R

when f is continuous on the rectangle R.

Theorem 9.3.2. (Fubini™s theorem for continuous functions.) Let

R = [a, b] — [c, d]. If f : R ’ R is continuous, then the functions F1 : [a, b] ’

R and F2 : [c, d] ’ R de¬ned by

d b

F1 (x) = f (x, s) ds and F2 (y) = f (t, y) dt

c a

are continuous and

b d

F1 (x) dx = F2 (y) dy = f (x) dA.

a c R

This result is more usually written as

b d d b

f (x, y) dy dx = f (x, y) dx dy = f (x) dA,

a c c a R

or, simply,

b d d b

f (x, y) dy dx = f (x, y) dx dy = f (x) dA.

a c c a [a,b]—[c,d]

(See also Exercises K.152, K.154 and K.155.)

We prove Theorem 9.3.2 in two exercises.

Exercise 9.3.3. (We use the notation of Theorem 9.3.2.) If |f (x, s) ’

f (w, s)| ¤ for all s ∈ [c, d] show that |F1 (x) ’ F1 (w)| ¤ (d ’ c). Use

the uniform continuity of f to conclude that F1 is continuous.

For the next exercise we recall the notion of an indicator function IE for a

set E. If E ⊆ R, then IE : R ’ R is de¬ned by IE (a) = 1 if a ∈ E, IE (a) = 0

otherwise.

Exercise 9.3.4. We use the notation of Theorem 9.3.2. In this exercise

interval will mean open, half open or closed interval (that is intervals of the

form, (±, β), [±, β), (±, β] or [±, β]) and rectangle will mean the product of

two intervals. We say that g satis¬es the Fubini condition if

b d d b

g(x, y) dy dx = g(x, y) dy dx = g(x) dA.

a c c a R

214 A COMPANION TO ANALYSIS

> 0, we can ¬nd rectangles Rj ⊆ R and »j ∈ R

(i) Show that, given

such that, writing

N

» j IR j ,

H=

j=1

we have H(x) ’ ¤ F (x) ¤ H(x) + for all x ∈ R.

(ii) Show by direct calculation that IB satis¬es the Fubini condition when-

ever B is a rectangle. Deduce that H satis¬es the Fubini condition and use

(i) (carefully) to show that F does.

All this looks very satisfactory, but our treatment hides a problem. If we

look at how mathematicians actually use integrals we ¬nd that they want

to integrate over sets which are more complicated than rectangles with sides

parallel to coordinate axes. (Indeed one of the guiding principles of this book

is that coordinate axes should not have a special role.) If you have studied

mathematical methods you will have come across the formula for change of

variables4

‚(u, v)

f (u, v) du dv = f (u(x, y), v(x, y)) dx dy,

‚(x, y)

E E

where

E = {(u(x, y), v(x, y)) : (x, y) ∈ E}.

Even if you do not recognise the formula, you should see easily that any

change of variable formula will involve changing not only the integrand but

the set over which we integrate.

It is not hard to come up with an appropriate de¬nition for integrals over

a set E.

De¬nition 9.3.5. Let E be a bounded set and f : E ’ R a bounded func-

tion. Choose a < b and c < d such that R = [a, b] — [c, d] contains E and

˜ ˜ ˜ ˜

de¬ne f : R ’ R by f (x) = f (x) if x ∈ E, f (x) = 0 otherwise. If R f (x) dA

exists, we say that E f (x) dA exists and

˜

f (x) dA = f (x) dA.

E R

Exercise 9.3.6. Explain brie¬‚y why the de¬nition is independent of the

choice of R.

4

This formula is included as a memory jogger only. It would require substantial sup-

porting discussion to explain the underlying conventions and assumptions.

215

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

The most important consequence of this de¬nition is laid bare in the next

exercise.

Exercise 9.3.7. Let R = [a, b] — [c, d] and E ⊆ R. Let R be the set of

functions f : R ’ R which are Riemann integrable. Then E f (x) dA exists

for all f ∈ R if and only if IE ∈ R.

I (x) dA

If we think about the meaning of we are led to the following

RE

de¬nition5 .

De¬nition 9.3.8. A bounded set E in R2 has Riemann area 1 dA if that

E

integral exists.

Recall that, if R = [a, b] — [c, d], we write |R| = (b ’ a)(d ’ c).

