Appel W. Mathematics for Physics and Physicists

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The integral according to Mr. Lebesgue 67

vector space of µ-integrable complex-valued functions on X , up to equality

almost everywh ere.

In particular, L

(RR

RR) or simply L

denotes the vector space of functions

integrable on R with respect to Lebesgue measure, up to equality almost ev-

erywhere.

For any interval [a, b] of R, L

[ a, b] is the space of functions on [a, b]

integrable w ith respect to the restriction of Lebesgue measure to [a, b]. This

integrability condition can also be expressed as saying that the function on R

obtained by extending f by 0 outside [a, b] is in L

(R).

The following important theorem justiﬁes all the previous theory:

THEOREM 2.31 (Riesz-Fischer) Let (X , T , µ) be a measure space. Then the space

(X , µ) of complex-valued integrable functions, with the norm kf k

|f | dµ,

is a complete normed vector space. In par ticular, the spaces L

(R) and L

[a, b] are

complete normed vector spaces.

Moreover, the space C ([a, b]) of continuous functions on [a, b] is dense in the

space L

[a, b], which is therefore the completion of C ([a, b]), and similarly for the

space C (R) of continuous functions on R and L

(R).

2.2.i And today?

Lebesgue’s theory of integration became ver y quickly the standard approach

at the beginning of the twentieth century. However, some other constructions

remain useful, such as that of Stieltjes

(see S idebar 8). Generalizations of the

integral, such as the It

o stochastic integral,

or recently the work of Cécile

DeWitt-Morette and Pierre Cartier on the Fe ynman path integrals [53], have

also appeared. There is also a “pedagogical”

deﬁnition of an integral which

combines advantages of the Lebesgue and Riemann integrals, due to Kurzweil

and Henstock (see, e.g., [93]).

The so-called Riemann-Stieltjes integ ral deﬁnes integration on R with respect to a “Borel

measure”; they can be used to avoid ap pealing to distributions in some situations, and are still

of great use in probabiliy theory.

This concerns Brownian motion in p articular.

This is not the only reason it is interesting.

68 Measure theory and the Lebesgue integral

EXERCISES

 Exercise 2.1 Let d ∈ N

∗

. Prove or disprove (wit h a counterexample) the following state-

ments:

i) An open set in R

is bounded if and only if its Lebesgue measure i s ﬁnite.

ii) A Borel set in R

has positive Lebesgue measure if and only if it contains a non-empty

open set.

iii) For any integer p and 0 < p < d, the set R

×{0 }

d−p

has Lebesgue measure zero in R

 Exercise 2.2 Find the σ-algebras on R generated by a single e l ement.

 Exercise 2.3 Show that each of the following subsets generate the Bo rel σ-algebra B(R):

1. closed sets in R;

2. intervals of the type ]a, b];

3. intervals of the type ]−∞, b ].

Show th at B(R

) is generated by

1. closed sets in R

;

2. “bricks,” that is, sets of the type {x ∈ R

| a

< x

¶ b

, i = 1, . . . ,d};

3. closed half-spaces {x ∈ R

| x

¶ b}, where 1 ¶ i ¶ d and b ∈ R.

 Exercise 2.4 Let (X , T , µ) be a measure space. Prove the following properties:

i) µ(∅) = 0;

ii) µ is increasing: if A, B ∈ T and A ⊂ B, then we have µ(A) ¶ µ(B);

iii) let (A

)

n∈N

be an increasing sequence of measurable sets in T , and let A =

Then we have µ(A

) −−→

n→∞

µ(A);

iv) let (A

)

n∈N

be a decreasing sequence of measurable sets in T such that µ(A

) < +∞,

and let A =

. Then we have µ(A

) −−→

n→∞

µ(A). Show also that this is not

necessarily true if µ(A

) = +∞;

v) let (A

)

n∈N

be any sequence of measurable sets in T . Then we have



[



µ(A

) (σ-subadditivity of the measure).

 Exercise 2.5 In this exercise, we denote by O the set of open subsets in R for the usual

topology, and (as usual) by B and L the Borel σ-algebra and th e Lebesgue σ-algebra. We have

O ⊂ B ⊂ L ⊂ P(R), and we will show that all inclusions are strict.

i) Show that O $ B.

ii) Let C be the Cantor triadic set, deﬁned by

C =

p¾1

, where F

def



x ∈ R ; ∃(k

)

n∈N

∈ E

, x =

n¾1



def

)

n∈N

∈ {0, 1, 2}

; ∀n ¶ p, k

6= 1

a) Show that C is a compact set with µ(C ) = 0.

