AN INTRODUCTION TO MEASURE THEORY AND THE … · AN INTRODUCTION TO MEASURE THEORY AND THE LEBESGUE INTEGRAL MARCO M. PELOSO Contents 1. First elements of measure theory 1 1.1. Measure

AN INTRODUCTION TO MEASURE THEORY AND THE LEBESGUEINTEGRAL

MARCO M. PELOSO

Contents

1. First elements of measure theory 11.1. Measure spaces 11.2. Measures 42. Abstract integration theory 72.1. Measurable functions 72.2. Integration of non-negative functions 112.3. Integration of complex-valued functions 162.4. The space L1(µ) 193. The Lebesgue measure on R 203.1. Outer measures 203.2. The Lebesgue measure on R. 263.3. Examples 323.4. The Cantor set and its generalizations 343.5. Integrals depending on a parameter 363.6. More on L1(m) 384. Product measure spaces and the Lebesgue integral in Rn 404.1. Product measure spaces 404.2. Integration on product measure spaces 434.3. The Lebesgue integral in Rn 494.4. Polar coordinates in Rn 535. Hausdorff measaures 565.1. A quick review of submanifolds in Rn 565.2. Hausdorff measures 575.3. Hausdorff measures in Rn 606. The fundamental theorem of calculus and the theorems of Gauss and Green 646.1. The fundamental theorem of calculus and Gauss theorem 646.2. Extesion to domains with lower regularities 706.3. Green’s Theorem and identities 71References 73

Notes for the course Analisi Matematica 4 per i corsi di Laurea presso Dipartimento di Matematicadell’Universita di Milano, a.a. 2015/16.

A † denotes the proofs and the parts that are not stricly required for the exam for the a.a. 2016/17.– June 12, 2017.

MEASURE THEORY AND LEBESGUE INTEGRAL 1

In these notes we present a concise introduction to abstract measure theory and to theLebesgue integral in euclidean spaces. These notes should be considered only as a supportfor the preparation for the exam, and not as detailed introduction to the subject.

1. First elements of measure theory

We would like to indroduce a notion of measure as a function µ that assigns to every subsetE of Rn a value µ(E) ∈ [0,+∞] in such a way the following conditions are satisfied:

(i) if E1, . . . , Ek . . . are disjoint subsets, then µ(∪kEk) =∑

k µ(Ek);

(ii) if F is congruent to E, that is, obtained from E by a translation, rotation or reflection,then µ(F ) = µ(E);

(iii) µ(Q) = 1, where Q = [0, 1)n denotes the unit cube.

Unfortunately, these three conditions are mutually incompatible, as the next example shows.

Example 1.1. Assume that n = 1. Let us introduce an equivalent relation ∼ in [0, 1) by settingx ∼ y if x − y is rational. Let N be a subset of [0, 1) containing exactly one element for eachequivalent class.1 Let R = Q ∩ [0, 1) and for r ∈ R define

Nr =x+ r : x ∈ N ∩ [0, 1− r)

∪x+ r : −1x ∈ N ∩ [1− r, 1)

.

Then Nr ⊂ [0, 1) and each x ∈ [0, 1) belongs to exactly one Nr. Indeed, let y ∈ N be the elementthat is in the same equivalent class as x. If x ≥ y, then x ∈ Nr, where x − y = r ∈ R, whileif x < y, then x ∈ Nr, where x − y + 1 = r ∈ R. Thus, x ∈ Nr for some r ∈ R. On the otherhand, if x ∈ Nr ∩Ns, then we would have x− r (or x− r+ 1) and x− s (or x− s+ 1) as distinctelements in N but belonging to the same equivalent class, against our choice of N .

Suppose then that µ : P(R)→ [0,+∞] satisfy (i)-(iii). By (i) and (ii) we have that

µ(x+ r : x ∈ N ∩ [0, 1− r)

)+ µ

(x+ r : −1x ∈ N ∩ [1− r, 1)

)= µ(Nr)

for every r ∈ R. Moreover, since [0, 1) is the disjoint union of the Nr and these sets are countablymany, by (ii) and (iii) we have

1 = µ([0, 1)

)= µ

(∪r∈R Nr

)=∑r∈R

µ(Nr) =∑r∈R

µ(N) .

But this is impossible, since the right-hand-side equals +∞ is µ(N) > 0, or it equals 0 ifµ(N) = 0.

This example easily generilizes to the case n > 1.

1.1. Measure spaces. Led by the previous example, we try to define such function µ on adomain which is strictly contained in P(Rn) but still satisfying (i)-(iii). To this end we introducethe following definitions.

Definition 1.2. Given a set X we call algebra a non-empty collection A of subsets of X thatis closed under finite unions and complements, that is, if the following conditions are satisfied:

(i) if E1, . . . , Em ∈ A, then ∪mk=1Ek ∈ A;

(ii) if E ∈ A, then cE ∈ A.

1To obtain this, we need to assume the validity of the axiom of choice.

2 M. M. PELOSO

Notice that, if E ∈ A, X = E ∪ cE ∈ A, so that also ∅ = cX ∈ A.A non-empty collection A of subsets of X is called a σ-algebra if it is an algebra and it is

closed under countable unions. In other words, if the following conditions are satisfied:

(i’) if E1, E2, . . . ,∈ A, then ∪+∞k=1Ek ∈ A;

(ii’) if E ∈ A, then cE ∈ A.

Notice that a σ-algebra A is closed under countable intersections, since(∩+∞j=1 Ej

)= c

(∪+∞j=1

cEj)∈ A.

We observe that an algebra A that is closed under countable unions of disjoint subsets of X ,is a σ-algebra. Indeed, given a sequence Ek of elements in A, define

F1 = E1, Fk = Ek \( k−1⋃j=1

Ej

)= Ek

⋂c( k−1⋃j=1

Ej

). (1)

Then the Fk are disjoint and ∪+∞k=1Ek = ∪+∞

k=1Fk, which belongs to A by assumption.

Example 1.3. Let X be any set.

(1) P(X ) is a σ-algebra;

(2) if ∅,X is a σ-algebra;

(3) if X is uncountable and we set A =E : E is countable, or cE is countable2

, then A

is a σ-algebra.

It is easy to see that the intersection of σ-algebras is again a σ-algebra. Therefore, thefollowing definition makes sense.

Definition 1.4. Given any subset E of P(X ), we call M(E) the σ-algebra generated by E asthe smallest σ-algebra containing E , that is, the intersection of all σ-algebras containing E .

Observe that P(X ) is a σ-algebra containing E , so the above intersection is not empty. Thefollowing lemma is elementary, but it deserves its own statement.

Lemma 1.5. If A is a σ-algebra, and E ⊆M(F) ⊆ A, then M(E) ⊆M(F).

Proof. SinceM(F) is a σ-algebra containing E , it contains the smallest σ-algebra containing E ,i.e. M(E).

The above definition leads us to the following fundamental notion.

Definition 1.6. If X is a topological space, we call the Borel σ-algebra in X , and we denote itby BX , the σ-algebra generated by the collections of open sets in X , and a set E ∈ BX a Borelset in X .

Observe that if F is a closed set, then F ∈ BX , so are countable unions of closed sets, theircomplements, countable unions of such sets, etc.

We call a countable intersection of open sets a Gδ-set, and countable union of closed sets anFσ-set.

If X = R is endowed with its natural topology, we denote by BR the σ-algebra of Borel sets inR (with respect to this topology). It is going to play a fundamental role in remeinder of thesenotes.

