# The Ross-Darboux-Stieltjes Integral

David Cruz-Uribe, OFS  Dept. of Mathematics
University of Alabama
Tuscaloosa, AL 35487, USA<EMAIL_ADDRESS>and  Jacob Glidewell  Dept. of
Mathematics
University of Alabama
Tuscaloosa, AL 35487, USA<EMAIL_ADDRESS>
(Date: May 27, 2024)

###### Abstract.

Motivated by the limitations of the traditional definitions of the Riemann-
Stieltjes and Darboux-Stieltjes integrals, we introduce a generalized Darboux-
Stieltjes integral that is equivalent to an earlier generalization by Ross
[19]. Our definition builds upon an approach to the Darboux-Stieltjes integral
recently introduced by the first author and Convertito [5]. We show that our
definition agrees with all previous definitions, but that the class of
integrable functions is much larger. We develop all the analogs of the classic
results for the Riemann integral, and rectify the problems inherent in the
definition of the Darboux-Stieltjes integral in [5]. In particular, we show
the Bounded Convergence Theorem holds for our definition and that it agrees
with the Lebesgue-Stieltjes integral.

###### Key words and phrases:

Stieltjes integral, Lebesgue-Stieltjes integral, functions of bounded
variation

###### 2010 Mathematics Subject Classification:

Primary 26A42

This paper is a revised version of the masters thesis written by the second
author under the direction of the first. The first author is partially
supported by a Simons Foundation Travel Support for Mathematicians Grant.

###### Contents

  1. 1 Introduction
  2. 2 The Darboux-Stieltjes Integral
  3. 3 The Ross-Darboux-Stieltjes Integral for Step functions
  4. 4 The Ross-Darboux-Stieltjes Integral
  5. 5 Consequences of the RDS Criterion
  6. 6 Convergence Theorems
  7. 7 RDS Integrability Theorems
  8. 8 The Bounded Convergence Theorem
  9. 9 RDS Integral and Related Integrals

##  1\. Introduction

The Stieltjes integral plays an important role in analysis and in
applications. However, since its introduction it has been recognized that its
original definition had serious drawbacks, and there have been several
attempts to improve it, none completely successful. The goal of this paper is
to introduce a new definition which we believe overcomes the drawbacks of
previous definitions. To support this claim and to provide context for our
results, we will first review the history of the Stieltjes integal. This
review is not intended to be comprehensive, but should suffice to illustrate
our point. Hereafter, $B[a,b]$ will be the set of bounded functions on the
interval $[a,b]$, $I[a,b]$, will be increasing functions, and $BV[a,b]$ will
be functions of bounded variation.

Stieltjes [22] first introduced the Riemann-Stieltjes integral by generalizing
the definition of the Riemann integral.

######  Definition 1.1.

Given $f\in B[a,b]$, $\alpha\in BV[a,b]$, a partition
$\mathcal{P}=\\{x_{i}\\}_{i=0}^{n}$ of $[a,b]$ and sample points
$x_{i}^{*}\in[x_{i-1},x_{i}]$, we form a Riemann-Stieltjes sum

$\sum_{i=1}^{n}f(x_{i}^{*})[\alpha(x_{i})-\alpha(x_{i-1})].$

If there exists $A$ such that for every $\epsilon>0$, there exists $\delta>0$
such that for every Riemann-Stieltjes sum with $|\mathcal{P}|<\delta$,

$\left|\sum_{i=1}^{n}f(x_{i}^{*})[\alpha(x_{i})-\alpha(x_{i-1})]-A\right|<\epsilon,$

then we define the Riemann-Stieltjes integral of $f$ with respect to $\alpha$
by

$(RS)\int_{a}^{b}f(x)\,d\alpha=A.$

Stieltjes introduced this integral, hereafter the RS-integral, to study
continued fractions. Later, M. Riesz [17] showed the importance of this
definition by using the RS-integral to characterize the space of linear
functionals on $C[a,b]$, the space of continuous functions on $[a,b]$. (See
also F. Riesz and Sz.-Nagy [16].) Both Stieltjes and Riesz were primarily
interested in this integral when the integrand $f$ was a continuous function.

Straightforward examples (see [5, Example 5.30]) show that the RS-integral
only exists if $f$ and $\alpha$ have no common points of discontinuity. To
overcome this weakness and widen the class of functions for which a Stieltjes-
type integral is defined, Pollard [14] gave two equivalent definitions: the
Pollard-Stieltjes integral (PS-integral) and what is now referred to as the
Darboux-Stieltjes integral (DS-integral). The PS-integral changes Definition
1.1 by replacing the convergence of the Riemann-Stieltjes sums (referred to as
"norm convergence" in [8]) with a weaker form of convergence (referred to as
"$\sigma$-convergence" in [8]). More precisely, for each $\epsilon>0$, it only
considers the Riemann-Stieltjes sums on the refinements of a specific
partition that depends on $\epsilon$, $\mathcal{P}_{\epsilon}$, instead of all
partitions with $|\mathcal{P}|<\delta$. (See Definition 9.10 for more
details.)

The DS-integral generalizes the definition of the classical Darboux integral
(often mistakenly referred to as the Riemann integral). Pollard only stated it
for $\alpha$ increasing; however, it can be extended to all $\alpha\in
BV[a,b]$ using linearity and the Jordan decomposition theorem.

######  Definition 1.2.

Given $f\in B[a,b]$, $\alpha\in I[a,b]$, and a partition
$\mathcal{P}=\\{x_{i}\\}_{i=1}^{n}$, define

$\overline{M}_{i}=\sup\\{f(x)\;:x\in[x_{i-1},x_{i}]\\},\quad\overline{m}_{i}=\inf\\{f(x)\;:x\in[x_{i-1},x_{i}]\\},$

and form the upper and lower sums

$\displaystyle U_{\alpha}^{e}(f,\mathcal{P})$
$\displaystyle=\sum_{i=1}^{n}\overline{M}_{i}[\alpha(x_{i})-\alpha(x_{i-1})],$
$\displaystyle L_{\alpha}^{e}(f,\mathcal{P})$
$\displaystyle=\sum_{i=1}^{n}\overline{m}_{i}[\alpha(x_{i})-\alpha(x_{i-1})].$

If $\inf U_{\alpha}^{e}(f,\mathcal{P})=\sup L_{\alpha}^{e}(f,\mathcal{P})=A$,
where the infimum and supremum are taken over all partitions of $[a,b]$, then
we define the Darboux-Stieltjes integral of $f$ with respect to $\alpha$ by

$(DS)\int_{a}^{b}f(x)\,d\alpha=A.$

For $\alpha\in I[a,b]$, Pollard showed the PS-integral and the DS-integral are
equivalent and agree with the RS-integral whenever the latter exists. However,
both of these integrals are more general than the RS-integral: there exist
functions which are DS-integrable but whose RS-integrals do not exist since
$f$ and $\alpha$ have common discontinuities. The correct necessary condition
in this case (see [13, Theorem 4.13]) is that the DS-integral exists if at
each common discontinuity $x\in[a,b]$, at least one of $f$ or $\alpha$ is
right continuous and at least one is left continuous at $x$.

Several other versions of the Stieltjes integral have appeared in the
literature. Hildebrandt [8] and Dushnik [6] studied generalizations of the PS-
integral. Pollard gave an “interior" version of the DS-integral where
$\overline{M}_{i}$ and $\overline{m}_{i}$ are replaced with $M_{i}$ and
$m_{i}$, the supremum and infimum of $f(x)$ when $x\in(x_{i-1},x_{i})$. To
distinguish this from Definition 1.2 we will sometimes refer to that one as
the "exterior" DS-integral. (This explains the superscript "$e$" that appears
above.) An equivalent version of the interior DS-integral was independently
discovered by Convertito and the first author and developed in [5]. It is
equivalent to the exterior DS-integral wherever the the latter is defined, but
is defined on a larger collection of functions: in [5, Theorem 5.19] they
showed that a necessary condition for the integral to exist is that at each
common discontinuity $x\in[a,b]$, $\alpha$ is right continuous or $f(x+)$
exists, and $\alpha$ is left continuous or $f(x-)$ exists.

The RS, DS, and PS-integrals were the definitions used for the Stieltjes
integral in many of the standard analysis books in the second half of the
twentieth century. The RS-integral was used by Taylor and Mann [23]. Apostol
[1] used the PS-integral and showed the equivalence of the DS-integral as
theorem. He noted in an exercise that the RS-integral is not equivalent.
Bartle [2] also used the PS-integral and gave the RS and DS-integrals as
exercises. From a pedagogic perspective, we note that neither author explains
the rationale for using $\sigma$-convergence in their definition. Rudin [21]
used the DS-integral, as did Burkhill and Burkhill [3], and Krantz [11].
Protter and Morrey [15] and Wheeden and Zygmund [24] defined both the RS and
DS-integrals and highlighted the fact that they are not equivalent.

Ross [19, 18], motivated by the restrictive necessary conditions for the RS
and DS-integrals to exist, proposed a modified version of the DS-integral.

######  Definition 1.3.

Given $f\in B[a,b]$, $\alpha\in I[a,b]$, and a partition
$\mathcal{P}=\\{x_{i}\\}_{i=1}^{n}$, define

$M_{i}=\sup\\{f(x)\;:x\in(x_{i-1},x_{i})\\},\quad
m_{i}=\inf\\{f(x)\;:x\in(x_{i-1},x_{i})\\},$

and

$J_{\alpha}(f,\mathcal{P})=\sum_{i=0}^{n}f(x_{i})[\alpha(x_{i}+)-\alpha(x_{i}-)].$

Form the upper and lower sums

$\displaystyle U_{\alpha}^{r}(f,\mathcal{P})$
$\displaystyle=J_{\alpha}(f,\mathcal{P})+\sum_{i=1}^{n}M_{i}[\alpha(x_{i}-)-\alpha(x_{i-1}+)],$
$\displaystyle L_{\alpha}^{r}(f,\mathcal{P})$
$\displaystyle=J_{\alpha}(f,\mathcal{P})+\sum_{i=1}^{n}m_{i}[\alpha(x_{i}-)-\alpha(x_{i-1}+)].$

If $\inf U_{\alpha}^{r}(f,\mathcal{P})=\sup L_{\alpha}^{r}(f,\mathcal{P})=A$,
where the infimum and supremum are taken over all partitions of $[a,b]$, then
we define the Ross-Darboux-Stieljtes (RDS) integral of $f$ with respect to
$\alpha$ by

$(RDS)\int_{a}^{b}f(x)\,d\alpha=A.$

Ross [19] also gave an equivalent definition in terms of a modified RS-
integral. He loosely motivated the additional term $J_{\alpha}$ used to define
the upper and lower sums by suggesting that it allows individual points to be
assigned a positive "weight." It is clear from other comments in [18] that he
was also motivated by the Lebesgue-Stieltjes integral.

######  Remark 1.4.

Another variant of the RS-integral was proposed by Ter Horst [9] which is a
generalization of the definition given by Ross. His definition is a hybrid
between the classical Stieltjes integrals and the more complex definition of
the Lebesgue integral. We will not discuss it in detail in this paper.

As noted above, in [5] the authors developed the interior DS-integral using an
equivalent definition in terms of approximation by step functions. (See
Definition 2.8 below.) Like Ross, they wanted to weaken the restrictive
necessary conditions associated with the standard definitions. They were also
motivated by pedagogical concerns: they felt that their definition was
intuitive and also created a bridge between classical undergraduate analysis
and graduate analysis and measure theory. However, as they pointed out in
their introduction, there are two significant problems with the interior DS-
integral. First, the Bounded Convergence Theorem (the analog of the Lebesgue
Dominated Convergence Theorem for the DS-integral) does not hold for this
definition of the Stieltjes integral unless $\alpha$ is continuous. See [5,
Example 4.49 and Theorem 4.50]. Second, and perhaps more problematic from an
advanced perspective, this definition does not always agree with the Lebesgue-
Stieltjes integral when $f$ and $\alpha$ have a common discontinuity.

The goal of this paper is to address the problems with the interior DS-
integral as defined in [5]. We give a definition which is equivalent to
Definition 1.3 but which adopts the perspective taken in [5]. In particular,
we draw a more explicit connection to the Lebesgue-Stieltjes integral. In the
definitions of the RS, PS and DS-integrals, as well as in Definition 2.6
below, the "$\alpha$-length" of a partition interval $[x_{i-1},x_{i}]$ is
given by $\alpha(x_{i})-\alpha(x_{i-1})$. The problem with this, as Ross
correctly noted, is that the Stieltjes integral should give weight to
individual points in the partition if they correspond to jump discontinuities
of $\alpha$. To accomplish this, we will replace this definition of
$\alpha$-length with the following: given an open interval $I=(c,d)$, we
define $\alpha(I)=\alpha(d-)-\alpha(c+)$, where $x\pm$ denotes the right and
left-sided limits at $x$. We also define the $\alpha$-length of a single point
$x$ by $\alpha(\\{x\\})=\alpha(x+)-\alpha(x-)$. This modest change lets us
correctly capture all the behavior of the Stieltjes integral when $f$ and
$\alpha$ have common discontinuities. This notion of $\alpha$-length is
implicit in the definition of $J_{\alpha}$ in Definition 1.3; it also
corresponds to the premeasure used to define the Lebesgue-Stieltjes measure.
(See [4, Section 4.1].)

Given our definition, we show that all of the standard properties of the
Stieltjes integral can be easily derived. These results can be thought of as
extending and complementing the development of the interior DS-integral in [5]
and the development of the RDS-integral in [19]. We then prove some more
advanced theorems which show that in some sense the RDS-integral is the right
definition to use. We completely characterize the functions which are RDS-
integrable by proving the precise analogue of the Lebesgue criterion for
Riemann-Darboux integrability.

######  Theorem 1.5 (RDS-Lebesgue Criterion).

Given $f\in B[a,b]$ and $\alpha\in BV[a,b]$, $f\in RDS_{\alpha}[a,b]$ if and
only if the subset of $[a,b]$ where $f$ is discontinuous has $\alpha$-measure
$0$.

As an immediate consequence of this theorem we have that if $f$ and $\alpha$
are functions of bounded variation, then the RDS-integral of $f$ with respect
to $\alpha$ exists. We also prove that the Bounded Convergence theorem holds
for the RDS-integral.

######  Theorem 1.6 (RDS Bounded Convergence Theorem).

Given $\alpha\in BV[a,b]$ and a uniformly bounded sequence of functions
$\\{f_{n}\\}_{n=1}^{\infty}$ on $[a,b]$, if

  * •

for all $n\in\mathbb{N}$, $f_{n}\in RDS_{\alpha}[a,b]$,

  * •

$f_{n}\to f$ pointwise as $n\to\infty$,

  * •

$f\in RDS_{\alpha}[a,b]$,

then

$\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,d\alpha=\int_{a}^{b}f(x)\,d\alpha.$

Finally, we prove that the RDS-integral agrees with the Lebesgue-Stieltjes
integral whenever the former is defined.

The remainder of this paper is organized as follows. In Section 2 we give some
preliminary definitions and sketch the approach in [5] to the Darboux-
Stieltjes integral. This will serve as a foundation for our definition of the
RDS-integral, which we give in Sections 3 and 4. In Section 4 we also prove
some basic properties of the RDS-integral, including the RDS criterion, an
analogue of the Darboux criterion for integrability. Finally, we give
sufficient conditions, in terms of continuity, for the RDS and DS-integrals to
agree. In Section 5, we use the RDS criterion to prove linearity, additivity,
monotonicity in the integrand and linearity of the integrator. In Section 6,
we prove two convergence theorems: uniform convergence in the integrand and
BV-norm convergence in the integrator. In Section 7 we consider necessary and
sufficient conditions for RDS-integrability and prove the RDS-Lebesgue
criterion. As an application we prove a generalization of integration by parts
for the RDS-integral. In Section 8 we prove the Bounded Convergence Theorem.
Finally, in Section 9, we compare various definitions of the Stieltjes
integral. We first give a characterization the functions whose RDS and DS-
integrals agree. We then give three alternative definitions of a generalized
Riemann-Stieltjes integral and discuss their relationship with the RDS-
integral. Finally, we prove that the RDS-integral and the Lebesgue-Stieltjes
integral agree.

Throughout, our notation will follow that used in [5]; similarly, in our
proofs we will draw upon a number of results proved in this work. In Sections
3–6 many of our proofs are very similar to proofs of the corresponding result
for the Darboux or DS-integrals, and we have tried to strike a balance between
being self-contained and repeating arguments that can be found elsewhere. The
results in Sections 7–9 are (with one exception) all new, and we give complete
proofs.

##  2\. The Darboux-Stieltjes Integral

In this section we briefly state the definition of the interior Darboux-
Stieltjes integral as defined in [5, Chapter 4]. This will help to motivate
the definition of the RDS-integral that we will give in Section 3 below. As
part of of this we define functions of bounded variation; additional
properties of functions of bounded variation will be called upon throughout
the text, and we refer to [5, Chapter 3] for complete information.

Fundamental to all of our definitions is the notion of a partition of an
interval.

######  Definition 2.1.

Given a closed interval $[a,b]$, a partition of $[a,b]$ is a finite set of
points $\mathcal{P}=\\{x_{i}\\}_{i=0}^{n}$ such that

$a=x_{0}<x_{1}<x_{2}<\dots<x_{n-1}<x_{n}=b.$

Given $\mathcal{P}$, define the disjoint partition intervals
$I_{i}=(x_{i-1},x_{i})$ for each $1\leq i\leq n$. The mesh size
$|\mathcal{P}|$ of a partition is the length of the longest partition
interval:

$|\mathcal{P}|=\max_{1\leq i\leq n}\\{x_{i}-x_{i-1}\\}.$

######  Definition 2.2.

Given partitions $\mathcal{P}=\\{x_{i}\\}_{i=0}^{n}$ and
$\mathcal{Q}=\\{y_{j}\\}_{j=0}^{m}$ of the interval $[a,b]$, we say
$\mathcal{Q}$ is a refinement of $\mathcal{P}$ if
$\mathcal{P}\subset\mathcal{Q}$. We say $\mathcal{R}$ is a common refinement
of $\mathcal{P}$ and $\mathcal{Q}$ if $\mathcal{R}$ is a refinement of both
$\mathcal{P}$ and $\mathcal{Q}$.

We will use the above notation throughout. Moreover, given a second partition
$\mathcal{Q}=\\{y_{j}\\}_{j=0}^{m}$, we will use $J_{j}=(y_{j-1},y_{j})$ for
each $1\leq j\leq m$ for the partition intervals of $\mathcal{Q}$.

We can now define the bounded variation of a function.

######  Definition 2.3.

Given a function $f\in B[a,b]$ and a partition $\mathcal{P}$ of $[a,b]$,
define the variation of $f$ on $\mathcal{P}$ by

$V(f,\mathcal{P})=\sum_{i=0}^{n}|f(x_{i})-f(x_{i-1})|.$

We say that $f$ is of bounded variation, and write $f\in BV[a,b]$, if

$V(f,[a,b])=\sup_{\mathcal{P}}V(f,\mathcal{P})<\infty,$

where the supremum is taken over all partitions $\mathcal{P}$ of $[a,b]$.

For the definition of the DS-integral, we recall the Jordan decomposition
theorem: if $\alpha\in BV[a,b]$, then there exist functions
$\alpha^{+},\,\alpha^{-}\in I[a,b]$ such that $\alpha=\alpha^{+}-\alpha^{-}$.

The definition of the DS-integral occurs in two stages. First, we define it
for step functions. The integrals of step functions will take the place of the
upper and lower Darboux sums in the traditional definition, e.g., Definition
1.2.

######  Definition 2.4.

A function $f\in B[a,b]$ is a step function if there exists a partition
$\mathcal{P}$ of $[a,b]$ and real numbers $\\{c_{i}\\}_{i=1}^{n}$ such that
$f(x)=c_{i}$ if $x\in I_{i},1\leq i\leq n$. The function f is said to be a
step function with respect to the partition $\mathcal{P}$.

######  Definition 2.5.

Given $\alpha\in BV[a,b]$ and an open interval $I=(c,d)\subset[a,b]$, we
define the (classical) $\alpha$-length of $I$ to be
$\alpha(I)=\alpha(d)-\alpha(c)$.

We refer to this as the classical $\alpha$-length because of the role it
played historically in the definition of the Stieltjes integral. Below in
Definition 3.1 we will redefine it to correspond to the premeasure used to
define the Lebesgue-Stieltjes integral.

######  Definition 2.6.

Given $f\in S[a,b]$ defined with respect to a partition $\mathcal{P}$, suppose
$f(x)=c_{i}$ for $x\in I_{i}$. Define the Stieltjes integral with respect to
$\alpha\in BV[a,b]$ by

$(DS)\int_{a}^{b}f(x)\,d\alpha=\sum_{i=1}^{n}c_{i}\alpha(I_{i}).$

As the second step, we define the DS-integral of any bounded function. To do
so, we define the collections of step functions that bracket $f$.

######  Definition 2.7.

Given $f\in B[a,b]$, define the sets

$\displaystyle S_{U}(f,[a,b])$ $\displaystyle=\\{u\in S[a,b]\;:\;f(x)\leq
u(x),x\in[a,b]\\},$ $\displaystyle S_{L}(f,[a,b])$ $\displaystyle=\\{v\in
S[a,b]\;:\;f(x)\geq v(x),x\in[a,b]\\}.$

We say $u,\,v\in S[a,b]$ bracket $f$ if $u\in S_{U}(f,[a,b])$ and $v\in
S_{L}(f,[a,b]).$

######  Definition 2.8.

Given $\alpha\in I[a,b]$ and $f\in B[a,b]$, define

$\displaystyle
L^{i}_{\alpha}(f,[a,b])=\sup\left\\{\int_{a}^{b}v(x)\,d\alpha\;:\;v\in
S_{L}(f,[a,b])\right\\},$ $\displaystyle
U^{i}_{\alpha}(f,[a,b])=\inf\left\\{\int_{a}^{b}u(x)\,d\alpha\;:\;u\in
S_{U}(f,[a,b])\right\\}.$

If $L^{i}_{\alpha}(f,[a,b])=U^{i}_{\alpha}(f,[a,b])$, then we say $f$ is
(interior) Darboux-Stieltjes integrable with respect to $\alpha$ and write

$(DS)\int_{a}^{b}f(x)\,d\alpha$

as their common value. Moreover, for $\alpha\in BV[a,b]$, if there exists a
decomposition $\alpha=\alpha^{+}-\alpha^{-}$ where $f\in DS_{\alpha^{+}}[a,b]$
and $f\in DS_{\alpha^{-}}[a,b]$, then we define $f\in DS_{\alpha}[a,b]$ and

$(DS)\int_{a}^{b}f(x)\,d\alpha=(DS)\int_{a}^{b}f(x)\,d\alpha^{+}-(DS)\int_{a}^{b}f(x)\,d\alpha^{-}.$

We denote the set of functions which are Darboux-Stieltjes integrable with
respect to $\alpha$ as $DS_{\alpha}[a,b]$.