Exercises 69

b) Show that C is in one-to-one correspondance with {0, 1}

, and in one-to-one corre-

spondence with R.

c) Show that L is in one-to-one correspondance with P(R).

d) One can show that there exists an injective map from B to R. Deduce that the

second inclusion i s strict.

iii) Use Sidebar 2 on the following page to conclude that the last inclusion is also strict.

 Exercise 2.6 Let f be a map from R to R. Is it true that the two statements below are

equivalent?

i) f is almost everywhere continuous;

ii) there exists a continuous function g such that f = g almost everywhere.

 Exercise 2.7 Let f be a measurable function on a measure space (X , T , µ), such that f (x) >

0 for x ∈ A with µ(A) > 0. Show that

f dµ > 0.

Deduce from this a proof of Theorem 2.29 on page 66.

 Exercise 2.8 (Egorov ’ s theorem) Let (X , T , µ) be a ﬁnite measure space, that is, a measure

space such that µ(X ) < +∞ (for instance, X = [a, b ] with th e Lebesgue measure). Let

( f

)

n∈N

be a sequence of complex-valued measurable functions on X converging pointwise to a

function f .

For n ∈ N and k ∈ N

∗

, let

(k)

def

p¾n

x ∈ X ;



(x) − f (x)



For ﬁxed k, show that the sequence



(k)



n¾1

is increasing (for inclusion) and that the measure

ofE

(k)

tends to µ(X ) as n tends to inﬁnity. Deduce the following th eorem:

THEOREM 2.32 (Egorov) Let (X , T , µ) be a ﬁnite measure space. Let ( f

)

n∈N

be a sequence of

complex-valued measurable functions on X converging pointwise to a function f . Then, for any ǫ > 0,

there exists a measurable set A

with measure µ(A

) < ǫ such that ( f

)

n∈N

converges uniformly to f on

the complement X \ A

This theorem shows that, on a ﬁnite measure space, pointwise convergence is not so

different from uniform convergence: one can ﬁnd a set of measure arbitrarily close to the

measure of X such that the convergence is uniform on this set.

70 Measure theory an d the Lebesgue integral

Sidebar 2 (A nonmeasurable set) We describe here the construction of a subset

of R which is not measurable (for the Lebesgue σ-algebra).

Consider the following relation on R: for x, y ∈ R, we say that x ∼ y if and

only if (x − y) ∈ Q.

This relation “∼” is of course an equivalence relation.

We now appeal to the Axiom of Ch oice of set theor y

to deﬁne a subset E ⊂ [0, 1]

containing exactly one element in each equivalence class of ∼

. Note that the use

of the Axiom of Choice means that this set E is not deﬁned constructively: we know

it exists, but it is impossible to give an explicit deﬁnition of it.

We now claim that E is not in the Lebesgue σ-algebra.

To prove this, for any r ∈ Q, let E

= {x + r ; x ∈ E}.

If r, s ∈ Q and r 6= s, the sets E

and E

are disjoint. Indeed, if there exists

x ∈ E

∩E

, then by deﬁnition there exist y, z ∈ E such that x = y + r = z + s,

which implies that z − y = r − s is a rational number, and so z ∼ y. But E

contains a single element in each equivalence class, so this means that y = z and

hence r = s.

Assuming now that E is measurable, let α = µ(E). Then for each r ∈ Q, E

is measurable with measure µ(E

) = α (because it is a translate of E).

Let S

def

[

r∈Q∩[−1,1]

As a countable union of measurable sets, this set is itself measurable. If α 6= 0, then

µ(S ) = +∞ since S is a disjoint union, so we can use σ-additivity to compute

its measure. But clearly S ⊂ [−1, 2] so µ(S ) ¶ 3. The assumption α > 0 is

therefore u ntenable, and w e must ha ve α = 0. Then µ(S ) = 0. But w e ca n also

easily check that

[0, 1] ⊂ S ⊂ [−1, 2] .

This implies that µ(S ) ¾ 1, contradicting this last assumption.

It follows that E is not measurable.

This means that, for all x, y, z :

• x ∼ x;

• x ∼ y if and only if y ∼ x;

• if x ∼ y and y ∼ z, then x ∼ z.