The following are elementary consequences of the definition.

2In this case we say that E is co-countable.


Proposition 1.7. The σ-algebra BR is generated by each of the following sets:

(1) the collection of open intervals E1 =

(a, b) : a < b

;

(2) the collection of closed intervals E2 =

[a, b] : a < b

;

(3) the collection of half-open intervals E3 =

(a, b] : a < b

, or E ′3 =

[a, b) : a < b

;

(4) the collection of open rays E4 =

(−∞, b) : b ∈ R

, or E ′4 =

(a,+∞) : a ∈ R

;

(5) the collection of closed rays E5 =

(−∞, b] : b ∈ R

, or E ′5 =

[a,+∞) : a ∈ R

.

Proof. The elements of the Ej and E ′j are all open, or closed, or intersection of open and closed

sets, so that each of the σ-algebras in (1)-(5) is contained in BR. Conversely, it is clear thatM(E1) ⊇ BR since each open set is countable union of elements of E1. In all remaining cases,it suffices to show that open sets are contained in M(Ej) or M(E ′j). By symmetry, we consider

only the cases of M(Ej), j = 1, . . . , 5. Clearly, (a, b) = ∪+∞k=1

[a− 1

k , b−1k

], so (2) follows. Next,

(a, b) = ∪+∞k=N (a, b− 1

k ], where N is chosen so that a < b− 1N . This shows (3). To prove (4) we

consider [b,+∞) = c(−∞, b) and (−∞, a) ∩ [b,+∞) = [b, a), if b < a. Thus, (4) follows fromthe case (3) (for E ′3). Case (5) is analogous and left to the reader.

A similar description can be given for the σ-algebra of Borel sets in Rn. If X1, . . . ,Xn is acollection of non-empty sets, X =

∏nj=1Xj , we denote by πj : X → Xj the coordinate maps. If

Mj is a σ-algebra on Xj , j = 1, . . . , n, we setn⊗j=1

Mj =M(π−1j (Ej) : Ej ∈Mj , j = 1, . . . , n

)that is, the σ-algebra generated by π−1

j (Ej) where Ej ∈Mj , j = 1, . . . , n.

Proposition 1.8. We have that

BRn =n⊗j=1

BR .

Proof. (†) Let Ej ∈ BR, j = 1, . . . , n. Then π−11 (E1) = E1 × R × · · ·R, and similarly for

π−1j (Ej). Then

∏nj=1Ej = ∩nj=1π

−1j (Ej). Hence, ⊗nj=1BR contains the σ-algebra generated

by∏n

j=1Ej : Ej open in R

=: F , i.e.⊗nj=1BR ⊇ M(F). On the other hand, for each

j = 1, . . . , n, the setEj ⊆ R : π−1

j (Ej) ∈M(cF )

is a σ-algebra, that contains BR. Hence,

n⊗j=1

BR =M( n∏

j=1

Ej : Ej open in R)

.

This implies that ⊗nj=1BR ⊆ BRn .To see the reverse inclusion, let E be any open set in Rn and consider its points with rational

coordinates, which is a dense subset in E. Take a ngbh, contained in E, of any such point thatis a cartesian product of open sets in R in each coordinate. Then E is union of open sets thatare of the form

∏nj=1Ej with Ej open in R. Thus,

BRn ⊆M( n∏

j=1

Ej : Ej open in R)

=n⊗j=1

BR .

This proves the proposition.

4 M. M. PELOSO

1.2. Measures. Let X be a set and M a σ-algebra in P(X ). The pair (X ,M) is called ameasurable space.

Definition 1.9. Given a measurable space (X ,M), a function µ : M → [0,+∞] is called ameasure on X if

(i) µ(∅) = 0;

(ii) if E1, E2, · · · ∈ M, are disjoint, then µ(+∞⋃j=1

Ej

)=

+∞∑j=1

µ(Ej).

The triple (X ,M, µ) is called a measure space.

Property (ii) is called countable additivity, in contrast with an analogous property

(ii’) if E1, E2, . . . , Em ∈M, are disjoint, then µ(∪mj=1 Ej

)=∑m

j=1 µ(Ej),

which will be called finite additivity. Notice that a measure µ satisfies also (ii’) since we maytake Ek = ∅ for all k ≥ 1.

Given a measure space (X ,M, µ), µ is said to be finite if µ(X ) < +∞ and is said to be σ-finiteif there exists a collection of sets Ej, with Ej ∈M, ∪+∞

j=1Ej = X and such that µ(Ej) < +∞.In other words, µ is said to be σ-finite if X is countable union of sets of finite measure. Ourtreatment will essentially concern only with σ-finite measures.

Here are some simple examples of measure spaces.

Example 1.10. (1) Let X be any set, and letM = P(X ). Define the counting measure onX by setting

µ(E) =

+∞ if E is infinite

m if E contains exactly m elements

0 if E = ∅ .

Then µ is a measure. Notice that µ(x) = 1 for all x ∈ X , and that it is σ-finite if Xis countable, and it is finite if X is finite.

(2) A particular case of (1) is when X is countable, or more in particular, equals N or Z.

(3) Instead, a generalization of (1) is the following. Let f : X → [0,+∞] be given and define

µ(E) =∑x∈E

f(x) .

Since f(x) is non-negative, there is no ambiguity in the above definition even if E isuncountable. For, if there exist uncountably many x ∈ E such that f(x) > 0, then thereexists n ∈ N for which En =

x ∈ E : f(x) > 1/n

is infinite. Then,

∑x∈E f(x) ≥∑

x∈Enf(x) = +∞. If there exist at most countably many x ∈ E such that f(x) > 0,

then the sum becomes a series, and the notion of convergence is the standard one. It iseasy to chech again that such µ is a measure.

A particular case is when f is the function that equals 1 at a given x0 ∈ X and it is0 anywhere else. Such measure, is called the Dirac delta at x0.

(4) if X is uncountable, let A be the σ-algebra of countable or co-countable sets. Define

µ(E) =

0 if E is countable

1 if E is co-countable .


Again, it is easy to check that such µ is a measure on A.

The first elementary properties of measures are given in the next result.

Proposition 1.11. Let (X ,M, µ) be a measure space. Then, the following properties hold true.

(i) Monotonicity: If E,F ∈M with E ⊆ F , then µ(E) ≤ µ(F ).

(ii) Subadditivity: If Ej ⊂ M, then

µ(+∞⋃j=1

Ej

)≤

+∞∑j=1

µ(Ej) .

(iii) Continuity from below: If Ej ⊂ M and E1 ⊆ E2 ⊆ · · · , then

limj→+∞

µ(Ej) = µ(+∞⋃j=1

Ej

).

(iv) Continuity from above: If Ej ⊂ M and E1 ⊇ E2 ⊇ · · · and µ(E1) < +∞ then

limj→+∞

µ(Ej) = µ(+∞⋂j=1

Ej

).

Proof. (i) Since F ⊇ E, we have that

µ(F ) = µ((F \ E) ∪ (F ∩ E)

)= µ

((F \ E) ∪ E

)= µ(F \ E) + µ(E) ≥ µ(E) .

(ii) Given Ej, we define Fj as in (1), that is, we set

F1 = E1, Fk = Ek \( k−1⋃j=1

Ej

)for k ≥ 2 .

Since the Fj are disjoint and ∪+∞j=1Fj = ∪+∞

j=1Ej , by (i) we have that

µ(+∞⋃j=1

Ej

)= µ

(+∞⋃j=1

Fj

)=

+∞∑j=1

µ(Fj) ≤+∞∑j=1

µ(Ej) .

