<< . .

. 20
( : 70)



. . >>

x∈[0,1)
(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
Please send corrections however trivial to twk@dpmms.cam.ac.uk

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

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

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

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

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

<< . .

. 20
( : 70)



. . >>