7.7. CHAPTER 4
57
Suppose that g is integrable on [a, b]. If so, then there is a Cousin cover
with the property that for any partition
of [a, b] from
Let
Define
00
=
00
= cfw_([ai , bi ], i ) : i = 1, 2, . . . , n
,
0
\
n
X
g(i )L([ai , bi ])
Z
b
g(x) dx
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
6.9. DESCRIPTIVE CHARACTERIZATION OF THE INTEGRAL
41
[Fine version] A set E is a null set if for every > 0 there is a fine cover
of E so that
n
X
(bi ai ) <
i=1
whenever = cfw_([ai , bi ], xi ) : i = 1, 2, . . . , n is a subpartition chosen from
.
L(E)
CHAPTER 6
Measure Theory
Warning: This is not an elementary chapter!
The measure theory in this chapter should not be considered in the same pedagogical light as the earlier four chapters. Up to now no techniques beyond simple
compactness arguments are in
7.7. CHAPTER 4
69
now a partition of the larger interval [a, b]. We note that t = a1 and 1 = a so that
f (1 ) = f (a) = 0. In particular then
X
X
f (x)L(I) =
f (x)L(I).
(I,x)21
(I,x)22
Now we recall, since 2 is a partition of [a, b] from , that
F (b)
F (a
6.3. LEBESGUES MEASURE
39
6.3. Lebesgues measure
The measure L is known as Lebesgues measure and assigns a natural length
to every set. There are numerous presentations of Lebesgues measure and any one
of these can be consulted. We oer no proofs of the st
7.6. CHAPTER 3
from
have
51
chosen in such a way that the intervals cfw_[ai , bi ] do not overlap, we must
n
X
(bi
ai ) < /2.
i=1
It is easy to check that
partition
is a Cousin cover of the interval [c, d]. Thus there is a
cfw_([ai , bi ], xi ) : i = 1, 2
iv
PREFACE
function is adequate justification in the correct theory of integration, but not at all
for the Riemann integral.
More distressingly would be their solution to this problem: Verify that
Z 1
1
xp dx =
( 1 < p < 0).
1
+
p
0
My students would use
7.7. CHAPTER 4
55
many steps, to handle all the extra points where the derivative might fail. This
suggests that we split into infinitely many parts:
= + + + + n.
2 4 8
2
Proof.
/8:
Let > 0. Construct
1
and
2
as before but using the choices /4) and
1
= c
5.10. COMPARING RIEMANN SUMS
29
Some rewriting will help in the comparison. First of all use
L([a, b]) = b
a
to denote the length of the interval [a, b]. We can even allow a = b so that [a, a]
is a degenerate interval of length L([a, a]) = 0. The length o
CHAPTER 7
PROOFS
The proofs assume the usual calculus background for definitions of continuity
and derivatives. We need the mean-value theorem of the calculus but little more
than that. It would be useful if the student knew that continuous functions on
c
5.18. INFINITE INTEGRALS
35
5.18.1. Summing inside the integral. We now can extend the formula we
have used before:
!
!
Z b X
Z b
1
1
X
fn (x) dx =
fn (x) dx .
a
n=1
n=1
a
Our assumptions previously (Theorem 5.22) required the integrals to exist as finite
7.7. CHAPTER 4
53
At the remaining points we take advantage of the derivative formula F 0 (x) = f (x)
to define
3
where
= cfw_([c, d], x) : c x d and |F (d)
=
F (c)
2(b
a)
f (x)(d
c)| < (d
c)
.
Combine the covering relations
=
1
[
[
2
3.
Simply check (poi
5.14. THE INDEFINITE INTEGRAL
31
5.14. The indefinite integral
The essential ingredient in the Newton definition of the integral of a function
f on a compact interval [a, b] is that there must be an indefinite integral, i.e., a
function F defined on [a, b
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
7.5. CHAPTER 2
49
7.4.2. Proof of Lemma 1.3. If F and G are continuous functions on an
interval [a, b] and if
F 0 (x) = G0 (x)
for all a < x < b with at most finitely many exceptions then
F (b)
Proof.
F (a) = G(b)
G(a).
Let us suppose that the exceptions
7.7. CHAPTER 4
67
case ([ai , bi ], i ) is from 2 or j0 belongs to N and so the corresponding element
([a0j , b0j ], j0 ) is from 2 . That sum is smaller than
X
f (i ) f (j0 ) L([ai , bi ] \ [a0j , b0j ]) < 2M [/8M ] + 2M [/8M ] = /2.
