Exercise 8.5.4. Justify each step in the chain of equalities and inequalities

which concluded the preceding proof.

Exercise 8.5.5. Show that the collection R of Riemann integrable functions

f : [a, b] ’ Rm forms a real vector space with the natural operations. If we

write

b

Tf = f (x) dx

a

and f ∞ = supt∈[a,b] f (t) , show that T : R ’ R is a linear map and

T f ¤ (b ’ a) f ∞ .

Chapter 9

Developments and limitations

of the Riemann integral ™

9.1 Why go further?

Let us imagine a conversation in the 1880™s between a mathematician opposed

to the ˜new rigour™ and a mathematician who supported it. The opponent

might claim that the de¬nition of the Riemann integral given in section 8.2

was dull and gave rise to no new theorems. The supporter might say, as

this book does, that de¬nitions are necessary in order that we know when

we have proved something and to understand what we have proved when we

have proved it. He would, however, have to admit both the dullness and the

lack of theorems. Both sides would regretfully agree that there was probably

little more to say about the matter.

Twenty years later, Lebesgue, building on work of Borel and others,

produced a radically new theory of integration. From the point of view

of Lebesgue™s theory, Riemann integration has a profound weakness. We

saw in Lemma 8.2.11 and Exercise 8.2.14 that we cannot leave the class of

Riemann integrable functions if we only perform algebraic operations (for

example the product of two Riemann integrable functions is again Riemann

integrable). However we can leave the class of Riemann integrable functions

by performing limiting operations.

Exercise 9.1.1. Let fn : [0, 1] ’ R be de¬ned by fn (r2’n ) = 1 if r is an

integer with 0 ¤ r ¤ 2n , fn (x) = 0, otherwise.

(i) Show that fn is Riemann integrable.

(ii) Show that there exists an f : [0, 1] ’ R, which you should de¬ne

explicitly, such that fn (x) ’ f (x) as n ’ ∞, for each x ∈ [0, 1].

(iii) Show, however, that f is not Riemann integrable.

205

206 A COMPANION TO ANALYSIS

[See also Exercise K.138.]

The class of Lebesgue integrable functions includes every Riemann inte-

grable function but behaves much better when we perform limiting opera-

tions. As an example, which does not give the whole picture but shows the

kind of result that can be obtained, contrast Exercise 9.1.1 with the following

lemma.

Lemma 9.1.2. Let fn : [a, b] ’ R be a sequence of Lebesgue integrable

functions with |fn (x)| ¤ M for all x ∈ [0, 1] and all n. If fn (x) ’ f (x) as

n ’ ∞ for each x ∈ [0, 1], then f is Lebesgue integrable and

b b

fn (x) dx ’ f (x) dx.

a a

It is important to realise that mathematicians prize the Lebesgue inte-

gral, not because it integrates more functions (most functions that we meet

explicitly are Riemann integrable), but because it gives rise to beautiful the-

orems and, at a deeper level, to beautiful theories way beyond the reach of

the Riemann integral.

Dieudonn´ dismisses the Riemann integral with scorn in [13], Chap-

e

ter VIII.

It may well be suspected that, had it not been for its pres-

tigious name, this [topic] would have been dropped long ago

[from elementary analysis courses], for (with due reverence to

Riemann™s genius) it is certainly clear to any working mathemati-

cian that nowadays such a ˜theory™ has at best the importance of

a mildly interesting exercise in the general theory of measure and

integration. Only the stubborn conservatism of academic tradi-

tion could freeze it into a regular part of the curriculum, long

after it had outlived its historical importance.

Stubborn academic conservatives like the present writer would reply that,

as a matter of observation, many working mathematicians1 do not use and

have never studied Lebesgue integration and its generalisation to measure

theory. Although measure theory is now essential for the study of all branches

of analysis and probability, it is not needed for most of number theory, alge-

bra, geometry and applied mathematics.

1

Of course, it depends on who you consider to be a mathematician. A particular French

academic tradition begins by excluding all applied mathematicians, continues by excluding

all supporters of the foreign policy of the United States and ends by restricting the title

´

to pupils of the Ecole Normale Sup´rieure.

e

207

Please send corrections however trivial to twk@dpmms.cam.ac.uk

It is frequently claimed that Lebesgue integration is as easy to teach as

Riemann integration. This is probably true, but I have yet to be convinced

that it is as easy to learn. Under these circumstances, it is reasonable to

introduce Riemann integration as an ad hoc tool to be replaced later by a

more powerful theory, if required. If we only have to walk 50 metres, it makes

no sense to buy a car.