(iii) Let now Ej ⊂ M and E1 ⊆ E2 ⊆ · · · . Setting E0 := ∅, we have that the sets Ej \Ej−1

are disjoint and ∪+∞j=1Ej \ Ej−1 = ∪+∞

j=1Ej . Then,

µ(+∞⋃j=1

Ej

)=

+∞∑j=0

µ(Ej \ Ej−1

)= lim

n→+∞

n∑j=0

µ(Ej \ Ej−1

)= lim

n→+∞µ(En) .

(iv) Set Fj = E1 \Ej . Then F1 ⊆ F2 ⊆ · · · , and µ(E1) = µ(Fj) + µ(Ej), since Ej and Fj aredisjoint, and ∪+∞

j=1Fj = E1 \ ∩+∞j=1Ej . Finally, using (iii) we have

µ(E1) = µ(+∞⋂j=1

Ej

)+ limj→+∞

µ(Fj) = µ(+∞⋂j=1

Ej

)+ limj→+∞

µ(E1)− µ(Ej) .

Since µ(E1) < +∞, we can substract it from both side of the equation to obtain (iv).

6 M. M. PELOSO

We remark that the assumption µ(E1) < +∞ could be replaced by µ(En) < +∞ for somen, but the finiteness of some µ(En) is necessary. For, consider the case of (N,P(N), µ), whereµ is the counting measure, and the sets Ej = n : n ≥ j. Then, µ(Ej) = +∞ for all j butµ(∩+∞j=1 Ej

)= µ(∅) = 0.

Let (X ,M, µ) be a measure space. A very important class of sets, is the class of sets of measurezero, also called null sets, that is, the sets E such that µ(E) = 0. By countable subadditivity,a countable union of null sets is again a null set. By monotonicity, if E ∈ M, F ⊆ E, F ∈ Mand µ(E) = 0, then also µ(F ) = 0. But, in general, it is not true that F ∈M.

Definition 1.12. A measure space (X ,M, µ) is said to be complete if M contains all subsetsof null sets, that is, for every E ∈M with µ(E) = 0 and F ⊆ E, we have F ∈M.

The next result shows that any measure space can be extended to a complete measure space.

Theorem 1.13. Let (X ,M, µ) be a measure space and let M = M∪ N , where N =E ∈

P(X ) : there exists F ∈M with µ(F ) = 0

. If E ∪N ∈M, with E ∈M and N ∈ N , define

µ(E ∪N

)= µ(E) .

Then (X ,M, µ) is a complete measure space, called the completion of (X ,M, µ).

Observe that the completion of (X ,M, µ) is uniquely determined, in the sense that once Mis constructed, there exists a unique measure µ on M that is complete and restricted to Mcoincides with µ.

Proof. We begin by showing that M is a σ-algebra. If Ej ⊆ M, then Ej = Ej ∪ Nj , whereEj ∈ M and Nj ∈ N , for all j. Since NJ ∈ N , there exists Fj ∈ M such that Nj ⊆ Fj and

µ(Fj) = 0. Therefore, ∪+∞j=1Nj ⊆ ∪+∞

j=1Fj =: F , where µ(F ) ≤∑+∞

j=1 µ(Fj) = 0. Therefore,

∪+∞j=1Nj ∈ N . Hence,

+∞⋃j=1

Ej =+∞⋃j=1

(Ej ∪Nj

)=(+∞⋃j=1

Ej

)∪(+∞⋃j=1

Nj

)∈M∪N =M .

Next, let E′ = E ∪N ∈ M∪N . Then there exists F ∈ M such that N ⊆ F and µ(F ) = 0.Then, E ∪N =

(E ∪ F

)∩(cF ∪N) and

c(E ∪N

)= c

(E ∪ F

)∪(F \N

).

Notice that c(E ∪ F

)∈ M, while F \N ⊆ F , with µ(F ) = 0, so that F \N ∈ N . This shows

that M is a σ-algebra.We now show that µ is well defined and is a complete measure. If E1 ∪ N1 = E2 ∪ N2

with Ej ∈ M and Nj ∈ N , with Nj ⊆ Fj and µ(Fj) = 0, j = 1, 2, then E1 ⊆ E2 ∪ F2 andµ(E1) ≤ µ(E2) +µ(F2) = µ(E2). Thus, µ(E1) ≤ µ(E2). Arguing in the same way we obtain thereverse inequality so that µ(E1) = µ(E2). Therefore, µ is well defined.

Next, let Ej ∪Nj be a sequence of disjoint sets in M. If Fj ∈M, Nj ⊆ Fj and µ(Fj) = 0,then

µ(+∞⋃j=1

(Ej ∪Nj))

= µ((+∞⋃

j=1

Ej)∪(+∞⋃j=1

Nj

))= µ

(+∞⋃j=1

Ej

)=

+∞∑j=1

µ(Ej) =+∞∑j=1

µ(Ej ∪Nj) .

It follows that µ is a measure, and it is clear that it is complete.


2. Abstract integration theory

In order to begin our approach to the theory of integration, we need to discuss the notion ofmeasurable functions.

2.1. Measurable functions. In analogy with the definition of continuous functions, as mor-phisms between topological spaces, we have the following definition.

Definition 2.1. Let (X ,M), (Y,N ) be measurable spaces, and f : X → Y be given. We saythat f is (M,N )-measurable if f−1(E) ∈M for all E ∈ N .

It is clear that if f : X → Y is (M,N )-measurable and g : Y → Z is (N ,O)-measurable,then g f : X → Z is (M,O)-measurable. Also, observe that if N is a σ-algebra in P(Y),then

f−1(E) : E ∈ N

is a σ-algebra in P(X ). In fact, ∪jf−1(Ej) = f−1

(∪j Ej

)and

c(f−1(E)

)= f−1( cE), and the conclusion follows.

Proposition 2.2. Let f : X → Y be (M,N )-measurable and suppose that N is generated by E.Then f is (M,N )-measurable if and only if f−1(E) ∈M for all E ∈ E.

Hence, if X ,Y are topological spaces, and f : X → Y continuous, then f is (BX ,BY)-measurable.

Proof. First of all, we observe that the implication “only if” is trivial. Conversely, suppose thatf−1(E) ∈ M for all E ∈ E . It is easy to check that

E ⊆ Y : f−1(E) ∈ M

is a σ-algebra in

P(Y) that contains E . Hence, it contains the σ-algebra generated by E , that is, it contains Nand f is (M,N )-measurable. The second conclusion now is obvious.

Definition 2.3. If (X ,M) is a measurable space, then f : X → R is said to be measurable if itis (M,BR)-measurable.

It is however convenient to consider functions that take value in the extended reals, that is,in R = R ∪ ±∞ = [−∞,+∞]. In this case, the σ-algebra of Borel sets BR is defined asE ⊆ R : E ∩R ∈ BR

. Then we say that f : X → R is measurable if it is (M,BR)-measurable.

Proposition 2.4. Let (X ,M) be a measurable space, and let f : X → R be given. Then thefollowing conditions are equivalent.

(i) f is measurable, that is, it is (M,BR)-measurable.

(ii) f−1([−∞, b)

)∈M for every b ∈ R.

(iii) f−1([−∞, b]

)∈M for every b ∈ R.

(iv) f−1((a,+∞]

)∈M for every a ∈ R.