1
The second part co
7.7. CHAPTER 4
But is also a subset of
n
X
i=1
59
, so we know that
Z b
g(i )L([ai , bi ])
g(x) dx < /2.
00
a
Putting these two inequalities together we deduce that
Z b
n
X
f (i )L([ai , bi ])
g(x) dx
a
i=1
n
X
n
X
f (i )L([ai , bi ])
i=1
g(i )L([ai , bi
7.7. CHAPTER 4
71
Fix an integer n and let > 0. There exists a Cousin cover
that
X
(7.1)
| F (I) f (x)L(I)| < /2,
1
of [a, b] such
(I,x)2
for every partition of [a, b] contained in 1 . Since Nn is a null set there is a Cousin
cover 2 so that
X
(7.2)
L(I)
6.13. LEBESGUES PROGRAM: CONSTRUCTION OF THE INTEGRAL
43
A function f : [a, b] ! R would be measurable if the corresponding set
E = cfw_x 2 [a, b] : f (x) < r
is a measurable set. An equally simple way to think of this is that a function f
defined on an i
5.6. IGNORING NULL SETS
27
.
PROOF IN SECTION 7.7.5
Lemma 5.4. Let e1 , e2 , e3 , . . . be a sequence of points from the interval [a, b]
and suppose that f and g are functions defined on [a, b] and for which f (x) = g(x)
for every point x not belonging to
7.8. CHAPTER 5
77
7.8.3. Step 1: Vitali covering theorem. For the first step in the proof
of the Vitali covering theorem we prove a preliminary covering theorem that is of
interest on its own.
Lemma 7.1 (Rad
o Covering Theorem). Let E be a set of finite L
8.3. CHAPTER 3
91
8.2.4. Solution for Exercise 13. Let F be dierentiable at a point x0 , let
> 0. Then we can choose > 0 so that
|F (x)
F (x0 )
F 0 (x0 )(x
x0 )| |x
x0 |/2
for all |x x0 | < . Suppose that [c, d] is an interval containing x0 and with leng
7.7. CHAPTER 4
61
7.7.11. Proof of Theorem 5.9. Our goal is to establish the useful formula
Z b
Z c
Z b
f (x) dx =
f (x) dx +
f (x) dx
a
a
c
for integrating over two adjacent intervals [a, c] and [c, b].
Proof. We begin by invoking our right to change the
7.7. CHAPTER 4
73
To prove this formula is not sophisticated. It merely uses the definition of the
integral. Given an > 0 can we find a Cousin cover so that the Riemann sums for
the cover are within of the value we require for the integral? The computatio
7.8. CHAPTER 5
Z
N
X
b
fj (x) dx + 2
j
a
j=1
!
75
Z
1
X
j=1
b
!
fj (x) dx
a
+ .
We have established this inequality for all t < 1:
!
!
Z b
1
X
X
t
tf (x)L(I)
fj (x) dx + .
a
j=1
This gives the upper bound for all Riemann sums from
00
that we wanted.
7.7.
x
D.R.I.P. HISTORY
of the success of his project. In fact Lebesgue felt rather naturally compelled to show
that his integral could be expressed as a limit of Riemann sums. He didnt, however,
find the right filter to express this. If he had found it and th
5.17. SUMMING INSIDE THE INTEGRAL
33
For the constructive definition of the integral we are not yet sure that F 0 (x0 ) =
f (x0 ) would have to be true at any point. The following theorem is useful in
calculus courses for providing a simple situation in w
7.8. CHAPTER 5
87
Set = 1 \ 2 ; this is a Cousin cover of [a, b]. We deduce from the inequalities
(7.12) and (7.13) that, for all Riemann sums from any partition of [a, b],
X
(7.14)
L(K)
K (x)L(I) < L(K) + .
(I,x)2
The identity of the theorem is now clea
7.8. CHAPTER 5
83
The two growth lemmas are used to check that each of the following sets is a
null set:
(7.10)
and
(7.11)
Apq = cfw_x 2 [a, b] : Df (x) < p < q < DF (x)
B = cfw_x 2 [a, b] : Df (x) = 1.
The first set (7.10) must satisfy
L(f (Epq ) = L(Epq