(v) if x, y ∈ [0, 1), then either [x] = [y] or [x] © [y] = ….

We now consider a set A which contains exactly one element from each

equivalence class.

Exercise 8.1.3. (This is easy.) Show that if t ∈ [0, 1) then the equation

t≡a+q mod 1

has exactly one solution with a ∈ A, q rational and q ∈ [0, 1).

[Here t ≡ x + q mod 1 means t ’ x ’ q ∈ Z.]

We are now in a position to produce our example. It will be easiest to

work in C identi¬ed with R2 in the usual way and to de¬ne

E = {r exp 2πia : 1 > r > 0, a ∈ A}.

Since Q is countable, it follows that its subset Q © [0, 1) is countable and we

can write

Q © [0, 1) = {qj : j ≥ 1}

with q1 , q2 , . . . all distinct. Set

Ej = {r exp 2πi(a + qj ) : 1 > r > 0, a ∈ A}.

171

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Exercise 8.1.4. Suppose that conditions (a) to (d) all hold.

(i) Describe the geometric relation of E and Ej . Deduce that |E| = |Ej |.

(ii) Use Exercise 8.1.3 to show that Ej © Ek = … if j = k.

(iii) Use Exercise 8.1.3 to show that

∞

Ej = U

j=1

where U = {z : 0 < |z| < 1}.

(iv) Deduce that

∞

|Ej | = |U |.

j=1

Show from Exercise 8.1.1 (ii) that 0 < |U |.

(v) Show that (i) and (iv) lead to a contradiction if |E| = 0 and if |E| > 0.

Thus (i) and (iv) lead to a contradiction whatever we assume. It follows that

conditions (a) to (d) cannot all hold simultaneously.

Exercise 8.1.5. De¬ne E and Eq as subsets of R2 without using complex

numbers.

The example just given is due to Vitali. It might be hoped that the

problem raised by Vitali™s example are due to the fact that condition (d)

involves in¬nite sums. This hope is dashed by the following theorem of

Banach and Tarski.

Theorem 8.1.6. The unit ball in R3 can be decomposed into a ¬nite number

of pieces which may be reassembled, using only translation and rotation, to

form 2 disjoint copies of the unit ball.

Exercise 8.1.7. Use Theorem 8.1.6 to show that the following four condi-

tions are not mutually consistent.

(a) Every bounded set E in R3 has an volume |E| with |E| ≥ 0.

(b) Suppose that E is a bounded set in R3 . If E is congruent to F (that

is E can be obtained from F by translation and rotation), then |E| = |F |.

(c) Any cube E of side a has volume |E| = a3 .

(d) If E1 and E2 are disjoint bounded sets in R3 , then |E1 ∪ E2 | = |E1 | +

|E2 |.

The proof of Theorem 8.1.6, which is a lineal descendant of Vitali™s ex-

ample, is too long to be given here. It is beautifully and simply explained in

172 A COMPANION TO ANALYSIS

a book [46] devoted entirely to ideas generated by the result of Banach and

Tarski1 .

The examples of Vitali and Banach and Tarski show that if we want a

well behaved notion of area we will have to say that only certain sets have

area. Since the notion of an integral is closely linked to that of area, (˜the

integral is the area under the curve™) this means that we must accept that

only certain functions have integrals. It also means that that we must make

sure that our de¬nition does not allow paradoxes of the type discussed here.

8.2 Riemann integration

In this section we introduce a notion of the integral due to Riemann. For

the moment we only attempt to de¬ne our integral for bounded functions on

bounded intervals.

Let f : [a, b] ’ R be a function such that there exists a K with |f (x)| ¤ K

for all x ∈ [a, b]. [To see the connection with ˜the area under the curve™ it is

helpful to suppose initially that 0 ¤ f (x) ¤ K. However, all the de¬nitions

and proofs work more generally for ’K ¤ f (x) ¤ K. The point is discussed

in Exercise K.114.] A dissection (also called a partition) D of [a, b] is a ¬nite

subset of [a, b] containing the end points a and b. By convention, we write

D = {x0 , x1 , . . . , xn } with a = x0 ¤ x1 ¤ x2 ¤ · · · ¤ xn = b.

