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Note that, although we have justi¬ed the concept of a ˜delta function™, we have not
justi¬ed the concept of ˜the derivative of the delta function™. 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 de¬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 de¬nition is greatly increased if we allow improper Riemann-Stieltjes
integrals somewhat along the lines of De¬nition 9.2.14.

De¬nition 9.4.13. Let G be as throughout this section. If f : R ’ R, and
R, S > 0 we de¬ne fRS : R ’ R by

if R ≥ f (t) ≥ ’S
fRS (t) = f (t)
fRS (t) = ’S if ’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 ¬nd an L such that, given > 0, we can ¬nd an R0 ( ) > 0 such that

fRS (x) dG(x) ’ 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. (Tchebychev™s inequality.) If P is associated with a
real-valued random variable X and EX 2 exists then
EX 2
Pr{X > a or ’ a ≥ X} ¤ 2 .
a
Proof. Observe that

x2 ≥ a2 IR\(’a,a] (x)

for all x and so

a2 IR\(’a,a] (x) dG(x).
x2 dG(x) ≥
R R
223
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Thus

IR\(’a,a] (x) dG(x).
x2 dG(x) ≥ a2
R R

In other words,
EX 2 ≥ a2 Pr{X ∈ (’a, a]},
/
which is what we want to prove.
Exercise 9.4.15. (i) In the proof of Tchebchyev™s theorem we used various
simple results on improper Riemann-Stieltjes integrals without proof. Identify
these results and prove them.
(ii) If P (t) = ( π ’ tan’1 x)/π, 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 ’ 2’n if 2n ¤ t < 2n+1 , n ≥ 0 an integer.
Exercise 9.4.16. (Probabilists call this result ˜Markov™s inequality™. Ana-
lysts simply call it a ˜Tchebychev type inequality™.) Suppose φ : [0, ∞) ’ R
is an increasing continuous positive function. If P is associated with a real-
valued random variable X and Eφ(X) exists, show that
Eφ(X)
Pr{X ∈ (’a, a]} ¤
/ .
φ(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 ˜density function™). 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 ¤ y} = 1 ’ e’y for y ≥ 0. The height is measured by a gauge which
registers X = min(Y, 1). Find Pr{X ¤ 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 ¤ x, Y ¤ 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 in¬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? ™
9.5
The topic of line integrals is dealt with quickly and e¬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 ˜learning de¬nitions™ for when she studies a more
e¬cient text.
The ¬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 de¬ning a curve
is that it is a continuous map γ : [a, b] ’ Rm . If we do this it is helpful to
consider the following examples.
: [0, 1] ’ R2 with γ 1 (t) = (cos 2πt, sin 2πt)
γ1
: [1, 2] ’ R2 with γ 2 (t) = (cos 2πt, sin 2πt)
γ2
: [0, 2] ’ R2 with γ 3 (t) = (cos πt, sin πt)
γ3
: [0, 1] ’ R2 γ 4 (t) = (cos 2πt2 , sin 2πt2 )
with
γ4
: [0, 1] ’ R2 γ 5 (t) = (cos 2πt, ’ sin 2πt)
with
γ5
: [0, 1] ’ R2 with γ 6 (t) = (cos 4πt, sin 4πt)
γ6


Exercise 9.5.1. Trace out the curves γ 1 to γ 6 . State in words how the
curves γ 1 , γ 4 , γ 5 and γ 6 di¬er.
Exercise 9.5.2. (i) Which of the curves γ 1 to γ 6 are equivalent and which
are not, under the following de¬nitions.
(a) Two curves „ 1 : [a, b] ’ R2 and „ 2 : [c, d] ’ R2 are equivalent
if there exist real numbers A and B with A > 0 such that Ac + B = a,
Ad + B = b and „ 1 (At + b) = „ 2 (t) for all t ∈ [c, d].
(b) Two curves „ 1 : [a, b] ’ R2 and „ 2 : [c, d] ’ R2 are equivalent if
there exists a strictly increasing continuous surjective function θ : [c, d] ’
[a, b] such that „ 1 (θ(t)) = „ 2 (t) for all t ∈ [c, d].
(c) Two curves „ 1 : [a, b] ’ R2 and „ 2 : [c, d] ’ R2 are equivalent
if there exists a continuous bijective function θ : [c, d] ’ [a, b] such that
„ 1 (θ(t)) = „ 2 (t) for all t ∈ [c, d].
225
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(d) Two curves „ 1 : [a, b] ’ R2 and „ 2 : [c, d] ’ R2 are equivalent if
„ 1 ([a, b]) = „ 2 ([c, d]).
(ii) If you know the de¬nition of an equivalence relation verify that con-
ditions (a) to (d) do indeed give equivalence relations.
Naturally we demand that ˜equivalent curves™ (that is curves which we
consider ˜identical™) should have the same length. I think, for example, that
a de¬nition which gave di¬erent lengths to the curves described by γ 1 and
γ 2 would be obviously unsatisfactory. However, opinions may di¬er as to
when two curves are ˜equivalent™. 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 γ 1 and γ 6 should have the same length.
Most of the time, most mathematicians9 would say that the curves γ 1 and
γ 6 are not equivalent and that, ˜since γ 6 is really γ 1 done twice™, γ 6 should
have twice the length of γ 1 . If the reader is dubious she should replace the
phrase ˜length of curve™ by ˜distance traveled along the curve™.
The following chain of ideas leads to a natural de¬nition of length. Sup-
pose γ : [a, b] ’ Rm is a curve (in other words γ is continuous). As usual,
we consider dissections