The Darboux-Stieltjes integral is well-defined; in particular, Definition 2.8
applied to a step function agrees with Definition 2.6. Moreover, this
definition is equivalent to the definition of the interior DS-integral
described in Section 1: details are left to the interested reader. See the
proof of the corresponding result for the RDS-integral, Corollary 4.8.

##  3\. The Ross-Darboux-Stieltjes Integral for Step functions

We now turn to the definition of the RDS-integral. Following the pattern of
the definition of the DS-integral described in Section 2, in this section we
give a new definition of $\alpha$-length and define the RDS-integral of step
functions. We will define the integral in general, and prove some of its basic
properties, in Section 4.

We first give our revised definition of $\alpha$-length, which takes more
careful account of the jump discontinuities of the integrator $\alpha$.

######  Definition 3.1.

Define the $\alpha$-length of an open interval $I=(c,d)$ as

$\mu_{\alpha}(I)=\alpha(d-)-\alpha(c+),$

of a closed interval $\overline{I}=[c,d]$ as

$\mu_{\alpha}(\;\overline{I}\;)=\alpha(d+)-\alpha(c-),$

and of a singleton set $\\{c\\}$ as

$\mu_{\alpha}(\\{c\\})=\alpha(c+)-\alpha(c-).$

We also extend this to half-open intervals

$\mu_{\alpha}((c,d])=\alpha(d+)-\alpha(c+)\quad\text{and}\quad\mu_{\alpha}([c,d))=\alpha(d-)-\alpha(c-).$

Clearly, $\mu_{\alpha}$ is finitely additive. Moreover, if $\alpha$ is
increasing, $\mu_{\alpha}$ is nonnegative.

######  Remark 3.2.

In dealing with a function $\alpha$ defined on a closed interval $[a,b]$, we
adopt the following conventions: $\alpha(a-)=\alpha(a)$ and
$\alpha(b+)=\alpha(b)$. This convention will be important in the proving the
additivity of the RDS-integral. See the proofs of Theorems 3.9 and 5.4.

We now define the integral of a step function. Notice the definition has two
terms: the first is analogous to Ross’s jump term $J_{\alpha}(f,\mathcal{P})$
in Definition 1.3, and the second is the classical step function integral as
in Definition 2.6.

######  Definition 3.3.

Given $f\in S[a,b]$ defined with respect to a partition $\mathcal{P}$, suppose
$f(x)=c_{i}$ for $x\in I_{i}$. Define the Ross-Stieltjes integral with respect
to $\alpha\in BV[a,b]$ by

$(\mathcal{P})\int_{a}^{b}f(x)\,d\alpha=\sum_{i=0}^{n}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}c_{i}\mu_{\alpha}(I_{i}).$

Clearly, a step function can be defined with respect to multiple paritions:
for instance, a constant function is a step function with respect to any
partition. Therefore, we have to show our definition is well-defined, i.e.,
that it is independent of the chosen partition.

######  Proposition 3.4.

Given $\alpha\in BV[a,b]$, if $f\in S[a,b]$ is defined with respect to both
partitions $\mathcal{P}$ and $\mathcal{Q}$, then

$(\mathcal{P})\int_{a}^{b}f(x)\,d\alpha=(\mathcal{Q})\int_{a}^{b}f(x)\,d\alpha.$

###### Proof.

First, suppose $\mathcal{Q}=\\{y_{j}\\}_{j=0}^{m}$ is a refinement of
$\mathcal{P}=\\{x_{i}\\}_{i=0}^{n}$ with only one additional point. That is,
$m=n+1$. Note there exists indices $1\leq\ell\leq n$, $1\leq k\leq n$ such
that

$y_{k-1}=x_{\ell-1}<y_{k}<y_{k+1}=x_{\ell}.$

Hence, $f(y_{k})=c_{\ell}=d_{k}=d_{k+1}$. Then,

$\displaystyle(\mathcal{Q})\int_{a}^{b}f(x)\,d\alpha$
$\displaystyle\qquad=\sum_{j=0}^{n+1}f(y_{j})\mu_{\alpha}(\\{y_{j}\\})+\sum_{j=1}^{n+1}d_{j}\mu_{\alpha}(J_{j})$
$\displaystyle\qquad=f(y_{k})\mu_{\alpha}(\\{y_{k}\\})+\sum_{i=0}^{n}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})$
$\displaystyle\qquad\quad+d_{k}\mu_{\alpha}(J_{k})+d_{k+1}\mu_{\alpha}(J_{k+1})-c_{\ell}\mu_{\alpha}(I_{\ell})+\sum_{i=1}^{n}c_{i}\mu_{\alpha}(I_{i})$
$\displaystyle\qquad=f(y_{k})\mu_{\alpha}(\\{y_{k}\\})+\sum_{i=0}^{n}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})$
$\displaystyle\qquad\quad+c_{\ell}[\alpha(y_{k}-)-\alpha(y_{k-1}+)+\alpha(y_{k+1}-)-\alpha(y_{k}+)-\alpha(x_{\ell}-)+\alpha(x_{\ell-1}+)]$
$\displaystyle\qquad\quad+\sum_{i=1}^{n}c_{i}\mu_{\alpha}(I_{i})$
$\displaystyle\qquad=f(y_{k})\mu_{\alpha}(\\{y_{k}\\})+c_{\ell}[\alpha(y_{k}-)-\alpha(y_{k}+)]+\sum_{i=0}^{n}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}c_{i}\mu_{\alpha}(I_{i})$
$\displaystyle\qquad=f(y_{k})\mu_{\alpha}(\\{y_{k}\\})-f(y_{k})\mu_{\alpha}(\\{y_{k}\\})+\sum_{i=0}^{n}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}c_{i}\mu_{\alpha}(I_{i})$
$\displaystyle\qquad=\sum_{i=0}^{n}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}c_{i}\mu_{\alpha}(I_{i})$
$\displaystyle\qquad=(\mathcal{P})\int_{a}^{b}f(x)\,d\alpha.$

Next, assume $\mathcal{Q}$ is a refinement of $\mathcal{P}$. Thus, we can
write $\mathcal{Q}=\mathcal{P}\cup\\{t_{1},\dots,t_{r}\\}$. Define
$\mathcal{P}_{0}=\mathcal{P}$,
$\mathcal{P}_{1}=\mathcal{P}_{0}\cup\\{t_{1}\\},\mathcal{P}_{2}=\mathcal{P}_{1}\cup\\{t_{2}\\},\dots,\mathcal{P}_{r}=\mathcal{P}_{r-1}\cup\\{t_{r}\\}=\mathcal{Q}$.
From the previous case, for each $0\leq i\leq r-1$,

$(\mathcal{P}_{i})\int_{a}^{b}f(x)\,d\alpha=(\mathcal{P}_{i+1})\int_{a}^{b}f(x)\,d\alpha.$

By combining these equalities, we have

$(\mathcal{P})\int_{a}^{b}f(x)\,d\alpha=(\mathcal{P}_{0})\int_{a}^{b}f(x)\,d\alpha=(\mathcal{P}_{r})\int_{a}^{b}f(x)\,d\alpha=(\mathcal{Q})\int_{a}^{b}f(x)\,d\alpha.$

Finally, we prove the general case. Let $\mathcal{R}$ be a common refinement
of $\mathcal{P}$ and $\mathcal{Q}$. Thus, by the previous case,

$(\mathcal{P})\int_{a}^{b}f(x)\,d\alpha=(\mathcal{R})\int_{a}^{b}f(x)\,d\alpha=(\mathcal{Q})\int_{a}^{b}f(x)\,d\alpha.$

∎

Given Proposition 3.4, we may drop the indication of the partition and simply
write

$(S)\int_{a}^{b}f(x)\,d\alpha$

for the RDS integral of a step function. In fact, for the rest of this
section, for brevity we will drop the "$(S)$" as well. We will use this
notation again in Section 4.

The canonical example of a step function is the Heaviside function; we will
use this function repeatedly throughout this paper.

######  Example 3.5.

Given $c\in\mathbb{R}$, let

$H_{c}(x)=\begin{cases}0,&x<0,\\\ c,&x=0,\\\ 1,&x>0.\end{cases}$

Then, for every $a,b\in\mathbb{R}$,

$\displaystyle(S)\int_{-1}^{1}H_{a}(x)\,dH_{b}=a.$

However,

$(DS)\displaystyle\int_{-1}^{1}H_{a}(x)\,dH_{b}=1-b.$

######  Remark 3.6.

In [5], the right continuous function $H_{1}$ is referred to as "the"
Heaviside function, and the left continuous function $H_{0}$ is referred to as
the "Jeaviside" function.

###### Proof.

We use the partition $\mathcal{P}=\\{-1,0,1\\}$. Note $H_{c}$ is a step
function with respect to this partition. Now,

$\displaystyle\displaystyle\int_{-1}^{1}H_{a}(x)\,dH_{b}$
$\displaystyle=H_{a}(-1)\mu_{H_{b}}(\\{-1\\})+H_{a}(0)\mu_{H_{b}}(\\{0\\})+H_{a}(1)\mu_{H_{b}}(\\{1\\})$
$\displaystyle\quad+(0)\mu_{H_{b}}((-1,0))+(1)\mu_{H_{b}}((0,1))$
$\displaystyle=(0)(0)+(a)(1)+(1)(0)+(0)(0)+(1)(0)$ $\displaystyle=a.$

However,

$\displaystyle(DS)\displaystyle\int_{-1}^{1}H_{a}(x)\,dH_{b}$
$\displaystyle=(0)H_{b}((-1,0))+(1)H_{b}((0,1))$
$\displaystyle=(0)(b-0)+(1)(1-b)$ $\displaystyle=1-b.$

∎

In Example 3.5, $a$ is the desired value: since $H_{b}$ is flat except at zero
where it jumps, the integral should only "see" $H_{a}$ at zero and be equal to
its value there. Moreover, this agrees with the Lebesgue-Stieltjes integral,
since this is what is gotten by integrating the function $H_{a}$ against a
point mass at the origin (modeled by integrating with respect to $H_{b})$. The
DS-integral, however, only correctly models integration against a point mass
when the integrand is continuous: see [5, Lemma 5.6].

Another application of Definition 3.3 we will continually use is the integral
of a constant.

######  Lemma 3.7.

Let $f:[a,b]\to\mathbb{R}$ be the constant function $f(x)=c$ for every
$x\in[a,b]$. Then

$\int_{a}^{b}f(x)\,d\alpha=c\mu_{\alpha}([a,b]).$

###### Proof.

Note that $f$ is a step function with respect to the trivial partition
$\\{a,b\\}$. Thus,

$\displaystyle\int_{a}^{b}f(x)\,d\alpha$
$\displaystyle=f(a)\mu_{\alpha}(\\{a\\})+f(b)\mu_{\alpha}(b)+c\mu_{\alpha}((a,b))$
$\displaystyle=c(\alpha(a+)-\alpha(a-)+\alpha(b+)-\alpha(b-)+\alpha(b-)-\alpha(a+))$
$\displaystyle=c(\alpha(b+)-\alpha(a-))$ $\displaystyle=c\mu_{\alpha}([a,b]).$

∎

We now show that the usual properties of the Stieltjes integral hold:
linearity, additivity, monotonicity, and linearity of integrator. The proofs
of these results are very similar to the proofs for the DS-integral, but with
some differences. To illustrate this we give the proofs of the first two
results, and refer to [5] for the others.

######  Theorem 3.8.

Given $f,g\in S[a,b]$, for any $r_{1},r_{2}\in\mathbb{R}$,

$\int_{a}^{b}[r_{1}f(x)+r_{2}g(x)]\,d\alpha=r_{1}\int_{a}^{b}f(x)\,d\alpha+r_{2}\int_{a}^{b}g(x)\,d\alpha.$

###### Proof.

We argue as in [5, Theorem 1.18, Theorem 4.7]. Given step functions $f,g$, by
passing to a common refinement, we may assume both are defined with respect to
the partition $\mathcal{R}=\\{x_{i}\\}_{i=0}^{n}$ where
$f(x)=c_{i},\;g(x)=d_{i}$ for $x\in I_{i}$. Then,

$\displaystyle\int_{a}^{b}[r_{1}f(x)+r_{2}g(x)]\,d\alpha$
$\displaystyle=\sum_{i=0}^{n}[r_{1}f(x_{i})+r_{2}g(x_{i})]\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}[r_{1}c_{i}+r_{2}d_{i}]\mu_{\alpha}(I_{i})$
$\displaystyle=r_{1}\sum_{i=0}^{n}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})+r_{2}\sum_{i=0}^{n}g(x_{i})\mu_{\alpha}(\\{x_{i}\\})$
$\displaystyle\quad+r_{1}\sum_{i=1}^{n}c_{i}\mu_{\alpha}(I_{i})+r_{2}\sum_{i=1}^{n}d_{i}\mu_{\alpha}(I_{i})$
$\displaystyle=r_{1}\int_{a}^{b}f(x)\,d\alpha+r_{2}\int_{a}^{b}g(x)\,d\alpha.$

∎

######  Theorem 3.9.

Given $f\in S[a,b]$, for any $m\in(a,b)$,

$\int_{a}^{m}f(x)\,d\alpha+\int_{m}^{b}f(x)\,d\alpha=\int_{a}^{b}f(x)\,d\alpha.$

The corresponding result for the DS-integral is proved in [5, Theorem 1.18,
Theorem 4.7] but there are some differences in the proof for the RDS-integral.

###### Proof.

By passing to a refinement, suppose $f$ is a step function on $[a,b]$ with
respect to a partition $\mathcal{P}=\\{x_{i}\\}_{i=0}^{n}$ which contains $m$,
so there exists $i_{0}$ where $x_{i_{0}}=m$. Then, $f$ is a step function on
$[a,m]$ with respect to the partition $\\{x_{i}\\}_{i=0}^{i_{0}}$ and on
$[m,b]$ with respect to the partition $\\{x_{i}\\}_{i=i_{0}}^{n}$. Then,

$\displaystyle\int_{a}^{b}f(x)\,d\alpha$
$\displaystyle=\sum_{i=0}^{n}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}c_{i}\mu_{\alpha}(I_{i})$
$\displaystyle=\sum_{i=0}^{i_{0}-1}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})+f(m)[\alpha(m+)-\alpha(m-)]+\sum_{i=i_{0}+1}^{n}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})$
$\displaystyle\quad+\sum_{i=1}^{i_{0}}c_{i}\mu_{\alpha}(I_{i})+\sum_{i=i_{0}+1}^{n}c_{i}\mu_{\alpha}(I_{i})$
$\displaystyle=\sum_{i=0}^{i_{0}-1}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})+f(m)[\alpha(m)-\alpha(m-)]+\sum_{i=1}^{i_{0}}c_{i}\mu_{\alpha}(I_{i})$
$\displaystyle\quad+f(m)[\alpha(m+)-\alpha(m)]+\sum_{i=i_{0}+1}^{n}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=i_{0}+1}^{n}c_{i}\mu_{\alpha}(I_{i}).$

Note that on $[a,m]$, $\alpha(m+)=\alpha(m)$ and similarly on $[m,b]$,
$\alpha(m-)=\alpha(m)$. Hence,

$\int_{a}^{b}f(x)\,d\alpha=\int_{a}^{m}f(x)\,d\alpha+\int_{m}^{b}f(x)\,d\alpha.$

∎

######  Theorem 3.10.

Given $\alpha\in I[a,b]$ and $f,g\in S[a,b]$, if for every $x\in[a,b]$,
$f(x)\geq g(x)$, then

$\int_{a}^{b}f(x)\,d\alpha\geq\int_{a}^{b}g(x)\,d\alpha.$

###### Proof.

The proof is the same as for the DS-integral: see[5, Theorem 1.18, Theorem
4.7]. ∎

######  Theorem 3.11.

Given $\alpha,\beta\in BV[a,b]$, let $r_{1},r_{2}\in\mathbb{R}$ and define
$\gamma=r_{1}\alpha+r_{2}\beta\in BV[a,b]$. Then, for every $f\in S[a,b]$,

$\int_{a}^{b}f(x)\,d\gamma=r_{1}\int_{a}^{b}f(x)\,d\alpha+r_{2}\int_{a}^{b}f(x)\,d\beta.$

###### Proof.

The proof is the same as for the DS-integral: see[5, Theorem 4.10]. ∎

##  4\. The Ross-Darboux-Stieltjes Integral

In this section we define the RDS-integral for an arbitrary bounded function
$f$ and prove some basic properties of the integral. Additional properties,
proved using the so-called RDS criterion, Theorem 4.5, will be given in
Section 5. We follow the approach in [5] as outlined in Section 2 and
approximate the integral of $f$ by the integral of step functions that bracket
it. For the next definition recall $S_{U}$ and $S_{L}$, given in Definition
2.7.

######  Definition 4.1.

Given $\alpha\in I[a,b]$ and $f\in B[a,b]$, define

$\displaystyle
L_{\alpha}(f,[a,b])=\sup\left\\{(S)\int_{a}^{b}v(x)\,d\alpha\;:\;v\in
S_{L}(f,[a,b])\right\\},$ $\displaystyle
U_{\alpha}(f,[a,b])=\inf\left\\{(S)\int_{a}^{b}u(x)\,d\alpha\;:\;u\in
S_{U}(f,[a,b])\right\\}.$

If $L_{\alpha}(f,[a,b])=U_{\alpha}(f,[a,b])$, then we say $f$ is Ross-Darboux-
Stieltjes integrable with respect to $\alpha$ and we denote their common value
by

$(RDS)\int_{a}^{b}f(x)\,d\alpha.$

Moreover, for $\alpha\in BV[a,b]$, if there exists a decomposition
$\alpha=\alpha^{+}-\alpha^{-}$ with $\alpha^{\pm}\in I[a,b]$ where $f\in
RDS_{\alpha^{+}}[a,b]$ and $f\in RDS_{\alpha^{-}}[a,b]$, then we define $f\in
RDS_{\alpha}[a,b]$ and

$(RDS)\int_{a}^{b}f(x)\,d\alpha=(RDS)\int_{a}^{b}f(x)\,d\alpha^{+}-(RDS)\int_{a}^{b}f(x)\,d\alpha^{-}.$

We denote the set of functions which are Ross-Darboux-Stieltjes integrable
with respect to $\alpha$ as $RDS_{\alpha}[a,b]$.

For brevity, unless we need to work with different definitions of the
integral, hereafter we will omit the notation "$(RDS)$" and simply write
$\int_{a}^{b}f\,d\alpha$.

To work with this definition, we need to establish three things. First, we
have to show that it is well-defined in the sense that all of the objects in
the definition exist.

######  Proposition 4.2.

Given $\alpha\in I[a,b]$ and $f\in B[a,b]$, $L_{\alpha}(f,[a,b])$ and
$U_{\alpha}(f,[a,b])$ always exist and $L_{\alpha}(f,[a,b])\leq
U_{\alpha}(f,[a,b])$.

The same questions needed to be resolved for the DS-integral as defined above.
This is done in the discussion after [5, Definition 4.12]. We include the
short proof for completeness.

###### Proof.

Fix $f\in B[a,b]$. Then, there exists $m,M$ such that $m\leq f(x)\leq M$ for
all $x\in[a,b].$ Hence, $u_{M}(x)=M$ is in $S_{U}(f,[a,b])$ and $v_{m}(x)=m$
is in $S_{L}(f,[a,b])$, so these sets are non-empty. Moreover, fix $u\in
S_{U}(f,[a,b])$ and $v\in S_{L}(f,[a,b])$. Since $v(x)\leq f(x)\leq u(x)$ for
all $x\in[a,b]$, we have by Theorem 3.10,

$(S)\int_{a}^{b}v(x)\,d\alpha\leq(S)\int_{a}^{b}u(x)\,d\alpha.$

Thus, the set of integrals of step functions in $S_{U}(f,[a,b])$ is bounded
below by $(S)\int_{a}^{b}v(x)\,d\alpha$ so the infimum $U_{\alpha}(f,[a,b])$
exists and

$(S)\int_{a}^{b}v(x)\,d\alpha\leq U_{\alpha}(f,[a,b]).$

From here, we see the set of $\alpha$-integrals of step functions in
$S_{L}(f,[a,b])$ is bounded above by $U_{\alpha}(f,[a,b])$ so the supremum
$L_{\alpha}(f,[a,b])$ exists as well and

$L_{\alpha}(f,[a,b])\leq U_{\alpha}(f,[a,b]).$

∎

Second, we must show that given a step function, the integrals defined by
Definitions 3.3 and 4.1 agree. Again a similar question is addressed after [5,
Definition 4.12] for the DS-integral.

######  Proposition 4.3.

Given $\alpha\in BV[a,b]$ and $f\in S[a,b]$, $f\in RDS_{\alpha}[a,b]$ and

$(S)\int_{a}^{b}f(x)\,d\alpha=\int_{a}^{b}f(x)\,d\alpha.$

###### Proof.

First, suppose $\alpha\in I[a,b]$. Fix $f\in S[a,b]$. Then we have that $f\in
S_{L}(f,[a,b])$ and $f\in S_{U}(f,[a,b])$. Hence, by Proposition 4.2,

$(S)\int_{a}^{b}f(x)\,d\alpha\leq L_{\alpha}(f,[a,b])\leq
U_{\alpha}(f,[a,b])\leq(S)\int_{a}^{b}f(x)\,d\alpha.$

Therefore,

$L_{\alpha}(f,[a,b])=U_{\alpha}(f,[a,b])=(S)\int_{a}^{b}f(x)\,d\alpha$

as desired.