Exercise 9.3.9. Show that a bounded set E has Riemann area |E| if and

only if, given any , we can ¬nd disjoint rectangles Ri = [ai , bi ] — [ci , di ]

[1 ¤ i ¤ N ] and (not necessarily disjoint) rectangles Rj = [aj , bj ] — [cj , dj ]

[1 ¤ j ¤ M ] such that

N M N M

Ri ⊆ E ⊆ |Ri | ≥ |E| ’ and |Rj | ¤ |E| + .

Rj ,

i=1 j=1 i=1 j=1

Exercise 9.3.10. Show that, if E has Riemann area and f is de¬ned and

Riemann integrable on some rectangle R = [a, b] — [c, d] containing E, then

f (x) dA exists and

E

f (x) dA ¤ sup |f (x)||E|.

x∈E

E

In other words

size of integral ¤ area — sup .

Our discussion tells us that in order to talk about

f (x) dA

E

we need to know not only that f is well behaved (Riemann integrable) but

that E is well behaved (has Riemann area). Just as the functions f which

occur in ˜¬rst mathematical methods™ courses are Riemann integrable, so the

sets E which appear in such courses have Riemann area, though the process

of showing this may be tedious.

5

Like most of the rest of this chapter, this is not meant to be taken too seriously. What

we call ˜Riemann area™ is traditionally called ˜content™. The theory of content is pretty

but was rendered obsolete by the theory of measure.

216 A COMPANION TO ANALYSIS

Exercise 9.3.11. (The reader may wish to think about how to do this exer-

cise without actually writing down all the details.)

(i) Show that a rectangle whose sides are not necessarily parallel to the

axis has Riemann area and that this area is what we expect.

(ii) Show that a triangle has Riemann area and that this area is what we

expect.

(iii) Show that a polygon has Riemann area and that this area is what

we expect. (Of course, the answer is to cut it up into a ¬nite number of

triangles, but can this always be done?)

However, if we want to go further, it becomes rather hard to decide which

sets are nice and which are not. The problem is already present in the

one-dimensional case, but hidden by our insistence on only integrating over

intervals.

De¬nition 9.3.12. A bounded set E in R has Riemann length if, taking any

[a, b] ⊇ E, we have IE ∈ R([a, b]). We say then that E has Riemann length

b

IE (t) dt.

|E| =

a

Exercise 9.3.13. (i) Explain why the de¬nition just given is independent of

the choice of [a, b].

(ii) Show that

IA∪B = IA + IB ’ IA IB .

Hence show that, if A and B have Riemann length, so does A ∪ B. Prove

similar results for A © B and A \ B.

(iii) By reinterpreting Exercise 9.1.1 show that we can ¬nd An ⊆ [0, 1]

such that An has Riemann length for each n but ∞ An does not.

n=1

(iv) Obtain results like (ii) and (iii) for Riemann area.

It also turns out that the kind of sets we have begun to think of as nice,

that is open and closed sets, need not have Riemann area.

Lemma 9.3.14. There exist bounded closed and open sets in R which do not

have Riemann length. There exist bounded closed and open sets in R2 which

do not have Riemann area.

The proof of this result is a little complicated so we have relegated it to

Exercise K.156.

Any belief we may have that we have a ˜natural feeling™ for how area

behaves under complicated maps is ¬nally removed by an example of Peano.

217

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

Theorem 9.3.15. There exists a continuous surjective map f : [0, 1] ’

[0, 1] — [0, 1].

Thus there exists a curve which passes through every point of a square!

A proof depending on the notion of uniform convergence is given in Exer-

cise K.224.

Fortunately all these di¬culties vanish like early morning mist in the light

of Lebesgue™s theory.

The Riemann-Stieltjes integral ™

9.4

In this section we discuss a remarkable extension of the notion of integral due

to Stieltjes. The reader should ¬nd the discussion gives an excellent revision

of many of the ideas of Chapter 8.

Before doing so, we must dispose of a technical point. When authors talk

about the Heaviside step function H : R ’ R they all agree that H(t) = 0

for t < 0 and H(t) = 1 for t > 0. However, some take H(0) = 0, some take

H(0) = 1 and some take H(0) = 1/2. Usually this does not matter but it is

helpful to have consistency.

De¬nition 9.4.1. Let E ⊆ R We say that a function f : E ’ R is a right

continuous function if, for all x ∈ E, f (t) ’ f (x) whenever t ’ x through

values of t ∈ E with t > x.

Exercise 9.4.2. Which de¬nition of the Heaviside step function makes H

right continuous?