On the other hand, as the distance to be traveled becomes longer, walking

becomes less attractive. We could walk from London to Cambridge but few

people wish to do so. This chapter contains a series of short sections showing

how the notion of the integral can be extended in various directions. I hope

that the reader will ¬nd them interesting and instructive but, for the reasons

just given, she should not invest too much time and e¬ort in their contents

which, in many cases, can be given a more elegant, inclusive and e¬cient

exposition using measure theory.

I believe that, provided it is not taken too seriously, this chapter will be

useful to those who do not go on to do measure theory by showing that

the theory of integration is richer than most elementary treatments would

suggest and to those who will go on to do measure theory by opening their

minds to some of the issues involved.

Improper integrals ™

9.2

We have de¬ned Riemann integration for bounded functions on bounded

intervals. However, the reader will already have evaluated, as a matter of

routine, so called ˜improper integrals™2 in the following manner

1 1

’1/2

x’1/2 dx = lim [2x1/2 ]1 = 2,

x dx = lim

’0+ ’0+

0

and

∞ R

’2

x’2 dx = lim [’x’1 ]R = 1.

x dx = lim 1

R’∞ R’∞

1 1

A full theoretical treatment of such integrals with the tools at our disposal

is apt to lead into into a howling wilderness of ˜improper integrals of the ¬rst

kind™, ˜Cauchy principal values™ and so on. Instead, I shall give a few typical

2

There is nothing particularly improper about improper integrals (at least, if they are

absolutely convergent, see page 211), but this is what they are traditionally called. Their

other traditional name ˜in¬nite integrals™ removes the imputation of moral obliquity but

is liable to cause confusion in other directions.

208 A COMPANION TO ANALYSIS

theorems, de¬nitions and counterexamples from which the reader should be

able to construct any theory that she needs to justify results in elementary

calculus.

De¬nition 9.2.1. If f : [a, ∞) ’ R is such that f |[a,X] ∈ R[a, X] for each

∞

X

X > a and a f (x) dx ’ L as X ’ ∞, then we say that a f (x) dx exists

with value L.

Lemma 9.2.2. Suppose f : [a, ∞) ’ R is such that f |[a,X] ∈ R[a, X] for

∞

each X > a. If f (x) ≥ 0 for all x, then a f (x) dx exists if and only if there

X

exists a K such that a f (x) dx ¤ K for all X.

Proof. As usual we split the proof into two parts dealing with ˜if™ and ˜only

if™ separately.

∞ X

Suppose ¬rst that a f (x) dx exists, that is to say a f (x) dx tends to

n

a limit as X ’ ∞. Let un = a f (x) dx when n is an integer with n ≥ a.

Since f is positive, un is an increasing sequence. Since un tends to a limit, it

must be bounded, that is to say, there exists a K such that un ¤ K for all

n ≥ a. If X ≥ a we choose an integer N ≥ X and observe that

X N

f (x) dx ¤ f (x) dx = uN ¤ K

a a

as required.

X

Suppose, conversely, that there exists a K such that a f (x) dx ¤ K for

n

all X ≥ a. De¬ning un = a f (x) dx as before, we observe that the un form

an increasing sequence bounded above by K. By the fundamental axiom it

follows that un tends to a limit L, say. In particular, given > 0, we can

¬nd an n0 ( ) such that L ’ < un ¤ L for all n ≥ n0 ( ).

If X is any real number with X > n0 ( ) + 1, we can ¬nd an integer n

with n + 1 ≥ X > n. Since n ≥ n0 ( ), we have

X

L ’ < un ¤ f (x) dx ¤ un+1 ¤ L

a

X X

and |L ’ f (x) dx ’ L as X ’ ∞, as required.

f (x) dx| < . Thus

a a

n

Exercise 9.2.3. Show that 0 sin(2πx) dx tends to a limit as n ’ ∞ through

X

integer values, but 0 sin(2πx) dx does not tend to a limit as X ’ ∞.

We use Lemma 9.2.2 to prove the integral comparison test.

