i \subset X_{i+1}$ for each $i$.
We prove that $f$ is $H_1$-integrable on~$X$.
We may assume that the result of Lemma~9
holds; that is, for every $i$
there exists an integer $n(i) \geq i$ such
that for any partial division
$D$ of $[a,b]$ with $u$ or $v$ in $X_i$, we
have
$$(D) \sum |F_{n(i)}(u,v)-F(u,v)| < \frac{1}{
2^i},$$
where $F$ is the primitive of $f$ on $[a,b]$
and $F_{n(i)}(x)$ is
as defined in the proof of Lemma~9.
Since $f$ is $H_1$-integrable on $X_{n(i)}$
for each $i$, $f_{n(i)}$
is also $H_1$-integrable on $X_{n(i)}$,
where $f_{n(i)}$ is as defined
in Lemma~9. There exists
$\delta_{n(i)}(x)>0$
such that for any $\delta_{n(i)}$-fine
division $D$ of $[a,b]$ we have
$$(D)\sum_{\xi \in
X_i}|f_{n(i)}(\xi)(v-u)-F_{n(i)}(u,v)|<\frac{1}{2^i}.$$
For $i=1,2,\ldots$, put
$\delta(x)=\delta_i(x)$ if $x \in
X_i-X_{i-1}$ with
$X_0=\emptyset$; otherwise, put
$\delta(x)>0$ arbitrary. We may modify
$\delta_i(x)$, if necessary, as in Lemma~4.
Given $\epsilon>0$, there exists a positive
integer $N=n(i_0)$ such that
$$\sum_{i=i_0+1}^{\infty}\frac{1}{2^i} <
\epsilon.$$
Further, there exists a $\delta$-fine
division $D_N$ of $[a,b]$ such that
for any $\delta$-fine division $D \geq D_N$
of $[a,b]$ we have
$$\Bigl|(D)\sum_{\xi\in
X_{i_0}}\{f(\xi)(v-u)-F_N(u,v)\}\Bigr|<\epsilon.
$$
For any $\delta$-fine division
$D=\{([u,v],\xi)\}$ of $[a,b]$
with $\xi\in X_n-X_{n-1}$ and $D\geq D_N$,
we have
\begin{align*}
\Bigl|(D)\sum_{\xi\in X}&\{f(\xi)(v-u)-F(u,v)\}\Bigr|
\leq
\Bigl|(D)\sum_{\xi\in
X_{i_0}}\{f(\xi)(v-u)-F_N(u,v)\}\Bigr|\\
& \quad +
\sum_{i=i_0+1}^{\infty}|(D)\sum_{\xi\in
X_i-X_{i-1}}
\{f(\xi)(v-u)-F_{n(i)}(u,v)\}|\\
& \quad + \Bigl|(D) \sum_{\xi\in
X_{i_0}}\{F_N(u,v)-F(u,v)\}\Bigr|\\
& \quad +
\sum_{i=i_0+1}^{\infty}\Bigl|(D)\sum_{\xi\in
X_i-X_{i-1}}
\{F_{n(i)}(u,v)-F(u,v)\}\Bigr|\\
&< \epsilon + \sum_{i=i_0+1}^{\infty}\frac{1}{2^i} + \frac{1}{2^{i_0}} +
\sum_{i=i_0+1}^{\infty}\frac{1}{2^i}\\
&< 4\epsilon.
\end{align*}
Therefore, $f$ is $H_1$-integrable on~$X$.
\qed
\begin{cor} A function $f$ is
Henstock integrable on $[a,b]$ if
and only if $f(x)=g(x)$ almost everywhere in
$[a,b]$ for some
$H_1$-integrable function $g$ on $[a,b]$.\end{cor}
It is not known whether every Henstock
integrable function on $[a,b]$ is also
$H_1$-integrable there. We conjecture that
it is not.
\begin{thebibliography}{9}
\bibitem{1} N. Dunford and J. T. Schwartz, {\it
Linear Operators} I, Interscience 1958.
\bibitem{2} C. Goffman, {\it Real Functions},
Rinehart 1953.
\bibitem{3}
P. Y. Lee, {\it Lanzhou Lectures in
Henstock Integration}, World Scientific
1989.
\bibitem{4}
S. Saks, {\it Theory of the Integral},
2nd ed, Hafner 1937.
\end{thebibliography}
\end{document}