(v) f−1([a,+∞]

)∈M for every a ∈ R.

Proof. If f : X → R, the conclusions follow at once from Prop.’s 1.7 and 2.2. If f takes valuesin [−∞,+∞], the conclusions are a simple consequence of Def. 2.3.

The following simple result will be needed later on.

Lemma 2.5. Let (X ,M) be a measurable space, and let f : X → C be given. Then f ismeasurable if and only if Re f and Im f are measurable.

8 M. M. PELOSO

Proof. (†) We identify f with (f1, f2) : X → R2. Observe that the coordinate maps πj :R2 → R, j = 1, 2 are measurable. Since composition of measurable functions is measurable andf1 = Re f = π1 f , f2 = Im f = π2 f , f measurable implies Re f, Im f are measurable.

Conversely, suppose Re f and Im f are measurable. By Prop. 1.8 we know that BC = BR2 =BR ⊗ BR. Then, if E ∈ BC, E = E1 × E2, with E1, E2 ∈ BR and f−1(E) = f−1

1 (E1) ∩ f−12 (E2)

that is in M by assumption.

We remark that when we require that f takes values in R we include the case of f havingfinite values.

Theorem 2.6. Let (X ,M) be a measurable space.

(i) If f, g : X → R are measurable, then f + g, fg are measurable.

(ii) If fj : X → R are measurable, j = 1, 2, . . . , then

g1 = supjfj , g2 = inf

jfj , g3 = lim sup

j→+∞fj , g4 = lim inf

jfj

are measurable.

(iii) f, g : X → R are measurable, then

max(f, g), min(f, g)

are measurable.

(iv) If fj : X → C are measurable, j = 1, 2, . . . , and g(x) = limj→+∞ fj(x) exists, then g ismeasurable.

Proof. (i) Observe that f + g = S F , where F (x) = (f(x), g(x)) and S : C × C → C is thesum-funciton, i.e. S(z + w) = z + w. Arguing as in Lemma 2.5 we see that F is measurable.Since S is continuous is measurable, and then f + g is measurable. Replacing S by P , whereP (z, w) = zw, we obtain the measurability of fg.

(ii) Notice that

g−11

((a,+∞]

)=x ∈ X : sup

jfj(x) > a

=

+∞⋃j=1

f−1j

((a,+∞]

),

while

g−12

([−∞, b)

)=x ∈ X : inf

jfj(x) < b

=

+∞⋃j=1

f−1j

([−∞, b)

).

The measurability of g1 and g2 follows from Prop. 2.4.Next, we observe that3

g3(x) = lim supj→+∞

fj(x) = infk

(supj≥k

fj(x))

Then hk(x) = supj<k fj(x) are measurable, so is infk hk(x) = g3(x). The argument for g4 isanalogous, since

g4(x) = lim infj→+∞

fj(x) = supk

(infj≥k

fj(x)).

3We recall that, given a sequence an, lim supn an = s∗, where s∗ is sup of the limit points of an. Then,s∗ = infk(supn≥k an) = limk→∞(supn≥k an). Analogously, if we set s∗ = lim infn an, then s∗ = supk(infn≥k an) =

limk→∞(infn≥k an).


Finally, (iii) and (iv) are trivial consequences of (ii)

Corollary 2.7. If f : X → R is measurable, then f+ = max(f, 0) and f− = −min(f, 0) aremeasurable. If g : X → C is measurable, then |g| and sgn g := g/|g| are measurable.

We point out that f = f+ − f−, |f | = f+ + f−, and g = sgn g · |g|.

Proof. We only need to prove the statements for |g| and sgn g. But these are elementary andwe leave the details to the reader.

Definition 2.8. Let (X ,M) be a measurable space, and let E ∈M. We define the characteristicfunction of E as the function

χE(x) =

1 if x ∈ E0 otherwise .

We call a simple function a finite linear combination with complex coefficients of characteristicfuncitons of measurable sets Ej

f(x) =n∑j=1

cjχEj (x) . (2)

Observe that simple functions can characterized as the measurable functions whose range isfinite (that is, they attain at most finitely many values). We remark that the representation (2)of a simple function f is not unique, but it becomes unique if we write

f(x) =

m∑j=1

djχFj (x) .

where Fj = f−1(dj), dj 6= 0, andd1, . . . , dm

= f

(X)\ 0 is the range of f taken away the

value 0.Notice also that finite sums and products of simple functions are again simple functions.

The main result of this section is that measurable can be suitable approximated with simplefunctions, as the next result shows.

Theorem 2.9. Let (X ,M) be a measurable space.(1) Let f : X → [0,+∞] be a measurable funtion. Then there exists an increasing sequence of

non-negative measurable functions 0 ≤ s1 ≤ s2 ≤ · · · ≤ f such that sn(x) → f(x) as n → +∞and the convergence is uniform on all sets where f is bounded.

(2) Let f : X → C be a measurable funtion. Then there exists a sequence of complex-valuedsimple measurable functions ϕn such that 0 ≤ |ϕ1| ≤ |ϕ2| ≤ · · · ≤ |f | such that ϕn(x)→ f(x)as n→ +∞ and the convergence is uniform on all sets where f is bounded.

Proof. (1) For n = 0, 1, 2, . . . and k integer, 0 ≤ k ≤ 22n − 1 we define the sets

Fn = f−1((2n,+∞]

)En,k = f−1

((k2−n, (k + 1)2−n]

),

and set

sn(x) = 2nχFn(x) +22n−1∑k=0

k2−nχEn,k(x) .

10 M. M. PELOSO

Clearly sn are measurable and non-negative. In order to check that sn is increasing, noticethat(

k2−n, (k + 1)2−n]

=(2k2−(n+1), (2k + 2)2−(n+1)

]=(2k2−(n+1), (2k + 1)2−(n+1)

]∪((2k + 1)2−(n+1), (2k + 2)2−(n+1)

].

This implies that

En,k = En+1,2k ∪ En+1,2k+1 . (3)

Moreover,

(2n, 2n+1

]=

22(n+1)−1⋃j=22n+1

(j2−(n+1), (j + 1)2−(n+1)

],

so that

Fn \ Fn+1 =22(n+1)−1⋃j=22n+1

En+1,j . (4)

Therefore, using (3) and (4),

sn(x) = 2nχFn(x) +

22n−1∑k=0

k2−n(χEn+1,2k

(x) + χEn+1,2k+1(x))

= 2n(χFn+1(x) + χFn\Fn+1

(x))

+22n−1∑k=0

2k2−(n+1)(χEn+1,2k

(x) + χEn+1,2k+1(x))

≤ 2n(χFn+1(x) + χFn\Fn+1

(x))

+22n−1∑k=0

2k2−(n+1)χEn+1,2k(x) +

22n−1∑k=0

(2k + 1)2−(n+1)χEn+1,2k+1(x)

= 2nχFn+1(x) + 2n22(n+1)−1∑j=22n+1

χEn+1,j (x) +22n+1−1∑`=0

`2−(n+1)χEn+1,`(x)

≤ 2n+1χFn+1(x) +22(n+1)−1∑j=22n+1

j2−(n+1)χEn+1,j (x) +22n+1−1∑`=0

`2−(n+1)χEn+1,`(x)

= sn+1(x) .

Hence, sn is monotone increasing. It is clear that sn(x) ≤ f(x) for all x. Finally, notice that forx /∈ Fn, 0 ≤ f(x)− sn(x) ≤ 2−n. Then, sm → f uniformly on each set cFn =

x : f(x) ≤ 2n

.