Lemma 9.2.4. Suppose f : [1, ∞) ’ R is a decreasing continuous positive

∞

function. Then ∞ f (n) exists if and only if 1 f (x) dx does.

n=1

209

Please send corrections however trivial to twk@dpmms.cam.ac.uk

∞

Just as with sums we sometimes say that ˜ 1 f (x) dx converges™ rather

∞

than ˜ 1 f (x) dx exists™. The lemma then says ˜ ∞ f (n) converges if and

n=1

∞

only if 1 f (x) dx does™.

The proof of Lemma 9.2.4 is set out in the next exercise.

Exercise 9.2.5. Suppose f : [1, ∞) ’ R is a decreasing continuous positive

function.

(i) Show that

n+1

f (n) ≥ f (x) dx ≥ f (n + 1).

n

(ii) Deduce that

N N +1

N +1

f (n) ≥ f (x) dx ≥ f (n).

1

1 2

(iii) By using Lemma 9.2.2 and the corresponding result for sums, deduce

Lemma 9.2.4.

Exercise 9.2.6. (i) Use Lemma 9.2.4 to show that ∞ n’± converges if

n=1

± > 1 and diverges if ± ¤ 1.

(ii) Use the inequality established in Exercise 9.2.5 to give a rough esti-

mate of the size of N required to give N n’1 > 100.

n=1

(iii) Use the methods just discussed to do Exercise 5.1.10.

Exercise 9.2.7. (Simple version of Stirling™s formula.) The ideas of

Exercise 9.2.5 have many applications.

(i) Suppose g : [1, ∞) ’ R is an increasing continuous positive function.

Obtain inequalities for g corresponding to those for f in parts (i) and (ii) of

Exercise 9.2.5.

(ii) By taking g(x) = log x in part (i), show that

N

log(N ’ 1)! ¤ log x dx ¤ log N !

1

and use integration by parts to conclude that

log(N ’ 1)! ¤ N log N ’ N + 1 ¤ log N ! .

(iii) Show that log N ! = N log N ’ N + θ(N )N where θ(N ) ’ 0 as

N ’ ∞.

[A stronger result is proved in Exercise K.141.]

210 A COMPANION TO ANALYSIS

We have a result corresponding to Theorem 4.6.12

Lemma 9.2.8. Suppose f : [a, ∞) ’ R is such that f |[a,X] ∈ R[a, X] for

∞ ∞

each X > a. If a |f (x)| dx exists, then a f (x) dx exists.

It is natural to state Lemma 9.2.8 in the form ˜absolute convergence of

the integral implies convergence™.

Exercise 9.2.9. Prove Lemma 9.2.8 by using the argument of Exercise 4.6.14 (i).

Exercise 9.2.10. Prove the following general principle of convergence for

integrals.

Suppose f : [a, ∞) ’ R is such that f |[a,X] ∈ R[a, X] for each X > a.

∞

Show that a f (x) dx exists if and only if, given any > 0, we can ¬nd an

X0 ( ) > a such that

Y

f (x) dx <

X

whenever Y ≥ X ≥ X0 ( ).

Exercise 9.2.11. (i) Following the ideas of this section and Section 8.5,

∞

provide the appropriate de¬nition of a f (x) dx for a function f : [a, ∞) ’

Rm .

(ii) By taking components and using Exercise 9.2.10, or otherwise, prove

a general principle of convergence for such integrals.

(iii) Use part (ii) and the method of proof of Theorem 4.6.12 to prove the

following generalisation of Lemma 9.2.8.

Suppose f : [a, ∞) ’ Rm is such that f |[a,X] ∈ R[a, X] for each X > a. If

∞ ∞

f (x) dx exists then a f (x) dx exists.

a

Exercise 9.2.12. Suppose f : [a, b) ’ R is such that f |[a,c] ∈ R[a, c] for

each a < c < b. Produce a de¬nition along the lines of De¬nition 9.2.1 of

b

what it should mean for a f (x) dx to exist with value L.

State and prove results analogous to Lemma 9.2.2 and Lemma 9.2.8.

Additional problems arise when there are two limits involved.

Example 9.2.13. If », µ > 0 then

»R

x »

dx ’ log

1 + x2 µ

’µR

as R ’ ∞.

211

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Proof. Direct calculation, which is left to the reader.