Now, suppose $\alpha\in BV[a,b]$. Then there exists a decomposition
$\alpha=\alpha^{+}-\alpha^{-}$. By the previous case, $f\in
RDS_{\alpha^{+}}[a,b]$ and $f\in RDS_{\alpha^{-}}[a,b]$. Hence, $f\in
RDS_{\alpha}[a,b]$ and by Theorem 3.11,

$\int_{a}^{b}f(x)\,d\alpha=\int_{a}^{b}f(x)\,d\alpha^{+}-\int_{a}^{b}f(x)\,d\alpha^{-}\\\
=(S)\int_{a}^{b}f(x)\,d\alpha^{+}-(S)\int_{a}^{b}f(x)\,d\alpha^{-}=(S)\int_{a}^{b}f(x)\,d\alpha.$

∎

######  Remark 4.4.

Given Proposition 4.3, hereafter we will use either definition when computing
the integral of a step function.

Third, we need to show that given $\alpha\in BV[a,b]$ the definition of the
RDS-integral is independent of the Jordan decomposition of $\alpha$ used. This
requires more work; the final proof is in Theorem 4.12 below. To prove this
result we need some preliminary results. The first is the following
characterization of RDS-integrability. This corresponds to the Darboux
criterion for the Darboux integral and the corresponding result for the DS-
integral (see [5, Theorems 1.27, 4.17]) and we will refer to it as the Ross-
Darboux-Stieltjes criterion or RDS criterion. The short proof is nearly
identical to the proofs of these results; we include the details for
completeness.

######  Theorem 4.5.

Given $\alpha\in I[a,b]$ and $f\in B[a,b]$, $f\in RDS_{\alpha}[a,b]$ if and
only if for every $\epsilon>0$ there exists $u,\,v\in S[a,b]$ which bracket
$f$, that is, $v(x)\leq f(x)\leq u(x)$ for each $x\in[a,b]$, and

$\int_{a}^{b}u(x)-v(x)\,d\alpha<\epsilon.$

###### Proof.

Suppose $f\in RDS_{\alpha}[a,b]$. Fix $\epsilon>0$. By Definition 4.1 and the
properties of infimum and supremum, there exists $u,\,v\in S[a,b]$ such that
$v(x)\leq f(x)\leq u(x)$ and

$\int_{a}^{b}u(x)\,d\alpha-\frac{\epsilon}{2}<\int_{a}^{b}f(x)\,d\alpha<\int_{a}^{b}v(x)\,d\alpha+\frac{\epsilon}{2}.$

We rearrange and use Theorem 3.8 to get the desired inequality.

Now for the converse, suppose for each $\epsilon>0$, there exists $u,\,v\in
S[a,b]$ such that $v(x)\leq f(x)\leq u(x)$ for each $x\in[a,b]$ and

$\int_{a}^{b}u(x)-v(x)\,d\alpha<\epsilon.$

Fix such $\epsilon>0$. Then,

$\int_{a}^{b}v(x)\,d\alpha\leq L_{\alpha}(f,[a,b])\leq
U_{\alpha}(f,[a,b])\leq\int_{a}^{b}u(x)\,d\alpha.$

Hence,

$U_{\alpha}(f,[a,b])-L_{\alpha}(f,[a,b])\leq\int_{a}^{b}u(x)\,d\alpha-\int_{a}^{b}v(x)\,d\alpha<\epsilon.$

Since $\epsilon$ is arbitrary, this is true for all $\epsilon>0$ and thus,
$U_{\alpha}(f,[a,b])=L_{\alpha}(f,[a,b])$. By Definition 4.1, we have the
result. ∎

Before continuing our proof of Theorem 4.12, we pause to give two corollaries
of the RDS criterion. The first will be used below in later sections. The
proof is identical to that for the DS-integral (see [5, Corollaries 1.28,
4.18]) and so we omit the details.

######  Corollary 4.6.

Given $\alpha\in I[a,b]$, if $f\in RDS_{\alpha}[a,b]$ and $|f(x)|\leq M$ for
every $x\in[a,b]$, then for any $\epsilon>0$, there exists step functions
$u,\,v$ such that

$-M\leq v(x)\leq f(x)\leq u(x)\leq M$

and

$\int_{a}^{b}u(x)-v(x)\,d\alpha<\epsilon.$

As our second corollary we prove that our definition of the RDS-integral in
Definition 4.1 in terms of step functions is equivalent to the definition by
Ross in Definition 1.3 in terms of upper and lower sums. To do so, we
introduce the notion of best-fit step functions (see [5, Definition 1.30]).

######  Definition 4.7.

Given $f\in B[a,b]$ and a partition $\mathcal{P}$ of $[a,b]$, define the best-
fit step functions of $f$ with respect to $\mathcal{P}$, denoted
$\hat{u},\hat{v}\in S[a,b]$, as follows: on each partition interval $I_{i}$,
let

$M_{i}=\sup\\{f(x)\;:\;x\in I_{i}\\},\quad m_{i}=\inf\\{f(x)\;:\;x\in
I_{i}\\},$

and define $\hat{u}(x)=M_{i}$ and $\hat{v}(x)=m_{i}$ for $x\in I_{i}$. At each
partition point $x_{i}\in\mathcal{P}$, let
$\hat{u}(x_{i})=\hat{v}(x_{i})=f(x_{i})$.

Observe that for any $\alpha\in I[a,b]$,

$\int_{a}^{b}\hat{u}(x)\,d\alpha=U_{\alpha}^{r}(f,\mathcal{P})\>\text{ and
}\;\int_{a}^{b}\hat{v}(x)\,d\alpha=L_{\alpha}^{r}(f,\mathcal{P}).$

Moreover, if $u,\,v$ are step functions with respect to the partition
$\mathcal{P}$ that bracket $f$, then we immediately have that for all
$x\in[a,b]$,

(4.1) $v(x)\leq\hat{v}(x)\leq f(x)\leq\hat{u}(x)\leq u(x).$

Using this observation, we can pass between the RDS criterion in Theorem 7.13
and the corresponding result proved by Ross [19, Theorem 35.6] using his
definition. As an immediate consequence we have that the two definitions are
equivalent. To distinguish them, temporarily denote the set of functions
integrable by Definition 1.3 by $RDS^{*}_{\alpha}[a,b]$.

######  Corollary 4.8.

Given $\alpha\in I[a,b]$, $f\in RDS_{\alpha}[a,b]$ if and only if for any
$\epsilon>0$, there exists a partition $\mathcal{P}$ and best-fit step
function $\hat{u},\hat{v}$ of $f$ with respect to $\mathcal{P}$ such that

(4.2)
$U_{\alpha}^{r}(f,\mathcal{P})-L_{\alpha}^{r}(f,\mathcal{P})=\int_{a}^{b}\hat{u}(x)-\hat{v}(x)\,d\alpha<\epsilon.$

As a consequence, $f\in RDS_{\alpha}[a,b]$ if and only if $f\in
RDS_{\alpha}^{*}[a,b]$ and the two integrals agree.

###### Proof.

We argue as for the Darboux integral in [5, Corollary 1.32]. Suppose first
that $f\in RDS_{\alpha}[a,b]$. Fix $\epsilon>0$. By the RDS criterion, Theorem
4.5, there exist $u,\,v\in S[a,b]$ which bracket $f$ such that

$\int_{a}^{b}u(x)-v(x)\;d\alpha<\epsilon.$

We may assume that $u$ and $v$ are step functions with respect to a common
partition $\mathcal{P}$. Let $\hat{u}$ and $\hat{v}$ be the best-fit step
functions of $f$ with respect to $\mathcal{P}$. Then by (4.1), we have that
$\hat{u}(x)-\hat{v}(x)\leq u(x)-v(x)$, and so by Theorem 3.10,

$\int_{a}^{b}\hat{u}(x)-\hat{v}(x)\,d\alpha\leq\int_{a}^{b}u(x)-v(x)\,d\alpha<\epsilon.$

Therefore, by [19, Theorem 35.6], $f\in RDS^{*}_{\alpha}[a,b]$.

Conversely, if (4.2) holds for every $\epsilon>0$, then, since $\hat{u}$ and
$\hat{v}$ are step functions that bracket $f$, by Theorem 4.5 we have $f\in
RDS_{\alpha}[a,b]$.

Finally, to see that the integrals agree, note that given $\epsilon>0$ and the
best fit step functions such that (4.2) holds, then by their respective
definitions we have that

$\bigg{|}(RDS)\int_{a}^{b}f(x)\,d\alpha-(RDS^{*})\int_{a}^{b}f(x)\,d\alpha\bigg{|}\leq\int_{a}^{b}\hat{u}(x)-\hat{v}(x)\,d\alpha<\epsilon.$

Since $\epsilon>0$ is arbitrary, the two integrals are equal. ∎

######  Lemma 4.9.

Given $\alpha\in BV[a,b]$ such that $\alpha(x)=c$ is constant. Then, for every
$f\in B[a,b]$, $f\in RDS_{\alpha}[a,b]$ and

(4.3) $\int_{a}^{b}f(x)\,d\alpha=0.$

###### Proof.

We argue as for the DS-integral in [5, Theorem 4.20]. Note $\alpha\in I[a,b]$.
Fix $f\in B[a,b]$. Fix $u\in S_{U}(f,[a,b])$ defined with respect to a
partition $\mathcal{P}=\\{x_{i}\\}_{i=0}^{n}$ of $[a,b]$ For each $i$,

$\mu_{\alpha}(\\{x_{i}\\})=\alpha(x_{i}+)-\alpha(x_{i}-)=c-c=0;$

similarly, $\mu_{\alpha}(I_{i})=0$. Hence, for every $u\in S_{U}(f,[a,b])$,

$\int_{a}^{b}u(x)\,d\alpha=0$

and so $U_{\alpha}(f,[a,b])=0$. By the same argument, $L_{\alpha}(f,[a,b])=0$.
Thus, $f\in RDS_{\alpha}[a,b]$ and (4.3) holds. ∎

The proof of the following lemma is identical to the proof for the DS-integral
[5, Prop 4.21], and we omit the details.

######  Lemma 4.10.

Given $\alpha,\beta\in I[a,b]$ and $r_{1},r_{2}\geq 0$, define
$\gamma=r_{1}\alpha+r_{2}\beta\in I[a,b]$. If $f\in RDS_{\alpha}[a,b]$ and
$f\in RDS_{\beta}[a,b]$, then $f\in RDS_{\gamma}[a,b]$ and

$\int_{a}^{b}f(x)\,d\gamma=r_{1}\int_{a}^{b}f(x)\,d\alpha+r_{2}\int_{a}^{b}f(x)\,d\beta.$

######  Corollary 4.11.

Given $\alpha\in I[a,b]$ and $c\in\mathbb{R}$, if $f\in RDS_{\alpha}[a,b]$,
then $f\in RDS_{\alpha+c}[a,b]$ and

$\int_{a}^{b}f(x)\,d\alpha=\int_{a}^{b}f(x)\,d[\alpha+c].$

###### Proof.

This follows immediately from Lemma 4.9 and Lemma 4.10. ∎

We can now prove the independence of the RDS-integral from the decomposition
of the integrator $\alpha$ as the difference of increasing functions. The
proof is essentially the same as for the DS-integral [5, Lemma 4.27], but we
include the short proof for completeness.

######  Theorem 4.12.

For $\alpha\in BV[a,b]$, given two different decompositions
$\alpha=\alpha^{+}-\alpha^{-}=\beta^{+}-\beta^{-},$ if $f\in
RDS_{\alpha}[a,b]$ with respect to both decompositions, then

$\int_{a}^{b}f(x)\,d\alpha^{+}-\int_{a}^{b}f(x)\,d\alpha^{-}=\int_{a}^{b}f(x)\,d\beta^{+}-\int_{a}^{b}f(x)\,d\beta^{-}.$

###### Proof.

Let $\gamma=\alpha^{+}+\beta^{-}=\alpha^{-}+\beta^{+}$. Then $\gamma\in
I[a,b]$ and by Lemma 4.10,

$\int_{a}^{b}f(x)\,d\alpha^{+}+\int_{a}^{b}f(x)\,d\beta^{-}=\int_{a}^{b}f(x)\,d\gamma=\int_{a}^{b}f(x)\,d\alpha^{-}+\int_{a}^{b}f(x)\,d\beta^{+}.$

If we rearrange terms, we get the desired equality. ∎

By Theorem 4.12, when computing the RDS-integral we can choose any
decomposition of $\alpha$ for which it is defined. Determining such a
decomposition may seem like a problem, but we will show that if a function is
RDS-integrable with respect to any decomposition of $\alpha$, then it is
always integrable with respect to a canonical decomposition of $\alpha$ in
terms of its positive and negative variation. The proof requires a definition
and two lemmas. For the first definition, recall the definition of the
variation $V(f,[a,b])$ given in Definition 2.3.

######  Definition 4.13.

Fix $\alpha\in BV[a,b]$, and define the variation function of $\alpha$,
$V\alpha$, by

$V\alpha(x)=V(\alpha,[a,x]).$

Further, define the positive and negative variations of $\alpha$ by

$P\alpha=\frac{1}{2}(V\alpha+\alpha-\alpha(a))\qquad\text{and}\qquad
N\alpha=\frac{1}{2}(V\alpha-\alpha+\alpha(a)).$

The functions $V\alpha,\,P\alpha,\,N\alpha$ are all increasing and yield a
Jordan decomposition of $f$. For a proof, see [5, Section 3.3].

######  Lemma 4.14.

Given $\alpha\in BV[a,b]$, $V\alpha,\,P\alpha,\,N\alpha\in I[a,b]$ and

$\alpha(x)=P\alpha(x)-N\alpha(x)+\alpha(a).$

The second lemma is a kind of converse to Lemma 4.10.

######  Lemma 4.15.

Given $\alpha,\beta\in I[a,b]$ and $r_{1},r_{2}>0$, define
$\gamma=r_{1}\alpha+r_{2}\beta\in I[a,b]$. If $f\in RDS_{\gamma}[a,b]$, then
$f\in RDS_{\alpha}[a,b]$ and $f\in RDS_{\beta}[a,b].$

###### Proof.

It suffices to show $f\in RDS_{\alpha}[a,b]$ since by a symmetric argument,
$f\in RDS_{\beta}[a,b]$. Fix $\epsilon>0$. By Theorem 4.5, there exist step
functions $u,\,v$ which bracket $f$ such that

$\int_{a}^{b}u(x)-v(x)\,d\gamma<r_{1}\epsilon.$

Then, since $u-v$ is nonnegative and $r_{2}\beta$ is increasing, by Theorems
3.10 and 3.11,

$\displaystyle\int_{a}^{b}u(x)-v(x)\,d\alpha$
$\displaystyle=\frac{1}{r_{1}}\int_{a}^{b}u(x)-v(x)\,d(r_{1}\alpha)$
$\displaystyle\leq\frac{1}{r_{1}}\left(\int_{a}^{b}u(x)-v(x)\,d(r_{1}\alpha)+\int_{a}^{b}u(x)-v(x)\,d(r_{2}\beta)\right)$
$\displaystyle=\frac{1}{r_{1}}\int_{a}^{b}u(x)-v(x)\,d\gamma$
$\displaystyle<\epsilon.$

Hence, by Theorem 4.5, $f\in RDS_{\alpha}[a,b].$ ∎

######  Theorem 4.16.

Given $\alpha\in BV[a,b]$, $f\in RDS_{\alpha}[a,b]$ if and only if $f\in
RDS_{P\alpha}[a,b]$, $f\in RDS_{N\alpha}[a,b]$. Moreover,

$\int_{a}^{b}f(x)\,d\alpha=\int_{a}^{b}f(x)\,dP\alpha-\int_{a}^{b}f(x)\,dN\alpha.$

###### Proof.

We argue as in [5, Theorem 4.28] for the DS-integral. Suppose first that $f\in
RDS_{\alpha}[a,b]$. By Lemma 4.14, $P\alpha,N\alpha\in I[a,b]$. Since $f\in
RDS_{\alpha}[a,b]$, there exists a decomposition
$\alpha=\alpha^{+}-\alpha^{-}$ such that $f\in RDS_{\alpha^{+}}[a,b]$ and
$f\in RDS_{\alpha^{-}}[a,b]$. Thus, for all $x\in[a,b]$

$\alpha(x)=[\alpha^{+}(x)-\alpha^{+}(a)]-[\alpha^{-}(x)-\alpha^{-}(a)]+\alpha(a).$

By [5, Theorem 3.21], there exists $\beta\in I[a,b]$ such that
$\alpha^{+}(x)-\alpha^{+}(a)=P\alpha(x)+\beta(x)$ and
$\alpha^{-}(x)-\alpha^{-}(a)=N\alpha(x)+\beta(x)$. By Lemma 4.10, $f\in
RDS_{\alpha^{+}-\alpha^{+}(a)}[a,b]=RDS_{P\alpha+\beta}[a,b]$. Thus, by Lemma
4.15, $f\in RDS_{P\alpha}[a,b]$, $f\in RDS_{N\alpha}[a,b]$, and $f\in
RDS_{\beta}[a,b]$.

Now suppose $f\in RDS_{P\alpha}[a,b]$ and $f\in RDS_{N\alpha}[a,b]$. Since
$P\alpha+\alpha(a)\in I[a,b]$. By Lemmas 4.9 and 4.10, $f\in
RDS_{P\alpha+\alpha(a)}[a,b]$. Then, by Definition 4.1, since
$\alpha=P\alpha-N\alpha+\alpha(a)$, $f\in RDS_{\alpha}[a,b].$ Moreover,

$\displaystyle\int_{a}^{b}f(x)\,d\alpha^{+}$
$\displaystyle=\int_{a}^{b}f(x)\,d[\alpha^{+}(a)]+\int_{a}^{b}f(x)\,dP\alpha+\int_{a}^{b}f(x)\,d\beta$
$\displaystyle=\int_{a}^{b}f(x)\,dP\alpha+\int_{a}^{b}f(x)\,d\beta,$

and

$\displaystyle\int_{a}^{b}f(x)\,d\alpha^{-}$
$\displaystyle=\int_{a}^{b}f(x)\,d[\alpha^{-}(a)]+\int_{a}^{b}f(x)\,dN\alpha+\int_{a}^{b}f(x)\,d\beta$
$\displaystyle=\int_{a}^{b}f(x)\,dN\alpha+\int_{a}^{b}f(x)\,d\beta.$

Hence,

$\int_{a}^{b}f(x)\,d\alpha=\int_{a}^{b}f(x)\,d\alpha^{+}-\int_{a}^{b}f(x)\,d\alpha^{-}=\int_{a}^{b}f(x)\,dP\alpha-\int_{a}^{b}f(x)\,dN\alpha.$

∎

We conclude this section by showing that the DS-integral given in Definition
2.8 and the RDS-integral agree if the integrator or integrand are continuous.
We will give a complete characterization of the relationship between these two
definitions of the integral in Section 9. To state our first result, let
$CBV[a,b]$ be the set of functions which are both continuous and of bounded
variation on $[a,b]$.

######  Theorem 4.17.

Given $\alpha\in CBV[a,b]$, $f\in RDS_{\alpha}[a,b]$ if and only if $f\in
DS_{\alpha}[a,b]$ and in this case,

(4.4) $\int_{a}^{b}f(x)\,d\alpha=(DS)\int_{a}^{b}f(x)\,d\alpha.$

###### Proof.

First, consider $\alpha\in I[a,b]$. Suppose $f\in RDS_{\alpha}[a,b]$. Fix
$u\in S[a,b]$. Since $\alpha\in C[a,b]$,

$\int_{a}^{b}u(x)\,d\alpha=(DS)\int_{a}^{b}u(x)\,d\alpha.$

By Definition 2.8 and Definition 4.1, we have

$L^{i}_{\alpha}(f,[a,b])=L_{\alpha}(f,[a,b])=U_{\alpha}(f,[a,b])=U^{i}_{\alpha}(f,[a,b]).$

Thus, $f\in DS[a,b]$. The reverse implication is proved in the same way, and
it is immediate that (4.4) holds.

Now, consider $\alpha\in BV[a,b]$. Suppose $f\in RDS_{\alpha}[a,b]$. Then, by
Theorem 4.16, $f\in RDS_{P\alpha}[a,b]$ and $f\in RDS_{N\alpha}[a,b]$. By [5,
Definition 3.16], it is easy to see that $P\alpha,N\alpha\in CBV[a,b]$. By the
previous case, $f\in DS_{P\alpha}[a,b]$ and $f\in DS_{N\alpha}[a,b]$. By [5,
Theorem 4.28], $f\in DS_{\alpha}[a,b]$. The converse follows similarly since
all these implications are reversible. By the previous case,

$\int_{a}^{b}f(x)\,dP\alpha=(DS)\int_{a}^{b}f(x)\,dP\alpha,$
$\int_{a}^{b}f(x)\,dN\alpha=(DS)\int_{a}^{b}f(x)\,dN\alpha.$

Therefore, by Theorem 4.16 and [5, Theorem 4.28], (4.4) holds. ∎

######  Theorem 4.18.

Suppose $f\in C[a,b]$ and $\alpha\in BV[a,b]$. Then, $f\in RDS_{\alpha}[a,b]$
if and only if $f\in DS_{\alpha}[a,b]$ and in this case, (4.4) holds.

###### Proof.

By Theorem 7.1 below and [5, Theorem 4.35], if $f\in C[a,b]$, then $f\in
RDS_{\alpha}[a,b]$ and $f\in DS_{\alpha}[a,b]$. It suffices to show the
integrals are equal.