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

n

S(f, D) = (xj ’ xj’1 ) sup f (x),

x∈[xj’1 ,xj ]

j=1

n

s(f, D) = (xj ’ xj’1 ) inf f (x)

x∈[xj’1 ,xj ]

j=1

b

[Observe that, if the integral a f (t) dt exists, then the upper sum ought to

provide an upper bound and the lower sum a lower bound for that integral.]

Exercise 8.2.1. (i) Suppose that a ¤ c ¤ b. If D = {a, b} and D =

{a, c, b}, show that

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

1

In more advanced work it is observed that our discussion depends on a principle

called the ˜axiom of choice™. It is legitimate to doubt this principle. However, anyone who

doubts the axiom of choice but believes that every set has volume resembles someone who

disbelieves in Father Christmas but believes in ¬‚ying reindeer.

173

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(ii) Let c = a, b. Show by examples that, in (i), we can have either

S(f, D) = S(f, D ) or S(f, D) > S(f, D ).

(iii) Suppose that a ¤ c ¤ b and D is a dissection. Show that, if D =

D ∪ {c}, then

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

(iv) Suppose that D and D are dissections with D ⊇ D. Show, using

(iii), or otherwise, that

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

The result of Exercise 8.2.1 (iv) is so easy that it hardly requires proof.

None the less it is so important that we restate it as a lemma.

Lemma 8.2.2. If D and D are dissections with D ⊇ D then

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

The next lemma is again hardly more than an observation but it is the

key to the proper treatment of the integral.

Lemma 8.2.3 (Key integration property). If f : [a, b] ’ R is bounded

and D1 and D2 are two dissections, then

S(f, D1 ) ≥ S(f, D1 ∪ D2 ) ≥ s(f, D1 ∪ D2 ) ≥ s(f, D2 ).

The inequalities tell us that, whatever dissection you pick and whatever

dissection I pick, your lower sum cannot exceed my upper sum. There is no

way we can put a quart into a pint pot2 and the Banach-Tarski phenomenon

is avoided.

Since S(f, D) ≥ ’(b ’ a)K for all dissections D we can de¬ne the upper

integral as I — (f ) = inf D S(f, D). We de¬ne the lower integral similarly as

I— (f ) = supD s(f, D). The inequalities tell us that these concepts behave

well.

Lemma 8.2.4. If f : [a, b] ’ R is bounded, then I — (f ) ≥ I— (f ).

b

[Observe that, if the integral a f (t) dt exists, then the upper integral

ought to provide an upper bound and the lower integral a lower bound for

that integral.]

2

Or a litre into a half litre bottle. Any reader tempted to interpret such pictures

literally is directed to part (iv) of Exercise K.171.

174 A COMPANION TO ANALYSIS

If I — (f ) = I— (f ), we say that f is Riemann integrable and we write

b

f (x) dx = I — (f ).

a

We write R[a, b] or sometimes just R for the set of Riemann integrable func-

tions on [a, b].

Exercise 8.2.5. If k ∈ R show that the constant function given by f (t) = k

for all t is Riemann integrable and

b

k dx = k(b ’ a).

a

The following lemma provides a convenient criterion for Riemann inte-

grability.

Lemma 8.2.6. (i) A bounded function f : [a, b] ’ R is Riemann integrable

if and only if, given any > 0, we can ¬nd a dissection D with

S(f, D) ’ s(f, D) < .

(ii) A bounded function f : [a, b] ’ R is Riemann integrable with integral

I if and only if, given any > 0, we can ¬nd a dissection D with

S(f, D) ’ s(f, D) < and |S(f, D) ’ I| ¤ .

Proof. (i) We need to prove necessity and su¬ciency. To prove necessity,

suppose that f is Riemann integrable with Riemann integral I (so that I =

I — (f ) = I— (f )). If > 0 then, by the de¬nition of I — (f ), we can ¬nd a

dissection D1 such that

I + /2 > S(f, D1 ) ≥ I.

Similarly, by the de¬nition of I— (f ), we can ¬nd a dissection D2 such that

I ≥ s(f, D2 ) > I ’ /2.

Setting D = D1 ∪ D2 and using Lemmas 8.2.2 and 8.2.3, we have

I + /2 > S(f, D1 ) ≥ S(f, D) ≥ s(f, D) ≥ s(f, D2 ) > I ’ /2,

so S(f, D) ’ s(f, D) < as required.

175

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To prove su¬ciency suppose that, given any > 0, we can ¬nd a dissection

D with

S(f, D) ’ s(f, D) < .