In the discussion that follows, G : R ’ R will be a right continuous

increasing function. (Exercise K.158 sheds some light on the nature of such

functions, but is not needed for our discussion.) We assume further that

there exist A and B with G(t) ’ A as t ’ ’∞ and G(t) ’ B as t ’ ∞.

Exercise 9.4.3. If F : R ’ R is an increasing function show that the fol-

lowing two statements are equivalent:-

(i) F is bounded.

(ii) F (t) tends to (¬nite) limits as t ’ ’∞ and as t ’ ∞.

We shall say that any ¬nite set D containing at least two points is a

dissection of R. By convention we write

D = {x0 , x1 , . . . , xn } with x0 < x1 < x2 < · · · < xn .

(Note that we now demand that the xj are distinct.)

218 A COMPANION TO ANALYSIS

Now suppose f : R ’ R is a bounded function. We de¬ne the upper

Stieltjes G sum of f associated with D by

n

SG (f, D) =(G(x0 ) ’ A) sup f (t) + (G(xj ) ’ G(xj’1 )) sup f (t)

t¤x0 t∈(xj’1 ,xj ]

j=1

+ (B ’ G(xn )) sup f (t)

t>xn

(Note that we use half open intervals, since we have to be more careful about

overlap than when we dealt with Riemann integration.)

Exercise 9.4.4. (i) De¬ne the lower Stieltjes G sum sG (f, D) in the appro-

priate way.

(ii) Show that, if D and D are dissections of R, then SG (f, D) ≥ sG (f, D ).

(iii) De¬ne the upper Stieltjes G integral by I — (G, f ) = inf D S(f, D).

Give a similar de¬nition for the lower Stieltjes G integral I— (G, f ) and show

that I — (G, f ) ≥ I— (G, f ).

If I — (G, f ) = I— (G, f ), we say that f is Riemann-Stieltjes integrable with

respect to G and we write

f (x) dG(x) = I — (G, f ).

R

Exercise 9.4.5. (i) State and prove a criterion for Riemann-Stieltjes inte-

grability along the lines of Lemma 8.2.6.

(ii) Show that the set RG of functions which are Riemann-Stieltjes in-

tegrable with respect to G forms a vector space and the integral is a linear

functional (i.e. a linear map from RG to R).

(iii) Suppose that f : R ’ R is Riemann-Stieltjes integrable with respect

to G, that K ∈ R and |f (t)| ¤ K for all t ∈ R. Show that

f (x) dG(x) ¤ K(B ’ A).

R

(iv) Show that, if f, g : R ’ R are Riemann-Stieltjes integrable with

respect to G, so is f g (the product of f and g).

(v) If f : R ’ R is Riemann-Stieltjes integrable with respect to G, show

that |f | is also and that

|f (x)| dG(x) ≥ f (x) dG(x) .

R R

(vi) Prove that, if f : R ’ R is a bounded continuous function, then f is

Riemann-Stieltjes integrable with respect to G. [Hint: Use the fact that f is

uniformly continuous on any [’R, R]. Choose R su¬ciently large.]

219

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

The next result is more novel, although its proof is routine (it resembles

that of Exercise 9.4.5 (ii)).

Exercise 9.4.6. Suppose that F, G : R ’ R are right continuous increasing

bounded functions and », µ ≥ 0. Show that, if f : R ’ R is Riemann-

Stieltjes integrable with respect to both F and G, then f is Riemann-Stieltjes

integrable with respect to »F + µG and

f (x) d(»F + µG)(x) = » f (x) dF (x) + µ f (x) dG(x).

R R R

Exercise 9.4.7. (i) If a ∈ R, show, by choosing appropriate dissections, that

I(’∞,a] is Riemann-Stieltjes integrable with respect to G and

I(’∞,a] (x) dG(x) = G(a) ’ A.

R

(ii) If a ∈ R, show that I(’∞,a) is Riemann-Stieltjes integrable with respect

to G if and only if G is continuous at a. If G is continuous at a show that

I(’∞,a) (x) dG(x) = G(a) ’ A.

R

(iii) If a < b, show that I(a,b] is Riemann-Stieltjes integrable with respect

to G and

I(a,b] (x) dG(x) = G(b) ’ G(a).

R

(iv) If a < b, show that I(a,b) is Riemann-Stieltjes integrable with respect

to G if and only if G is continuous at b.