On F = ∩nFn, f(x) = +∞ and sn(x)→ +∞ for x ∈ F . This proves (1).(2) now follows easily from (1). If f : X → C, we write f = u+ iv, u = u+−u−, v = v+− v−.

Then, there exists increasing sequences s±n and σ±n of non-negative simple functions suchthat

s±n → u±, and σ±n → v± ,

as in (1). Then, setting

ϕn = s+n − s−n + i

(σ+n − σ−n

)


we have that ϕn → u+ − u− + i(v+ − v−) = f , pointwise, and uniformly on all sets where f isbounded. Finally, it also holds that

|ϕn| =[(s+n − s−n )2 + (σ+

n − σ−n )2]1/2

=[(s+n )2 + (s−n )2 + (σ+

n )2 + (σ−n )2]1/2

≤[(s+n+1)2 + (s−n+1)2 + (σ+

n+1)2 + (σ−n+1)2]1/2

≤ |ϕn+1|

for all n, where we have used the obvious fact that g+g− = 0 for all g.

Definition 2.10. Given a measure space (X ,M, µ), we say that a property (P) is valid µ-a.e.(or, simply, a.e. if the measure µ is understood by the context), if there exists a set F ∈ M,with µ(F ) = 0 such that the property (P) holds on cF .

One convenience of working with complete measures is the following result.

Proposition 2.11. Let (X ,M, µ) be a measure space The following implications are true if andonly if the measure µ is complete.

(1) If f is measurable and g = f µ-a.e., then g is measurable.

(2) If fn are measurable, n = 1, 2, . . . and fn → f pointwise µ-a.e., then f is measurable.

Proof. (†) Exercise.

On the other hand, we have the following result.

Proposition 2.12. Let (X ,M, µ) be a measure space and let (X ,M, µ) be its completion. If fis M-measurable, then there exists g M-measurable function such that f = g µ-a.e.

Proof. The proof is indicative of a method that will be often used. Suppose first f is a charac-teristic function of an M-measurable set E = E ∪ N , with E ∈ M, N ⊆ F , µ(F ) = 0. Thenf = χE = χE µ-a.e. and the conclusion holds true in this case. In the same way we see that theconclusion follows in the case of a simple function f =

∑nj=1 cjχEj

.

In the general case, given f an M-measurable function, by Thm. 2.9 we can find a sequenceof M-measurable functions ϕn pointwise converging to f . For each n, let ψn be an M-measurable function, ψn = ϕn except on a set Nn ∈ M with µ(Nn) = 0. Then, there existFn ∈ M, Fn ⊇ Nn and µ(Fn) = 0. Set F = ∪nFn and define g = limn→+∞ χ cFψn. Then g isM-measurable by the previous Prop. 2.11 (2), and g = f on cF , that is, µ-a.e.

2.2. Integration of non-negative functions. We begin the construction of the integral withthe simple functions.

Definition 2.13. Let (X ,M, µ) be a measure space. We define

L+ =f : X → [0,+∞], M−measurable

. (5)

If s ∈ L+ is a simple function s =∑n

j=1 cjχEj we define the integral of s with respect to µ,∫s dµ =

n∑j=1

cjµ(Ej) ,

12 M. M. PELOSO

(with the standard convention that 0 · ∞ = 0). More generally, if E ∈M we also define∫Es dµ =

∫χEs dµ =

n∑j=1

cjµ(E ∩ Ej) .

We will use the following notation to denote the integral of s w.r.t. µ:∫s dµ =

∫s =

∫Xs dµ =

∫Xs(x) dµ(x)

and similarly in the case of∫E s dµ.

The following properties are easy to check.

Proposition 2.14. Let (X ,M, µ) be a measure space and s, σ be non-negative simple functionson X . Then, the following properties hold true.

(i) If c ≥ 0, then

∫cs dµ = c

∫s dµ ;

(ii)

∫(s+ σ) dµ =

∫s dµ+

∫σ dµ;

(iii) if s ≤ σ, then

∫s dµ ≤

∫σ dµ;

(iv) the map M3 E 7→∫Es dµ is a measure.

We are now in the position to define the integral of non-negative measurable functions.

Definition 2.15. Let (X ,M, µ) be a measure space. We define

L+ =f : X → [0,+∞], M−measurable

. (6)

If f ∈ L+ we set ∫f dµ = sup

∫s dµ : 0 ≤ s ≤ f, s simple

. (7)

For any E ∈M we also define ∫Ef dµ =

∫fχE dµ .

It follows at once that, if c ≥ 0 and f ∈ L+, then cf ∈ L+ and∫cfdµ = c

∫fdµ. It also

follows at once that if f, g ∈ L+ and f ≤ g, then∫f dµ ≤

∫g dµ . (8)

The next result is a fundamental building block of this theory.

Theorem 2.16. (Monotone Convergence Theorem.) Let (X ,M, µ) be a measure spacefn be given such that fn : X → [0,+∞], 0 ≤ f1 ≤ f2 ≤ · · · . Setting f(x) = limn→+∞ fn(x),we have ∫

f dµ = limn→+∞

∫fn dµ .

Notice that the conclusion can we written in the suggestive form∫ (lim

n→+∞fn

)dµ = lim

n→+∞

∫fn dµ .


Proof. Since 0 ≤ f1(x) ≤ f2(x) ≤ · · · is a monotone sequence the limit function f exists, and itis measurable by Thm. 2.6 (iv). Moreover, by the mononicity of the integral, property (8),∫

f1 dµ ≤∫f2 dµ ≤ · · · ,

that is, the sequence ∫

fndµ

is increasing, so that it has a limit and since∫fndµ ≤

∫fdµ for

all n4, by the comparison test

limn→+∞

∫fn dµ ≤

∫f dµ .

In order to prove the reverse inequality, let s be a simple function such that 0 ≤ s ≤ f andlet 0 < α < 1. Since limn→+∞ fn(x) = f(x), setting

En =x : fn(x) ≥ αs(x)

,

we have that E1 ⊆ E2 ⊆ · · · , and ∪+∞n=1En = X . Therefore, by the monotonicity of the integral,

α

∫En

s dµ ≤∫En

fn dµ =

∫χEnfn dµ ≤

∫fn dµ . (9)

By Prop. 2.14 (iv), since En is an increasing sequence of measurable sets whose union is X ,limn→+∞

∫Ens dµ =

∫s dµ, therefore, passing to the limit in (9) we obtain

α

∫s dµ ≤ lim

n→+∞

∫fn dµ .

This holds for all 0 < α < 1 so that∫s dµ ≤ lim

n→+∞

∫fn dµ .

Passing to the supremum on the left hand side for all simple functions s with 0 ≤ s ≤ f weobtain ∫

f dµ ≤ limn→+∞

∫fn dµ ,

and we are done.

Corollary 2.17. (1) Let f, g ∈ L+. Then∫(f + g) dµ =

∫f dµ+

∫g dµ .

(2) If fn ∈ L+, n = 1, 2, . . . , then∫ ( +∞∑n=1

fn

)dµ =

+∞∑n=1

∫fn dµ

Notice that part (1) is just the statement of the additivity of the integral. In the presenttheory, in order to obtain such property we had to recourse to the Monotone ConvergenceTheorem (that we will abbreviate as MCT in what follows).