Using the de¬nition of the upper and lower integrals I — (f ) and I— (f ) together

with the fact that I — (f ) ≥ I— (f ) (a consequence of our key Lemma 8.2.3), we

already know that

S(f, D) ≥ I — (f ) ≥ I— (f ) ≥ s(f, D),

so we may conclude that ≥ I — (f ) ’ I— (f ) ≥ 0. Since is arbitrary, we have

I — (f ) ’ I— (f ) = 0 so I — (f ) = I— (f ) as required.

(ii) Left to the reader.

Exercise 8.2.7. Prove part (ii) of Lemma 8.2.6.

Many students are tempted to use Lemma 8.2.6 (ii) as the de¬nition of

the Riemann integral. The reader should re¬‚ect that, without the inequality

, it is not even clear that such a de¬nition gives a unique value for I. (This

is only the ¬rst of a series of nasty problems that arise if we attempt to

develop the theory without ¬rst proving , so I strongly advise the reader

not to take this path.) We give another equivalent de¬nition of the Riemann

integral in Exercise K.113.

It is reasonably easy to show that the Riemann integral has the properties

which are normally assumed in elementary calculus.

Lemma 8.2.8. If f, g : [a, b] ’ R are Riemann integrable, then so is f + g

and

b b b

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

a a a

b b

Proof. Let us write I(f ) = a f (x) dx and I(g) = a g(x) dx. Suppose > 0

is given. By the de¬nition of the Riemann integral, we can ¬nd dissections

D1 and D2 of [a, b] such that

I(f ) + /4 >S(f, D1 ) ≥ I(f ) > s(f, D1 ) ’ /4 and

I(g) + /4 >S(g, D2 ) ≥ I(g) > s(g, D2 ) ’ /4.

and the de¬nition of I — (f )

If we set D = D1 ∪ D2 , then our key inequality

tell us that

I(f ) + /4 > S(f, D1 ) ≥ S(f, D) ≥ I(f ).

176 A COMPANION TO ANALYSIS

Using this and corresponding results, we obtain

I(f ) + /4 >S(f, D) ≥ I(f ) > s(f, D) ’ /4 and

I(g) + /4 >S(g, D) ≥ I(g) > s(g, D) ’ /4.

Now

n

S(f + g, D) = (xj ’ xj’1 ) sup (f (x) + g(x))

x∈[xj’1 ,xj ]

j=1

n

¤ (xj ’ xj’1 )( sup f (x) + sup g(x))

x∈[xj’1 ,xj ] x∈[xj’1 ,xj ]

j=1

= S(f, D) + S(g, D)

and similarly s(f +g, D) ≥ s(f, D)+s(g, D). Thus, using the ¬nal inequalities

of the last paragraph,

I(f ) + I(g) + /2 > S(f, D) + S(g, D) ≥ S(f + g, D)

≥ s(f + g, D) ≥ s(f, D) + s(g, D) > I(f ) + I(g) ’ /2.

Thus S(f + g, D) ’ s(f + g, D) < and |S(f + g, D) ’ (I(f ) + I(g))| < .

Exercise 8.2.9. How would you explain (NB explain, not prove) to someone

who had not done calculus but had a good grasp of geometry why the result

b b b

f (x) + g(x) dx = f (x) dx + g(x) dx

a a a

is true for well behaved functions. (I hope that you will agree with me that,

obvious as this result now seems to us, the ¬rst mathematicians to grasp this

fact had genuine insight.)

Exercise 8.2.10. (i) If f : [a, b] ’ R is bounded and D is a dissection of

[a, b], show that S(’f, D) = ’s(f, D).

(ii) If f : [a, b] ’ R is Riemann integrable, show that ’f is Riemann

integrable and

b b

(’f (x)) dx = ’ f (x) dx.

a a

(iii) If » ∈ R, » ≥ 0, f : [a, b] ’ R is bounded and D is a dissection of

[a, b], show that S(»f, D) = »S(f, D).

177

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(iv) If » ∈ R, » ≥ 0 and f : [a, b] ’ R is Riemann integrable, show that

»f is Riemann integrable and

b b

»f (x) dx = » f (x) dx.

a a

(v) If » ∈ R and f : [a, b] ’ R is Riemann integrable, show that »f is

Riemann integrable and

b b

»f (x) dx = » f (x) dx.

a a

Combining Lemma 8.2.8 with Exercise 8.2.10, we get the following result.