First, consider $\alpha\in I[a,b]$ . If
$\mu_{\alpha}([a,b])=\alpha(b)-\alpha(a)=0$, then $\alpha$ is constant. Hence,
by Lemma 4.9 and [5, Lemma 4.20], we have both integrals are $0$. Otherwise,
suppose $\mu_{\alpha}([a,b])=\alpha(b)-\alpha(a)>0$. Fix $\epsilon>0$. Since
$f$ is continuous on $[a,b]$, it is uniformly continuous. Thus, there exists
$\delta>0$ such that for every $x,y\in[a,b]$ with $|x-y|<\delta$,

$|f(x)-f(y)|<\frac{\epsilon}{\mu_{\alpha}([a,b])}.$

Fix $N\in\mathbb{N}$ such that $N>\frac{b-a}{\delta}.$ Moreover, let
$\mathcal{P}_{N}=\\{x_{i}\\}_{i=1}^{N}=\\{a+\frac{b-a}{N}i\\}_{i=0}^{N}$ be a
regular partition of $[a,b]$. Hence, $\overline{I_{i}}=[x_{i-1},x_{i}]$ is
compact so $f$ attains its extrema on each interval. For each $1\leq i\leq N$,
let

$M_{i}=\sup\\{f(x)\;:\;x\in\overline{I_{i}}\\}=f(x_{i}^{M})$

and

$m_{i}=\inf\\{f(x)\;:\;x\in\overline{I_{i}}\\}=f(x_{i}^{m}).$

Define step functions $u,\,v$ by

(4.5) $u(x)=\begin{cases}M_{i},&x\in I_{i},\\\
f(x_{i}),&x=x_{i},\end{cases}\quad\text{and}\quad
v(x)=\begin{cases}m_{i},&x\in I_{i},\\\ f(x_{i}),&x=x_{i}.\end{cases}$

Observe that

$\displaystyle\int_{a}^{b}u(x)\,d\alpha-(DS)\int_{a}^{b}v(x)\,d\alpha$
$\displaystyle\qquad\qquad=\sum_{i=0}^{N}f(x_{i})\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{N}M_{i}\mu_{\alpha}(I_{i})-\sum_{i=1}^{N}m_{i}\alpha(I_{i})$
$\displaystyle\qquad\qquad=\sum_{i=0}^{N-1}f(x_{i})[\alpha(x_{i}+)-\alpha(x_{i})]+\sum_{i=1}^{N}f(x_{i})[\alpha(x_{i})-\alpha(x_{i}-)]$
$\displaystyle\qquad\qquad\quad+\sum_{i=1}^{N}M_{i}\mu_{\alpha}(I_{i})-\sum_{i=1}^{N}m_{i}\alpha(I_{i})$
$\displaystyle\qquad\qquad\leq\sum_{i=0}^{N-1}M_{i+1}[\alpha(x_{i}+)-\alpha(x_{i})]+\sum_{i=1}^{N}M_{i}[\alpha(x_{i})-\alpha(x_{i}-)]$
$\displaystyle\qquad\qquad\quad+\sum_{i=1}^{N}M_{i}[\alpha(I_{i})+\alpha(x_{i}-)-\alpha(x_{i})-\alpha(x_{i-1}+)+\alpha(x_{i-1})]$
$\displaystyle\qquad\qquad\quad-\sum_{i=1}^{N}m_{i}\alpha(I_{i})$
$\displaystyle\qquad\qquad=\sum_{i=1}^{N}M_{i}\alpha(I_{i})-\sum_{i=1}^{N}m_{i}\alpha(I_{i})$
$\displaystyle\qquad\qquad<\frac{\epsilon}{\mu_{\alpha}([a,b])}\sum_{i=1}^{N}\alpha(I_{i})$
$\displaystyle\qquad\qquad=\frac{\epsilon}{\mu_{\alpha}([a,b])}(\alpha(b)-\alpha(a))$
$\displaystyle\qquad\qquad=\epsilon.$

A similar argument yields

$(DS)\int_{a}^{b}u(x)\,d\alpha-\int_{a}^{b}v(x)\,d\alpha<\epsilon.$

Hence,

$\left|(DS)\int_{a}^{b}f(x)\,d\alpha-\int_{a}^{b}f(x)\,d\alpha\right|<\epsilon.$

Since $\epsilon>0$ is arbitrary, (4.4) holds.

Now, we consider the case $\alpha\in BV[a,b]$. By the previous case,

$\int_{a}^{b}f(x)\,dP\alpha=(DS)\int_{a}^{b}f(x)\,dP\alpha,$
$\int_{a}^{b}f(x)\,dN\alpha=(DS)\int_{a}^{b}f(x)\,dN\alpha.$

Therefore, by Theorem 4.16 and [5, Theorem 4.28], (4.4) holds. ∎

##  5\. Consequences of the RDS Criterion

In this section we develop the basic properties of the RDS-integral as a
consequence of the RDS Criterion, Theorem 4.5. In particular, we show that the
RDS-integral is linear, additive, monotonic, and linear in the integrator. In
several cases the proofs are essentially identical to the proofs of the
corresponding results for the Darboux integral and DS-integral given in [5],
and so we omit the details. We will instead concentrate on those results where
some additional care is needed because of the additional term in the
definition of the RDS-integral of step functions.

######  Theorem 5.1.

(Linearity) Given $\alpha\in BV[a,b]$ and $f,g\in RDS_{\alpha}[a,b]$, for any
$r_{1},r_{2}\in\mathbb{R}$, $r_{1}f+r_{2}g\in RDS_{\alpha}[a,b]$ and

$\int_{a}^{b}[r_{1}f(x)+r_{2}g(x)]\,d\alpha=r_{1}\int_{a}^{b}f(x)\,d\alpha+r_{2}\int_{a}^{b}g(x)\,d\alpha.$

The proof of Theorem 5.1 is an immediate consequence of Lemmas 5.2 and 5.3.
The proof of the first is identical to the proof for the DS-integral in [5,
Theorem 4.19, Theorem 4.30]; the proof of the second is identical to the
proofs in [5, Theorem 1.39, Theorem 4.28, Theorem 4.30].

######  Lemma 5.2.

Given $\alpha\in BV[a,b]$ and $f,g\in RDS_{\alpha}[a,b]$, $f+g\in
RDS_{\alpha}[a,b]$ and

$\int_{a}^{b}f(x)+g(x)\,d\alpha=\int_{a}^{b}f(x)\,d\alpha+\int_{a}^{b}g(x)\,d\alpha.$

######  Lemma 5.3.

Given $\alpha\in BV[a,b]$ and $f\in RDS_{\alpha}[a,b]$, for any
$c\in\mathbb{R}$, $cf\in RDS_{\alpha}[a,b]$ and

$\int_{a}^{b}cf(x)\,d\alpha=c\int_{a}^{b}f(x)\,d\alpha.$

######  Theorem 5.4.

(Additivity) Given $\alpha\in BV[a,b]$, for any $m\in(a,b)$, $f\in
RDS_{\alpha}[a,b]$ if and only if $f\in RDS_{\alpha}[a,m]$ and $f\in
RDS_{\alpha}[m,b]$. Furthermore,

(5.1)
$\int_{a}^{m}f(x)\,d\alpha+\int_{m}^{b}f(x)\,d\alpha=\int_{a}^{b}f(x)\,d\alpha.$

###### Proof.

The proof of the first part is similar to that for the DS-integral in [5,
Theorem 1.39, Theorem 4.19, Theorem 4.30], but the second part requires more
care. First, consider $\alpha\in I[a,b]$. Fix $m\in(a,b)$ and suppose $f\in
RDS_{\alpha}[a,b]$. Fix $\epsilon>0$. Then by Theorem 4.5, there exists
$u,\,v\in S[a,b]$ which bracket $f$ such that

$\int_{a}^{b}u(x)-v(x)\,d\alpha<\epsilon.$

Thus, $u|_{[a,m]},v|_{[a,m]}\in S[a,b]$ and they bracket $f$ on $[a,m]$.
Moreover, by Theorem 3.10, we have

$\int_{m}^{b}u(x)-v(x)\,d\alpha\geq 0,$

and by Theorem 3.9,

$\int_{a}^{m}u(x)-v(x)\,d\alpha\\\
\leq\int_{a}^{m}u(x)-v(x)\,d\alpha+\int_{m}^{b}u(x)-v(x)\,d\alpha=\int_{a}^{b}u(x)-v(x)\,d\alpha<\epsilon.$

Hence, by Theorem 4.5, $f\in RDS_{\alpha}[a,m]$. By a similar arguement, $f\in
RDS_{\alpha}[m,b]$.

Now, fix $f$ such that $f\in RDS_{\alpha}[a,m]$ and $f\in RDS_{\alpha}[m,b]$.
Fix $\epsilon>0$. By Theorem 4.5, there exists $u_{1},v_{1}\in S[a,m]$ and
$u_{2},v_{2}\in[m,b]$ such that both pairs bracket $f$ and

$\int_{a}^{m}u_{1}(x)-v_{1}(x)\,d\alpha<\frac{\epsilon}{2},\quad\int_{m}^{b}u_{2}(x)-v_{2}(x)\,d\alpha<\frac{\epsilon}{2}.$

We combine $u_{1}$ and $u_{2}$ as follows: let

$u(x)=\begin{cases}u_{1}(x),&x\in[a,m),\\\ f(m),&x=m,\\\
u_{2}(x),&x\in(m,b].\end{cases}$

Similarly, we combine $v_{1}$ and $v_{2}$ as follows: let

$v(x)=\begin{cases}v_{1}(x),&x\in[a,m),\\\ f(m),&x=m,\\\
v_{2}(x),&x\in(m,b].\end{cases}$

It is easy to see $v(x)\leq f(x)\leq u(x)$ for all $x\in[a,b]$. Also, observe
that on $[a,m]$, $u(x)-v(x)\leq u_{1}(x)-v_{1}(x)$ and on $[m,b]$,
$u(x)-v(x)\leq u_{2}(x)-v_{2}(x).$ Thus, by Theorems 3.9 and 3.10,

$\int_{a}^{b}u(x)-v(x)\,d\alpha=\int_{a}^{m}u(x)-v(x)\,d\alpha+\int_{m}^{b}u(x)-v(x)\,d\alpha\\\
\leq\int_{a}^{m}u_{1}(x)-v_{1}(x)\,d\alpha+\int_{m}^{b}u_{2}(x)-v_{2}(x)\,d\alpha<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$

Hence, by Theorem 4.5, $f\in RDS_{\alpha}[a,b]$. Further, by Definition 4.1,

$\int_{a}^{b}v(x)\,d\alpha\leq\int_{a}^{b}f(x)\,d\alpha\leq\int_{a}^{b}u(x)\,d\alpha$

and

$\int_{a}^{m}v(x)\,d\alpha+\int_{m}^{b}v(x)\,d\alpha\leq\int_{a}^{m}f(x)\,d\alpha+\int_{m}^{b}f(x)\,d\alpha\leq\int_{a}^{m}u(x)\,d\alpha+\int_{m}^{b}u(x)\,d\alpha.$

Therefore,

$\left|\int_{a}^{m}f(x)\,d\alpha+\int_{m}^{b}f(x)\,d\alpha-\int_{a}^{b}f(x)\,d\alpha\right|\leq\epsilon$

for all $\epsilon>0$, and so (5.1) holds.

Now, consider $\alpha\in BV[a,b]$. Fix $m\in(a,b)$. Suppose $f\in
RDS_{\alpha}[a,b]$. Then by Theorem 4.16, $f\in RDS_{P\alpha}[a,b]$ and $f\in
RDS_{N\alpha}[a,b]$. By the previous case, $f\in RDS_{P\alpha}[a,m],f\in
RDS_{P\alpha}[m,b],f\in RDS_{N\alpha}[a,m]$, and $f\in RDS_{N\alpha}[m,b].$
Then by Theorem 4.16, $f\in RDS_{\alpha}[a,m]$ and $f\in RDS_{\alpha}[m,b]$.
The reverse implications hold as well. Moreover,

$\displaystyle\int_{a}^{m}f(x)\,d\alpha+\int_{m}^{b}f(x)\,d\alpha$
$\displaystyle=\int_{a}^{m}f(x)\,dP\alpha+\int_{m}^{b}f(x)\,dP\alpha$
$\displaystyle\quad\quad-\int_{a}^{m}f(x)\,dN\alpha-\int_{m}^{b}f(x)\,dN\alpha$
$\displaystyle=\int_{a}^{b}f(x)\,dP\alpha-\int_{a}^{b}f(x)\,dN\alpha$
$\displaystyle=\int_{a}^{b}f(x)\,d\alpha.$

∎

The proof of the following result is identical to the proof for the DS-
integral in [5, Theorem 1.39, Theorem 4.19, Theorem 4.30] and so is omitted.

######  Theorem 5.5.

(Monotonicity) Given $\alpha\in I[a,b]$ and $f,g\in RDS_{\alpha}[a,b]$, if for
every $x\in[a,b]$, $f(x)\geq g(x)$, then

$\int_{a}^{b}f(x)\,d\alpha\geq\int_{a}^{b}g(x)\,d\alpha.$

######  Theorem 5.6.

(Linearity of the Integrator) Given $\alpha,\beta\in BV[a,b]$ and
$r_{1},r_{2}\in\mathbb{R}$, define $\gamma=r_{1}\alpha+r_{2}\beta\in BV[a,b]$.
If $f\in RDS_{\alpha}[a,b]$ and $f\in RDS_{\beta}[a,b]$, then $f\in
RDS_{\gamma}[a,b],$ and

$\int_{a}^{b}f(x)\,d\gamma=r_{1}\int_{a}^{b}f(x)\,d\alpha+r_{2}\int_{a}^{b}f(x)\,d\beta.$

The proof of Theorem 5.6 is an immediate consequence of Lemmas 5.7 and 5.8.
The proofs of both lemmas are the same as that for the DS-integral [5, Theorem
4.30], and so omitted.

######  Lemma 5.7.

Given $\alpha,\beta\in BV[a,b]$, define $\gamma=\alpha+\beta\in BV[a,b]$. If
$f\in RDS_{\alpha}[a,b]$ and $f\in RDS_{\beta}[a,b]$, then $f\in
RDS_{\gamma}[a,b],$ and

$\int_{a}^{b}f(x)\,d\gamma=\int_{a}^{b}f(x)\,d\alpha+\int_{a}^{b}f(x)\,d\beta.$

######  Lemma 5.8.

Given $\alpha\in BV[a,b]$ and $c\in\mathbb{R}$, define $\gamma=c\alpha\in
BV[a,b]$. If $f\in RDS_{\alpha}[a,b]$, then $f\in RDS_{\gamma}[a,b],$ and

$\int_{a}^{b}f(x)\,d\gamma=c\int_{a}^{b}f(x)\,d\alpha.$

We conclude this section with two lemmas, the first of which uses the RDS
Criterion in its proof, and which we will use below.

######  Lemma 5.9.

Given $\alpha\in BV[a,b]$ and $f\in RDS_{\alpha}[a,b]$, if we define
$f^{+}(x)=\max(f(x),0)$, then $f^{+}\in RDS_{\alpha}[a,b].$ Similarly, if we
set $f^{-}(x)=\max(-f(x),0)$, then $f^{-}\in RDS_{\alpha}[a,b]$.

###### Proof.

We will prove this for $f^{+}$; the proof for $f^{-}$ is essentially the same.
First, consider $\alpha\in I[a,b]$. Fix $\epsilon>0$. By Theorem 4.5, there
exists step functions $u_{f},v_{f}$ on some common partition
$\mathcal{P}=\\{x_{i}\\}_{i=0}^{n}$ of $[a,b]$ which bracket $f$ such that

$\int_{a}^{b}u_{f}(x)-v_{f}(x)\,d\alpha<\epsilon.$

Let $u_{f}(x)=c_{i}$ and $v_{f}(x)=d_{i}$ for $x\in I_{i}$. Define $u,\,v$ by

$u(x)=\begin{cases}\max(c_{i},0),&x\in I_{i},\\\
f^{+}(x),&x=x_{i},\end{cases}$

and

$v(x)=\begin{cases}\max(d_{i},0),&x\in I_{i},\\\
f^{+}(x),&x=x_{i}.\end{cases}$

Then, for $x\in I_{i}$,

$v(x)=\max(d_{i},0)\leq\max(f(x),0)\leq\max(c_{i},0)=u(x)$

and for $x=x_{i}$, $v(x)=f^{+}(x)=u(x)$. Moreover, if $\max(c_{i},0)=c_{i}$,
then

$\max(c_{i},0)-\max(d_{i},0)=c_{i}-\max(d_{i},0)\leq c_{i}-d_{i}$

and if $\max(c_{i},0)=0$, then

$0\leq\max(c_{i},0)-\max(d_{i},0)=-\max(d_{i},0)\leq 0.$

Hence, since $c_{i}-d_{i}\geq 0$, we have

$\max(c_{i},0)-\max(d_{i},0)\leq\max(c_{i}-d_{i},0)=c_{i}-d_{i}.$

Now by Definition 3.3,

$\displaystyle\int_{a}^{b}u(x)-v(x)\,d\alpha$
$\displaystyle=\sum_{i=0}^{n}[u(x_{i})-v(x_{i})]\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}(\max(c_{i},0)-\max(d_{i},0))\mu_{\alpha}(I_{i})$
$\displaystyle=\sum_{i=1}^{n}(\max(c_{i},0)-\max(d_{i},0))\mu_{\alpha}(I_{i})$
$\displaystyle\leq\sum_{i=1}^{n}(c_{i}-d_{i})\mu_{\alpha}(I_{i})$
$\displaystyle\leq\sum_{i=0}^{n}[u_{f}(x_{i})-v_{f}(x_{i})]\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}(c_{i}-d_{i})\mu_{\alpha}(I_{i})$
$\displaystyle=\int_{a}^{b}u_{f}(x)-v_{f}(x)\,d\alpha$
$\displaystyle<\epsilon.$

So by Theorem 4.5, $g\in RDS_{\alpha}[a,b]$.

Now, consider $\alpha\in BV[a,b]$. Then by Theorem 4.16, $f\in
RDS_{P\alpha}[a,b]$ and $f\in RDS_{N\alpha}[a,b]$. By the previous case,
$f^{+}\in RDS_{P\alpha}[a,b]$ and $f^{+}\in RDS_{N\alpha}[a,b]$. Hence, by
Theorem 4.16, $f^{+}\in RDS_{\alpha}[a,b]$ as desired. ∎

######  Remark 5.10.

Given $f\in B[a,b]$, $f(x)=f^{+}(x)-f^{-}(x)$ and $|f(x)|=f^{+}(x)+f^{-}(x)$.

The proof of the second lemma, which relies on Lemma 5.9, is identical to that
for the DS-integral in [5, Theorem 4.34], and so is omitted.

######  Lemma 5.11.

Given $\alpha\in BV[a,b]$ and $f\in RDS_{\alpha}[a,b]$, $|f|\in
RDS_{\alpha}[a,b]$ and $f\in RDS_{V\alpha}[a,b]$, and

$\left|\int_{a}^{b}f(x)\,d\alpha\right|\leq\int_{a}^{b}|f(x)|\,dV\alpha.$

##  6\. Convergence Theorems

In this section, we begin to consider limits and the RDS-integral. Our
ultimate goal is to prove the bounded convergence theorem, but we defer that
proof to Section 8 as we will need results from Sections 6 and 7 to prove it.
Here we prove that the limit and the integral can be exchanged in two cases:
first when the integrands converge uniformly, and second when the integrators
converge in bounded variation norm.

######  Theorem 6.1.

Given $\alpha\in BV[a,b]$ and a sequence $\\{f_{k}\\}_{k=1}^{\infty}$ of
bounded functions on $[a,b]$, suppose $f_{k}\in RDS_{\alpha}[a,b]$ for each
$k\in\mathbb{N}$ and there exists a function $f$ on $[a,b]$ such that
$f_{k}\to f$ uniformly. Then, $f\in RDS_{\alpha}[a,b]$ and

(6.1)
$\lim_{k\to\infty}\int_{a}^{b}f_{k}(x)\,d\alpha=\int_{a}^{b}f(x)\,d\alpha.$

###### Proof.

First, we consider the case when $\alpha\in I[a,b]$. Suppose
$\mu_{\alpha}([a,b])=\alpha(b)-\alpha(a)=0$. Then, $\alpha$ is constant. By
Lemma 4.9, $f\in RDS_{\alpha}[a,b]$ and for each $k\in\mathbb{N},$

$\int_{a}^{b}f_{k}(x)\,d\alpha=0=\int_{a}^{b}f(x)\,d\alpha.$

Hence, (6.1) holds.

Now suppose $\mu_{\alpha}([a,b])=\alpha(b)-\alpha(a)>0.$ Fix $\epsilon>0$. By
uniform convergence, there exists $K\in\mathbb{N}$ such that for all
$x\in[a,b]$,

$|f_{K}(x)-f(x)|<\frac{\epsilon}{3(\alpha(b)-\alpha(a))}.$

Since $f_{K}\in RDS_{\alpha}[a,b]$, by Theorem 4.5, there exists $u_{K},v_{K}$
step functions with respect to a common partition $\\{x_{i}\\}_{i=0}^{n}$
which bracket $f_{K}$ such that

$\int_{a}^{b}u_{K}(x)-v_{K}(x)\,d\alpha<\frac{\epsilon}{3}.$

Let $u_{K}(x)=c_{i}$ and $v_{K}(x)=d_{i}$ for $x\in I_{i}$.

Define step functions $u,\,v$ as

$u(x)=u_{K}(x)+\frac{\epsilon}{3(\alpha(b)-\alpha(a))}$

and

$v(x)=v_{K}(x)-\frac{\epsilon}{3(\alpha(b)-\alpha(a))}$

Hence, for $x\in[a,b]$,

$f(x)<f_{K}(x)+\frac{\epsilon}{3(\alpha(b)-\alpha(a))}\leq
u_{K}(x)+\frac{\epsilon}{3(\alpha(b)-\alpha(a))}=u(x)$

and

$f(x)>f_{K}(x)-\frac{\epsilon}{3(\alpha(b)-\alpha(a))}\geq
v_{K}(x)-\frac{\epsilon}{3(\alpha(b)-\alpha(a))}=v(x).$

Moreover,

$\displaystyle\int_{a}^{b}u(x)-v(x)\,d\alpha$
$\displaystyle=\sum_{i=0}^{n}[u(x_{i})-v(x_{i})]\mu_{\alpha}(\\{x_{i}\\})$
$\displaystyle\quad\quad+\sum_{i=1}^{n}(c_{i}+\frac{\epsilon}{3(\alpha(b)-\alpha(a))}-d_{i}+\frac{\epsilon}{3(\alpha(b)-\alpha(a))})\mu_{\alpha}(I_{i})$
$\displaystyle=\sum_{i=0}^{n}[u_{K}(x_{i})-v_{K}(x_{i})]\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}(c_{i}-d_{i})\mu_{\alpha}(I_{i})$
$\displaystyle\quad\quad+\frac{2\epsilon}{3(\alpha(b)-\alpha(a))}\left[\sum_{i=0}^{n}\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}\mu_{\alpha}(I_{i})\right]$
$\displaystyle=\int_{a}^{b}u_{K}(x)-v_{K}(x)\,d\alpha+\frac{2\epsilon}{3(\alpha(b)-\alpha(a))}\int_{a}^{b}1\,d\alpha$
$\displaystyle<\frac{\epsilon}{3}+\frac{2\epsilon}{3(\alpha(b)-\alpha(a))}(\alpha(b)-\alpha(a))$
$\displaystyle=\epsilon.$

So by Theorem 4.5, $f\in RDS_{\alpha}[a,b].$

To show that (6.1) holds, fix $\epsilon>0$. Then, there exists $K$ such that
for all $k\geq K^{\prime}$,

$|f_{k}(x)-f(x)|<\frac{\epsilon}{2(\alpha(b)-\alpha(a))}.$

Hence, for $k\geq K^{\prime}$,

$\left|\int_{a}^{b}f_{k}(x)\,d\alpha-\int_{a}^{b}f(x)\,d\alpha\right|=\left|\int_{a}^{b}f_{k}(x)-f(x)\,d\alpha\right|\leq\int_{a}^{b}|f_{k}(x)-f(x)|\,d\alpha\\\
\leq\int_{a}^{b}\frac{\epsilon}{2(\alpha(b)-\alpha(a))}\,d\alpha=\frac{\epsilon}{2(\alpha(b)-\alpha(a))}(\alpha(b)-\alpha(a))=\frac{\epsilon}{2}<\epsilon.$

Now consider $\alpha\in BV[a,b]$. Then by Theorem 4.16, for each
$k\in\mathbb{N}$, $f_{k}\in RDS_{P\alpha}[a,b]$ and $f_{k}\in
RDS_{N\alpha}[a,b]$. By the previous case, $f\in RDS_{P\alpha}[a,b]$ and $f\in
RDS_{N\alpha}[a,b]$. Thus, $f\in RDS_{\alpha}[a,b]$. Moreover,

$\lim_{k\to\infty}\int_{a}^{b}f_{k}(x)\,d\alpha=\lim_{k\to\infty}\left[\int_{a}^{b}f_{k}(x)\,dP\alpha-\int_{a}^{b}f_{k}(x)\,dN\alpha\right]\\\
=\int_{a}^{b}f(x)\,dP\alpha-\int_{a}^{b}f(x)\,dN\alpha=\int_{a}^{b}f(x)\,d\alpha.$

∎

To state our result on convergence of the integrators, we first give the
definition of the bounded variation norm, following [5, Proposition 3.43];
recall the definition of the variation of a BV function in Definition 2.3.

