0

Show, however, that f is not Riemann integrable.

Exercise K.156. [9.3, H] The object of this exercise is to construct a

bounded open set in R which does not have Riemann length. Our con-

struction will take place within the closed interval [’2, 2] and all references

to dissections and upper sums and so forth will refer to this interval.

(i) Enumerate the rationals in [0, 1] as a sequence y1 , y2 , . . . . Let Uj =

, yk + 2’k’4 ) and U = ∞ (yk ’ 2’k’4 , yk + 2’k’4 ). Explain

j ’k’4

k=1 (yk ’ 2 k=1

—

why U is open and I (IU ) ≥ 1.

(ii) We wish to show that I— (IU ) < 1, so that IU is not Riemann integrable

and U has no Riemann length. To this end, suppose, if possible, that I— (IU ) ≥

1. Explain why this means that we can ¬nd disjoint closed intervals Ir =

[ar , br ] [1 ¤ r ¤ N ] lying inside [’2, 2] such that

N N

(br ’ ar ) ≥ 15/16 yet [ar , br ] ‚ U.

r=1 r=1

(iii) Let Kj = ([’2, 2] \ Uj ) © N [ar , br ]. Explain why Kj is a closed

r=1

bounded set. By considering the total length of the intervals making up Uj ,

show that Kj = …. Use Exercise 4.3.8 to show that

N

([’2, 2] \ U ) © [ar , br ] = ….

r=1

Deduce, by reductio ad absurdum, that I— (IU ) < 1 and U has no Riemann

length.

(iv) Show that there is a closed bounded set in R which does not have

Riemann length. (This is a one line argument.)

514 A COMPANION TO ANALYSIS

(v) Show that there are bounded closed sets and bounded open sets in

R2 which do not have Riemann area.

[The ¬rst example of this type was found by Henry Smith. Smith was a

major pure mathematician at a time and place (19th century Oxford) not

particularly propitious for such a talent. He seems to have been valued more

as a good College and University man than for anything else10 .]

Exercise K.157. [9.3, H, ‘ ] (This continues with the ideas of Exercise K.156

above.)

(i) If (a, b) ⊆ [’2, 2], ¬nd a sequence of continuous functions fn : [’2, 2] ’

R such that 0 ¤ fn (x) ¤ I(a,b) (x) for all n and fn (x) ’ I(a,b) (x) as n ’ ∞

for all x ∈ [a, b].

(ii) If Uj and U are as in Exercise K.156, ¬nd a sequence of continuous

functions fn : [’2, 2] ’ R such that 0 ¤ fn (x) ¤ IUn (x) for all n and

fn (x) ’ IU (x) as n ’ ∞ for all x ∈ [a, b].

(iii) Show that it is possible to ¬nd a sequence of bounded continuous

functions hn : [0, 1] ’ R and a function h : [0, 1] ’ R which is not Riemann

integrable such that hn (x) ’ h(x) as n ’ ∞, for each x ∈ [0, 1].

Exercise K.158. [9.4, T] Let f : [a, b] ’ R be an increasing function.

(i) If tj ∈ [a, b] and t1 ¤ t2 ¤ t3 ¤ . . . , explain why f (tj ) tends to a limit.

(ii) Suppose x ∈ [a, b]. If tj , sj ∈ [a, b],

t1 ¤ t2 ¤ t3 ¤ . . . , and s1 ¤ s2 ¤ s3 ¤ . . . ,

tj ’ x, sj ’ x and tj , sj < x for all j show that limj’∞ f (tj ) = limj’∞ f (sj ).

Show that the condition tj , sj < x for all j can not be omitted.

(iii) Suppose x ∈ [a, b]. Show that, if x ∈ (a, b] and t ’ x through values

of t with a ¤ t < x, then f (t) tends to a ˜left limit™ f (x’).

(iv) State and prove the appropriate result on ˜right limits™.

(v) If we write J(x) = f (x+) ’ f (x’) for the ˜jump™ at x, show that

J(x) ≥ 0. Show that, if a ¤ x1 < x2 < · · · < xN ’1 < xN ¤ b, then

N

J(xr ) ¤ f (b) ’ f (a).

r=1

(vi) Let us write Ek = {x ∈ [a, b] : J(x) ≥ 1/k}. Use (v) to show that

Ek is ¬nite. Deduce that E = ∞ Ek is countable and conclude that an

k=1

increasing function is continuous except at a countable set of points.

10

He even gave extra teaching on Sunday afternoon, telling his students that ˜It was

lawful on the Sabbath day to pull an ass out of the ditch™.

515

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

(vi) Let us write Ek = {x ∈ [a, b] : J(x) ≥ 1/k}. Use (v) to show that

Ek is ¬nite. Deduce that E = ∞ Ek is countable and conclude that an

k=1

increasing function is continuous except at a countable set of points.

˜ ˜

(vii) If we de¬ne f (x) = f (x+) when x ∈ E, f (x) = f (x) otherwise, show

˜

that f : [a, b] ’ R is a right continuous increasing function.

(viii) If g : R ’ R is an increasing function, show that g is continuous

except at a countable set of points E, say. If we de¬ne g (x) = f (x+) when

˜

x ∈ E, g (x) = g(x) otherwise, show that g : R ’ R is a right continuous

˜ ˜

increasing function.

(ix) Not all discontinuities are as well behaved as those of increasing

functions. If g : [’1, 1] ’ R is de¬ned by g(x) = sin(1/x) for x = 0

show that there is no choice of value for g(0) which will make g left or right

continuous at 0.

Exercise K.159. [9.4, P] De¬ne H : R ’ R by H(t) = 0 if t < 0, H(t) = 1

if t ≥ 0.

(i) If f (t) = sin(1/t) for t > 0, and f (t) = 0 for t < 0, show that, whatever

value we assign to f (0), the function f is not Riemann-Stieltjes integrable

with respect to H.

(ii) If f (t) = sin(1/t) for t < 0, and f (t) = 0 for t > 0, show that f is

Riemann-Stieltjes integrable with respect to H if and only if f (0) = 0.

(iii) By re¬‚ecting on parts (i) and (ii), or otherwise, formulate and prove

a necessary and su¬cient condition for a bounded function f : R ’ R to be

Riemann-Stieltjes integrable with respect to H.

Exercise K.160. [9.4, P, ‘ ] (i) Let H be the Heaviside function discussed

in the previous exercise. Give an example of continuous function f : R ’ R

such that f (t) ≥ 0 for all t and f is not identically zero but

f (x) dH(x) = 0.

(a,b]

(ii) By re¬‚ecting on part (i) and the proof of Exercise 8.3.3, formulate a

necessary and su¬cient condition on G so that the following theorem holds.

If f : R ’ R is continuous and bounded, f (t) ≥ 0 for all t and

f (x) dG(x) = 0,

R

then f (t) = 0 for all t.

Prove that your condition is, indeed, necessary and su¬cient.

(iii) By re¬‚ecting on part (ii), formulate a necessary and su¬cient condi-

tion on G so that the following theorem holds.

516 A COMPANION TO ANALYSIS

If f : R ’ R is continuous and bounded and

f (x)g(x) dG(x) = 0,

R

whenever g : R ’ R is continuous and bounded, it follows that f (t) = 0 for

all t.

Prove that your condition is, indeed, necessary and su¬cient.

Exercise K.161. [9.4, P, ‘ ] (i) Show that, if G : R ’ R is bounded,

increasing and right continuous and f : R ’ R is bounded, increasing and

left continuous, then f is Riemann-Stieltjes integrable with respect to G.

(ii) Suppose G : R ’ R is bounded, increasing and right continuous and

f : R ’ R is bounded, increasing and right continuous. Find necessary and

su¬cient conditions for f to be Riemann-Stieltjes integrable with respect to

G and prove that your statement is correct.

Exercise K.162. (Bounded variation.) [9.4, T] We say that a function

F : [a, b] ’ R is of bounded variation if there exists a constant K such that

whenever a = x0 ¤ x1 ¤ · · · ¤ xn = b, we have

n

|f (xj ) ’ f (xj’1 )| ¤ K.

j=1

(i) Show that any bounded increasing function G : [a, b] ’ R is of

bounded variation.

(ii) Show that the sum of two functions of bounded variation is of bounded

variation. Deduce, in particular, that the di¬erence F ’ G of two increasing

functions F, G : [a, b] ’ R is of bounded variation. What modi¬cations, if

any, would you need to make to obtain a similar result for functions F, G :

R ’ R?

(iii) Show that any function of bounded variation is bounded.

(iv) Let f : [’1, 1] ’ R be de¬ned by f (x) = x sin(1/x) for x = 0,

f (0) = 0. Show that f is continuous everywhere on [’1, 1] but not of bounded

variation. [For variations on this theme consult Exercise K.165.]

Exercise K.163. [9.4, T, ‘ ] (This continues the previous question.) In this

exercise it will be useful to remember that, to prove A ¤ B, it is su¬cient

to prove that A ¤ B + for all > 0. Suppose F : [a, b] ’ R is of bounded

variation.

(i) Explain why we can de¬ne VF : [a, b] ’ R by taking VF (t) to be the

supremum of all sums

n

|f (xj ) ’ f (xj’1 )|,

j=1

517

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

where a = x0 ¤ x1 ¤ · · · ¤ xn = t and n ≥ 1.

Explain why we can de¬ne F+ : [a, b] ’ R by taking F+ (t) to be the

supremum of all sums

n

(f (yj ) ’ f (xj )),

j=0

where a = x0 ¤ y0 ¤ x1 ¤ y1 ¤ x2 ¤ y2 ¤ · · · ¤ xn ¤ yn = t and n ≥ 0.

We de¬ne F’ : [a, b] ’ R by taking F’ (t) to be the supremum of all sums

n

(f (yj’1 ) ’ f (xj )),

j=1

where a = x0 ¤ y0 ¤ x1 ¤ y1 ¤ x2 ¤ y2 ¤ · · · ¤ xn ¤ yn = t and n ≥ 1.

(ii) Show that F+ and F’ are increasing functions and

VF = F + + F ’ , F = F + ’ F ’ .

In particular, we have shown that every function of bounded variation on

[a, b] is the di¬erence of two increasing functions. We call function VF (t) the

total variation of F on [a, t].

(iii) Suppose that we have two increasing functions G+ , G’ : [a, b] ’ R

such that G+ (a) = G’ (a) = 0 and F = G+ ’ G’ . Show that G+ (t) ≥ F+ (t)

and G’ (t) ≥ F’ (t) for all t ∈ [a, b].

Show, more precisely, that G+ ’F+ and G’ ’F’ are increasing functions.

Exercise K.164. [9.4, T, ‘ ] We use the notation and hypotheses of the

previous question. We need the results of Exercise K.158 which the reader

should either do or reread before continuing.

(i) By Exercise K.158 there is a countable (possibly ¬nite or empty) set

E+ such that

lim F+ (t) ’ lim F+ (t) > 0

t’e, t>e t’e, t<e

for e ∈ E+ , and f is continuous at each x ∈ E+ . The same result holds

/

with F+ and E+ replaced by F’ and E’ . By using Exercise K.162 (iii), or

otherwise, show that E+ © E’ = ….

(ii) Use part (i) to show that F+ and F’ are right continuous if F is.

Show that F+ and F’ are continuous if F is.

(iii) Suppose that F is continuously di¬erentiable. Show that

t t

max(F (x), 0) dx, F’ (t) = ’

F+ (t) = min(F (x), 0) dx.

a a

518 A COMPANION TO ANALYSIS

[In setting out your proof, remember that a continuous function may change

sign in¬nitely often in an interval.]

Conclude that F+ and F’ are continuously di¬erentiable if F is.

Exercise K.165. [9.4, P, ‘ ] (This exercise is not required for later parts

of the sequence.) If ± and β are real let f±β : [’1, 1] ’ R be de¬ned by

f±β (x) = x± sin(xβ ) for x = 0, f±β (0) = 0.

(i) For which values of ± and β is f±β of bounded variation?

(ii) For which values of ± and β is f±β everywhere continuous?

(iii) For which values of ± and β is f±β everywhere di¬erentiable?

(iv) For which values of ± and β is f±β everywhere di¬erentiable with

continuous derivative?

Exercise K.166. [9.4, T, ‘ ] We say that a function G : R ’ R is of

bounded variation if there exists a constant K such that, whenever x0 <

x1 < · · · < xn , we have

n

|G(xj ) ’ G(xj’1 )| ¤ K.

j=1

(i) Show that a function G : R ’ R is of bounded variation if and only if

it is the di¬erence of two bounded increasing functions.

(ii) Consider a function G : R ’ R. Show that its restriction G|[a,b] is

of bounded variation for every closed interval [a, b] if and only if it is the

di¬erence of two increasing functions.

(iii) Show that if a function G : R ’ R is of bounded variation then

there exist unique bounded increasing functions G+ , G’ : R ’ R such that

G+ (t), G’ (t) ’ 0 as t ’ ’∞ and G = G+ ’ G’ with the property that, if

F+ , F’ : R ’ R are increasing functions with G = F+ ’ F’ , then F+ ’ G+

and F’ ’ G’ increasing.

(iv) Suppose that G, G+ , G’ are as in (iii). Show that G+ and G’ are

right continuous if G is. Show that G+ and G’ are continuous if G is. Show

that G+ and G’ are continuously di¬erentiable if G is.

(v) We write GV = G+ + G’ . Identify GV as the supremum of a certain

set of sums and prove that your identi¬cation is correct.

Exercise K.167. [9.4, T, ‘ ] (i) Suppose that F, G : R ’ R are bounded

increasing right continuous functions. Show that a bounded function f :

R ’ R is Riemann-Stieltjes integrable with respect to both F and G if and

only if it is Riemann-Stieltjes integrable with respect to F + G.

(ii) Let F1 , F2 , G1 , G2 : R ’ R be bounded increasing right continuous

functions with F1 ’ G1 = F2 ’ G2 . If a bounded function f : R ’ R is

519

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

Riemann-Stieltjes integrable with respect to F1 , F2 , G1 and G2 , show that

f (x) dF1 (x) ’ f (x) dF2 (x) ’

f (x) dG1 (x) = f (x) dG2 (x).

R R R R

Exercise K.168. [9.4, T, ‘ ] We use the notation and results of Exer-

cise K.166. Although we do not use the results of Exercise K.167 directly,

they show that the path chosen is a reasonable one.

If G : R ’ R is right continuous of bounded variation, then we know that

G+ and G’ are increasing bounded right continuous functions. We say that

a bounded function f : R ’ R is Riemann-Stieltjes integrable with respect

to G if it is Riemann-Stieltjes integrable with respect to both G+ and G’ .

We write

f (x) dG+ (x) ’

f (x) dG(x) = f (x) dG’ (x).

R R R

Develop the theory of this extended integral along the lines of Section 9.4,

starting at Exercise 9.4.5 (ii) and ending at Exercise 9.4.11. Note that,

though this is easy, you must be careful to make the right adjustments. Thus,

for example, the conclusion of the result corresponding to Exercise 9.4.5 (iii)

is

f (x) dG(x) ¤ K lim GV (t),

t’∞

R

and, in Exercise 9.4.11 (ii), we can choose »j positive or negative.

Exercise K.169. (The trigonometric functions via arc length.) [9.5,

T] In section 5.5 we developed the theory of the trigonometric functions via

their di¬erential equations. In that development, the trigonometric functions

come ¬rst and angle is de¬ned in terms of those functions (see Exercise 5.5.6).

Here we reverse the process.

(i) Consider the map γ : [’1, 1] ’ R2 given by

γ(y) = ((1 ’ y 2 )1/2 , y)

where we take the positive square root. Convince yourself that this represents

an arc of the unit circle. If 1 > y ≥ 0, we de¬ne the angle θ(y) subtended

at the origin by γ(0) and γ(y) to be length of the curve γ between those

two points. If 0 ≥ y > ’1, we de¬ne the angle θ(y) subtended at the origin

by γ(0) and γ(y) to be minus the length of the curve γ between those two

points.

520 A COMPANION TO ANALYSIS

Show that

y

1

θ(y) = dt

(1 ’ t2 )1/2

0

for ’1 < y < 1. Show that there is a real number ω such that θ(y) ’ ω as

y ’ 1 through values of y < 1, and θ(y) ’ ’ω as y ’ ’1 through values of

y > ’1. We de¬ne θ(1) = ω and θ(’1) = ’ω.

(ii) Show that θ is a strictly increasing continuous function on [’ω, ω].

We may thus de¬ne a function sin : [’ω, ω] ’ R by

sin t = θ ’1 (t).

Show that sin is once di¬erentiable on (’ω, ω) with sin t = (1 ’ (sin t)2 )1/2 .

Show that sin is twice di¬erentiable on (’ω, ω) with sin t = ’ sin t.

(iii) We now de¬ne sin t = sin(2ω ’ t) for t ∈ [ω, 3ω]. Show (paying

particular attention to behaviour at ω) that sin : [’ω, 3ω] ’ R is a well

de¬ned continuous function and that sin is twice di¬erentiable on (’ω, 3ω)

with sin t = ’ sin t.

(iv) Show that we can extend the de¬nition of sin to obtain a twice

di¬erentiable function sin : R ’ R with sin t = ’ sin t.

(v) Show that, if we de¬ne cos t = sin(t ’ ω), then cos2 t + sin2 t = 1 for

all t and 0 ¤ cos s ¤ 1 for all 0 ¤ s ¤ ω. Explain why this gives the ˜right

geometrical meaning™ to cos t for 0 ¤ t ¤ ω. Show also that, if we set π = 2ω,

this gives the right geometrical meaning for π in terms of the circumference

formula for the circle.

Exercise K.170. (The trigonometric functions via area.) [9.5, T, ‘ ]

We might argue that it is better to base our de¬nition of angle on area rather

than the more delicate concept of length. Consider the part of the unit circle

shown in Figure K.3 with A the point (1, 0), B the point (x, (1 ’ x2 )1/2 ) and

C the point (x, 0). Explain why the area of the area of the sector OAB is

1

x(1 ’ x2 )1/2

(1 ’ t2 )1/2 dt.

θ(x) = +

2 x

Use the de¬nition of θ just given to obtain an appropriate de¬nition of

cos t in an appropriate range [0, ω]. Show that cos has the properties we

should expect. Extend your de¬nition so that cos is de¬ned on the whole

real line and has the properties we expect.

Exercise K.171. [9.5, G, M!] (You should treat this exercise informally.)

Here are some simple examples of a phenomenon which has to be accepted

in any reasonably advanced account of length, area and volume.

521

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

Figure K.3: The area of a sector

(i) We work in R2 . Let

Dn = {(x, y) : (x ’ n)2 + y 2 ¤ (n + 1)’2 }.

Sketch E = ∞ Dn and convince yourself that the area of E ought to be

n=1

∞ ∞ ’2

n=1 area Dn = π n=1 (n + 1) , which is ¬nite, but that the length of

the boundary of E ought to be ∞ circumference Dn = 2π ∞ (n + 1)’1 ,

n=1 n=1

which is in¬nite.

(ii) Construct a similar example in R3 involving volume and area.

(iii) We work in C. Consider

∞ n

{z : |z ’ 2’n exp(2πir/n)| ¤ 2’n’3 }.

E = {0} ∪

n=1 r=1

Convince yourself that it is reasonable to say that E has ¬nite area but that

E has boundary of in¬nite length.

(iv) (Torricelli™s trumpet, often called Gabriel™s horn.) This example goes

back historically to the very beginning of the calculus. Consider the volume

of revolution E obtained by revolving the curve y = 1/x for x ≥ 1 around

the x axis. Thus

E = {(x, y, z) ∈ R3 : y 2 + z 2 ¤ x’2 , x ≥ 1}.

∞

We write f (x) = 1/x. Convince yourself that the volume of E is π 1 f (x)2 dx,

∞

which is ¬nite11 but the curved surface of E has area 2π 1 f (x)(1+f (x)2 )1/2 dx,

which is in¬nite.

11

This result is due to Torricelli. In modern terms, this was the ¬rst in¬nite integral

ever to be considered. Torricelli ¬rst evaluated the integral by ˜slicing™ and then gave a

˜proof by exhaustion™ which met Greek (and thus modern) standards of rigour The result

created a sensation because it showed that a solid could have ˜in¬nite extent™ but ¬nite

volume. Thomas Hobbes, a political philosopher who fancied himself as mathematician,

wrote of Torricelli™s result: ˜To understand this for sense, it is not required that a man

should be a geometrician or a logician but that he should be mad™[36].

522 A COMPANION TO ANALYSIS

Thus, if we have a trumpet in the form of the curved surface of E, we

would ˜clearly™ require an in¬nite amount of paint to paint its inside (since