Lemma 8.2.11. If », µ ∈ R and f, g : [a, b] ’ R are Riemann integrable,

then »f + µg is Riemann integrable and

b b b

»f (x) + µg(x) dx = » f (x) dx + µ g(x) dx.

a a a

In the language of linear algebra, R[a, b] (the set of Riemann integrable

functions on [a, b]) is a vector space and the integral is a linear functional

(i.e. a linear map from R[a, b] to R).

Exercise 8.2.12. (i) If E is a subset of [a, b], we de¬ne the indicator func-

tion IE : [a, b] ’ R by IE (x) = 1 if x ∈ E, IE (x) = 0 otherwise. Show

directly from the de¬nition that, if a ¤ c ¤ d ¤ b, then I[c,d] is Riemann

integrable and

b

I[c,d] (x) dx = d ’ c.

a

(ii) If a ¤ c ¤ d ¤ b, we say that the intervals (c, d), (c, d], [c, d), [c, d] all

have length d ’ c. If I(j) is a subinterval of [a, b] of length |I(j)| and » j ∈ R

show that the step function n »j II(j) is Riemann integrable and

j=1

n n

b

»j II(j) dx = »j |I(j)|.

a j=1 j=1

Exercise 8.2.13. (i) If f, g : [a, b] ’ R are bounded functions with f (t) ≥

g(t) for all t ∈ [a, b] and D is a dissection of [a, b], show that S(f, D) ≥

S(g, D).

178 A COMPANION TO ANALYSIS

(ii) If f, g : [a, b] ’ R are Riemann integrable functions with f (t) ≥ g(t)

for all t ∈ [a, b], show that

b b

f (x) dx ≥ g(x) dx.

a a

(iii) Suppose that f : [a, b] ’ R is a Riemann integrable function, K ∈ R

and f (t) ≥ K for all t ∈ [a, b]. Show that

b

f (x) dx ≥ K(b ’ a).

a

State and prove a similar result involving upper bounds.

(iv) Suppose that f : [a, b] ’ R is a Riemann integrable function, K ∈ R,

K ≥ 0 and |f (t)| ¤ K for all t ∈ [a, b]. Show that

b

f (x) dx ¤ K(b ’ a).

a

Although part (iv) is weaker than part (iii), it generalises more easily and

we shall use it frequently in the form

|integral| ¤ length — sup.

Exercise 8.2.14. (i) Let M be a positive real number and f : [a, b] ’ R

a function with |f (t)| ¤ M for all t ∈ [a, b]. Show that |f (s)2 ’ f (t)2 | ¤

2M |f (s) ’ f (t)| and deduce that

sup f (x)2 ’ inf f (x)2 ¤ 2M ( sup f (x) ’ inf f (x)).

x∈[a,b] x∈[a,b]

x∈[a,b] x∈[a,b]

(ii) Let f : [a, b] ’ R be a bounded function. Show that, if D is a

dissection of [a, b],

S(f 2 , D) ’ s(f 2 D) ¤ 2M (S(f, D) ’ s(f, D)).

Deduce that, if f is Riemann integrable, so is f 2 .

(iii) By using the formula f g = 1 ((f + g)2 ’ (f ’ g)2 ), or otherwise,

4

deduce that that if f, g : [a, b] ’ R are Riemann integrable, so is f g (the

product of f and g). (Compare Exercise 1.2.6.)

Exercise 8.2.15. (i) Consider a function f : [a, b] ’ R. We de¬ne f+ , f’ :

[a, b] ’ R by

if f (t) ≥ 0

f+ (t) = f (t), f’ (t) = 0

f’ (t) = ’f (t) if f (t) ¤ 0.

f+ (t) = 0,

179

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Check that f (t) = f+ (t) ’ f’ (t) and |f (t)| = f+ (t) + f’ (t).

(ii) If f : [a, b] ’ R is bounded and D is a dissection of [a, b], show that

S(f, D) ’ s(f, D) ≥ S(f+ , D) ’ s(f+ , D) ≥ 0.

(iii) If f : [a, b] ’ R is Riemann integrable, show that f+ and f’ are

Riemann integrable.

(iv) If f : [a, b] ’ R is Riemann integrable, show that |f | is Riemann

integrable and

b b