The Axiom of Choice states for any non-empty set A, there exists a map f : P(A) →

A, called a choice function, such that f (X ) ∈ X for any non-empty subset X ⊂ A. In

other words, th e choice functio n select s, in an arbitrary way, an ele ment in any non-empty

subset of X . It is interesting to know that a more restricted form of the Axiom of Choice

(the Axiom of Countable Choice) is compatible with the assertion th at any subset of R is

Lebesgue-measurable. (But is not sufﬁcient to prove this statement). The interested reader

may look in books suc h as [51] for a very intuitive approach, or [45] whi ch i s very clear.

This means that for any x ∈ R, there is a unique y ∈ E with x ∼ y.

Solutions of exercises 71

SOLUTIONS

 Solution of exercise 2 .1

i) Wrong. Take, for instance, d = 1 and the unbounded open set

U =

[

k¾1



k, k +



which has Lebesgue measure µ(U ) =

k¾1

−k

= 1.

ii) Wrong. For d = 1, consider for example the Borel set (R\Q)∩[0, 1], which has measure

1 but does not contain any non-empty open set.

iii) True. It is easy to describe for any ǫ > 0 a subset of R

containing R

×{0}

d−p

with

measure less than ǫ.

 Solution of exercise 2 .6. Wrong. Let f = D be the Dirichlet function, equal to 1 on Q

and zero on R \Q. Then f is nowhere continuous, but f = 0 almost everywhere.

 Solution of exercise 2 .7. For any n ¾ 1, let

def



x ∈ X ; f (x) > 1 /n



The sequence (E

)

n∈N

∗

is an increasing se quence of measurable sets, and its union is



x ∈ X ; f (x) > 0



which has positi ve measure since it contains A. Using the σ-additivity, there exists n

such

that E

has positive measure. Then the integral of f is at least equal to µ(E

)/n

> 0.

 Solution of exercise 2 .8. For any k ∈ N, we have µ



(k)



−−→

n→∞

µ(X ). It is the n possible

to construct a strictly increasing sequence (n

)

k∈N

such that µ



X − E

(k)



< ǫ/2

for all k ∈ N,

and then conclude.

Chapter

Integral calculus

3.1

Integrability in pr actice

A physicist is much more likely to encounter “ concrete” functions than

abstract functions for which integrability might depend on subtle theoretical

arguments. Hence it is important to remember the usual tricks that can be

used to prove that a function is integrable.

The standard method involves two steps:

• ﬁrst, ﬁnd some general criteria proving that some “standard” functions

are integrable;

• second, prove and use comparison theorems, which show how to reduce

the integrability of a “complicated” function to t he integrability of one

of the standard ones.

3.1.a Standard functions

The situation we consider is that of functions f deﬁned on a half-open in-

terval [a, b[, where b may be either a real number or +∞, and where the

functions involved are assumed to be integrable on any interval [a, c] wh ere

a < c is a real number (for instance, f may be a continuous function on

[a, b[). This is a fairly general situation, and integrability on R may be re-

duced to integrability on [0, +∞[ and ] −∞, 0].

74 Integral calculus

PROPOSITION 3.1 (Integrability of a non-negative function) Let a be a real

and let b ∈ R ∪ {±∞}. Let f : [a, b[ → R

be a non-negative measurable

function which has a primitive F . Then f is integrable on [a, b[ if and only if F h as

a ﬁnite limit in b.

Example 3.2 The function log is integrable on ]0, 1]. The functions x 7→ e

−ax

are i ntegrable

on [0, +∞[ if and only if a > 0.

PROPOSITION 3.3 (Riemann functions t

) The function f

: t 7→ t

with α ∈

R is integrable on [1, +∞[ if and only if α < −1. It is integra ble on ]0, 1] if and

only if α > −1.

PROPOSITION 3.4 (Bertrand functions t

log

t) Th e function t 7→ t

(log t)

with α and β real numbers is integrable on [1, +∞[ if and only if

α < −1 or (α = 1 a nd β < −1).

3.1.b Comparison theorems

Once we know some integrable functions, the following simple result gives

information concerning the integrability of many more functions.

PROPOSITION 3.5 (Comparison theorem) Let g : [a, b[ → C be an integrable

function, and let f : [a, b[ → C be a measurable function. Then

i) if |f | ¶ |g|, then f is integrable on [a, b[;

ii) if f = O

(g) or if f = o

(g) and if f is integrable on any interval [ a, c] with

c < b, then f is integrable on [a, b[;

iii) if g is non-negative, f ∼

g, and f is integrable on any interval [a, c] with

c < b, then f is integrable on [a, b[.

 Exercise 3.1 Study the integrability of the following functions (in increasing order of difﬁ-

culty):

i) t 7→ cos(t) log(tan t) on ]0, π/2[;

ii) t 7→ (sin t)/t on ]0, +∞[;

iii) t 7→ 1/(1 + t

sin

t) on [0, +∞[, where α > 0.

PROPOSITION 3.6 (Asymptotic comparison) Let g be a non-negative measur-

able function on [a, b[, and let f be a measurable function on [a, b[. Assume that

g is not integrable and that f and g are both integrable on any interval [a, c] with

c < b. Then asymptotic comparison relations between f and g e xtend to their integrals

as follows:

f ∼

g =⇒

f ∼

x→b

g and f = o

(g) =⇒

f = o

x→b

R



Exchanging integrals and limits or series 75

If g and h are integrable on [a, b[, then asymptotic comparison relations between

f and h extend to the “remainders” as follows:

f ∼

h =⇒

f ∼

x→b

h and f = o

(h) =⇒

f = o

x→b





 Exercise 3.2 Show that

+∞

−x t

1 + t

dt ∼

x→0

−log x.

3.2

Exchanging integrals and limits or series

Even when dealing with “concrete” functions, Lebesgue’s theory is ex-

tremely useful because of the power of its general statements about exchanging

limits and integrals. T he main theorem is Lebesgue’s dominated convergence

theorem.

THEOREM 3.7 (Lebesgue’s dominated convergence theorem) Let (X , T , µ)

be a measure space, for instance, R or R

with the Borel σ-algebra and the Lebesgue

measure. Let ( f

)

n∈N

be a sequence of complex-valued measurable functions on X . As-

sume that ( f

)

n∈N

converges pointwise almost everywh ere to a function f : X → R,

and moreover assume that there exists a µ-integrable function g : X → R

such that

| ¶ g almost everywhere on X , for all n ∈ N . Then

i) f is µ-integrable;

ii) for any measura ble subset A ⊂ X , w e ha ve

lim

n→∞

dµ =

f dµ.

This result is commonly used both to prove the integrability of a function

obtained by a limit process, and as a way to compute the integral of such

functions.

Example 3.8 Suppose we want to compute lim

n→∞

+∞

−∞

n e

−x

cos x

1 + n

dx. By the change of vari-

able y = nx, we are led to

+∞

−∞

−y

cos( y/n)

1 + y

dy.

76 Integral calculus

The sequence of functions f

: y 7→ e

−y

cos( y/n)/(1 + y

) converges pointwise on R to

the function f : x 7→ 1/(1 + x

). Moreover, |f

| ¶ f , which is integrable on R (by comparison

with t 7→ 1/t

at ±∞).

Therefore, by Lebesgue’s theorem, the limit exists and is equal to

+∞

−∞

1 + x

dx =



arctan x



+∞

−∞

= 1.

Another useful result does not require the “domination” assumption, but

instead is restricted to increasing sequences. This condition is more restrictive,

but often very easy to check, in particular for series of non-negative terms.

THEOREM 3.9 (Beppo Levi’s monotone convergence theorem) Let (X , T , µ)

be a measure space, for instance, R or R

with the Borel σ-algebra and the Lebesgue

measure. L et ( f

)

be a sequence of non-negative measurable functions on X . As-

sume that the sequences



(x)



are increasing for all x ∈ X . Then the function

f : X → R ∪{+∞}, deﬁned by

f (x) = lim

n→∞

(x) for all x ∈ X ,

is measurable and we ha ve

lim

n→∞

dµ =

f dµ

(this quantity may be either ﬁnite or inﬁnite).

This result is often used as a ﬁrst step to prove that a function is integrable.

Here ﬁnally is the version of the dominated convergence theorem adapted

to a series of functions:

THEOREM 3.10 (Term by term integration of a series) Let (X , T , µ) be a mea-

sure space, for instance, R or R

with the Borel σ-algebra and the Lebesgue measure.

Let

be a series of complex-valued measurable functions on X such that

∞

n=0

| dµ < +∞.

Then the series of functions

converges absolutely almost everywhere on X and

its sum is µ-integrable. Moreover, its integral may be computed by term by term

integration: we have

∞

n=0

dµ =

∞

n=0

dµ.