14 M. M. PELOSO

Proof. Let sn, σn be sequences of non-negative simple functions, monotonically convergingto f and g, respectively. Then, sn + σn is a sequence of non-negative simple functions,monotonically converging to f + g. By the MCT,∫

(f + g) dµ = limn→+∞

∫(sn + σn) dµ = lim

n→+∞

∫sn dµ+ lim

n→+∞

∫σn dµ =

∫f dµ+

∫g dµ .

This proves (1). In order to prove (2), notive that by (1) and induction we have that∫ ( N∑n=1

fn

)dµ =

N∑n=1

∫fn dµ ,

for every N . Observe that∑N

n=1 fn

is a sequence of functions in L+ that converges mono-

tonically a∑+∞

n=1 fn. Hence, applying the MCT again we obtain∫ ( +∞∑n=1

fn

)dµ = lim

n→+∞

∫ ( N∑n=1

fn

)dµ = lim

n→+∞

N∑n=1

∫fn dµ =

+∞∑n=1

∫fn dµ ,

as we wished to show.

Corollary 2.18. Let f ∈ L+. Then∫fdµ = 0 if and only if f = 0 µ-a.e. Therefore, if

f, g ∈ L+ and f = g a.e., then ∫f dµ =

∫g dµ .

Proof. Assume first the f is simple, f =∑

j cjχEj . Then f = 0 clearly implies∫fdµ = 0, while

if 0 =∫fdµ =

∑j cjµ(Ej), it must be either cj = 0 or µ(Ej) = 0, for each j. In any case, f = 0

µ-a.e.Let now f ∈ L+ and assume first f = 0 µ-a.e. If s is simple, 0 ≤ s ≤ f , then s = 0 µ-a.e.,∫sdµ = 0, and passing to the supremum of such functions we obtain that

∫fdµ = 0. Finally,

suppose∫fdµ = 0. Then

x : f(x) > 0

=

+∞⋃n=1

x : f(x) > 1/n

=:

+∞⋃n=1

En ,

Since E1 ⊆ E2 ⊆ · · · ,µ(x : f(x) > 0

)= lim

n→+∞µ(En) .

If f is not equal to 0 a.e., then it must be µ(En) > 0 for some n, but then∫f dµ ≥

∫En

f dµ ≥∫En

1

ndµ =

1

nµ(En) > 0 ,

a contradiction. Then,∫fdµ = 0 implies f = 0 µ-a.e., and we are done.

The conclusion of the MCT holds true also if we just assume that fn is a sequence in L+

monotonically convergent a.e.

Corollary 2.19. Let fn is a sequence in L+ monotonically convergent a.e. to f ∈ L+, then∫f dµ = lim

n→+∞

∫fn dµ .


Proof. Let E ∈ M be such that µ(E) = 0, and f1(x) ≤ f2(x) ≤ · · · and limn→+∞ fn(x) = f(x)for x ∈ cE. Setting gn = χ cEfn and g = χ cEf , we observe that gn = fn a.e. and g = f a.e. sothat

∫gndµ =

∫fndµ, for all n and

∫gdµ =

∫fdµ, by Cor. 2.18.

We may apply the MCT to gn and obtain that∫f dµ =

∫g dµ = lim

n→+∞

∫gn dµ = lim

n→+∞

∫fn dµ

as we wish to show.

In the MCT the key assumption, besides the non-negativity of the functions, was the mono-tonicity of the sequence. Such monotonicity guaranteed the existence of the limits. In thegeneral case, we have the following.

Theorem 2.20. (Fatou’s Lemma.) Let fn ⊆ L+, then∫ (lim infn→+∞

fn

)dµ ≤ lim inf

n→+∞

∫fn dµ .

Proof. As in the proof of Thm. 2.6 (iv), if we set gk(x) = infj≤k fj(x), for k = 1, 2, . . . , we havethat gk is an increasing sequence in L+ and limk gk = lim infn→+∞ fn. Moreover, gk ≤ fk sothat

limk→+∞

∫gk dµ ≤ lim inf

n→+∞

∫fn dµ .

Therefore, by the MCT,∫ (lim infn→+∞

fn

)dµ =

∫ (lim

k→+∞gk

)dµ = lim

k→+∞

∫gk dµ ≤ lim inf

n→+∞

∫fn dµ ,


The following result follows at once from Fatou’s Lemma and Cor. 2.19.

Corollary 2.21. Let fn ⊆ L+ and suppose fn → f a.e. Then∫f dµ ≤ lim inf

n→+∞

∫fn dµ .

Proof. Let E ∈M be a set such that fn → f in E and µ( cE) = 0. Set gn = χEfn and g = χEf .Then gn → g, fn = gn a.e. and f = g a.e. By Fatou’s Lemma,∫

f dµ =

∫g dµ ≤ lim inf

n→+∞

∫gn dµ = lim inf

n→+∞

∫fn dµ .

Remark 2.22. We observe that the previous result cannot be improved to have equality evenif we assume that the limit of the sequence

∫fndµ

exists.

Indeed, consider the measure space (N,P(N), µ), where µ is the counting measure. Let snbe the numerical sequence sn,k that is equal to 1 if k = n and equals 0 otherwise. Then thesequence of functions on N sn converges to 0 pointwise, but∫

Nsn dµ =

+∞∑k=1

sn,k = 1

for all n. Therefore,

0 =

∫Ns dµ ≤ lim

n→+∞

∫Nsn dµ = 1 .

16 M. M. PELOSO

Remark 2.23. We also observe that there exists a version of the MCT for decreasing sequencesfn, but one needs to assume that

∫f1dµ < +∞. Precisely, if fn ⊆ L+, f1 ≥ f2 ≥ · · · , and∫

f1dµ < +∞, then ∫f dµ = lim

n→+∞

∫fn dµ .

Proof. We first observe that if 0 ≤ g ≤ f , then f − g ≥ 0 and∫ (f − g) dµ =

∫f dµ−

∫g dµ .

Next, since fn is monotone and non-negative, the limit f exists it is non-negative, andfn ≥ f for each n. Set gn = f1 − fn. Then gn is a sequence of non-negative functions suchthat gn ≤ gn+1 for all n and converging to f1 − f . Applying the MCT we obtain∫

f1 dµ−∫f dµ =

∫ (f1 − f) dµ =

∫g dµ = lim

n→+∞

∫gn dµ = lim

n→+∞

(∫f1 dµ−

∫fn dµ

).

Therefore, since∫f1dµ < +∞ we can subtractive from both sides to obtain∫

f dµ = limn→+∞

∫fn dµ ,

as we wished to prove.

2.3. Integration of complex-valued functions. We now pass to consider generic real, orcomplex, valued functions. If (X ,M, µ) is a measure space, f : X → C is measurable, andu = Re f , v = Im f , we write

f = u+ − u− + i(v+ − v−) .

Then, u±, v± : X → [0,+∞). If f is real-valued, we simply write f = f+ − f−.

Definition 2.24. Given a measure space (X ,M, µ), given f : X → C is measurable, we saythat f is absolutely integrable, or simply integrable, if∫

|f | dµ < +∞;

equivalently, if

0 ≤∫u± dµ,

∫v± dµ < +∞ .

For f integrable we set∫f dµ =

∫u+ dµ−

∫u− dµ+ i

(∫v+ dµ−

∫v− dµ

).

In particular, if f is real-valued, we have∫f dµ =

∫f+ dµ−

∫f− dµ .

Remark 2.25. Notice that, if f is integrable, f = u+ iv, then 0 ≤ u±, v± ≤ |f | so that

0 ≤∫u± dµ,

∫v± dµ ≤

∫|f | dµ < +∞ .