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

with a = t0 ¤ t1 ¤ t2 ¤ · · · ¤ tn = b. We write
n
L(γ, D) = γ(tj’1 ) ’ γ(tj ) ,
j=1

where a ’ b is the usual Euclidean distance between a and b.
Exercise 9.5.3. (i) Explain why L(γ, D) may be considered as the ˜length
of the approximating curve obtained by taking straight line segments joining
each γ(tj’1 ) to γ(tj )™.
(ii) Show that, if D1 and D2 are dissections with D1 ⊆ D2 ,

L(γ, D2 ) ≥ L(γ, D1 ).

Deduce that, if D3 and D4 are dissections, then

L(γ, D3 ∪ D4 ) ≥ max(L(γ, D3 ), L(γ, D4 )).

The two parts of Exercise 9.5.3 suggest the following de¬nition.
9
But not all mathematicians and not all the time. One very important de¬nition of
length associated with the name Hausdor¬ agrees with the school level view.
226 A COMPANION TO ANALYSIS

De¬nition 9.5.4. We say that a curve γ : [a, b] ’ Rm is recti¬able if
there exists a K such that L(γ, D) ¤ K for all dissections D. If a curve is
recti¬able, we write

length(γ) = sup L(γ, D)
D

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

Not all curves are recti¬able.

Exercise 9.5.5. (i) Let f : [0, 1] ’ R be the function given by the conditions
f (0) = 0, f is linear on [2’n’2 3, 2’n ] with f (2’n’2 3) = 0 and f (2’n ) =
(n + 1)’1 , f is linear on [2’n’1 , 2’n’2 3] with f (2’n’2 3) = 0 and f (2’n’1 ) =
(n + 2)’1 [n ≥ 0].
Sketch the graph of f and check that f is continuous. Show that the curve
γ : [0, 1] ’ R2 given by γ(t) = (t, f (t)) is not recti¬able.
(ii) Let g : [’1, 1] ’ R be the function given by the conditions g(0) = 0,
g(t) = t2 sin |t|± for t = 0, where ± is real. Show that g is di¬erentiable
everywhere, but that, for an appropriate choice of ±, the curve „ : [’1, 1] ’
R2 given by „ (t) = (t, g(t)) is not recti¬able.

Exercise 9.5.6. (i) By using the intermediate value theorem, show that a
continuous bijective function θ : [c, d] ’ [a, b] is either strictly increasing or
strictly decreasing.
(ii) Suppose that γ : [a, b] ’ Rm is a recti¬able curve and θ : [c, d] ’ [a, b]
is a continuous bijection. Show that γ —¦ θ (where —¦ denotes composition) is
a recti¬able curve and

length(γ —¦ θ) = length(γ).

(iii) Let „ : [’1, 1] ’ R2 given by „ (t) = (sin πt, 0). Show that length(„ ) =
4. Comment brie¬‚y.

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

Exercise 9.5.7. Suppose that γ : [a, b] ’ Rm is recti¬able. Show that, if
a ¤ t ¤ b, then the restriction γ|[a,t] : [a, t] ’ Rm is recti¬able. If we write

lγ (t) = length(γ|[a,t] ),

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

With a little extra e¬ort we can say rather more about lγ .
227
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Exercise 9.5.8. We use the hypotheses and notation of Exercise 9.5.7.
(i) Suppose that γ has length L and that

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

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

L(γ, D) ≥ L ’ .

