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|f (x)| dx ≥ f (x) dx .
a a

Exercise 8.2.16. In each of Exercises 8.2.10, 8.2.14 and 8.2.15 we used
a roundabout route to our result. For example, in Exercise 8.2.10 we ¬rst
proved that if f 2 is Riemann integrable whenever f is and then used this
result to prove that f g is Riemann integrable whenever f and g are. It is
natural to ask whether we can give a direct proof in each case. The reader
should try to do so. (In my opinion, the direct proofs are not much harder,
though they do require more care in writing out.)
Exercise 8.2.17. (i) Suppose that a ¤ c ¤ b and that f : [a, b] ’ R is a
bounded function. Consider a dissection D1 of [a, c] given by
D1 = {x0 , x1 , . . . , xm } with a = x0 ¤ x1 ¤ x2 ¤ · · · ¤ xm = c,
and a dissection D2 of [c, b] given by
D2 = {xm+1 , xm+2 , . . . , xn } with c = xm+1 ¤ xm+2 ¤ xm+3 ¤ · · · ¤ xn = b.
If D is the dissection of [a, b] given by
D = {x0 , x1 , . . . , xn },
show that S(f, D) = S(f |[a,c] , D1 ) + S(f |[c,b] , D2 ). (Here f |[a,c] means the
restriction of f to [a, c].)
(ii) Show that f ∈ R[a, b] if and only if f |[a,c] ∈ R[a, c] and f |[c,b] ∈ R[c, b].
Show also that, if f ∈ R[a, b], then
b c b
f |[a,c] (x) dx + f |[c,b] (x) dx.
f (x) dx =
a a c

In a very slightly less precise and very much more usual notation we write
b c b
f (x) dx = f (x) dx + f (x) dx.
a a c
180 A COMPANION TO ANALYSIS

There is a standard convention that we shall follow which says that, if
b ≥ a and f is Riemann integrable on [a, b], we de¬ne
a b
f (x) dx = ’ f (x) dx.
b a

Exercise 8.2.18. Suppose β ≥ ± and f is Riemann integrable on [±, β].
Show that if a, b, c ∈ [±, β] then
b c b
f (x) dx = f (x) dx + f (x) dx.
a a c

[Note that a, b and c may occur in any of six orders.]
However, this convention must be used with caution.
Exercise 8.2.19. Suppose that b ≥ a, », µ ∈ R, and f and g are Riemann
integrable. Which of the following statements are always true and which are
not? Give a proof or counterexample. If the statement is not always true,
¬nd an appropriate correction and prove it.
a a a
(i) »f (x) + µg(x) dx = » f (x) dx + µ g(x) dx.
b b b
a a
(ii) If f (x) ≥ g(x) for all x ∈ [a, b], then f (x) dx ≥ g(x) dx.
b b

Riemann was unable to show that all continuous functions were integrable
(we have a key concept that Riemann did not and we shall be able to ¬ll
this gap in the next section). He did, however, have the result of the next
exercise. (Note that an increasing function need not be continuous. Consider
the Heaviside function H : R ’ R given by H(x) = 0 for x < 0, H(x) = 1
for x ≥ 0.)
Exercise 8.2.20. Suppose f : [a, b] ’ R is increasing. Let N be a strictly
positive integer and consider the dissection
D = {x0 , x1 , . . . , xN } with xj = a + j(b ’ a)/N .
Show that
N
S(f, D) = f (xj )(b ’ a)/N,
j=1

