106
ESSAY: WHY THE RIEMANN INTEGRAL IS MORE DIFFICULT
Definition 8.6. A sequence of functions cfw_fn converges uniformly to a function f on an interval [a, b] if for every > 0 there is an integer N so that
|fn (x)
for all n
f (x)| <
N and all x in [a, b

Contents
Preface
Course
0.1.
0.2.
0.3.
iii
Design
Step 1: Use Newtons integral
Step 2: Replace the Riemann integral by the natural integral
Step 3: Prepare the way for the measure theory
D.R.I.P. HISTORY
v
v
vi
vii
ix
Chapter 1. Newtons Original Integral

8.8. LIMITS AND LIMITATIONS
105
and when that happens i and j0 are inside an interval of length smaller than 2 .
Consequently
f (i )
f (j0 ) < /(b
a).
The proof is easy enough then to write up. Compare it with the version for the
natural integral.
Theorem

90
8. SOLUTIONS FOR SELECTED EXERCISES
for all |x x0 | < .
Note that if an interval [c, d] contains the point x0 and if d
x0 c < and d x0 < . But
|F (d)
|F (d)
F (c)| = |[F (d)
F (x0 )] + [F (x0 )
F (x0 )| + |F (x0 )
c < , then
F (c)]|
F (c)| < /2 + /2 =

Index
The Lebesgue integral is dead., xi
ignoring negligible sets, 27
indefinite integral, 31
infinite integral, 34
intersection of Cousin covers, 13
absolutely continuous function, 22, 94
additivity, 28
Arzel`
a, Cesare, 107
Kurzweil integral, iii
Bartle

1.6. THE GENERAL NEWTON INTEGRAL
7
In any of these cases we can be assured that f has an indefinite integral F , that
Z d
f (x) dx = F (d) F (c) (a c < d b)
c
and that
F 0 (x) = f (x)
at every point x at which f is continuous.
Example. One of the most imp

2.3. RIEMANN SUMS CONSTRUCTED FROM THE DERIVATIVE
11
2.3.1. A full cover. The idea is a good one if we approach it dierently. Let
us assume that
F 0 (x) = f (x) (a x b)
(which is a somewhat stronger assumption than we need for Newtons integral).
Let > 0 a

CHAPTER 2
Covering Relations
The full definition of the integral will require the following notions:
Covering relations.
Full covers.
Partitions of a compact interval [a, b].
Riemann sums.
Cousin covering lemma.
We motivate these concepts in this cha

1.4. INTEGRATION FORMULAS
5
allowing us to apply Lemma 1.5. From that lemma we deduce that F and G dier
by a constant and that F (b) F (a) = G(b) G(a). Thus the definite integral is
unambiguously defined no matter what indefinite integral we choose to use

8.4. CHAPTER 4
97
By Exercise 24 it follows that F cannot grow on the union of the sequence of
sets cfw_N \ Ek . But this union is all of N .
8.4. Chapter 4
8.4.1. Solution for Exercise 43. Let f be the function defined to be 1
at every irrational number

8.3. CHAPTER 3
93
so that k (k = 1, 2, 3, . . . ) just contains elements ([ai , bi ], i ) from that happen
also to belong to k . Most of these new subpartitions are empty: in fact since
contains only n elements, there are at most n of the k that are nonem

94
8. SOLUTIONS FOR SELECTED EXERCISES
and !F ([c, d]) < 2 j 1 . Note that there are, at most, two choices of i with the
same j. It follows that
n
X
!F ([ai , bi ]) < 2
i=1
1
X
j 1
2
= .
j=1
By definition, then, F does not grow on E.
8.3.7. Solution for E

`
8.9. ARZELAS
BOUNDED CONVERGENCE THEOREM
107
8.9. Arzel`
as bounded convergence theorem
As a test of the Riemann integral consider the following problem which is
a slight, but critical, variant of the problem that we discussed in the preceding
section.

8.6. IMPROVING THE RIEMANN VERSION
103
Theorem 8.6.1. Let F be a uniformly dierentiable function on an interval
[a, b]. Then the function f (x) = F 0 (x) is Riemann integrable on [a, b] and
Z
b
f (x) dx = F (b)
F (a).
a
The proof is obvious, also elementa

104
ESSAY: WHY THE RIEMANN INTEGRAL IS MORE DIFFICULT
8.7. Continuous functions are integrable
Any modestly ambitious course on the Riemann integral will prove that continuous functions are integrable. More accurately such a course will prove that
uniform