Explain why

lγ (tj ) ’ lγ (tj’1 ) ¤

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

lγ (t ) ’ lγ (t) ¤

for all 1 ¤ j ¤ n.
(ii) Use part (i) to show that lγ : [a, b] ’ 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 γ. Show that, if γ : [a, b] ’ Rm is injective, then setting θ = lγ we
have θ : [a, b] ’ [0, L] a bijective continuous map. Explain why this means
that θ’1 is continuous (see the proof of Lemma 5.6.7 if necessary). If we set
„ = γ —¦ θ’1 , show that

l„ (s) = s

for 0 ¤ s ¤ L. We say that the curve „ is the curve γ ˜reparameterised by
arc length™.

If we de¬ne Lγ : R ’ R by

for t ¤ a,
Lγ (t) = lγ (a)
Lγ (t) = lγ (t) for a < t < b,
for b ¤ t,
Lγ (t) = lγ (b)

then, since Lγ is an increasing function, we can de¬ne the Riemann-Stieltjes
integral
b
g(t) dLγ (t)
a
228 A COMPANION TO ANALYSIS

for any continuous function g : [a, b] ’ R. It follows that, if f : Rm ’ R is
continuous, we can de¬ne the ˜integral along the curve™ by
b
f (x) ds = f (γ(t)) dLγ (t).
a
γ

Note that, if the curve „ is the curve γ ˜reparameterised by arc length™ in
the sense of Example 9.5.9, then
L
f (x) ds = f („ (t)) dt.
0


We have de¬ned an integral along a curve, but we have not shown how
to calculate it. If γ is su¬ciently smooth, we can proceed as follows.
Exercise 9.5.10. Suppose x, y : [a, b] ’ R are continuous functions, >0
and A, B are real numbers such that
|(x(t) ’ x(s)) ’ A(t ’ s)|, |(y(t) ’ y(s)) ’ B(t ’ s)| ¤
for all t, s ∈ [a, b]. Show that, if γ : [a, b] ’ R2 is the curve given by
γ(t) = (x(t), y(t)), then
(min(|A| ’ , 0))2 + (min(|B| ’ , 0))2 (t ’ s)2 ¤ γ(t) ’ γ(s) 2

¤ (|A| + )2 + (|B| + )2 (t ’ s)2
for all t, s ∈ [a, b]. Deduce that γ is recti¬able and
1/2
(min(|A| ’ , 0))2 + (min(|B| ’ , 0))2 (b ’ a) ¤ length(γ)
1/2
¤ (|A| + )2 + (|B| + )2 (b ’ 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] ’ R
have continuous derivatives (with the usual conventions about left and right
derivatives at end points). Show that the curve γ : [a, b] ’ R2 given by
γ(t) = (x(t), y(t)) is recti¬able and, by considering the behaviour of
lγ (s) ’ lγ (t)
s’t
as s ’ t, show that lγ is everywhere di¬erentiable on [a, b] with
lγ (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
γ 1 to γ 6 de¬ned on page 224.

Using Exercise 9.4.10, we see that, if x, y : [a, b] ’ R2 have continuous
derivatives (with the usual conventions about left and right derivatives at end
points), and we consider the curve γ : [a, b] ’ R2 given by γ(t) = (x(t), y(t)),
then, if f : R2 ’ R is continuous,
b
f (x(t), y(t))(x (t)2 + y (t))1/2 dt,
f (x) ds =
a
γ


a result coinciding with that we would expect from mathematical methods
courses.

Exercise 9.5.13. Extend this result to smooth curves γ : [a, b] ’ 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 ˜approximating by polygonal paths™ as in
Figure 9.1.
In 1890, H. A. Schwarz10 published an example showing that any naive at-
tempt to ¬nd the area of a surface by ˜approximating by polyhedral surfaces™
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 Schwarz™s polyhedral approximation




Figure 9.3: Two more views of Schwarz

Exercise 9.5.14. (Schwarz™s counterexample.) Split a 1 by 2π rectangle
into mn rectangles each m’1 by 2πn’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 2π, and use the vertices of the 4mn
triangles as the vertices of an inscribed polyhedron with 4mn ¬‚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)’1 sin(π/n). Show that Y W has length 2(sin(π/2n))2 and
deduce, or prove otherwise, that the triangle XAB has area sin(π/n)((2m) ’1 +
4(sin(π/2n))2 ). Conclude that
1/2
π π 2
2
A(m, n) = n sin 1 + 1 + 16m sin .
n 2n
11
The last two sentences are copied directly from Billingsley™s splendid Probability and
Measure [6], since I cannot see how to give a clearer description than his.
231
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