Conversely, if 0 ≤∫u± dµ,

∫v± dµ < +∞, then |f | ≤ |u|+ |v| = u+ + u− + v+ + v−, so that∫

|f | dµ ≤∫u+ dµ+

∫u− dµ+

∫v+ dµ+

∫v− dµ < +∞ .

Moreover, the set of integrable functions is a complex vector space, as it is easy to check, andfor integrable f, g and a, b ∈ C we have∫ (

af + bg) dµ = a

∫f dµ+ b

∫g dµ .

Proposition 2.26. The following properties hold true.

(i) If f is integrable, then∣∣∣ ∫ f dµ

∣∣∣ ≤ ∫ |f |dµ .(ii) If f, g are integrable, then

∫Ef dµ =

∫Eg dµ for every E ∈M if and only if |f − g| = 0

a.e. if and only if

∫|f − g| dµ = 0.

Proof. (i) If∫f = 0 the result is trivial. Next, if f is real∣∣∣ ∫ f dµ

∣∣∣ =∣∣∣ ∫ f+ dµ−

∫f− dµ

∣∣∣ ≤ ∫ f+ dµ+

∫f− dµ =

∫|f | dµ .

If f is complex-valued, and∫f 6= 0, from Cor. 2.7 that sgn f, |f | are measurable and we set

α = |∫fdµ|/(

∫fdµ). Then ∣∣∣ ∫ f dµ

∣∣∣ = α

∫f dµ =

∫αf dµ

so that, in particular∫αf dµ is non-negative. Therefore,∣∣∣ ∫ f dµ∣∣∣ = Re

∫αf dµ =

∫Re(αf) dµ ≤

∫|αf | dµ =

∫|f | dµ ,

since |α| = 1.(ii) If f, g ∈ L1(µ), from Cor. 2.18 we know that

∫|f − g| dµ = 0 if and only if |f − g| = 0

a.e., that is, if and only if f = g a.e. Now, if∫|f − g| dµ = 0, then, for every E ∈M,∣∣∣ ∫

Ef dµ−

∫Eg dµ

∣∣∣ ≤ ∫ χE |f − g| dµ ≤∫|f − g| dµ = 0 ,

so that ∫Ef dµ =

∫Eg dµ for every E ∈M .

Finally, suppose that the above condition holds, and assume for simplicity that f and g arereal-valued. Consider the function f − g = m and its decomposition m = m+ −m−. Then, byassumption we have that

∫Emdµ = 0 for all E ∈M. In particular if we take E+ =

x : m(x) ≤

0

, then we have

0 =

∫E+

mdµ =

∫E+

m+ dµ =

∫m+ dµ ,

since m+ = 0 on cE+. Thus, m+ = 0 a.e. and, with a similar argument we obtain also m− = 0a.e. Hence, m = 0 a.e., i.e. f = g a.e., as we wished to show.

18 M. M. PELOSO

From now on we will identify functions that differ only on a set of measure 0. Indeed, wehave the following definition.

Definition 2.27. Given a measure space (X ,M, µ), we consider the equivalent relation ∼ onthe space of integrable saying that f ∼ g if f = g a.e. We define the space L1(µ) as the spaceof equivalent classes of integrable function modulo the relation ∼ and we define the norm

‖f‖L1(µ) =

∫|f | dµ .

Remark 2.28. We need to check that the above definition gives in fact a norm. It is clear that‖f‖L1(µ) ≥ 0 and that f = 0 in L1(µ) (that is, f = 0 a.e.) implies ‖f‖L1(µ) = 0. Conversely, if

0 = ‖f‖L1(µ) =∫|f |dµ, then Cor. 2.18 implies that |f | = 0 a.e., ie. f = 0 a.e. and f = 0 in L1.

The symmetry and triangular inequality of the norm follow easily.In Thm. 2.31 we will show that L1 is a complete normed space, that is, a Banach space.

Therefore, we have a norm on L1(µ), hence a metric, given by the expression

d(f, g) = ‖f − g‖L1(µ) .

Then, given fn ⊆ L1(µ) we say that fn → f in L1(µ) (or, simply in L1 if the measure µ isunderstood) if

‖fn − f‖L1 → 0 as n→ +∞ .

Theorem 2.29. (Dominated Convergence Theorem.) Let (X ,M, µ) be a measure spaceand fn ⊆ L1(µ). Suppose that fn → f pointwise a.e. and that there exists a non-negativeg ∈ L1(µ) such that g ≥ |fn| for all n. Then, fn → f in L1(µ) and∫

f dµ = limn→+∞

∫fn dµ .

Proof. Since |fn| ≤ g for all n, since fn → f pointwise a.e., it follows that f is measurable and,by the comparison theorem, that |f | ≤ g. Therefore, f ∈ L1(µ).

Observing that |fn − f | ≤ 2g, we apply Fatou’s Lemma to the sequence 2g − |fn − f | andobtain ∫

2g dµ ≤ lim infn→+∞

∫2g − |fn − f | dµ =

∫2g dµ+ lim inf

n→+∞

(−∫|fn − f | dµ

)=

∫2g dµ− lim sup

n→+∞

∫|fn − f | dµ .

Since∫

2g dµ < +∞, we can subtract it from both sides of the inequality and obtain that

lim supn→+∞

∫|fn − f | dµ ≤ 0 .

This implis that limn→+∞∫|fn − f | dµ = 0, that is, fn → f in L1(µ). Finally,∣∣∣ ∫ fn dµ−

∫f dµ

∣∣∣ ≤ ∫ |fn − f | dµ→ 0

and the last conclusion follows.


2.4. The space L1(µ). Our next goal is to show that L1(µ) is a Banach space, that is it iscomplete in its norm. We recall that a normed space (X, ‖ · ‖X) is complete if it is complete as ametric space w.r.t the metric d(f, g) = ‖f − g‖X . We also recall that a series

∑n fn of elements

in X is said to be absolutely convergent in X if the numerical series∑

n ‖fn‖X is convergent.We have the following result.

Theorem 2.30. A normed space (X, ‖ · ‖X) is a Banach space if and only if every absolutelyconvergent series

∑n fn is convergent in X.

Proof. Suppose first X is a Banach space and that∑

n fn is an absolutely convergent series. Let

sN be the sequence of partial sums, that is, sN =∑+∞

n=1 fn. Then, given ε > 0, there existsnε such that for nε < M < N ,

d(sN , sM ) = ‖sN − sM‖X =∥∥ N∑n=M+1

fn∥∥X≤

N∑n=M+1

‖fn‖X < ε .

Hence, sN is a Cauchy sequence in X, that converges, since X is complete.Conversely, suppose every absolutely convergent series is convergent in X. Let gn be a

Cauchy sequence in X. Hence, for every k = 1, 2, . . . , there exists an integer Nk such that ifm,n ≥ Nk, d(gn, gm) = ‖gn − gm‖X < 2−k. Notice that we may assume that Nk+1 > Nk for allk. Then, the subsequence gNk

is such that ‖gNk+1−gNk

‖X < 2−k. Then,∑+∞

k=1 ‖gNk+1−gNk

‖Xconverges, so that by assumption,

∑+∞k=1 gNk+1

− gNkconverges to an element g ∈ X. But, the

series∑+∞

k=1 gNk+1− gNk

is telescopic and the partial sum

M∑k=1

gNk+1− gNk

= gNM+1− gN1 ,

so that gNk− gN1 → g as k → +∞. Hence, the original sequence must converge too, and the

conclusion follows.