¬nd s(f, D) and deduce that
S(f, D) ’ s(f, D) = (f (b) ’ f (a))(b ’ a)/N.
Conclude that f is Riemann integrable.
181
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Using Lemma 8.2.11 this gives the following result.
Lemma 8.2.21. If f : [a, b] ’ R can be written as f = f1 ’ f2 with f1 , f2 :
[a, b] ’ R increasing, then f is Riemann integrable.
At ¬rst sight, Lemma 8.2.21 looks rather uninteresting but, in fact, it
covers most of the functions we normally meet.
Exercise 8.2.22. (i) If
f1 (t) = 0, f2 (t) = ’t2 if t < 0
f1 (t) = t2 , f2 (t) = 0 if t ≥ 0,
show that f1 and f2 are increasing functions with t2 = f1 (t) ’ f2 (t).
(ii) Show that, if f : [a, b] ’ R has only a ¬nite number of local maxima
and minima, then it can be written in the form f = f1 ’ f2 with f1 , f2 :
[a, b] ’ R increasing.
Functions which are the di¬erence of two increasing functions are dis-
cussed in Exercise K.158, Exercises K.162 to K.166 and more generally in
the next chapter as ˜functions of bounded variation™. We conclude this sec-
tion with an important example of Dirichlet.
Exercise 8.2.23. If f : [0, 1] ’ R is given by
f (x) = 1 when x is rational,
f (x) = 0 when x is irrational,
show that, whenever D is a dissection of [0, 1], we have S(f, D) = 1 and
s(f, D) = 0. Conclude that f is not Riemann integrable.
Exercise 8.2.24. (i) If f is as in Exercise 8.2.23, show that
N N
1 1
f (r/N ) ’ 1 as N ’ ∞.
f (r/N ) = 1 and so
N N
r=1 r=1

(ii) Let g : [0, 1] ’ R be given by
g(r/2n ) = 1 when 1 ¤ r ¤ 2n ’ 1, n ≥ 1, and r and n are integers,
g(s/3n ) = ’1 when 1 ¤ s ¤ 3n ’ 1, n ≥ 1, and s and n are integers,
g(x) = 0 otherwise.
Discuss the behaviour of
N
1
g(r/N )
N r=1

as N ’ ∞ in as much detail as you consider desirable.
182 A COMPANION TO ANALYSIS

8.3 Integrals of continuous functions
The key to showing that continuous functions are integrable, which we have
and Riemann did not, is the notion of uniform continuity and the theo-
rem (Theorem 4.5.5) which tells us that a continuous function on a closed
bounded subset of Rn , and so, in particular, on a closed interval, is uniformly
continuous3 .

Theorem 8.3.1. Any continuous function f : [a, b] ’ R is Riemann inte-
grable.

Proof. If b = a the result is obvious, so suppose b > a. We shall show that f
is Riemann integrable by using the standard criterion given in Lemma 8.2.6.
To this end, suppose that > 0 is given. Since a continuous function on a
closed bounded interval is uniformly continuous, we can ¬nd a δ > 0 such
that

|f (x) ’ f (y)| ¤ whenever x, y ∈ [a, b] and |x ’ y| < δ.
b’a
Choose an integer N > (b ’ a)/δ and consider the dissection

D = {x0 , x1 , . . . , xN } with xj = a + j(b ’ a)/N .

If x, y ∈ [xj , xj+1 ], then |x ’ y| < δ and so

|f (x) ’ f (y)| ¤ .
b’a
It follows that

f (x) ’ f (x) ¤
sup inf
b’a
x∈[xj ,xj+1 ]
x∈[xj ,xj+1 ]


for all 0 ¤ j ¤ N ’ 1 and so
N ’1
S(f, D) ’ s(f, D) = (xj+1 ’ xj ) f (x) ’
sup inf f (x)
x∈[xj ,xj+1 ]
x∈[xj ,xj+1 ]
j=0
N ’1
b’a
¤ =,
N b’a
j=0


as required.
3
This is a natural way to proceed but Exercise K.118 shows that it is not the only one.
183
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Slight extensions of this result are given in Exercise I.11. In Exercise K.122
we consider a rather di¬erent way of looking at integrals of continuous func-
tions.
Although there are many functions which are integrable besides the con-
tinuous functions, there are various theorems on integration which demand
that the functions involved be continuous or even better behaved. Most of
the results of this section have this character.

Lemma 8.3.2. If f : [a, b] ’ R is continuous, f (t) ≥ 0 for all t ∈ [a, b] and
b
f (t) dt = 0,
a

it follows that f (t) = 0 for all t ∈ [a, b].

Proof. If f is a positive continuous function which is not identically zero, then
we can ¬nd an x ∈ [a, b] with f (x) > 0. Setting = f (x)/2, the continuity
of f tells us that there exists a δ > 0 such that |f (x) ’ f (y)| < whenever
|x ’ y| ¤ δ and y ∈ [a, b]. We observe that

f (y) ≥ f (x) ’ |f (x) ’ f (y)| > f (x) ’ = f (x)/2

whenever |x ’ y| ¤ δ and y ∈ [a, b]. If we de¬ne h : [a, b] ’ R by h(y) =
f (x)/2 whenever |x ’ y| ¤ δ and y ∈ [a, b] and h(y) = 0 otherwise, then
f (t) ≥ h(t) for all t ∈ [a, b] and so
b b
f (t) dt ≥ h(t) dt > 0.
a a




Exercise 8.3.3. (i) Let a ¤ c ¤ b. Give an example of a Riemann integrable
function f : [a, b] ’ R such that f (t) ≥ 0 for all t ∈ [a, b] and
b
f (t) dt = 0,
a

but f (c) = 0.
(ii) If f : [a, b] ’ R is Riemann integrable, f (t) ≥ 0 for all t ∈ [a, b] and
b
f (t) dt = 0,
a

show that f (t) = 0 at every point t ∈ [a, b] where f is continuous.
184 A COMPANION TO ANALYSIS

(iii) We say that f : [a, b] ’ R is right continuous at t ∈ [a, b] if f (s) ’
f (t) as s ’ t through values of s with b ≥ s > t. Suppose f is Riemann
integrable and is right continuous at every point t ∈ [a, b]. Show that if
f (t) ≥ 0 for all t ∈ [a, b] and
b
f (t) dt = 0,
a

it follows that f (t) = 0 for all t ∈ [a, b] with at most one exception. Give an
example to show that this exception may occur.

The reader should have little di¬culty in proving the following useful
related results.

Exercise 8.3.4. (i) If f : [a, b] ’ R is continuous and
b
f (t)g(t) dt = 0,
a

whenever g : [a, b] ’ R is continuous, show that f (t) = 0 for all t ∈ [a, b].
(ii) If f : [a, b] ’ R is continuous and
b
f (t)g(t) dt = 0,
a

whenever g : [a, b] ’ R is continuous and g(a) = g(b) = 0, show that f (t) = 0
for all t ∈ [a, b]. (We prove a slightly stronger result in Lemma 8.4.7.)

We now prove the fundamental theorem of the calculus which links the
processes of integration and di¬erentiation. Since the result is an important
one it is worth listing the properties of the integral that we use in the proof.

Lemma 8.3.5. Suppose », µ ∈ R, f, g : [±, β] ’ R are Riemann integrable
and a, b, c ∈ [±, β]. The following results hold.
b
1 dt = b ’ a.
(i)
a
b b b
(ii) »f (t) + µg(t) dt = » f (t) dt + µ g(t) dt.
a a a
b c c
(iii) f (t) dt + f (t) dt = f (t) dt.
a b a
b
f (t) dt ¤ |b ’ a| sup |f (a + θ(b ’ a))|.
(iv)
0¤θ¤1
a
185
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The reader should run through these results in her mind and make sure
that she can prove them (note that a, b and c can be in any order).
Theorem 8.3.6. (The fundamental theorem of the calculus.) Sup-
pose that f : (a, b) ’ R is a continuous function and that u ∈ (a, b). If we
set
t
F (t) = f (x) dx,
u

then F is di¬erentiable on (a, b) and F (t) = f (t) for all t ∈ (a, b).
Proof. Observe that, if t + h ∈ (a, b) and h = 0 then
t+h t
F (t + h) ’ F (t) 1
’ f (t) = f (x) dx ’ f (x) dx ’ hf (t)
h h u u
t+h t+h
1
f (x) dx ’
= f (t) dx
h t t
t+h
1
(f (x) ’ f (t)) dx
=
|h| t
¤ sup |f (t + θh) ’ f (t)| ’ 0
0¤θ¤1

as h ’ 0 since f is continuous at t. (Notice that f (t) remains constant as x
varies.)
Exercise 8.3.7. (i) Using the idea of the integral as the area under a curve,
draw diagrams illustrating the proof of Theorem 8.3.6.
(ii) Point out, explicitly, each use of Lemma 8.3.5 in our proof of Theo-
rem 8.3.6.
(iii) Let H be the Heaviside function H : R ’ R given by H(x) = 0 for
t
x < 0, H(x) = 1 for x ≥ 0. Calculate F (t) = 0 H(x) dx and show that F is
not di¬erentiable at 0. Where does our proof of Theorem 8.3.6 break down?
t
(iv) Let f (0) = 1, f (t) = 0 otherwise. Calculate F (t) = 0 f (x) dx and
show that F is di¬erentiable at 0 but F (0) = f (0). Where does our proof of
Theorem 8.3.6 break down?
Exercise 8.3.8. Suppose that f : (a, b) ’ R is a function such that f is
Riemann integrable on every interval [c, d] ⊆ (a, b). Let u ∈ (a, b) If we set
t
F (t) = f (x) dx
u

show that F is continuous on (a, b) and that, if f is continuous at some point
t ∈ (a, b), then F is di¬erentiable at t and F (t) = f (t).
186 A COMPANION TO ANALYSIS

Sometimes we think of the fundamental theorem in a slightly di¬erent way.

Theorem 8.3.9. Suppose that f : (a, b) ’ R is continuous, that u ∈ (a, b)
and c ∈ R. Then there is a unique solution to the di¬erential equation
g (t) = f (t) [t ∈ (a, b)] such that g(u) = c.

Exercise 8.3.10. Prove Theorem 8.3.9. Make clear how you use Theo-
rem 8.3.6 and the mean value theorem. Reread section 1.1.

We call the solutions of g (t) = f (t) inde¬nite integrals (or, simply, inte-
grals) of f .
Yet another version of the fundamental theorem is given by the next
theorem.

Theorem 8.3.11. Suppose that g : (±, β) ’ R has continuous derivative
and [a, b] ⊆ (±, β). Then
b
g (t) dt = g(b) ’ g(a).
a

Proof. De¬ne U : (±, β) ’ R by
t
g (x) dx ’ g(t) + g(a).
U (t) =
a

By the fundamental theorem of the calculus and earlier results on di¬erenti-
ation, U is everywhere di¬erentiable with

U (t) = g (t) ’ g (t) = 0

so, by the mean value theorem, U is constant. But U (a) = 0, so U (t) = 0
for all t and, in particular, U (b) = 0 as required.

[Remark: In one dimension, Theorems 8.3.6, 8.3.9 and 8.3.11 are so closely
linked that mathematicians tend to refer to them all as ˜The fundamental
theorem of the calculus™. However they generalise in di¬erent ways.
(1) Theorem 8.3.6 shows that, under suitable circumstances, we can re-
cover a function from its ˜local average™ (see Exercise K.130).
(2) Theorem 8.3.9 says that we can solve a certain kind of di¬erential
equation. We shall obtain substantial generalisations of this result in Sec-
tion 12.2.
(3) Theorem 8.3.11 links the value of the derivative f on the whole of
[a, b] with the value of f on the boundary (that is to say, the set {a, b}). If
187
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you have done a mathematical methods course you will already have seen a
similar idea expressed by the divergence theorem

· u dV = u · dS.
V ‚V

This result and similar ones like Stokes™ theorem turn out to be special cases
of a master theorem4 which links the behaviour of the derivative of a certain
mathematical object over the whole of some body with the behaviour of the
object on the boundary of that body.]
Theorems 8.3.6 and 8.3.11 show that (under appropriate circumstances)
integration and di¬erentiation are inverse operations and the the theories
of di¬erentiation and integration are subsumed in the greater theory of the
calculus. Under appropriate circumstances, if the graph of F has tangent
with slope f (x) at x
area under the graph of slope of tangent of F
= area under the graph of f
b b
F (x) dx = F (b) ’ F (a).
= f (x) dx =
a a

Exercise 8.3.12. Most books give a slightly stronger version of Theorem 8.3.11
in the following form.
If f : [a, b] ’ R has continuous derivative, then
b
f (t) dt = f (b) ’ f (a).
a

Explain what this means (you will need to talk about ˜left™ and ˜right™ deriva-
tives) and prove it.
Recalling the chain rule (Lemma 6.2.10) which tells us that (¦ —¦ g) (t) =
g (t)¦ (g(t)), the same form of proof gives us a very important theorem.
Theorem 8.3.13. (Change of variables for integrals.) Suppose that f :
(±, β) ’ R is continuous and g : (γ, δ) ’ R is di¬erentiable with continuous
derivative. Suppose further that g (γ, δ) ⊆ (±, β). Then, if c, d ∈ (γ, δ), we
have
g(d) d

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