######  Definition 6.2.

The function $\|\cdot\|_{BV}:BV[a,b]\to[0,\infty)$ defined for $\alpha\in
BV[a,b]$ by

$\|\alpha\|_{BV}=\sup_{x\in[a,b]}|\alpha(x)|+V\alpha(b).$

######  Remark 6.3.

For a proof that $BV[a,b]$ is a normed space with respect to this norm, and
for properties of this space, see [5, Section 3.6].

######  Theorem 6.4.

Given a sequence of functions $\\{\alpha_{k}\\}_{k=1}^{\infty}$, suppose for
each $k\in\mathbb{N}$, $\alpha_{k}\in BV[a,b]$ and $\alpha_{k}\to\alpha$ in BV
norm. Moreover, suppose $f\in RDS_{\alpha_{k}}[a,b]$ for all $k\in\mathbb{N}$.
Then, $f\in RDS_{\alpha}[a,b]$ and

(6.2)
$\lim_{k\to\infty}\int_{a}^{b}f(x)\,d\alpha_{k}=\int_{a}^{b}f(x)\,d\alpha.$

###### Proof.

First, we consider the case where all $\alpha_{k}\in I[a,b]$. Fix
$\epsilon>0$. Note $f\in B[a,b]$ so there exists $M>0$ such that $|f(x)|\leq
M$ for all $x\in[a,b]$. Further, there exists $K\in\mathbb{N}$ such that

$||\alpha-\alpha_{k}||_{BV}<\frac{\epsilon}{3M}.$

Since $f\in RDS_{\alpha_{k}}[a,b]$, by Corollary 4.6, there exist step
functions $u_{K},v_{K}$ such that

$-M\leq v_{K}(x)\leq f(x)\leq u_{K}(x)\leq M$

and

$\int_{a}^{b}u_{K}(x)-v_{K}(x)\,d\alpha_{k}<\frac{\epsilon}{3}.$

Suppose $u_{K},v_{K}$ are defined with respect to the common partition
$\\{x_{i}\\}_{i=0}^{n}$ where $u_{K}(x)=c_{i},v_{K}(x)=d_{i}$ for $x\in
I_{i}$. Define step functions $u,\,v$ by

$u(x)=\begin{cases}u_{K}(x),&x\in I_{i},\\\ f(x_{i}),&x=x_{i},\\\
\end{cases}\quad\text{ and }\quad v(x)=\begin{cases}v_{K}(x),&x\in I_{i},\\\
f(x_{i}),&x=x_{i}.\\\ \end{cases}$

So,

$\displaystyle\int_{a}^{b}u(x)-v(x)\,d\alpha_{k}$
$\displaystyle=\sum_{i=0}^{n}(u(x_{i})-v(x_{i}))\mu_{\alpha_{k}}(\\{x_{i}\\})+\sum_{i=1}^{n}(c_{i}-d_{i})\mu_{\alpha_{k}}(I_{i})$
$\displaystyle=\sum_{i=1}^{n}(c_{i}-d_{i})\mu_{\alpha_{k}}(I_{i})$
$\displaystyle\leq\sum_{i=0}^{n}(u_{K}(x_{i})-v_{K}(x_{i}))\mu_{\alpha_{k}}(\\{x_{i}\\})+\sum_{i=1}^{n}(c_{i}-d_{i})\mu_{\alpha_{k}}(I_{i})$
$\displaystyle=\int_{a}^{b}u_{K}(x)-v_{K}(x)\,d\alpha_{k}$
$\displaystyle<\frac{\epsilon}{3}.$

Then by Lemma 5.11,

$\displaystyle\int_{a}^{b}u(x)-v(x)\,d\alpha$
$\displaystyle=\int_{a}^{b}u(x)-v(x)\,d\alpha_{k}+\int_{a}^{b}u(x)-v(x)\,d(\alpha-\alpha_{k})$
$\displaystyle\leq\int_{a}^{b}u(x)-v(x)\,d\alpha_{k}+\left|\int_{a}^{b}u(x)-v(x)\,d(\alpha-\alpha_{k})\right|$
$\displaystyle<\frac{\epsilon}{3}+\int_{a}^{b}|u(x)-v(x)|\,dV(\alpha-\alpha_{k})$
$\displaystyle\leq\frac{\epsilon}{3}+2M\int_{a}^{b}\,dV(\alpha-\alpha_{k})$
$\displaystyle=\frac{\epsilon}{3}+2MV(\alpha-\alpha_{k})(b)$
$\displaystyle\leq\frac{\epsilon}{3}+2M\|\alpha-\alpha_{k}\|_{BV}$
$\displaystyle<\frac{\epsilon}{3}+2M\frac{\epsilon}{3M}$
$\displaystyle=\epsilon.$

Therefore, by Theorem 4.5, $f\in RDS_{\alpha}[a,b]$.

To show that (6.2) holds, fix $\epsilon>0$. Since $\alpha_{k}\to\alpha$ in BV
norm, there exists $K^{\prime}$ such that for all $k\geq K^{\prime}$,

$\|\alpha_{k}-\alpha\|_{BV}<\frac{\epsilon}{M}.$

Then, for all $k\geq K^{\prime}$,

$\left|\int_{a}^{b}f(x)\,d\alpha_{k}-\int_{a}^{b}f(x)\,d\alpha\right|=\left|\int_{a}^{b}f(x)\,d(\alpha_{k}-\alpha)\right|\leq\int_{a}^{b}|f(x)|\,dV(\alpha_{k}-\alpha)\\\
\leq M\int_{a}^{b}dV(\alpha_{k}-\alpha)=MV(\alpha_{k}-\alpha)(b)\leq
M\|\alpha_{k}-\alpha\|_{BV}<\epsilon.$

Now, consider $\alpha_{k}\in BV[a,b]$. Then by Theorem 4.16, $f\in
RDS_{P\alpha_{k}}[a,b]$ and $f\in RDS_{N\alpha_{k}}[a,b]$ for every
$k\in\mathbb{N}$. By [5, Proposition 3.47], $P\alpha_{k}\to
P\alpha,N\alpha_{k}\to N\alpha$ in $BV$ norm. Thus, $f\in RDS_{P\alpha}[a,b]$
and $f\in RDS_{N\alpha}[a,b]$ by the previous case. Therefore, $f\in
RDS_{\alpha}[a,b]$. Moreover,

$\lim_{k\to\infty}\int_{a}^{b}f(x)\,d\alpha_{k}=\lim_{k\to\infty}\left[\int_{a}^{b}f(x)\,dP\alpha_{k}-\int_{a}^{b}f(x)\,dN\alpha_{k}\right]\\\
=\int_{a}^{b}f(x)\,dP\alpha-\int_{a}^{b}f(x)\,dN\alpha=\int_{a}^{b}f(x)\,d\alpha.$

∎

Finally, we note that in Theorem 6.4, convergence in BV norm is necessary, in
the sense that it cannot be replaced with the weaker condition of uniform
convergence (which is implied by convergence in norm). This was shown for the
interior DS-integral in [5, Example 4.52]. In this example all the integrators
are continuous, so by Theorem 4.17 we conclude that the same example holds for
the RDS-integral as well.

##  7\. RDS Integrability Theorems

In this section we turn to the following problem: given $\alpha\in BV[a,b]$,
characterize the functions $f\in B[a,b]$ such that $f\in RDS_{\alpha}[a,b]$.
Our goal is to prove a version of the Lebesgue criterion for the RDS-integral,
Theorem 7.13; this will let us completely characterize the $f\in
RDS_{\alpha}[a,b]$ in terms of the discontinuities of $f$ and $\alpha$. To
accomplish this, we need to work with the so-called saltus decomposition of
functions in $BV[a,b]$: this will allow us to study the behavior of the RDS-
integral at a discontinuity of $\alpha$ by considering a Heaviside function
that corresponds to this discontinuity.

Finally, as an application of the results in this section, we prove a
generalized version of integration by parts for the RDS-integral. For the
classical definitions of the RS or DS-integrals, integration by parts can be
stated as an integrability theorem: see, for example, [2, Theorem 29.7]. While
we do not prove such a version here, it still seemed natural to include
integration by parts in this section.

We begin with a classical result that also holds for all definitions of the
Stieltjes integral.

######  Theorem 7.1.

Suppose $\alpha\in BV[a,b]$ and $f\in C[a,b]$. Then, $f\in RDS_{\alpha}[a,b]$.

###### Proof.

We will first consider the case when $\alpha\in I[a,b]$. The proof is very
similar to the proof of Theorem 4.18. Suppose
$\mu_{\alpha}([a,b])=\alpha(b)-\alpha(a)=0$. Then, $\alpha$ is constant, and
so by Lemma 4.9, $f\in RDS_{\alpha}[a,b]$.

Now suppose $\mu_{\alpha}([a,b])=\alpha(b)-\alpha(a)>0$. Fix $\epsilon>0$.
Since $f$ is continuous on the closed interval $[a,b]$, it must also be
uniformly continuous on $[a,b]$. Fix $\delta$, $N$ and the regular partition
$\mathcal{P}_{N}=\\{x_{i}\\}_{i=1}^{N}=\\{a+\frac{b-a}{N}i\\}_{i=0}^{N}$.
Define step functions $u,\,v$ as in (4.5). Then $u,\,v$ bracket $f$ on
$[a,b]$, and, since the partition intervals have lengths less than $\delta$,
if $x\in I_{i}$, then

$u(x)-v(x)=M_{i}-m_{i}=f(x_{i}^{M})-f(x_{i}^{m})=|f(x_{i}^{M})-f(x_{i}^{m})|<\frac{\epsilon}{\alpha(b)-\alpha(a)},$

and if $x=x_{i}$, $u(x)-v(x)=0$. Hence, by Definition 3.3,

$\displaystyle\int_{a}^{b}u(x)-v(x)\,d\alpha$
$\displaystyle=\sum_{i=0}^{N}[u(x_{i})-v(x_{i})]\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{N}[M_{i}-m_{i}]\mu_{\alpha}(I_{i})$
$\displaystyle<\sum_{i=1}^{N}\frac{\epsilon}{\alpha(b)-\alpha(a)}\mu_{\alpha}(I_{i})$
$\displaystyle=\frac{\epsilon}{\alpha(b)-\alpha(a)}\sum_{i=1}^{N}\mu_{\alpha}(I_{i})$
$\displaystyle\leq\frac{\epsilon}{\alpha(b)-\alpha(a)}\left[\sum_{i=0}^{N}\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{N}\mu_{\alpha}(I_{i})\right]$
$\displaystyle=\frac{\epsilon}{\alpha(b)-\alpha(a)}\int_{a}^{b}1\,d\alpha$
$\displaystyle=\epsilon.$

Therefore, by Theorem 4.5, $f\in RDS_{\alpha}[a,b]$.

Finally, if $\alpha\in BV[a,b]$, then by the previous case, $f\in
RDS_{P\alpha}[a,b]$ and $f\in RDS_{N\alpha}[a,b]$. Thus, by Theorem 4.16,
$f\in RDS_{\alpha}[a,b]$.

∎

In our next result we show that if the integrand is a function of bounded
variation, then it is integrable with respect to any integrator in $BV[a,b]$.
Already with this result, we see that the RDS-integral is much more general
than the classical definitions of the Stieltjes integral: the RS-integral does
not exist if $f$ and $\alpha$ have a common discontinuity (see [5, Theorem
5.31]), and the classical exterior DS-integral only exists in certain cases
(see [13, Theorem 4.13]). This result also holds for the interior DS-integral
(see [5, Theorem 4.38]) and was one of the motivations for using that
definition.

To prove this result we introduce the saltus decomposition which decomposes a
function of bounded variation into a continuous piece and an infinite sum of
Heaviside functions. This decomposition is developed in detail in [5, Section
3.5]; here we refine that definition slightly to meet our needs.

######  Definition 7.2.

A saltus set in $[a,b]$ is a sequence of ordered pairs $\\{(x_{i},a_{i})\\}$
such that the points $x_{i}$ are a collection of disjoint points in $[a,b]$,
and the $a_{i}\in\mathbb{R}$, $a_{i}\neq 0$, are such that

$\sum_{i=1}^{\infty}|a_{i}|<\infty.$

Associated with any saltus set is a saltus function,

$f(x)=\sum_{i=1}^{\infty}a_{i}H_{c}(x-x_{i}),$

where $c\in\mathbb{R}$. If $c=0$, $f$ is a left continuous saltus function,
and if $c=1$, it is a right continuous saltus function. A saltus function is
reduced if when $c=1$, $x_{i}\neq a$ for all $i\in\mathbb{N}$, or if when
$c=0$, $x_{i}\neq b$ for all $i\in\mathbb{N}$.

######  Proposition 7.3.

Given $f\in BV[a,b]$, there exists a saltus decomposition into functions of
bounded variation

(7.1) $f=Gf+S_{L}f+S_{R}f,$

where $Gf$ is continuous, $S_{L}f$ is a reduced, left continuous saltus
function, and $S_{R}f$ is a reduced, right continuous saltus function.

###### Proof.

Let $\\{x_{i}\\}_{i=1}^{\infty}$ be the set of discontinuities of $f$. By [5,
Theorem 3.38], we have a saltus decomposition (7.1), where $Gf$ is a
continuous function of bounded variation, and

$S_{L}f=\sum_{i=1}^{\infty}\big{(}f(x_{i}+)-f(x_{i})\big{)}H_{0}(x-x_{i})\;\text{and}\;S_{R}f=\sum_{i=1}^{\infty}\big{(}f(x_{i})-f(x_{i}-)\big{)}H_{1}(x-x_{i})$

are left and right continuous saltus functions (which are functions of bounded
variation: see [5, Proposition 3.37]). It follows from this definition (also
see [5, Remark 3.42]) that we may assume that $b$ is not in the saltus set of
$S_{L}f$ since $f(b+)-f(b)=0$ so even if $f$ is discontinuous at $b$, the
corresponding term in the definition of $S_{L}f$ is $0$. Similarly, we may
assume that $a$ is not in the saltus set of $S_{R}f$. Thus, both $S_{L}f$ and
$S_{R}f$ are reduced saltus functions. ∎

######  Remark 7.4.

The reduced saltus functions $S_{L}f$ and $S_{R}f$ satisfy
$S_{L}f(a)=S_{R}f(a)=0$.

######  Theorem 7.5.

Suppose $\alpha\in BV[a,b]$ and $f\in BV[a,b]$. Then, $f\in
RDS_{\alpha}[a,b]$.

###### Proof.

By Proposition 7.3, form the saltus decomposition of $f$. By Theorem 7.1,
$Gf\in RDS_{\alpha}[a,b]$. By [5, Lemma 3.36], there exists a sequence of step
functions $\\{S_{L}^{n}f\\}_{n=1}^{\infty}$ such that $S_{L}^{n}f\to S_{L}f$
uniformly. Since these are step functions, $S_{L}^{n}f\in RDS_{\alpha}[a,b]$
by Proposition 4.3. Thus, by Theorem 6.1, $S_{L}f\in RDS_{\alpha}[a,b]$.
Similarly, $S_{R}f\in RDS_{\alpha}[a,b]$. Therefore, by Theorem 5.1, $f\in
RDS_{\alpha}[a,b].$ ∎

We now consider the RDS-integral when the integrand $\alpha$ is a saltus
function. We will show that every bounded function is RDS-integrable when the
integrator is a saltus function that arises in the saltus decomposition of a
function of bounded variation. As a first step we give two key results for
integrators that are Heaviside functions. These results capture the fact that
such integrators act like Lebesgue-Stieltjes measures that are point masses.
This behavior highlights where our definition improves on the historic ones.
(Compare this result to the behavior of the interior DS-integral in [5, Lemma
5.6].)

######  Proposition 7.6.

Suppose $\alpha\in BV[a,b]$ is the Heaviside function
$\alpha(x)=H_{c}(x-x_{0})$ for some $x_{0}\in(a,b),c\in\mathbb{R}$ and $f\in
B[a,b]$. Then, $f\in RDS_{\alpha}[a,b]$ and

(7.2) $\int_{a}^{b}f(x)\,d\alpha=f(x_{0}).$

###### Proof.

Since $f\in B[a,b]$, there exists $M>0$ such that $|f(x)|\leq M$ for every
$x\in[a,b]$. Define step functions $u,\,v$ by

$u(x)=\begin{cases}M,&x\in[a,b]\backslash\\{x_{0}\\},\\\
f(x_{0}),&x=x_{0},\end{cases}\quad\text{and}\quad
v(x)=\begin{cases}-M,&x\in[a,b]\backslash\\{x_{0}\\},\\\
f(x_{0}),&x=x_{0}.\end{cases}$

Then $v(x)\leq f(x)\leq u(x)$ and, by Definition 3.3,

$\displaystyle\int_{a}^{b}u(x)\,d\alpha$
$\displaystyle=u(a)\mu_{\alpha}(\\{a\\})+u(x_{0})\mu_{\alpha}(\\{x_{0}\\})+u(b)\mu_{\alpha}(\\{b\\})$
$\displaystyle\quad\quad+M\cdot\mu_{\alpha}((a,x_{0}))+M\cdot\mu_{\alpha}((x_{0},b))$
$\displaystyle=M\cdot 0+f(x_{0})\cdot 1+M\cdot 0+M\cdot 0+M\cdot 0$
$\displaystyle=f(x_{0}).$

Similarly,

$\displaystyle\int_{a}^{b}v(x)\,d\alpha$
$\displaystyle=v(a)\mu_{\alpha}(\\{a\\})+v(x_{0})\mu_{\alpha}(\\{x_{0}\\})+v(b)\mu_{\alpha}(\\{b\\})$
$\displaystyle\quad\quad-M\cdot\mu_{\alpha}((a,x_{0}))-M\cdot\mu_{\alpha}((x_{0},1))$
$\displaystyle=-M\cdot 0+f(x_{0})\cdot 1-M\cdot 0-M\cdot 0-M\cdot 0$
$\displaystyle=f(x_{0}).$

Thus by Definition 4.1 and Proposition 4.2,

$f(x_{0})\leq L_{\alpha}(f,[a,b])\leq U_{\alpha}(f,[a,b])\leq f(x_{0}),$

which proves that $f\in RDS_{\alpha}[a,b]$ and (7.2) holds. ∎

In Proposition 7.6 we assumed that the discontinuity of the Heaviside function
was in the interior of the interval $(a,b)$. We now consider when the
discontinuity lies at $x_{0}=a$ or $x_{0}=b$. If $c=1$ and $x_{0}=a$ then
$H_{c}(x-x_{0})=1$, a constant function on $[a,b]$ and so, the result would
not be true by Lemma 4.9. A similar situation happens if $c=0$ and $x_{0}=b$.
Thus, we consider only the combinations $c=0,x_{0}=a$ or $c=1,x_{0}=b$.

######  Proposition 7.7.

Suppose $\alpha\in BV[a,b]$ is the Heaviside function
$\alpha(x)=H_{c}(x-x_{0})$ with either $x_{0}=a,c=0$ or $x_{0}=b,c=1$. If
$f\in B[a,b]$, then $f\in RDS_{\alpha}[a,b]$ and

$\int_{a}^{b}f(x)\,d\alpha=f(x_{0}).$

###### Proof.

Suppose $x_{0}=a$ and $c=0$. Define $u,\,v$ as in the proof of Proposition
7.6. Then,

$\displaystyle\int_{a}^{b}u(x)\,d\alpha$
$\displaystyle=u(a)\mu_{\alpha}(\\{a\\})+u(b)\mu_{\alpha}(\\{b\\})+M\cdot\mu_{\alpha}((a,b))$
$\displaystyle=f(a)\cdot(1-c)+M\cdot 0+M\cdot 0$ $\displaystyle=f(a),$

and

$\displaystyle\int_{a}^{b}v(x)\,d\alpha$
$\displaystyle=v(a)\mu_{\alpha}(\\{a\\})+v(b)\mu_{\alpha}(\\{b\\})-M\cdot\mu_{\alpha}((a,b))$
$\displaystyle=f(a)\cdot(1-c)-M\cdot 0-M\cdot 0$ $\displaystyle=f(a).$

Thus, as in the proof of Proposition 7.6, we are done. The other case
$x_{0}=b,c=1$, follows similarly. ∎

######  Theorem 7.8.

Suppose $\alpha\in BV[a,b]$ is a reduced saltus function of the form

$\alpha(x)=\sum_{i=1}^{\infty}a_{i}H_{c}(x-x_{i}).$

If $f\in B[a,b]$, then $f\in RDS_{\alpha}[a,b]$ and

$\int_{a}^{b}f(x)\,d\alpha=\sum_{i=1}^{\infty}a_{i}f(x_{i}).$

###### Proof.

Let $\alpha_{n}$ denote the partial sums of $\alpha$: that is,

$\alpha_{n}(x)=\sum_{i=1}^{n}a_{i}H_{c}(x-x_{i}).$

By [5, Proposition 3.48], $\alpha_{n}\to\alpha$ in BV norm. Moreover, for
every $n\in\mathbb{N}$, $\alpha_{n}$ is a linear combination of Heaviside
functions. Therefore, by Propositions 7.6 and 7.7, and Theorem 5.1, $f\in
RDS_{\alpha_{n}}[a,b]$ for every $n\in\mathbb{N}$. Hence, by Theorem 6.4,
$f\in RDS_{\alpha}[a,b]$ and

$\int_{a}^{b}f(x)\,d\alpha=\lim_{n\to\infty}\int_{a}^{b}f(x)\,d\alpha_{n}\\\
=\lim_{n\to\infty}\sum_{i=1}^{n}a_{i}\int_{a}^{b}f(x)\,dH_{c}(x-x_{i})=\lim_{n\to\infty}\sum_{i=1}^{n}a_{i}f(x_{i})=\sum_{i=1}^{\infty}a_{i}f(x_{i}).$

∎

We have two immediate corollaries of Theorem 7.8. First, we have that the
Dirichlet function,

$D(x)=\begin{cases}1,&x\in\mathbb{Q},\\\ 0,&x\not\in\mathbb{Q},\end{cases}$

is always integrable against a saltus function. This stands in contrast to
both the interior and exterior DS-integrals.

######  Corollary 7.9.

Let $\alpha\in BV[a,b]$ be a reduced saltus function. Then $D\in
RDS_{\alpha}[a,b]$.

Second, we can show that the values of $\alpha$ at its discontinuities in the
interior of $[a,b]$ do not affect the integral, so we can normalize the
integrand by assuming it is left or right continuous.

######  Corollary 7.10.

Given $\alpha\in BV[a,b]$, let $\\{x_{i}\\}_{i=1}^{\infty}$ be its set of
discontinuities and suppose $\alpha$ is not discontinuous at $b$. Suppose
$\beta\in BV[a,b]$ is such that $\alpha(x)=\beta(x)$ for all $x\neq x_{i}$ and
$\beta(x_{i})=\beta(x_{i}-)$ for every $i$. (That is, $\beta$ is left
continuous.) Then, $f\in RDS_{\alpha}[a,b]$ if and only if $f\in
RDS_{\beta}[a,b]$ and in this case,

(7.3) $\int_{a}^{b}f(x)\,d\alpha=\int_{a}^{b}f(x)\,d\beta.$

The same is true if we assume $\alpha$ is not discontinuous at $a$ and we take
$\beta$ to be right continuous.

###### Proof.

We will prove this for $\beta$ left continuous; the proof if it is right
continuous is essentially the same. Let $c_{i}=\alpha(x_{i})-\alpha(x_{i}-)$.
Then by Proposition 7.3, we have that $\\{(x_{i},c_{i}):c_{i}\neq 0\\}$ forms
the saltus set for $S_{R}\alpha$. Since $S_{R}\alpha$ is reduced, we have that
$a\not\in\\{x_{i}\\}_{i=1}^{\infty}$. by assumption,
$b\not\in\\{x_{i}\\}_{i=1}^{\infty}$.

Since $\alpha(x_{i}-)=\beta(x_{i}-)=\beta(x_{i})$,

$\alpha(x)-\beta(x)=\begin{cases}0,&x\neq x_{i},\\\
c_{i},&x=x_{i}.\end{cases}$

By Definition 7.2,

$\sum_{i=1}^{\infty}|c_{i}|<\infty,$

so we can rewrite this as

$\alpha(x)-\beta(x)=\sum_{i=1}^{\infty}c_{i}H_{1}(x-x_{i})-\sum_{i=1}^{\infty}c_{i}H_{0}(x-x_{i}).$

The righthand side is the difference of two reduced saltus functions.
Therefore, by Theorems 7.8 and 5.6, for any $f\in B[a,b]$, we have $f\in
RDS_{\alpha-\beta}[a,b]$.

To complete the proof, first suppose $f\in RDS_{\alpha}[a,b]$. Then, by
Theorem 5.6, since $\beta=-(\alpha-\beta)+\alpha$, we have $f\in
RDS_{\beta}[a,b]$. Alternatively, suppose $f\in RDS_{\beta}[a,b]$. Again, by
Theorem 5.6, we have $f\in RDS_{\alpha}[a,b]$. To see that (7.3) holds, note
that by Theorems 7.8 and 5.6,

$\int_{a}^{b}f(x)\,d\alpha-\int_{a}^{b}f(x)\,d\beta=\int_{a}^{b}f(x)\,d(\alpha-\beta)=\sum_{i=1}^{\infty}c_{i}f(x_{i})-\sum_{i=1}^{\infty}c_{i}f(x_{i})=0.$

∎

We now give a complete characterization of those functions $f$ which are
integrable with respect to a given function of bounded variation $\alpha$. To
put our result into context, recall that the Lebesgue criterion (see [5,
Theorem 2.2]) states that a bounded function $f$ is Darboux integrable if and
only if the set of discontinuities of $f$ has Lebesgue measure $0$. Lebesgue
measure corresponds to the Lebesgue-Stieltjes measure induced by the
continuous integrator $\alpha(x)=x$. Our result is the exact analogue of this,
and does not require the additional hypotheses needed for either the interior
or exterior DS-integral. To state it we need a definition.

######  Definition 7.11.

Given $\alpha\in I[a,b]$, a set $E\subset[a,b]$ has $\alpha$-measure $0$ if
for every $\epsilon>0$, there exists a collection $\\{I_{i}\\}_{i=1}^{\infty}$
of open intervals such that

$E\subset\bigcup_{i=1}^{\infty}I_{i}\quad\text{ and
}\quad\sum_{i=1}^{\infty}\mu_{\alpha}(I_{i})<\epsilon.$

######  Remark 7.12.

When $\alpha$ is continuous, this definition of $\alpha$-measure $0$ agrees
with [5, Definition 5.17]. If $\alpha$ has a non-trivial saltus component,
Definition 7.11 detects point masses, while the definition in [5] does not.

######  Theorem 7.13.

(RDS-Lebesgue Criterion) Given $f\in B[a,b]$ and $\alpha\in BV[a,b]$, $f\in
RDS_{\alpha}[a,b]$ if and only if the subset of $[a,b]$ where $f$ is
discontinuous has $G(P\alpha)$-measure $0$ and $G(N\alpha)$-measure $0$ where
$G(P\alpha),G(N\alpha)$ are the continuous pieces of the saltus decomposition
of $P\alpha,N\alpha$ respectively.

######  Remark 7.14.

An equivalent version of Theorem 7.13, with a very different proof couched in
terms of his definition of a generalized RS-integral, was given by Ter Horst
[9, Theorem D].

The following lemma let us takes advantage of the saltus decomposition to
reduce to only considering the $G\alpha$-measure of the discontinuities of
$f$.

######  Lemma 7.15.

Given $\alpha\in BV[a,b]$ and $f\in B[a,b]$, $f\in RDS_{\alpha}[a,b]$ if and
only if $f\in RDS_{G\alpha}[a,b]$.

###### Proof.

Suppose first that $f\in RDS_{\alpha}[a,b]$. By Theorem 7.8, $f\in
RDS_{S_{L}\alpha}[a,b]$ and $f\in RDS_{S_{R}\alpha}[a,b]$. Since
$G\alpha=\alpha-S_{L}\alpha-S_{R}\alpha$, by Theorem 5.6, $f\in
RDS_{G\alpha}[a,b]$. Conversely, suppose $f\in RDS_{G\alpha}[a,b]$. Since
$f\in RDS_{S_{L}\alpha}[a,b]$ and $f\in RDS_{S_{R}\alpha}[a,b]$, again by
Theorem 5.6 we have that $f\in RDS_{\alpha}[a,b]$. ∎

###### Proof of Theorem 7.13.

We first suppose that $\alpha\in I[a,b]$. Let $D\subset[a,b]$ be the of
discontinuities of $f$. We will show $f\in RDS_{\alpha}[a,b]$ if and only if
$D$ has $G\alpha$-measure $0$. If $f\in RDS_{\alpha}[a,b]$, then by Lemma
7.15, $f\in RDS_{G\alpha}[a,b]$. Since $G\alpha$ is continuous, by Theorem
4.18, $f\in DS_{G\alpha}[a,b]$. We can now use the characterization of the DS-
integral in [5, Proposition 5.25], which gives that this is the case if and
only if $D$ has $G\alpha$-measure $0$. (Note by Remark 7.12, the definitions
of $G\alpha$ agree.)

Conversely, suppose $D$ has $G\alpha$-measure $0$. By [5, Proposition 5.25],
$f\in RDS_{G\alpha}[a,b]$. By Theorem 7.8, $f\in RDS_{S_{L}\alpha}[a,b]$ and
$f\in RDS_{S_{R}\alpha}[a,b]$. Hence, by Theorem 5.6, $f\in
RDS_{\alpha}[a,b]$.

Now suppose that $\alpha\in BV[a,b]$. If $f\in RDS_{\alpha}[a,b]$, then by
Theorem 4.16, $f\in RDS_{P\alpha}[a,b]$ and $f\in RDS_{N\alpha}[a,b]$. From
the previous case, $D$ has $G(P\alpha)$-measure $0$ and $G(N\alpha)$-measure
$0$. The converse follows since this argument is reversible. ∎

We conclude this section by proving a generalization of integration by parts.
As with the DS-integral (see [5, Theorem 5.3]), the classical formula for
integration by parts need not hold. Fix $c,\,d\in\mathbb{R}$ and $t\in(a,b)$;
then we have that (see Lemma 3.5)

$\int_{a}^{b}H_{c}(x-t)\,dH_{d}(x-t)+\int_{a}^{b}H_{d}(x-t)\,dH_{c}(x-t)=c+d;$

On the other hand, $H_{d}(b)H_{c}(b)-H_{d}(a)H_{c}(a)=1$. Consequently, we
need to introduce a term, $A(t)$, into the formula for integration by parts
that compensates for a common discontinuity at a point $t$. This term is the
same as the the error term introduced in [4, Theorem 6.2.2] for the Lebesgue-
Stieltjes integral.

######  Theorem 7.16.

Fix $\alpha,\beta\in BV[a,b]$. If $S$ is the set of common discontinuities of
$\alpha$ and $\beta$, then

$\int_{a}^{b}\alpha(x)\,d\beta+\int_{a}^{b}\beta(x)\,d\alpha=\alpha(b)\beta(b)-\alpha(a)\beta(a)+\sum_{t\in
S}A_{\alpha,\beta}(t),$

where

$A_{\alpha,\beta}(t)=[\alpha(t)-\tfrac{1}{2}(\alpha(t+)+\alpha(t-))]\mu_{\beta}(\\{t\\})+[\beta(t)-\tfrac{1}{2}(\beta(t+)+\beta(t-))]\mu_{\alpha}(\\{t\\}).$

######  Remark 7.17.

Arguing as in [5, Lemma 5.2], we see that $A(t)=0$ if the discontinuities of
$\alpha$ and $\beta$ at $t$ are balanced; that is,

$\alpha(t)=\frac{\alpha(t-)+\alpha(t+)}{2},\qquad\beta(t)=\frac{\beta(t-)+\beta(t+)}{2}.$

The proof of Theorem 7.16 requires three lemmas whose proofs we defer until
after the main proof.

######  Lemma 7.18.

Given $\alpha,\beta\in BV[a,b]$, with reduced saltus decompositions

$\alpha=G\alpha+S_{L}\alpha+S_{R}\alpha,\quad\beta=G\beta+S_{L}\beta+S_{R}\beta,$

we have that

$A_{\alpha,\beta}(t)=A_{S_{L}\alpha,S_{L}\beta}(t)+A_{S_{R}\alpha,S_{R}\beta}(t).$

######  Lemma 7.19.

Given $\alpha,\beta\in BV[a,b]$, if $\alpha$ is continuous and $\beta$ is a
reduced, left or right continuous saltus function, then

$\int_{a}^{b}\alpha(x)\,d\beta+\int_{a}^{b}\beta(x)\,d\alpha=\alpha(b)\beta(b)-\alpha(a)\beta(a).$

######  Lemma 7.20.

Given $\alpha,\beta\in BV[a,b]$, suppose they are both reduced, left or right
continuous saltus functions. If $S$ is the set of common discontinuities of
$\alpha$ and $\beta$, then

$\int_{a}^{b}\alpha(x)\,d\beta+\int_{a}^{b}\beta(x)\,d\alpha=\alpha(b)\beta(b)-\alpha(a)\beta(a)+\sum_{t\in
S}A_{\alpha,\beta}(t).$

###### Proof of Theorem 7.16.

We first consider three special cases: (i) both $\alpha,\beta$ are continuous;
(ii) $\alpha$ is continuous, $\beta$ is a reduced, left or right continuous
saltus function; (iii) both $\alpha,\beta$ are reduced, left or right
continuous saltus functions. In Case (i), since both $\alpha$ and $\beta$ are
continuous, by Theorems 4.17 and 4.18 the RDS and DS-integrals agree, and the
desired identity follows from [5, Proposition 5.5]. Case (ii) follows from
Lemma 7.19. Case (iii) follows from Lemma 7.20.

We now argue in general. Fix $\alpha,\,\beta\in BV[a,b]$. By Proposition 7.3,
we have the saltus decompositions

$\alpha=G\alpha+S_{L}\alpha+S_{R}\alpha,\quad\beta=G\beta+S_{L}\beta+S_{R}\beta,$

where the saltus functions are reduced. Then by Theorems 5.1 and 5.6, and by
Lemma 7.18 we can decompose the integrals and argue as follows:

$\displaystyle\int_{a}^{b}\alpha(x)\,d\beta+\int_{a}^{b}\beta(x)\,d\alpha$
$\displaystyle=\int_{a}^{b}G\alpha(x)+S_{L}\alpha(x)+S_{R}\alpha(x)\,d[G\beta+S_{L}\beta+S_{R}\beta]$
$\displaystyle\quad+\int_{a}^{b}G\beta(x)+S_{L}\beta(x)+S_{R}\beta(x)\,d[G\alpha+S_{L}\alpha+S_{R}\alpha]$
$\displaystyle=\alpha(b)\beta(b)-\alpha(a)\beta(a)+\sum_{t\in
S}A_{S_{L}\alpha,S_{L}\beta}(t)+\sum_{t\in S}A_{S_{R}\alpha,S_{R}\beta}(t)$
$\displaystyle=\alpha(b)\beta(b)-\alpha(a)\beta(a)+\sum_{t\in
S}A_{\alpha,\beta}(t),$

where to get the term $\alpha(b)\beta(b)-\alpha(a)\beta(a)$ we have used the
saltus decomposition to recombine terms gotten from Cases (i) and (ii). ∎

###### Proof of Lemma 7.18.

Observe that

$\displaystyle\mu_{\alpha}(\\{t\\})=\alpha(t+)-\alpha(t-)$
$\displaystyle=G\alpha(t)+S_{L}\alpha(t+)+S_{R}\alpha(t)$
$\displaystyle\quad-G\alpha(t)-S_{L}\alpha(t)-S_{R}\alpha(t-)$
$\displaystyle=\mu_{S_{L}\alpha}(\\{t\\})+\mu_{S_{R}\alpha}(\\{t\\}).$

Hence,

$\displaystyle A_{\alpha,\beta}(t)$
$\displaystyle=[\alpha(t)-\tfrac{1}{2}(\alpha(t+)+\alpha(t-))]\mu_{\beta}(\\{t\\})+[\beta(t)-\tfrac{1}{2}(\beta(t+)+\beta(t-))]\mu_{\alpha}(\\{t\\})$
$\displaystyle=\tfrac{1}{2}[S_{L}\alpha(t)+S_{R}\alpha(t)-S_{L}\alpha(t+)-S_{R}\alpha(t-)][\mu_{S_{L}\beta}(\\{t\\})+\mu_{S_{R}\beta}(\\{t\\})]$
$\displaystyle\quad+\tfrac{1}{2}[S_{L}\beta(t)+S_{R}\beta(t)-S_{L}\beta(t+)-S_{R}\beta(t-)][\mu_{S_{L}\alpha}(\\{t\\})+\mu_{S_{R}\alpha}(\\{t\\})]$
$\displaystyle=\tfrac{1}{2}[\mu_{S_{R}\alpha}(\\{t\\})-\mu_{S_{L}\alpha}(\\{t\\})][\mu_{S_{L}\beta}(\\{t\\})+\mu_{S_{R}\beta}(\\{t\\})]$
$\displaystyle\quad+\tfrac{1}{2}[\mu_{S_{R}\beta}(\\{t\\})-\mu_{S_{L}\beta}(\\{t\\})][\mu_{S_{L}\alpha}(\\{t\\})+\mu_{S_{R}\alpha}(\\{t\\})]$
$\displaystyle=\mu_{S_{R}\alpha}(\\{t\\})\mu_{S_{R}\beta}(\\{t\\})-\mu_{S_{L}\alpha}(\\{t\\})\mu_{S_{L}\beta}(\\{t\\})$
$\displaystyle=A_{S_{R}\alpha,S_{R}\beta}(t)+A_{S_{L}\alpha,S_{L}\beta}(t).$

∎

###### Proof of Lemma 7.19.

Let $\\{(x_{i},b_{i})\\}_{i=1}^{\infty}$ be the saltus set associated to
$\beta$; then

$\beta(x)=\sum_{i=1}^{\infty}b_{i}H_{c}(x-x_{i}),$

where $c\in\\{0,1\\}$. By Theorem 7.8,

$\int_{a}^{b}\alpha(x)\,d\beta=\sum_{i=1}^{\infty}b_{i}\alpha(x_{i}).$

Let $\beta_{n}(x)=\sum_{i=1}^{n}b_{i}H_{c}(x-x_{i})$. By [5, Lemma 3.36],
$\beta_{n}\to\beta$ uniformly. By Theorem 6.1,

$\int_{a}^{b}\beta(x)\,d\alpha=\lim_{n\to\infty}\int_{a}^{b}\beta_{n}(x)\,d\alpha=\lim_{n\to\infty}\sum_{i=1}^{n}b_{i}\int_{a}^{b}H_{c}(x-x_{i})\,d\alpha\\\
=\lim_{n\to\infty}\sum_{i=1}^{n}b_{i}(\alpha(b)-\alpha(x_{i}))=\alpha(b)\beta(b)-\sum_{i=1}^{\infty}b_{i}\alpha(x_{i}).$

By Proposition 7.3, $\beta(a)=0$. Hence,

$\int_{a}^{b}\alpha(x)\,d\beta+\int_{a}^{b}\beta(x)\,d\alpha=\alpha(b)\beta(b)=\alpha(b)\beta(b)-\alpha(a)\beta(a).$

∎

###### Proof of Lemma 7.20.

Let

$\alpha(x)=\sum_{i=1}^{\infty}a_{i}H_{c}(x-x_{i}),\quad\beta(x)=\sum_{j=1}^{\infty}b_{j}H_{d}(x-t_{j}).$

First, suppose $c=d=0$. Then by left continuity,

$A(t)=-(\alpha(t+)-\alpha(t-))(\beta(t+)-\beta(t-)).$

By Theorem 7.8, we have

$\displaystyle\int_{a}^{b}\alpha(x)\,d\beta+\int_{a}^{b}\beta(x)\,d\alpha$
$\displaystyle=\sum_{j=1}^{\infty}b_{j}\alpha(t_{j})+\sum_{i=1}^{\infty}a_{i}\beta(x_{i})$
$\displaystyle=\sum_{j=1}^{\infty}b_{j}\sum_{i\;:\;x_{i}<t_{j}}a_{i}+\sum_{i=1}^{\infty}a_{i}\sum_{j\;:\;t_{j}<x_{i}}b_{j}$
$\displaystyle=\sum_{i=1}^{\infty}\sum_{j\;:\;t_{j}>x_{i}}a_{i}b_{j}+\sum_{i=1}^{\infty}\sum_{j\;:\;t_{j}<x_{i}}a_{i}b_{j}$
$\displaystyle=\sum_{i=1}^{\infty}\sum_{j\;:\;t_{j}\neq x_{i}}a_{i}b_{j}.$

Now fix $i\in\mathbb{N}$. If $x_{i}\not\in\\{t_{j}\\}_{j=1}^{\infty}$, then

$\displaystyle\sum_{j\;:\;t_{j}\neq x_{i}}a_{i}b_{j}$
$\displaystyle=a_{i}\sum_{j=1}^{\infty}b_{j}=a_{i}\beta(b).$

Next, suppose $x_{i}\in\\{t_{j}\\}_{j=1}^{\infty}$. That is, there exists
$j_{0}$ such that $x_{i}=t_{j_{0}}$. Then,

$\displaystyle\sum_{j\;:\;t_{j}\neq x_{i}}a_{i}b_{j}$
$\displaystyle=a_{i}\sum_{j=1}^{\infty}b_{j}-a_{i}b_{j_{0}}=a_{i}\beta(b)-a_{i}b_{j_{0}}.$

Hence,

$\displaystyle\int_{a}^{b}\alpha(x)\,d\beta+\int_{a}^{b}\beta(x)\,d\alpha$
$\displaystyle=\alpha(b)\beta(b)-\sum_{t\in
S}(\alpha(t+)-\alpha(t-))(\beta(t+)-\beta(t-))$
$\displaystyle=\alpha(b)\beta(b)-\alpha(a)\beta(a)+\sum_{t\in
S}A_{\alpha,\beta}(t)$

The case $c=d=1$ follows similarly as the previous case.

Finally, suppose $c=0,d=1$. Observe that $A_{\alpha,\beta}(t)=0$. Then, by
Theorem 7.8, we have

$\displaystyle\int_{a}^{b}\alpha(x)\,d\beta+\int_{a}^{b}\beta(x)\,d\alpha$
$\displaystyle=\sum_{j=1}^{\infty}b_{j}\alpha(t_{j})+\sum_{i=1}^{\infty}a_{i}\beta(x_{i})$
$\displaystyle=\sum_{j=1}^{\infty}b_{j}\sum_{i\;:\;x_{i}<t_{j}}a_{i}+\sum_{i=1}^{\infty}a_{i}\sum_{j\;:\;t_{j}\leq
x_{i}}b_{j}$
$\displaystyle=\sum_{i=1}^{\infty}\sum_{j\;:\;t_{j}>x_{i}}a_{i}b_{j}+\sum_{i=1}^{\infty}\sum_{j\;:\;t_{j}\leq
x_{i}}a_{i}b_{j}$
$\displaystyle=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i}b_{j}$
$\displaystyle=\alpha(b)\beta(b)-\alpha(a)\beta(b).$

∎

##  8\. The Bounded Convergence Theorem

In this section, we prove the Bounded Convergence Theorem for the RDS
integral. As we noted before, this theorem is not true for the interior DS-
integral defined in [5] unless $\alpha$ is continuous; our new definition
therefore overcomes this technical problem. For the convenience of the reader
we restate it from Section 1.

######  Theorem 8.1.

Given $\alpha\in BV[a,b]$ and a uniformly bounded sequence of functions
$\\{f_{n}\\}_{n=1}^{\infty}$ on $[a,b]$, if

  * •

for all $n\in\mathbb{N}$, $f_{n}\in RDS_{\alpha}[a,b]$,

  * •

$f_{n}\to f$ pointwise as $n\to\infty$,

  * •

$f\in RDS_{\alpha}[a,b]$,

then

(8.1)
$\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,d\alpha=\int_{a}^{b}f(x)\,d\alpha.$

Our proof of Theorem 8.1 is modeled on the proof of the bounded convergence
theorem for the Riemann integral due to Luxemburg [12]. We have organized our
proof as follows. In the main part of the proof, we show that we can reduce to
the case when our sequence of functions $\\{f_{n}\\}_{n=1}^{\infty}$ is
nonnegative and decreasing, and the integrator $\alpha$ is continuous; this
special case is given in Lemma 8.2. To prove this lemma we need to approximate
this decreasing sequence by a sequence of continuous functions; this is done
in Lemma 8.3. Given this lemma, to prove Lemma 8.2 we apply Dini’s Theorem to
get uniform convergence, and finally use Theorem 6.1.

######  Lemma 8.2.

Given $\alpha\in I[a,b]$ continuous and a sequence of functions
$\\{f_{n}\\}_{n=1}^{\infty}$ on $[a,b]$, if

  * •

$\\{f_{n}\\}_{n=1}^{\infty}$ is pointwise decreasing,

  * •

$f_{n}(x)\geq 0$ for every $x\in[a,b]$ and $n\in\mathbb{N}$,

  * •

$f_{n}\to 0$ pointwise as $n\to\infty$,

then

$\lim_{n\to\infty}L_{\alpha}(f_{n},[a,b])=0.$

We defer the proof of Lemma 8.2 to below.

###### Proof of Theorem 8.1.

Suppose first that $\alpha\in I[a,b]$. We will first prove the theorem for the
special case when $f_{n}(x)\geq 0$ for all $x\in[a,b],$ $n\in\mathbb{N}$, and
$f_{n}\to 0$ pointwise. By Proposition 7.3, we can decompose

$\alpha=G\alpha+S_{L}\alpha+S_{R}\alpha$

where $G\alpha$ is continuous and $S_{L}\alpha$ and $S_{R}\alpha$ are reduced
saltus functions. By Theorem 7.8 and Lemma 7.15, we have $f_{n}\in
RDS_{G\alpha}[a,b]$, $f_{n}\in RDS_{S_{L}\alpha}[a,b]$ and $f_{n}\in
RDS_{S_{R}\alpha}[a,b]$ for every $n\in\mathbb{N}$. By Theorem 5.6,

$\displaystyle\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,d\alpha$
$\displaystyle=\lim_{n\to\infty}\left[\int_{a}^{b}f_{n}(x)\,dG\alpha+\int_{a}^{b}f_{n}(x)\,dS_{L}\alpha+\int_{a}^{b}f_{n}(x)\,dS_{R}\alpha\right],$

and so to prove (8.1) it will suffice to show that each of the integrals on
the righthand side converges to $0$ and $n\rightarrow\infty$.

We first consider $G\alpha$. For each $n\in\mathbb{N}$, define
$p_{n}=\sup_{k\geq n}f_{k}.$ Then, $0\leq f_{n}(x)\leq p_{n}(x)$ for
$x\in[a,b]$ and $\\{p_{n}\\}_{n=1}^{\infty}$ decreases pointwise. In
particular, for $x\in[a,b]$,

$\lim_{n\to\infty}p_{n}(x)=\limsup_{n\to\infty}f_{n}(x)=\lim_{n\to\infty}f_{n}(x)=0.$

Hence, by Lemma 8.2,

$0\leq\liminf_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dG\alpha\leq\limsup_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dG\alpha\\\
=\limsup_{n\to\infty}L_{G\alpha}(f_{n},[a,b])\leq\limsup_{n\to\infty}L_{G\alpha}(p_{n},[a,b])=\lim_{n\to\infty}L_{G\alpha}(p_{n},[a,b])=0.$

Therefore,

$\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dG\alpha=0.$

We next consider $S_{L}\alpha$. Let $\\{(x_{i},a_{i})\\}$ be the saltus set
for $S_{L}\alpha$. Then we can write
$S_{L}\alpha=\sum_{i=1}^{\infty}a_{i}H_{c}(x-x_{i})$ where $c=0$. By Lemma
7.8, for each $n\in\mathbb{N}$,

$\int_{a}^{b}f_{n}(x)\,dS_{L}\alpha=\sum_{i=1}^{\infty}a_{i}f_{n}(x_{i}).$

Let $M$ be the uniform bound for $\\{f_{n}\\}_{n=1}^{\infty}$. By Definition
7.2,

$\sum_{i=1}^{\infty}|a_{i}||f_{n}(x_{i})|\leq\sum_{i=1}^{\infty}M|a_{i}|<\infty.$

Thus, for $n\in\mathbb{N}$,

$\sum_{i=1}^{\infty}a_{i}f_{n}(x_{i})$

converges absolutely.

Fix $\epsilon>0$. Let $K\in\mathbb{N}$ be such that

$\sum_{i=K+1}^{\infty}M|a_{i}|<\frac{\epsilon}{2}.$

For each $1\leq i\leq K$, let $N_{i}\in\mathbb{N}$ be such that for $n\geq
N_{i}$, $f_{n}(x_{i})<\frac{\epsilon}{2|a_{i}|K}.$ If we set $N=\max_{1\leq
i\leq K}\\{N_{i}\\}$, then for $n\geq N$,

$\left|\sum_{i=1}^{\infty}a_{i}f_{n}(x_{i})\right|\leq\left|\sum_{i=1}^{K}a_{i}f_{n}(x_{i})\right|+\left|\sum_{i=K+1}^{\infty}a_{i}f_{n}(x_{i})\right|\\\
\leq\sum_{i=1}^{K}|a_{i}|\frac{\epsilon}{2|a_{i}|K}+\sum_{i=K+1}^{\infty}M|a_{i}|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$

Hence,

$\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dS_{L}\alpha=0.$

Finally, for $S_{R}\alpha$, we can apply the same argument with $c=1$ to get

$\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dS_{R}\alpha=0.$

This completes the proof of this special case.

Now we will prove the theorem for the general case of
$\\{f_{n}\\}_{n=1}^{\infty}$. Define for each $n\in\mathbb{N},$

$g_{n}(x)=|f_{n}(x)-f(x)|.$

Note that $g_{n}\in RDS_{\alpha}[a,b]$ and $\\{g_{n}\\}_{n=1}^{\infty}$ is a
uniformly bounded sequence of functions. By the previous case,

$\lim_{n\to\infty}\int_{a}^{b}|f_{n}(x)-f(x)|\,d\alpha=\lim_{n\to\infty}\int_{a}^{b}g_{n}(x)\,d\alpha=0.$

By Lemma 5.11,

$0\leq\left|\int_{a}^{b}f_{n}(x)-f(x)\,d\alpha\right|\leq\int_{a}^{b}|f_{n}(x)-f(x)|\,d\alpha,$

and so, by the squeeze theorem,

$\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)-f(x)\,d\alpha=0,$

which yields (8.1).

Finally, we prove the theorem for $\alpha\in BV[a,b]$. By Theorem 4.16,
$f_{n}\in RDS_{P\alpha}[a,b]$ for every $n\in\mathbb{N}$ and $f\in
RDS_{P\alpha}[a,b]$. Therefore, by the previous case applied to $P\alpha$ and
$N\alpha$,

$\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,d\alpha=\lim_{n\to\infty}\bigg{(}\int_{a}^{b}f_{n}(x)\,dP\alpha-\int_{a}^{b}f_{n}(x)\,dN\alpha\bigg{)}\\\
=\int_{a}^{b}f(x)\,dP\alpha-\int_{a}^{b}f(x)\,dN\alpha=\int_{a}^{b}f(x)\,d\alpha.$

∎

To prove Lemma 8.2 we need another lemma, whose proof we give below.

######  Lemma 8.3.

Given $\alpha\in I[a,b]$ continuous and $f\in B[a,b]$ nonnegative, for each
$\epsilon>0$, there exists $g\in C[a,b]$ such that $0\leq g(x)\leq f(x)$ for
every $x\in[a,b]$ and

$L_{\alpha}(f,[a,b])<\int_{a}^{b}g(x)\,d\alpha+\epsilon.$

###### Proof of Lemma 8.2.

Fix $\epsilon>0$. By Lemma 8.3, for each $n\in\mathbb{N}$, there exists a
continuous function $g_{n}$ such that $0\leq g_{n}(x)\leq f_{n}(x)$ for
$x\in[a,b]$ and

(8.2)
$L_{\alpha}(f_{n},[a,b])<\int_{a}^{b}g_{n}(x)\,d\alpha+\frac{\epsilon}{2^{n}}.$

For each $n\in\mathbb{N}$, define $h_{n}=\min\\{g_{1},\dots,g_{n}\\}$. Then,
$0\leq h_{n}\leq g_{n}\leq f_{n}$ and $h_{n}\in C[a,b]$. Thus,
$\\{h_{n}\\}_{n=1}^{\infty}$ decreases to $0$ pointwise. Therefore, by Dini’s
Theorem [2, Exercise 24.H], the sequence $\\{h_{n}\\}_{n=1}^{\infty}$
converges uniformly to $0$ on $[a,b]$. Hence, by Theorem 6.1,

$\lim_{n\to\infty}\int_{a}^{b}h_{n}(x)\,d\alpha=0.$

Fix $n\in\mathbb{N}$ and $x\in[a,b]$. Then, $h_{n}(x)=g_{j}(x)$ for some
$1\leq j\leq n$. Thus,

(8.3) $\displaystyle
h_{n}(x)+\sum_{i=1}^{n}\big{(}\max\\{g_{i}(x),\dots,g_{n}(x)\\}-g_{i}(x)\big{)}$
$\displaystyle\qquad\qquad=g_{j}(x)+\sum_{i=1}^{n}(\max\\{g_{i}(x),\dots,g_{n}(x)\\}-g_{i}(x))$
$\displaystyle\qquad\qquad\geq
g_{j}(x)+\max\\{g_{j}(x),\dots,g_{n}(x)\\}-g_{j}(x)$
$\displaystyle\qquad\qquad=\max\\{g_{j}(x),\dots,g_{n}(x)\\}$
$\displaystyle\qquad\qquad\geq g_{n}(x).\ $

Since $x$ is arbitrary, this is true for all $x\in[a,b]$.

For $1\leq i\leq n$, $\max\\{g_{i},\dots,g_{n}\\}$ is continuous, so by
Theorem 7.1, $\max\\{g_{i},\dots,g_{n}\\}\in RDS_{\alpha}[a,b]$. Moreover,
since $\max\\{g_{i},\dots,g_{n}\\}\leq\max\\{f_{i},\dots,f_{n}\\}=f_{i}$ as
$\\{f_{i}\\}_{n=1}^{\infty}$ is pointwise decreasing, for $1\leq i\leq n$ we
have that

$\displaystyle L_{\alpha}(f_{i},[a,b])$ $\displaystyle\geq
L_{\alpha}(\max\\{g_{i},\dots,g_{n}\\},[a,b])$
$\displaystyle=\int_{a}^{b}\max\\{g_{i}(x),\dots,g_{n}(x)\\}\,d\alpha$
$\displaystyle=\int_{a}^{b}\max\\{g_{i}(x),\dots,g_{n}(x)\\}-g_{i}(x)\,d\alpha+\int_{a}^{b}g_{i}(x)\,d\alpha.$

If we combine this estimate with (8.2), we get

(8.4) $\int_{a}^{b}\max\\{g_{i}(x),\dots,g_{n}(x)\\}-g_{i}(x)\,d\alpha\leq
L_{\alpha}(f_{i},[a,b])-\int_{a}^{b}g_{i}(x)\,d\alpha<\frac{\epsilon}{2^{i}}.$

Therefore, by Theorem 5.5 and by (8.3) and (8.4),

$\displaystyle 0\leq L_{\alpha}(f_{n},[a,b])$
$\displaystyle<\frac{\epsilon}{2^{n}}+\int_{a}^{b}g_{n}(x)\,d\alpha$
$\displaystyle\leq\frac{\epsilon}{2^{n}}+\int_{a}^{b}h_{n}(x)\,d\alpha+\sum_{i=1}^{n}\int_{a}^{b}\max\\{g_{i}(x),\dots,g_{n}(x)\\}-g_{i}(x)\,d\alpha$
$\displaystyle\leq\frac{\epsilon}{2^{n}}+\int_{a}^{b}h_{n}(x)\,d\alpha+\sum_{i=1}^{n}\frac{\epsilon}{2^{i}}$
$\displaystyle=\epsilon+\int_{a}^{b}h_{n}(x)\,d\alpha.$

Thus, for every $\epsilon>0$,

$0\leq\liminf_{n\to\infty}L_{\alpha}(f_{n},[a,b])\leq\limsup_{n\to\infty}L_{\alpha}(f_{n},[a,b])\leq\epsilon+\lim_{n\to\infty}\int_{a}^{b}h_{n}(x)\,d\alpha=\epsilon,$

which gives

$\limsup_{n\to\infty}L_{\alpha}(f_{n},[a,b])=\limsup_{n\to\infty}L_{\alpha}(f_{n},[a,b])=0.$

Hence,

$\lim_{n\to\infty}L_{\alpha}(f_{n},[a,b])=0.$

∎

###### Proof of Lemma 8.3.

Fix $\epsilon>0$. By Definition 4.1, there exists a step function $v(x)\leq
f(x)$ defined with respect to the partition $\\{x_{i}\\}_{i=1}^{N}$ such that
$v(x)=c_{i}$ for $x\in I_{i}$ and

$L_{\alpha}(f,[a,b])\leq\int_{a}^{b}v(x)\,d\alpha+\frac{\epsilon}{2}.$

Without loss of generality, by Corollary 4.6 we may assume that $v(x)\geq 0$
for $x\in[a,b]$. Let $M=\max_{1\leq i\leq N}\\{c_{i}\\}+1$. Since $\alpha$ is
continuous on $[a,b]$, it is uniformly continuous on $[a,b]$. Thus, there
exists a $\delta_{1}>0$ such that for every $x,y\in[a,b]$ with
$|x-y|<\delta_{1}$,

$|\alpha(x)-\alpha(y)|<\frac{\epsilon}{4MN}.$

Let $\delta=\min(\frac{1}{2}\delta_{1},\min_{1\leq i\leq
N}\\{\frac{x_{i}-x_{i-1}}{4}\\}).$

We now define the function $g$ on $[a,b]$ by redefining $f$ in each
neighborhood $(x_{i}-\delta,x_{i}+\delta)$, $1\leq i\leq n-1$, if $x_{i}$ is a
discontinuity of $f$. (If $x_{0}$ or $x_{n}$ is a discontinuity, this argument
immediately adapts to the intervals $[x_{0},x_{0}+\delta)$,
$(x_{n}-\delta,x_{n}]$.) For each $i$, if
$v(x_{i})\leq\min\\{c_{i},c_{i+1}\\}$, let $g$ be the linear function such
that $g(x_{i}-\delta)=c_{i},g(x_{i})=v(x_{i}),g(x_{i}+\delta)=c_{i+1}$. On the
other hand, if $v(x_{i})>\min\\{c_{i},c_{i+1}\\}$, let $g$ be the piecewise
linear function with
$g(x_{i}-\delta)=c_{i},g(x_{i})=\min\\{c_{i},c_{i+1}\\},g(x_{i}+\delta)=c_{i+1}$.
Since the pieces of $g$ are continuous and agree at the endpoints, $g$ is
continuous. Moreover,

$0\leq g(x)\leq v(x)\leq f(x).$

We now estimate as follows: since $\alpha$ is continuous and increasing, by
the definition of the RDS-integral of a step function,

$\displaystyle\int_{a}^{b}g(x)\,d\alpha$
$\displaystyle=\sum_{i=1}^{N}\int_{x_{i-1}}^{x_{i}}g(x)\,d\alpha$
$\displaystyle\geq\sum_{i=1}^{N}\int_{x_{i-1}+\delta}^{x_{i}-\delta}g(x)\,d\alpha$
$\displaystyle=\sum_{i=1}^{N}c_{i}\mu_{\alpha}((x_{i-1}+\delta,x_{i}-\delta))$
$\displaystyle=\sum_{i=1}^{N}c_{i}\mu_{\alpha}((x_{i-1},x_{i}))-\sum_{i=1}^{N}c_{i}[\mu_{\alpha}((x_{i-1},x_{i-1}+\delta])+\mu_{\alpha}([x_{i}-\delta,x_{i}))]$
$\displaystyle=\int_{a}^{b}v(x)\,d\alpha-\sum_{i=1}^{N}c_{i}[|\alpha(x_{i-1}+\delta)-\alpha(x_{i-1})|+|\alpha(x_{i})-\alpha(x_{i}-\delta)|]$
$\displaystyle>\int_{a}^{b}v(x)\,d\alpha-\sum_{i=1}^{N}c_{i}\frac{\epsilon}{2NM}$
$\displaystyle\geq\int_{a}^{b}v(x)\,d\alpha-\sum_{i=1}^{N}M\frac{\epsilon}{2NM}$
$\displaystyle=\int_{a}^{b}v(x)\,d\alpha-\frac{\epsilon}{2}.$

Therefore, we have

$L_{\alpha}(f,[a,b])\leq\int_{a}^{b}v(x)\,d\alpha+\frac{\epsilon}{2}<\int_{a}^{b}g(x)\,d\alpha+\epsilon.$

∎

##  9\. RDS Integral and Related Integrals

In this section we continue our discussion of alternative definitions of the
Stieltjes integral, begun in Section 1, and compare them to the RDS-integral.
As we showed above in Corollary 4.8, our definition of the RDS-integral is
equivalent to the original definition by Ross, Definition 1.3. Further, as
shown in [5], the interior DS-integral is more general than the Riemann-
Stieltjes and exterior Darboux-Stieltjes integrals, and agrees with them when
they exist. Therefore, our first step is to characterize when the RDS-integral
and the interior DS-integral agree. We began this in Theorems 4.17 and 4.18,
but here we will use Theorem 7.13 to give a complete characterization. Then we
will define Riemann and Pollard-type integrals corresponding to the RDS-
integral. The Pollard-type integral is equivalent to the RDS-integral while
the Riemann-type integral is still weaker. We will also briefly consider a
variant of the RS-integral introduced by Ross that is equivalent to the RDS-
integral. Finally, we show the RDS-integral agrees with the Lebesgue-Stieltjes
integral whenever the former is defined.

To compare the RDS-integral and the interior DS-integral, we will use the
following characterization of interior DS-integrability proved in [5, Theorem
5.19].

######  Theorem 9.1.

Given $\alpha\in I[a,b]$ and $f\in B[a,b]$, $f\in DS_{\alpha}[a,b]$ if and
only if:

  1. (1)

the subset of $[a,b]$ where $f$ is discontinuous has $G\alpha$-measure $0$,
where $G\alpha$ is the continous part of the saltus decomposition of $\alpha$;

  2. (2)

If $\\{x_{i}\\}_{i=1}^{\infty}$ is the set of discontinuities of $\alpha$,
then for each $i$,

    1. (a)

$\alpha$ is right continuous at $x_{i}$ or $f(x_{i}+)$ exists, and

    2. (b)

$\alpha$ is left continuous at $x_{i}$ or $f(x_{i}-)$ exists.

As an immediate consequence of Theorems 7.13 and 9.1, we have the following
generalization of Theorems 4.17 and 4.18. Note that the assumption that $f$
and $\alpha$ have no common discontinuities is necessary for the RS-integral
to exist and be equal to the DS-integral, and so this is a natural sufficient
condition to consider.

######  Theorem 9.2.

Given $\alpha\in BV[a,b]$ and $f\in B[a,b]$, suppose $f$ and $\alpha$ have no
common discontinuities. Then $f\in RDS_{\alpha}[a,b]$ if and only if $f\in
DS_{\alpha}[a,b]$. Moreover,

$\int_{a}^{b}f(x)\,d\alpha=(DS)\int_{a}^{b}f(x)\,d\alpha.$

###### Proof.

Since $f$ and $\alpha$ have no common discontinuities, $f(x\pm)=f(x)$ at each
discontinuity $x$ of $\alpha$, so the second condition in Theorem 9.1 is
automatically fulfilled. By Remark 7.12, since $G\alpha$ is continuous, the
definition of $G\alpha$-measure $0$ used in Theorem 7.13 agrees with that used
in Theorem 7.13. Hence, $f\in DS_{\alpha}[a,b]$ if and only if $f\in
RDS_{\alpha}[a,b]$.

We now prove that the two integrals agree. By Proposition 7.3 we have the
reduced saltus decomposition $\alpha=G\alpha+S_{L}\alpha+S_{R}\alpha$. Let
$\\{(x_{i},a_{i})\\}$ be the reduced saltus set for $S_{L}\alpha$ and
$\\{(y_{i},b_{i})\\}$ be the reduced saltus set for $S_{R}\alpha$. By Theorems
5.6, 4.17, and 7.8 we have

$\displaystyle\int_{a}^{b}f(x)\,d\alpha$
$\displaystyle=\int_{a}^{b}f(x)\,dG\alpha+\int_{a}^{b}f(x)\,dS_{R}\alpha+\int_{a}^{b}f(x)\,dS_{L}\alpha$
$\displaystyle=(DS)\int_{a}^{b}f(x)\,dG\alpha+\sum_{i=1}^{\infty}a_{i}f(x_{i})+\sum_{i=1}^{\infty}b_{i}f(y_{i})$
$\displaystyle=(DS)\int_{a}^{b}f(x)\,dG\alpha+\sum_{i=1}^{\infty}a_{i}f(x_{i}+)+\sum_{i=1}^{\infty}b_{i}f(y_{i}-);$
by the corresponding results for the DS-integral, [5, Proposition 5.9] and [5,
Theorem 4.30c],
$\displaystyle=(DS)\int_{a}^{b}f(x)\,dG\alpha+(DS)\int_{a}^{b}f(x)\,dS_{R}\alpha+(DS)\int_{a}^{b}f(x)\,dS_{L}\alpha$
$\displaystyle=(DS)\int_{a}^{b}f(x)\,d\alpha.$

∎

Arguing as we did in the proof of Theorem 9.2, it is clear from Theorems 7.13
and 9.1 that if a function $f$ is interior DS-integrable, it is RDS-
integrable. However, as we showed in Lemma 3.5, the two integrals may not
agree if they share a common discontinuity. By carefully considering the
additional continuity conditions in the second part of Theorem 9.1, we can
give a complete characterization of when the two integrals agree in terms of a
condition reminiscent of the error term introduced for integration by parts,
Theorem 7.16.

######  Theorem 9.3.

Given $\alpha\in I[a,b]$, let $G\alpha+S_{L}\alpha+S_{R}\alpha$ be its reduced
saltus decomposition, with $\\{(x_{i},a_{i})\\}$ and $\\{(y_{i},b_{i})\\}$ the
reduced saltus sets for $S_{L}\alpha$ and $S_{R}\alpha$. Then for any $f\in
B[a,b]$, $f\in DS_{\alpha}[a,b]$ and

(9.1) $\int_{a}^{b}f(x)\,d\alpha=(DS)\int_{a}^{b}f(x)\,d\alpha$

if and only if $f\in RDS_{\alpha}[a,b]$, $f(x_{i}+)$ and $f(y_{i}-)$ exist for
all $i$ and

(9.2)
$\sum_{i=1}^{\infty}a_{i}[f(x_{i})-f(x_{i}+)]+\sum_{i=1}^{\infty}b_{i}[f(y_{i})-f(y_{i}-)]=0.$

###### Proof.

First suppose $f\in DS_{\alpha}[a,b]$ (and so $f\in RDS_{\alpha}[a,b]$) and

$\int_{a}^{b}f(x)\,d\alpha=(DS)\int_{a}^{b}f(x)\,d\alpha.$

We will prove that (9.2) holds. By [5, Proposition 4.22], $f\in
DS_{S_{L}\alpha}[a,b]$ and $f\in DS_{S_{R}\alpha}[a,b]$. Then, by Theorem 9.1,
$S_{L}\alpha$ is not right continuous for any $x_{i}$ and so $f(x_{i}+)$
exists. Similarly, $S_{R}\alpha$ is not left continuous for any $y_{i}$ and so
$f(y_{i}-)$ exists. Further, if we apply Theorem 5.6 and [5, Theorem 4.30c] to
each side of (9.1), we get that

$\int_{a}^{b}f(x)\,dG\alpha+\int_{a}^{b}f(x)\,dS_{L}\alpha+\int_{a}^{b}f(x)\,dS_{R}\alpha\\\
=(DS)\int_{a}^{b}f(x)\,dG\alpha+(DS)\int_{a}^{b}f(x)\,dS_{L}\alpha+(DS)\int_{a}^{b}f(x)\,dS_{R}\alpha.$

Moreover, by Theorem 4.17, we have

$\int_{a}^{b}f(x)\,dG\alpha=(DS)\int_{a}^{b}f(x)\,dG\alpha.$

Therefore, by Theorem 7.8 and [5, Proposition 5.9],

$\displaystyle\sum_{i=1}^{\infty}a_{i}f(x_{i})+\sum_{i=1}^{\infty}b_{i}f(y_{i})$
$\displaystyle=\int_{a}^{b}f(x)\,dS_{L}\alpha+\int_{a}^{b}f(x)\,dS_{R}\alpha$
$\displaystyle=(DS)\int_{a}^{b}f(x)\,dS_{L}\alpha+(DS)\int_{a}^{b}f(x)\,dS_{R}\alpha$
$\displaystyle=\sum_{i=1}^{\infty}a_{i}f(x_{i}+)+\sum_{i=1}^{\infty}b_{i}f(y_{i}-).$

If we rearrange terms, we get (9.2).

Now suppose that $f\in RDS_{\alpha}[a,b]$, $f(x_{i}+)$ and $f(y_{i}-)$ exist,
and (9.2) holds. Then, by Theorem 9.1, $f\in DS_{S_{L}\alpha}[a,b]$ and $f\in
DS_{S_{R}\alpha}[a,b]$. Moreover, since $f\in RDS_{\alpha}[a,b]$, by Lemma
7.15, $f\in RDS_{G\alpha}[a,b]$ and so by Theorem 4.17, $f\in
DS_{G\alpha}[a,b]$. Therefore, by [5, Theorem 4.30c], $f\in DS_{\alpha}[a,b]$.
Finally, if we argue exactly as we did at the end of the proof of Theorem 9.2,
we have that (9.1) holds. ∎

The hypothesis (9.2) in Theorem 9.3 is often not easy to check. However, as a
corollary we give a practical sufficient condition that is weaker than that in
Theorem 9.2. In particular, if $\alpha$ is not discontinuous at the endpoints
of $[a,b]$, then by Corollary 7.10 we may assume that $\alpha$ is either left
or right continuous, depending on $f$.

######  Corollary 9.4.

Given $\alpha\in I[a,b]$ and $f\in B[a,b]$, suppose for every common point of
discontinuity, either:

  1. (1)

$\alpha$ is right continuous and $f$ is left continuous, or

  2. (2)

$\alpha$ is left continuous and $f$ is right continuous.

Then $f\in RDS_{\alpha}[a,b]$ if and only if $f\in DS_{\alpha}[a,b]$.
Moreover,

$\int_{a}^{b}f(x)\,d\alpha=(DS)\int_{a}^{b}f(x)\,d\alpha.$

###### Proof.

First given our hypothesis on $f$ and $\alpha$, by Theorems 7.13 and 9.1, and
arguing as we did in the proof of Theorem 9.2 we have that $f\in
DS_{\alpha}[a,b]$ if and only if $f\in RDS_{\alpha}[a,b]$.

It remains to show that integrals are equal. Let $\\{(x_{i},a_{i})\\}$ be the
reduced saltus set for $S_{L}\alpha$ and $\\{(y_{i},b_{i})\\}$ be the reduced
saltus set for $S_{R}\alpha$. We will show that (9.2) holds by showing
$f(x_{i}+)=f(x_{i})$ and $f(y_{i}-)=f(y_{i})$ for all $i$. If $x_{i}$ is a
common point of discontinuity between $f$ and $\alpha$, then $\alpha$ is not
right continuous at $x_{i}$ as it is in the saltus set for $S_{L}\alpha$.
Thus, by assumption, $\alpha$ is left continuous and $f$ is right continuous;
that is, $f(x_{i}+)=f(x_{i})$. Otherwise, if $x_{i}$ is not a common point of
discontinuity, then $f$ must be continuous at $x_{i}$ and again
$f(x_{i}+)=f(x_{i})$. A similar argument shows that $f(y_{i}-)=f(y_{i})$ for
all $i$. ∎

We now consider Riemann-type integral corresponding to the RDS-integral. In
our definition we will follow the approach used in [5, Sections 2.2 and 5.3].
We will reinterpret the Riemann sums used to define the RS-integral (see
Definition 1.1) as the integral of a step function from a restricted class we
call Riemann step functions However, for our definition we will use the RDS-
integral of the step function, resulting in a more generalized Riemann sum.

######  Definition 9.5.

A tagged partition $\mathcal{P}^{*}$ of $[a,b]$ is a partition
$\mathcal{P}=\\{x_{i}\\}_{i=0}^{n}$ of $[a,b]$ together with a collection of
sample points $\\{x_{i}^{*}\\}_{i=0}^{n}$ where
$x_{i}^{*}\in\overline{I}_{i}.$ Given $f\in B[a,b]$ and a tagged partition
$\mathcal{P}^{*}$ of $[a,b]$, we say $r\in S[a,b]$ is a Riemann step function
of $f$ with respect to $\mathcal{P}^{*}$ if $r(x)=f(x_{i}^{*})$ for $x\in
I_{i}$ and $f(x_{i})=f(x_{i})$ for each $1\leq i\leq n$. The collection of all
Riemann step functions of $f$ defined with respect to all tagged partitions
$\mathcal{P}^{*}$ is denoted by $R^{*}(f,\mathcal{P})$.

######  Definition 9.6.

Given $\alpha\in BV[a,b]$ and $f\in B[a,b]$, we say $f$ is modified Riemann-
Stieltjes integrable on $[a,b]$ with respect to $\alpha$ if there exists
$A\in\mathbb{R}$ such that for every $\epsilon>0$ there exists $\delta>0$ so
that for any partition $\mathcal{P}$ with $|\mathcal{P}|<\delta$ and any $r\in
R^{*}(f,\mathcal{P})$,

$\left|\int_{a}^{b}r(x)\,d\alpha-A\right|<\epsilon.$

In this case we define the value of the modified Riemann-Stieltjes integral of
$f$ with respect to $\alpha$ by

$(mRS)\int_{a}^{b}f(x)\,d\alpha=A$

and the collection of all such $f$ by $mRS_{\alpha}[a,b]$.

######  Remark 9.7.

It might seem more natural to refer to this integral as the Ross-Riemann-
Stieltjes integral or the RRS-integral. However, we have reserved that
notation for Riemann-type integral introduced by Ross: see Definition 9.13
below.

However, as is the case with the RS and DS-integrals, this definition does not
agree with the definition of the RDS-integral, as the set of integrable
functions is smaller. In fact, we can use the same example as for those
integrals: see [5, Example 5.30].

######  Example 9.8.

Given $\alpha=H_{1}\in BV[-1,1]$, there exists a function $f$, namely $H_{1}$,
where $f\in RDS_{\alpha}[-1,1]$ while $f\not\in mRS_{\alpha}[-1,1]$.

###### Proof.

Since $H_{1}\in S[-1,1]$, we have by Proposition 4.3 that $H_{1}\in
RDS_{\alpha}[-1,1].$ To show that $H_{1}$ is not mRS-integrable, fix
$A\in\mathbb{R}$ and let $\epsilon=\frac{1}{2}$. Fix $\delta>0$ and let
$\mathcal{P}=\\{x_{i}\\}_{i=0}^{n}$ be a partition such that
$|\mathcal{P}|<\delta$ and $0$ is not a partition point. Then, there exists
$1\leq i_{0}\leq n$ where $0\in I_{i_{0}}$. Thus, $\mu_{\alpha}(I_{i_{0}})=1$
while $\mu_{\alpha}(I_{i})=0$ for all $i\neq i_{0}$. Let $r,s\in
R^{*}(f,\mathcal{P})$ such that $x_{i_{0}}^{*}$ lies in $(0,x_{i_{0}})$ for
$r$ and $x_{i_{0}}^{*}$ lies in $(x_{i_{0}-1},0)$ for $s$. Hence, $r(x)=1$ and
$s(x)=0$ for $x\in I_{i_{0}}.$ Therefore,

$\int_{-1}^{1}r(x)\,d\alpha=\sum_{i=0}^{n}r(x_{i})\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}r(x_{i}^{*})\mu_{\alpha}(I_{i})=r(x_{i_{0}}^{*})=1,$

and the same argument shows that

$\int_{-1}^{1}s(x)\,d\alpha=0.$

Thus,

$1=\left|\int_{-1}^{1}r(x)\,d\alpha-\int_{-1}^{1}s(x)\,d\alpha\right|\leq\left|\int_{-1}^{1}r(x)\,d\alpha-A\right|+\left|\int_{-1}^{1}s(x)\,d\alpha-A\right|.$

Hence, one of the terms on the right is at least $\epsilon=\frac{1}{2}$ and so
we have the desired step function in $R^{*}(f,\mathcal{P}).$ Therefore,
$H_{1}\not\in RS_{\alpha}[-1,1]$. ∎

######  Remark 9.9.

We leave it as an open problem to determine if the mRS-integral and RDS-
integrals agree whenever the former is defined. Moreover, there is also a
modified definition for which the same questions can be asked. Since in
Definition 9.5 the sample point can be taken as one of the endpoints of the
partition interval, we might think of this as an "exterior" modified RS-
integral. It is possible to define an "interior" modified RS-integral, by
restricting the sample points to be in the interior of the partition interval.
We leave it as on open problem to define and characterize the interior
modified RS-integral and determine its relationship to the exterior version we
define and to the RDS-integral. Similar questions exist for the RS and DS
integrals: see [5, Theorem 5.31, Exercises 5.20, 5.29].

As Example 9.8 makes clear, the problem arises because a common point of
discontinuity can be an interior point of a partition interval, no matter how
small the mesh size. As we noted in the introduction, to avoid this problem
for the RS-integral, Pollard introduced the notion of "$\sigma$-convergence"
which leads to a modified definition of the limit in the RS-integral. Here we
show that the same technique can be used to define a Ross-Pollard-Stieltjes
integral that is equivalent to the RDS-integral.

######  Definition 9.10.

Given $\alpha\in BV[a,b]$ and $f\in B[a,b]$, we say $f$ is Ross-Pollard-
Stieltjes integrable on $[a,b]$ with respect to $\alpha$ if there exists
$A\in\mathbb{R}$ such that for every $\epsilon>0$, there exists a partition
$\mathcal{P}_{\epsilon}$ so that for any refinement $\mathcal{Q}$ and any
$r\in R^{*}(f,\mathcal{Q})$,

$\left|\int_{a}^{b}r(x)\,d\alpha-A\right|<\epsilon.$

In this case we define the value of the Ross-Pollard-Stieltjes integral of $f$
with respect to $\alpha$ by

$(RPS)\int_{a}^{b}f(x)\,d\alpha=A$

and the collection of all such $f$ by $RPS_{\alpha}[a,b]$.

For brevity, we will only prove that the RPS-integral and the RDS-integral are
equivalent for $\alpha\in I[a,b]$. To prove the general case when $\alpha\in
BV[a,b]$ we would need to prove that the RPS-integral satisfied a canonical
decomposition of the integrator analogous to Theorem 4.16 for the RDS
integral. Such a result is true for the RS-integral: see [5, Lemma 5.38].

######  Theorem 9.11.

Given $\alpha\in I[a,b]$ and $f\in B[a,b]$, $f\in RDS_{\alpha}[a,b]$ if and
only if $f\in RPS_{\alpha}[a,b]$. In this case,

(9.3) $\int_{a}^{b}f(x)\,d\alpha=(RPS)\int_{a}^{b}f(x)\,d\alpha.$

###### Proof.

If $\mu_{\alpha}([a,b])=0$, then $\alpha$ is constant and so $f$ is integrable
in both definitions with common value is $0$. Indeed, by Lemma 4.9, we have
this for the RDS-integral. To see this for the RPS-integral, fix $\epsilon>0$
and let $\mathcal{P}_{\epsilon}$ be the trivial partition. Then, for any
refinement $\mathcal{Q}$ and $r\in R^{*}(f,\mathcal{Q})$,

$\left|\int_{a}^{b}r(x)\,d\alpha\right|=0<\epsilon.$

Hence, without loss of generality, $\mu_{\alpha}([a,b])>0$.

Suppose first that $f\in RDS_{\alpha}[a,b]$. Let

$A=\int_{a}^{b}f(x)\,d\alpha.$

Fix $\epsilon>0$. By Definition 4.1, there exists $u,\,v\in S[a,b]$ which
bracket $f$ and

$\int_{a}^{b}u(x)\,d\alpha<A+\epsilon,\quad\int_{a}^{b}v(x)\,d\alpha>A-\epsilon.$

Without loss of generality, suppose $u,\,v$ are step functions with respect to
a common partition $\mathcal{P}$. Let $\mathcal{P}_{\epsilon}=\mathcal{P}$.
Let $\mathcal{Q}$ be any refinement of $\mathcal{P}$ and fix any $r\in
R^{*}(f,\mathcal{Q})$. Since $u,\,v$ are also step functions with respect to
$\mathcal{Q}$ and bracket $f$ we have that $v(x)\leq r(x)\leq u(x)$ for all
$x\in[a,b]$. Thus, by Theorem 3.10,

$A-\epsilon<\int_{a}^{b}v(x)\,d\alpha\leq\int_{a}^{b}r(x)\,d\alpha\leq\int_{a}^{b}u(x)\,d\alpha<A+\epsilon.$

Since $\epsilon>0$ is arbitrary, by Definition 9.10, $f\in RPS_{\alpha}[a,b]$
and (9.3) holds.

Now suppose $f\in RPS_{\alpha}[a,b]$ and let

$A=(RPS)\int_{a}^{b}f(x)\,d\alpha.$

Fix $\epsilon>0$. By Definition 9.10, there exists a partition
$\mathcal{P}_{\epsilon}=\\{x_{i}\\}_{i=0}^{n}$ such that for any refinement
$\mathcal{Q}$ of $\mathcal{P}_{\epsilon}$ and $r\in R^{*}(f,\mathcal{Q})$,

$\left|\int_{a}^{b}r(x)\,d\alpha-A\right|<\frac{\epsilon}{4}.$

It will actually be enough to consider $\mathcal{Q}=\mathcal{P}_{\epsilon}$.
Denote the partition intervals of $\mathcal{P}_{\epsilon}$ by $I_{i}$. Let
$c_{i}=\sup_{x\in\overline{I_{i}}}f(x)$ and
$d_{i}=\inf_{x\in\overline{I_{i}}}f(x)$. Since $f$ is bounded, these are well-
defined. Note that there exist $y_{i},z_{i}\in\overline{I_{i}}$ such that

$f(y_{i})+\frac{\epsilon}{4\cdot\mu_{\alpha}([a,b])2^{i}}>c_{i},\quad
f(z_{i})-\frac{\epsilon}{4\cdot\mu_{\alpha}([a,b])2^{i}}<d_{i}.$

Define

$u(x)=\begin{cases}c_{i},&x\in I_{i},\\\ f(x),&x=x_{i},\end{cases}\quad
v(x)=\begin{cases}d_{i},&x\in I_{i},\\\ f(x),&x=x_{i}.\end{cases}$

Observe that $u,\,v$ bracket $f$. Now define

$r_{u}(x)=\begin{cases}f(y_{i}),&x\in I_{i},\\\ f(x),&x=x_{i},\end{cases}\quad
r_{v}(x)=\begin{cases}f(z_{i}),&x\in I_{i},\\\ f(x),&x=x_{i}.\end{cases}$

Notice that $r_{u},r_{v}\in R^{*}(f,\mathcal{P}_{\epsilon})$ and

$\displaystyle\int_{a}^{b}u(x)\,d\alpha$
$\displaystyle=\sum_{i=0}^{n}u(x_{i})\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}c_{i}\mu_{\alpha}(I_{i})$
$\displaystyle<\sum_{i=0}^{n}r_{u}(x_{i})\mu_{\alpha}(\\{x_{i}\\})+\sum_{i=1}^{n}f(y_{i})\mu_{\alpha}(I_{i})+\sum_{i=1}^{n}\frac{\epsilon}{4\cdot\mu_{\alpha}([a,b])2^{i}}\mu_{\alpha}(I_{i})$
$\displaystyle\leq\int_{a}^{b}r_{u}(x)\,d\alpha+\sum_{i=1}^{n}\frac{\epsilon}{4\cdot
2^{i}}$ $\displaystyle<\int_{a}^{b}r_{u}(x)\,d\alpha+\frac{\epsilon}{4}.$

Similarly,

$\int_{a}^{b}v(x)\,d\alpha>\int_{a}^{b}r_{v}(x)\,d\alpha-\frac{\epsilon}{4}.$

Therefore,

$\int_{a}^{b}u(x)-v(x)\,d\alpha<\int_{a}^{b}r_{u}(x)\,d\alpha-\int_{a}^{b}r_{v}(x)\,d\alpha+\frac{\epsilon}{2}\\\
=\left[\int_{a}^{b}r_{u}(x)\,d\alpha-A\right]+\left[A-\int_{a}^{b}r_{v}(x)\,d\alpha\right]+\frac{\epsilon}{2}<\epsilon.$

Hence, by Theorem 4.5, $f\in RDS_{\alpha}[a,b]$. By the previous argument,
(9.3) holds. ∎

Ross [19, Definition 35.24,Theorem 35.25] gave an alternative definition of a
Riemann-type integral associated to the RDS-integral. His idea was to replace
the mesh size of a partition with a mesh size defined using the
$\alpha$-lengths of the interval.

######  Definition 9.12.

Given $\alpha\in BV[a,b]$, the $\alpha$-mesh size of a partition
$\mathcal{P}=\\{x_{i}\\}_{i=0}^{n}$ is

$|\mathcal{P}|_{\alpha}=\max_{i}\mu_{\alpha}(I_{i}).$

Using this he defined what we refer to as the Ross-Riemann-Stieltjes integral,
and showed that it was equivalent to the RDS-integral. We refer the reader to
[19] for the proof.

######  Definition 9.13.

Given $\alpha\in BV[a,b]$ and $f\in B[a,b]$, we say $f$ is Ross-Riemann-
Stieltjes integrable on $[a,b]$ with respect to $\alpha$ if there exists
$A\in\mathbb{R}$ such that for every $\epsilon>0$ there exists $\delta>0$ so
that for any partition $\mathcal{P}$ with $|\mathcal{P}|_{\alpha}<\delta$ and
any $r\in R^{*}(f,\mathcal{P})$,

$\left|\int_{a}^{b}r(x)\,d\alpha-A\right|<\epsilon.$

We define the value of the Ross-Riemann-Stieltjes integral of $f$ with respect
to $\alpha$ by

$(RRS)\int_{a}^{b}f(x)\,d\alpha=A$

and denote the collection of all such $f$ by $RRS_{\alpha}[a,b]$.

######  Theorem 9.14.

Given $\alpha\in I[a,b]$ and $f\in B[a,b]$, $f\in RDS_{\alpha}[a,b]$ if and
only if $f\in RRS_{\alpha}[a,b]$. In this case,

$\int_{a}^{b}f(x)\,d\alpha=(RRS)\int_{a}^{b}f(x)\,d\alpha.$

######  Remark 9.15.