Combining the results of Exercise 9.4.7 with Exercise 9.4.5, we see that,

if f is Riemann-Stieltjes integrable with respect to G, we may de¬ne

I(a,b] (x)f (x) dG(x)

f (x) dG(x) =

R

(a,b]

and make similar de¬nitions for integrals like f (x) dG(x)

(’∞,a]

Exercise 9.4.8. Show that, if G is continuous and f is Riemann-Stieltjes

integrable with respect to G, then we can de¬ne [a,b] f (x) dG(x) and that

f (x) dG(x) = f (x) dG(x).

(a,b] [a,b]

220 A COMPANION TO ANALYSIS

Remark: When we discussed Riemann integration, I said that, in mathe-

matical practice, it was unusual to come across a function that was Lebesgue

integrable but not Riemann integrable. In Exercise 9.4.7 (iv) we saw that the

function I(a,b) , which we come across very frequently in mathematical prac-

tice, is not Riemann-Stieltjes integrable with respect to any right continuous

increasing function G which has a discontinuity at b. In the Lebesgue-Stieltjes

theory, I(a,b) is always Lebesgue-Stieltjes integrable with respect to G. (Ex-

ercise K.161 extends Exercise 9.4.7 a little.)

The next result has an analogous proof to the fundamental theorem of

the calculus (Theorem 8.3.6).

Exercise 9.4.9. Suppose that G : R ’ R is an increasing function with con-

tinuous derivative. Suppose further that f : R ’ R is a bounded continuous

function. If we set

I(t) = f (x) dG(x),

(’∞,t]

show that then I is di¬erentiable and I (t) = f (t)G (t) for all t ∈ R.

Using the mean value theorem, in the form which states that the only

function with derivative 0 is a constant, we get the following result.

Exercise 9.4.10. Suppose that G : R ’ R is an increasing function with

continuous derivative. If f : R ’ R is a bounded continuous function, show

that

b

f (x) dG(x) = f (x)G (x) dx.

(a,b] a

Show also that

∞

f (x) dG(x) = f (x)G (x) dx,

R ’∞

explaining carefully the meaning of the right hand side of the equation.

However, there is no reason why we should restrict ourselves even to

continuous functions when considering Riemann-Stieltjes integration.

Exercise 9.4.11. (i) If c ∈ R, de¬ne Hc : R ’ R by Hc (t) = 0 if t < c,

Hc (t) = 1 if t ≥ c. Show, by ¬nding appropriate dissections, that, if f : R ’

R is a bounded continuous function, we have

f (x) dHc (x) = f (c)

(a,b]

221

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

when c ∈ (a, b]. What happens if c ∈ (a, b]

/

(ii) If a < c1 < c2 < · · · < cm < b and »1 , »2 , . . . , »m ≥ 0, ¬nd a right

continuous function G : [a, b] ’ R such that, if f : (a, b] ’ R is a bounded

continuous function, we have

m

f (x) dG(x) = »j f (cj ).

(a,b] j=1

Exercise 9.4.11 shows that Riemann-Stieltjes integration provides a frame-

work in which point masses may be considered along with continuous densi-

ties6 .

The reader may agree with this but still doubt the usefulness of Riemann-

Stieltjes point of view. The following discussion may help change her mind.

What is a real-valued random variable? It is rather hard to give a proper

mathematical de¬nition with the mathematical apparatus available in 18807 .

However any real-valued random variable X is associated with a function

P (x) = Pr{X ¤ x}.

Exercise 9.4.12. Convince yourself that P : R ’ R is a right continuous

increasing function with P (t) ’ 0 as t ’ ’∞ and P (t) ’ 1 as t ’ ∞.

(Note that, as we have no proper de¬nitions, we can give no proper proofs.)

Even if we have no de¬nition of a random variable, we do have a de¬nition

of a Riemann-Stieltjes integral. So, in a typical mathematician™s trick, we

turn everything upside down.

Suppose P : R ’ R is a right continuous increasing function with P (t) ’

0 as t ’ ’∞ and P (t) ’ 1 as t ’ ∞. We say that P is associated with a

real-valued random variable X if

IE (x) dP (x)

Pr{X ∈ E} =

R

when IE is Riemann-Stieltjes integrable with respect to P . (Thus, for exam-

ple, E could be (’∞, a] or (a, b].) If the reader chooses to read Pr{X ∈ E}

as ˜the probability that X ∈ E™ that is up to her. So far as we are concerned,

Pr{X ∈ E} is an abbreviation for R IE (x) dP (x).

6