102
ESSAY: WHY THE RIEMANN INTEGRAL IS MORE DIFFICULT
(8.2)
n
X
[F (ai )
F (ai
1 )]
=
i=1
and
f (i )[ai
ai
1 ],
i=1
n
X
(8.3)
n
X
f (i )[ai
ai
i=1
1]
Z
b
f (x) dx.
a
The first identity (8.1) is just some algebra. The second identity (8.2) calls on
the me

Preface
These informal calculus notes are oered to those instructors who may wish to
design a calculus sequence that drops the Riemann integral in favor of the more
natural integral on the real line that has been called, at diverse times, by many
dierent

The Quiz
This quiz is designed for graduate students who have just completed their study
of the Lebesgue integral and are expected to remember their freshman calculus
study of the Riemann integral. They will be unable to use the natural integral,
of cours

1.1. BEYOND THE ORIGINAL NEWTON INTEGRAL
3
.
This means that, while a function can have many dierent indefinite integrals,
it has just one definite integral
Z b
f (x) dx
PROOF IN SECTION 7.4.1
a
that can be computed by taking any single indefinite integra

CHAPTER 1
Newtons Original Integral
Integration as it was originally conceived of by Newton is the process inverse
to dierentiation. Such a process, he saw, would have numerous applications in
geometry and physics.
In modern language we can describe the s

92
8. SOLUTIONS FOR SELECTED EXERCISES
8.3.2. Solution for Exercise 29. Suppose that E1 and E2 are null sets and
let E = E1 [ E2 be their union. Let > 0. Choose a full cover of E1 so that
n
X
(bi ai ) < /2
i=1
whenever the collection
cfw_([ai , bi ], xi )

100
ADDITIONAL READING
Washek Pfeer, The Riemann approach to integration, Cambridge University Press, Cambridge, 1993.
Washek Pfeer, Derivation and integration, Cambridge University Press,
Cambridge, 2001.
Charles Swartz, Introduction to gauge integral

8.11. MEASURE THEORY
111
8.11. Measure theory
Some of the rudiments of Lebesgues theory of measure can easily be introduced
in a first real analysis course. But when the only integration theory available is the
Riemann integral there hardly seems any poin

D.R.I.P. HISTORY
DUMP THE RIEMANN INTEGRAL PROJECT
The genuine history of ideas is not for amateurs: it requires skill,
dedication, integrity, and certainly training in the discipline of the
academic historian. There is another kind of history, wherein
th

Essay: Why the Riemann integral is more difficult
Most instructors of the calculus are likely to hold the mistaken belief that the
natural integral on the real line (presented in these notes) is a more difficult choice
for a theoretical calculus course th

96
8. SOLUTIONS FOR SELECTED EXERCISES
Most of these are empty: in fact since contains only n elements, there are at most
n of the k that are nonempty. Choose a large enough integer K so that k = ; for
k K
Now, since k is a subset of k , we have
X
!F ([ai

Additional Reading
The texts listed here, all of them except for the first, use the so-called gauge
version of the natural integral. In this text we have chosen to express this idea using
covering relations. I see no advantages in the gauge version except

8.10. THE BOUNDED CONVERGENCE THEOREM FOR THE NATURAL INTEGRAL 109
Although this simple argument can be used to defend the Riemann integral we
still remain concerned that the requirement that the limit function be assumed to
be Riemann integrable, which c

114
THE QUIZ
8.13. Problem #1
Z
1
x2 dx?
0
8.13.1. The integral: Observe that, with f (x) = x2 and F (x) = x3 /3, we
have F 0 (x) = f (x) at every point of the interval [0, 1].
Consequently we can use the following theorem of integration theory (known
som

108
ESSAY: WHY THE RIEMANN INTEGRAL IS MORE DIFFICULT
By using just the lower integral the problem gets reduced in complexity. Indeed
once we have expressed the problem in this manner, it should be transparent that
it is rather less ambitious an undertaki

CHAPTER 8
SOLUTIONS FOR SELECTED EXERCISES
8.1. Chapter 1
8.1.1. Solution for Exercise 5. To evaluate
Z 1
1
p dx
|x|
0
it is enough to find explicitly an indefinite integral. (Not always possible but here
it is.)
p
1
The function F (x) = 2 x = 2x 2 has f

CHAPTER 3
Growth of a Function on a Set
The calculus is concerned to a great extent with the growth of a function either
at a point or on a set. The derivative measures the growth of a function at a point.
A dierent concept is needed for the growth of a f

Course Design
The DRIP program is best viewed as a thought experiment for academic mathematicians. There is certainly no chance of the early calculus courses changing.
Any textbook writer who decided to alter material for a subsequent edition would
receiv