A pure mathematician gets round this problem by making a de¬nition along

these lines.

De¬nition 9.2.14. If f : R ’ R is such that f |[’X,Y ] ∈ R[’X, Y ] for each

∞

X, Y > 0, then ’∞ f (x) dx exists with value L if and only if the following

condition holds. Given > 0 we can ¬nd an X0 ( ) > 0 such that

Y

f (x) dx ’ L < .

’X

for all X, Y > X0 ( ).

Exercise 9.2.15. Let f : R ’ R be such that f |[’X,Y ] ∈ R[X, Y ] for

∞ ∞

each X, Y > 0. Show that ’∞ f (x) dx exists if and only if 0 f (x) dx =

R 0 0

limR’∞ 0 f (x) dx and ’∞ f (x) dx = limS’∞ ’S f (x) dx exist. If the inte-

grals exist, show that

∞ ∞

0

f (x) dx = f (x) dx + f (x) dx.

’∞ ’∞ 0

The physicist gets round the problem by ignoring it. If she is a real

physicist with correct physical intuition this works splendidly3 but if not,

not.

Speaking broadly, improper integrals E f (x) dx work well when they are

absolutely convergent, that is to say, E |f (x)| dx < ∞, but are full of traps

for the unwary otherwise. This is not a weakness of the Riemann integral but

inherent in any mathematical situation where an object only exists ˜by virtue

of the cancellation of two in¬nite objects™. (Recall Littlewood™s example on

page 81.)

Example 9.2.16. Suppose we de¬ne the PV (principle value) integral by

∞ R

PV f (x) dx = lim f (x) dx

R’∞

’∞ ’R

whenever the right hand side exists. Show, by considering Example 9.2.13, or

otherwise, that the standard rule for change of variables fails for PV integrals.

3

In [8], Boas reports the story of a friend visiting the Princeton common room ˜ . . .

where Einstein was talking to another man, who would shake his head and stop him;

Einstein then thought for a while, then started talking again; was stopped again; and so

on. After a while, . . . my friend was introduced to Einstein. He asked Einstein who the

other man was. “Oh,” said Einstein, “that™s my mathematician.” ™

212 A COMPANION TO ANALYSIS

Integrals over areas ™

9.3

At ¬rst sight, the extension of the idea of Riemann integration from functions

de¬ned on R to functions de¬ned on Rn looks like child™s play. We shall do

the case n = 2 since the general case is a trivial extension.

Let R = [a, b] — [c, d] and consider f : R ’ R such that there exists a K

with |f (x)| ¤ K for all x ∈ R. We de¬ne a dissection D of R to be a ¬nite

collection of rectangles Ij = [aj , bj ] — [cj , dj ] [1 ¤ j ¤ N ] such that

N

(i) Ij = R,

j=1

(ii) Ii © Ij is either empty or consists of a segment of a straight line

[1 ¤ j < i ¤ N ].

If D = {Ij : 1 ¤ j ¤ N } and D = {Ik : 1 ¤ k ¤ N } are dissections we

write D § D for the set of non-empty rectangles of the form Ij © Ik . If every

Ik ∈ D is contained in some Ij ∈ D we write D D.

We de¬ne the upper sum and lower sum associated with D by

N

S(f, D) = |Ij | sup f (x),

x∈Ij

j=1

N

s(f, D) = |Ij | inf f (x)

x∈Ij

j=1

where |Ij | = (bj ’ aj )(dj ’ cj ), the area of Ij .

Exercise 9.3.1. (i) Suppose that D and D are dissections with D D.

Show, using the method of Exercise 8.2.1, or otherwise, that

S(f, D) ≥ S(f, D ) ≥ s(f, D ) ≥ s(f, D).

(ii) State and prove a result corresponding to Lemma 8.2.3.

(iii) Explain how this enables us to de¬ne upper and lower integrals and

hence complete the de¬nition of Riemann integration. We write the integral

as

f (x) dA

R

when it exists.

(iv) Develop the theory of Riemann integration on R as far as you can.

(You should be able to obtain results like those in Section 8.2 as far as the

end of Exercise 8.2.15.) You should prove that if f is continuous on R then

it is Riemann integrable.

213

Please send corrections however trivial to twk@dpmms.cam.ac.uk

We can do rather more than just prove the existence of