Theorem 2.31. Let fn be a sequence of functions in L1(µ) such that∑+∞

n=1 ‖f‖L1(µ) < +∞.

Then, there exists f ∈ L1(µ) such∑+∞

n=1 fn converges to f in L1(µ) and∫f dµ =

+∞∑n=1

∫fn dµ .

As a consequence, L1(µ) is a Banach space. Finally, the simple functions are dense in L1(µ).

Proof. By Cor. 2.17 (ii) we know that∫ ∑n

|fn|dµ =∑n

∫|fn|dµ ≤

+∞∑n=1

‖fn‖L1(µ) < +∞ .

Then, setting g =∑+∞

n=1 |fn|, we have g ∈ L1(µ), and hence it is finite a.e. This implies that∑+∞n=1 fn converges but on a set of measure 0, call f such limit. We apply the DCT to the

sequence of partial sums sn =∑n

k=1 fk. Clearly, sn → f pointwise, and also

|sn| =∣∣∣ n∑k=1

fk

∣∣∣ ≤ n∑k=1

|fk| = g .

20 M. M. PELOSO

Hence, the DCT applies to give that sn → f in L1(µ) and limn→+∞∫sndµ =

∫fdµ, that is∫

f dµ = limn→+∞

∫ ( n∑k=1

fk dµ)

= limn→+∞

n∑k=1

∫fk dµ =

+∞∑n=1

∫fn dµ .

In particular, every absolutely convergent series also converges in L1(µ), hence L1(µ) is complete.Finally, let f ∈ L1(µ). Then, |f | is finite a.e. and the sequence of simple functions ϕn in

Thm. 2.9 (2), is such that ϕn → f pointwise, and |ϕn| ≤ |f |. Therefore, we can apply the DCTto obtain that ϕn → f in L1(µ).

We conclude this part by comparing L1(µ) with L1(µ). If f ∈ L1(µ) and g is given by by Prop.2.12, then g = f µ-a.e. and g ∈ L1(µ). Conversely, if g ∈ L1(µ), then g is also M-measurableand

∫|g|dµ =

∫|g|dµ. Then g ∈ L1(µ). Therefore, we have shown that the spaces L1(µ) and

L1(µ) can be identified and we will do in what follows.

3. The Lebesgue measure on R

Having established the definition and main properties of abstract integrals, we now go backto the construction of measures and in particular of the Lebesgue measure and of other Borelmeasures in R, and eventually in Rn.

3.1. Outer measures. Given a set X , an outer measure on X is a set function µ∗ defined onthe set of parts of X , P(X ) such that

(i) µ∗(∅) = 0;

(ii) if A ⊆ B ⊆ X , then µ∗(A) ≤ µ∗(B);

(iii) µ∗(∪+∞j=1 Aj

)≤∑+∞

j=1 µ∗(Aj), for all Aj ⊆ X .

The reason for this definition is that outer measures arise naturally when one has a family ofelementary sets and a notion of measure for such sets.

Proposition 3.1. Let E be a collection of sets in P(X ) such that ∅,X ∈ E, and let ρ : E →[0,+∞] such that ρ(∅) = 0 be given. Define

µ∗(A) = inf +∞∑j=1

ρ(Ej) : A ⊆+∞⋃j=1

Ej , Ej ∈ E.

Then µ∗ is an outer measure.

The key example to keep in mind is when E is the collection of (all, or a subclass of) intervalsI in R, and ρ(I) is the length of I, which could be +∞ if I is a ray.

Proof. First of all we observe that the definition makes sense, since for any A ⊆ X there exists acovering of A with sets in E , e.g. X , ∅, ∅, . . . , where ∅,X ∈ E . Also, it is clear that µ∗(∅) = 0,since ∅ can be covered by ∅, ∅, . . . . Property (ii) in the definition of an outer measure issatisfied since, if A ⊆ B, and Ej is a covering of B with sets in E , Ej is also a coveringof A. Thus, the infimum in the definition of µ∗(A) is taken with respect a larger collection ofconverings than in the case of µ∗(B); hence (ii) holds.


Finally, let Aj ⊆ P(X ). Fix ε > 0. Then, by the properties of the infimum of real numbers,for each j, there exists a covering Ej,k of sets in E of Aj such that

+∞∑k=1

ρ(Ej,k) < µ∗(Aj) + ε2−j .

Therefore, Ej,k for j, k = 1, 2, . . . is a covering of ∪+∞j=1Aj by sets in E such that

µ∗(+∞⋃j=1

Aj

)= inf

∑n

ρ(En) : En ∈ E ,+∞⋃j=1

Aj ⊆+∞⋃n=1

En

≤

+∞∑j,k=1

ρ(Ek,j)

≤+∞∑j=1

(µ∗(Aj) + ε2−j

)=

+∞∑j=1

µ∗(Aj) + ε .

Since ε > 0 was arbitrary, (iii) follows.

The key step to pass from an outer measure to an actual measure is based on the followingdefinition.

Definition 3.2. Let µ∗ be an outer maesure on a set X . We say that a set E ⊆ X is µ∗-maesurable if

µ∗(A) = µ∗(A ∩ E) + µ∗(A ∩ cE) ,

for all A ⊆ X .

Observe that we always have the inequality

µ∗(A) ≤ µ∗(A ∩ E) + µ∗(A ∩ cE) ,

by subadditivity. Then, in order to prove the µ∗-maesurability of a set E, we only need to provethe reverse inequality

µ∗(A) ≥ µ∗(A ∩ E) + µ∗(A ∩ cE) ,

for all sets A ⊆ X . This inequality is trivial if µ∗(A) = +∞. Therefore, E ⊆ X is µ∗-maesurableif and only if

µ∗(A) ≥ µ∗(A ∩ E) + µ∗(A ∩ cE) for all A with µ∗(A) < +∞ . (10)

The key step to pass from an outer measure to an actual measure is the following result,known as Caratheodory’s theorem.

Theorem 3.3. (Caratheodory’s Theorem.) Let X be a set and µ∗ an outer measure on X .Let M be the collection of µ∗-maesurable sets, and µ the restriction of µ∗ to M. Then M is aσ-algebra, and µ is a complete measure on M.

Proof. We first observe that E ∈ M implies that also cE ∈ M since the condition (10) issymmetric in E and cE. In order to show that M is a σ-algebra, it suffices to show that it is

22 M. M. PELOSO

closed under countable unions of disjoint sets – see the comment regarding formula (1). To thisend, we first show that M is closed under finite unions. Let E,F ∈M and let A ⊆ X . Then

µ∗(A) = µ∗(A ∩ E) + µ∗(A ∩ cE)

= µ∗((A ∩ E) ∩ F

)+ µ∗

((A ∩ E) ∩ cF

)+ µ∗

((A ∩ cE) ∩ F

)+ µ∗

((A ∩ cE) ∩ cF

).

We observe that

E ∪ F = (E ∩ F ) ∪ (E ∩ cF ) ∪ ( cE ∩ F )

so that, using subadditivity

µ∗(A ∩ E) ∩ F

)+ µ∗

((A ∩ E) ∩ cF

)+ µ∗

((A ∩ cE) ∩ F

)≥ µ∗

(A ∩ (E ∪ F )

)Since A ∩ cE) ∩ cF = A ∩ c(E ∪ F we then have

µ∗(A) ≥ µ∗(A ∩ (E ∪ F )

)+ µ∗

(A ∩ c(E ