ńņš. 26 |

justiļ¬ed the concept of ā˜the derivative of the delta functionā™. This requires a further

generalisation of our point of view to that of distributions.

7

The Holy Roman Empire was neither holy nor Roman nor an empire. A random

variable is neither random nor a variable.

222 A COMPANION TO ANALYSIS

In the same way we deļ¬ne the expectation Ef (X) by

Ef (X) = f (x) dP (x)

R

when f is Riemann-Stieltjes integrable with respect to P . The utility of

this deļ¬nition is greatly increased if we allow improper Riemann-Stieltjes

integrals somewhat along the lines of Deļ¬nition 9.2.14.

Deļ¬nition 9.4.13. Let G be as throughout this section. If f : R ā’ R, and

R, S > 0 we deļ¬ne fRS : R ā’ R by

if R ā„ f (t) ā„ ā’S

fRS (t) = f (t)

fRS (t) = ā’S if ā’S > f (t),

fRS (t) = R if f (t) > R.

If fRS is Riemann-Stieltjes integrable with respect to G for all R, S > 0, and

we can ļ¬nd an L such that, given > 0, we can ļ¬nd an R0 ( ) > 0 such that

fRS (x) dG(x) ā’ L < .

R

for all R, S > R0 ( ), then we say that f is Riemann-Stieltjes integrable with

respect to G with Riemann-Stieltjes integral

f (x) dG(x) = L.

R

|f (x)| dG(x)

(As before, we add a warning that care must be exercised if R

fails to converge.)

Lemma 9.4.14. (Tchebychevā™s inequality.) If P is associated with a

real-valued random variable X and EX 2 exists then

EX 2

Pr{X > a or ā’ a ā„ X} ā¤ 2 .

a

Proof. Observe that

x2 ā„ a2 IR\(ā’a,a] (x)

for all x and so

a2 IR\(ā’a,a] (x) dG(x).

x2 dG(x) ā„

R R

223

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

Thus

IR\(ā’a,a] (x) dG(x).

x2 dG(x) ā„ a2

R R

In other words,

EX 2 ā„ a2 Pr{X ā (ā’a, a]},

/

which is what we want to prove.

Exercise 9.4.15. (i) In the proof of Tchebchyevā™s theorem we used various

simple results on improper Riemann-Stieltjes integrals without proof. Identify

these results and prove them.

(ii) If P (t) = ( Ļ ā’ tanā’1 x)/Ļ, show that EX 2 does not exist. Show that

2

this is also the case if we choose P given by

P (t) = 0 if t < 1

P (t) = 1 ā’ 2ā’n if 2n ā¤ t < 2n+1 , n ā„ 0 an integer.

Exercise 9.4.16. (Probabilists call this result ā˜Markovā™s inequalityā™. Ana-

lysts simply call it a ā˜Tchebychev type inequalityā™.) Suppose Ļ : [0, ā) ā’ R

is an increasing continuous positive function. If P is associated with a real-

valued random variable X and EĻ(X) exists, show that

EĻ(X)

Pr{X ā (ā’a, a]} ā¤

/ .

Ļ(a)

In elementary courses we deal separately with discrete random variables

(typically, in our notation, P is constant on each interval [n, n + 1)) and con-

tinuous random variables8 (in our notation, P has continuous derivative, this

derivative is the ā˜density functionā™). It is easy to construct mixed examples.

Exercise 9.4.17. The height of water in a river is a random variable Y with

Pr{Y ā¤ y} = 1 ā’ eā’y for y ā„ 0. The height is measured by a gauge which

registers X = min(Y, 1). Find Pr{X ā¤ x} for all x.

Are there real-valued random variables which are not just a simple mix

of discrete and continuous? In Exercise K.225 (which depends on uniform

convergence) we shall show that there are.

The Riemann-Stieltjes formalism can easily be extended to deal with two

random variables X and Y by using a two dimensional Riemann-Stieltjes

integral with respect to a function

P (x, y) = Pr{X ā¤ x, Y ā¤ y}.

8

See the previous footnote on the Holy Roman Empire.

224 A COMPANION TO ANALYSIS

In the same way we can deal with n random variables X1 , X2 , . . . , Xn . How-

ever, we cannot deal with inļ¬nite sequences X1 , X2 , . . . of random variables

in the same way. Modern probability theory depends on measure theory.

In the series of exercises starting with Exercise K.162 and ending with

Exercise K.168 we see that the Riemann-Stieltjes integral can be generalised

further.

How long is a piece of string? ā™„

9.5

The topic of line integrals is dealt with quickly and eļ¬ciently in many texts.

The object of this section is to show why the texts deal with the matter in

the way they do. The reader should not worry too much about the details

and reserve such matters as ā˜learning deļ¬nitionsā™ for when she studies a more

eļ¬cient text.

The ļ¬rst problem that meets us when we ask for the length of a curve

is that it is not clear what a curve is. One natural way of deļ¬ning a curve

is that it is a continuous map Ī³ : [a, b] ā’ Rm . If we do this it is helpful to

consider the following examples.

: [0, 1] ā’ R2 with Ī³ 1 (t) = (cos 2Ļt, sin 2Ļt)

Ī³1

: [1, 2] ā’ R2 with Ī³ 2 (t) = (cos 2Ļt, sin 2Ļt)

Ī³2

: [0, 2] ā’ R2 with Ī³ 3 (t) = (cos Ļt, sin Ļt)

Ī³3

: [0, 1] ā’ R2 Ī³ 4 (t) = (cos 2Ļt2 , sin 2Ļt2 )

with

Ī³4

: [0, 1] ā’ R2 Ī³ 5 (t) = (cos 2Ļt, ā’ sin 2Ļt)

with

Ī³5

: [0, 1] ā’ R2 with Ī³ 6 (t) = (cos 4Ļt, sin 4Ļt)

Ī³6

Exercise 9.5.1. Trace out the curves Ī³ 1 to Ī³ 6 . State in words how the

curves Ī³ 1 , Ī³ 4 , Ī³ 5 and Ī³ 6 diļ¬er.

Exercise 9.5.2. (i) Which of the curves Ī³ 1 to Ī³ 6 are equivalent and which

are not, under the following deļ¬nitions.

(a) Two curves Ļ„ 1 : [a, b] ā’ R2 and Ļ„ 2 : [c, d] ā’ R2 are equivalent

if there exist real numbers A and B with A > 0 such that Ac + B = a,

Ad + B = b and Ļ„ 1 (At + b) = Ļ„ 2 (t) for all t ā [c, d].

(b) Two curves Ļ„ 1 : [a, b] ā’ R2 and Ļ„ 2 : [c, d] ā’ R2 are equivalent if

there exists a strictly increasing continuous surjective function Īø : [c, d] ā’

[a, b] such that Ļ„ 1 (Īø(t)) = Ļ„ 2 (t) for all t ā [c, d].

(c) Two curves Ļ„ 1 : [a, b] ā’ R2 and Ļ„ 2 : [c, d] ā’ R2 are equivalent

if there exists a continuous bijective function Īø : [c, d] ā’ [a, b] such that

Ļ„ 1 (Īø(t)) = Ļ„ 2 (t) for all t ā [c, d].

225

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(d) Two curves Ļ„ 1 : [a, b] ā’ R2 and Ļ„ 2 : [c, d] ā’ R2 are equivalent if

Ļ„ 1 ([a, b]) = Ļ„ 2 ([c, d]).

(ii) If you know the deļ¬nition of an equivalence relation verify that con-

ditions (a) to (d) do indeed give equivalence relations.

Naturally we demand that ā˜equivalent curvesā™ (that is curves which we

consider ā˜identicalā™) should have the same length. I think, for example, that

a deļ¬nition which gave diļ¬erent lengths to the curves described by Ī³ 1 and

Ī³ 2 would be obviously unsatisfactory. However, opinions may diļ¬er as to

when two curves are ā˜equivalentā™. At a secondary school level, most people

would say that the appropriate notion of equivalence is that given as (d) in

Exercise 9.5.2 and thus the curves Ī³ 1 and Ī³ 6 should have the same length.

Most of the time, most mathematicians9 would say that the curves Ī³ 1 and

Ī³ 6 are not equivalent and that, ā˜since Ī³ 6 is really Ī³ 1 done twiceā™, Ī³ 6 should

have twice the length of Ī³ 1 . If the reader is dubious she should replace the

phrase ā˜length of curveā™ by ā˜distance traveled along the curveā™.

The following chain of ideas leads to a natural deļ¬nition of length. Sup-

pose Ī³ : [a, b] ā’ Rm is a curve (in other words Ī³ is continuous). As usual,

we consider dissections

D = {t0 , t1 , t2 , . . . , tn }

with a = t0 ā¤ t1 ā¤ t2 ā¤ Ā· Ā· Ā· ā¤ tn = b. We write

n

L(Ī³, D) = Ī³(tjā’1 ) ā’ Ī³(tj ) ,

j=1

where a ā’ b is the usual Euclidean distance between a and b.

Exercise 9.5.3. (i) Explain why L(Ī³, D) may be considered as the ā˜length

of the approximating curve obtained by taking straight line segments joining

each Ī³(tjā’1 ) to Ī³(tj )ā™.

(ii) Show that, if D1 and D2 are dissections with D1 ā D2 ,

L(Ī³, D2 ) ā„ L(Ī³, D1 ).

Deduce that, if D3 and D4 are dissections, then

L(Ī³, D3 āŖ D4 ) ā„ max(L(Ī³, D3 ), L(Ī³, D4 )).

The two parts of Exercise 9.5.3 suggest the following deļ¬nition.

9

But not all mathematicians and not all the time. One very important deļ¬nition of

length associated with the name Hausdorļ¬ agrees with the school level view.

226 A COMPANION TO ANALYSIS

Deļ¬nition 9.5.4. We say that a curve Ī³ : [a, b] ā’ Rm is rectiļ¬able if

there exists a K such that L(Ī³, D) ā¤ K for all dissections D. If a curve is

rectiļ¬able, we write

length(Ī³) = sup L(Ī³, D)

D

the supremum being taken over all dissections of [a, b].

Not all curves are rectiļ¬able.

Exercise 9.5.5. (i) Let f : [0, 1] ā’ R be the function given by the conditions

f (0) = 0, f is linear on [2ā’nā’2 3, 2ā’n ] with f (2ā’nā’2 3) = 0 and f (2ā’n ) =

(n + 1)ā’1 , f is linear on [2ā’nā’1 , 2ā’nā’2 3] with f (2ā’nā’2 3) = 0 and f (2ā’nā’1 ) =

(n + 2)ā’1 [n ā„ 0].

Sketch the graph of f and check that f is continuous. Show that the curve

Ī³ : [0, 1] ā’ R2 given by Ī³(t) = (t, f (t)) is not rectiļ¬able.

(ii) Let g : [ā’1, 1] ā’ R be the function given by the conditions g(0) = 0,

g(t) = t2 sin |t|Ī± for t = 0, where Ī± is real. Show that g is diļ¬erentiable

everywhere, but that, for an appropriate choice of Ī±, the curve Ļ„ : [ā’1, 1] ā’

R2 given by Ļ„ (t) = (t, g(t)) is not rectiļ¬able.

Exercise 9.5.6. (i) By using the intermediate value theorem, show that a

continuous bijective function Īø : [c, d] ā’ [a, b] is either strictly increasing or

strictly decreasing.

(ii) Suppose that Ī³ : [a, b] ā’ Rm is a rectiļ¬able curve and Īø : [c, d] ā’ [a, b]

is a continuous bijection. Show that Ī³ ā—¦ Īø (where ā—¦ denotes composition) is

a rectiļ¬able curve and

length(Ī³ ā—¦ Īø) = length(Ī³).

(iii) Let Ļ„ : [ā’1, 1] ā’ R2 given by Ļ„ (t) = (sin Ļt, 0). Show that length(Ļ„ ) =

4. Comment brieļ¬‚y.

The next exercise is a fairly obvious but very useful observation.

Exercise 9.5.7. Suppose that Ī³ : [a, b] ā’ Rm is rectiļ¬able. Show that, if

a ā¤ t ā¤ b, then the restriction Ī³|[a,t] : [a, t] ā’ Rm is rectiļ¬able. If we write

lĪ³ (t) = length(Ī³|[a,t] ),

show that lĪ³ : [a, b] ā’ R is an increasing function with lĪ³ (a) = 0.

With a little extra eļ¬ort we can say rather more about lĪ³ .

227

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

Exercise 9.5.8. We use the hypotheses and notation of Exercise 9.5.7.

(i) Suppose that Ī³ has length L and that

D = {t0 , t1 , t2 , . . . , tn }

with a = t0 ā¤ t1 ā¤ t2 ā¤ Ā· Ā· Ā· ā¤ tn = b is a dissection such that

L(Ī³, D) ā„ L ā’ .

Explain why

lĪ³ (tj ) ā’ lĪ³ (tjā’1 ) ā¤

and deduce that, if tjā’1 ā¤ t ā¤ t ā¤ tj , then

lĪ³ (t ) ā’ lĪ³ (t) ā¤

for all 1 ā¤ j ā¤ n.

(ii) Use part (i) to show that lĪ³ : [a, b] ā’ R is continuous.

Exercise 9.5.9. [This a just a simple observation obscured by notation.]

We use the hypotheses and notation of Exercise 9.5.7. Let L be the length

of Ī³. Show that, if Ī³ : [a, b] ā’ Rm is injective, then setting Īø = lĪ³ we

have Īø : [a, b] ā’ [0, L] a bijective continuous map. Explain why this means

that Īøā’1 is continuous (see the proof of Lemma 5.6.7 if necessary). If we set

Ļ„ = Ī³ ā—¦ Īøā’1 , show that

lĻ„ (s) = s

for 0 ā¤ s ā¤ L. We say that the curve Ļ„ is the curve Ī³ ā˜reparameterised by

arc lengthā™.

If we deļ¬ne LĪ³ : R ā’ R by

for t ā¤ a,

LĪ³ (t) = lĪ³ (a)

LĪ³ (t) = lĪ³ (t) for a < t < b,

for b ā¤ t,

LĪ³ (t) = lĪ³ (b)

then, since LĪ³ is an increasing function, we can deļ¬ne the Riemann-Stieltjes

integral

b

g(t) dLĪ³ (t)

a

228 A COMPANION TO ANALYSIS

for any continuous function g : [a, b] ā’ R. It follows that, if f : Rm ā’ R is

continuous, we can deļ¬ne the ā˜integral along the curveā™ by

b

f (x) ds = f (Ī³(t)) dLĪ³ (t).

a

Ī³

Note that, if the curve Ļ„ is the curve Ī³ ā˜reparameterised by arc lengthā™ in

the sense of Example 9.5.9, then

L

f (x) ds = f (Ļ„ (t)) dt.

0

Ļ„

We have deļ¬ned an integral along a curve, but we have not shown how

to calculate it. If Ī³ is suļ¬ciently smooth, we can proceed as follows.

Exercise 9.5.10. Suppose x, y : [a, b] ā’ R are continuous functions, >0

and A, B are real numbers such that

|(x(t) ā’ x(s)) ā’ A(t ā’ s)|, |(y(t) ā’ y(s)) ā’ B(t ā’ s)| ā¤

for all t, s ā [a, b]. Show that, if Ī³ : [a, b] ā’ R2 is the curve given by

Ī³(t) = (x(t), y(t)), then

(min(|A| ā’ , 0))2 + (min(|B| ā’ , 0))2 (t ā’ s)2 ā¤ Ī³(t) ā’ Ī³(s) 2

ā¤ (|A| + )2 + (|B| + )2 (t ā’ s)2

for all t, s ā [a, b]. Deduce that Ī³ is rectiļ¬able and

1/2

(min(|A| ā’ , 0))2 + (min(|B| ā’ , 0))2 (b ā’ a) ā¤ length(Ī³)

1/2

ā¤ (|A| + )2 + (|B| + )2 (b ā’ a).

Exercise 9.5.11. This exercise uses the ideas of the previous exercise to-

gether with the mean value inequality. Suppose that x, y : [a, b] ā’ R

have continuous derivatives (with the usual conventions about left and right

derivatives at end points). Show that the curve Ī³ : [a, b] ā’ R2 given by

Ī³(t) = (x(t), y(t)) is rectiļ¬able and, by considering the behaviour of

lĪ³ (s) ā’ lĪ³ (t)

sā’t

as s ā’ t, show that lĪ³ is everywhere diļ¬erentiable on [a, b] with

lĪ³ (t) = (x (t)2 + y (t)2 )1/2 .

Explain why your proof does not apply to the counterexample in part (ii)

of Exercise 9.5.5.

229

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Figure 9.1: Arc length via polygonal paths

Exercise 9.5.12. Use Exercise 9.4.10 to compute the lengths of the curves

Ī³ 1 to Ī³ 6 deļ¬ned on page 224.

Using Exercise 9.4.10, we see that, if x, y : [a, b] ā’ R2 have continuous

derivatives (with the usual conventions about left and right derivatives at end

points), and we consider the curve Ī³ : [a, b] ā’ R2 given by Ī³(t) = (x(t), y(t)),

then, if f : R2 ā’ R is continuous,

b

f (x(t), y(t))(x (t)2 + y (t))1/2 dt,

f (x) ds =

a

Ī³

a result coinciding with that we would expect from mathematical methods

courses.

Exercise 9.5.13. Extend this result to smooth curves Ī³ : [a, b] ā’ Rm , giving

as much (or as little) detail as seems to you desirable.

This seems highly satisfactory until we read more advanced texts than

this and discover that, instead of using the sophisticated general ideas of

this section, these advanced texts use a rigorised version of the mathematical

methods approach. Why do they do this?

There are various reasons, but one of the most important is that the

approaches developed here do not apply in higher dimensions. More specif-

ically, we obtained arc length by ā˜approximating by polygonal pathsā™ as in

Figure 9.1.

In 1890, H. A. Schwarz10 published an example showing that any naive at-

tempt to ļ¬nd the area of a surface by ā˜approximating by polyhedral surfacesā™

must fail. The example provides a good exercise in simple three dimensional

geometry.

10

The Schwarz of the Cauchy-Schwarz inequality.

230 A COMPANION TO ANALYSIS

Figure 9.2: Part of Schwarzā™s polyhedral approximation

Figure 9.3: Two more views of Schwarz

Exercise 9.5.14. (Schwarzā™s counterexample.) Split a 1 by 2Ļ rectangle

into mn rectangles each mā’1 by 2Ļnā’1 , and split each of these into four

triangles by means of the two diagonals. Bend the large rectangle into a

cylinder of height 1 and circumference 2Ļ, and use the vertices of the 4mn

triangles as the vertices of an inscribed polyhedron with 4mn ļ¬‚at triangular

faces11 . We call the area of the resulting polyhedron A(m, n).

In Figure 9.2 we show one of the nm rectangles ABCD with diagonals

meeting at X before and after bending. In our discussion, we shall refer only

to the system after bending. Let W be the mid point of the arc AB, Y the

mid point of the chord AB and Z the mid point of the line BD as shown in

Figure 9.3.

By observing that W A = XZ, or otherwise, show that the area of the tri-

angle AXC is (2m)ā’1 sin(Ļ/n). Show that Y W has length 2(sin(Ļ/2n))2 and

deduce, or prove otherwise, that the triangle XAB has area sin(Ļ/n)((2m) ā’1 +

4(sin(Ļ/2n))2 ). Conclude that

1/2

Ļ Ļ 2

2

A(m, n) = n sin 1 + 1 + 16m sin .

n 2n

11

The last two sentences are copied directly from Billingsleyā™s splendid Probability and

Measure [6], since I cannot see how to give a clearer description than his.

231

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ńņš. 26 |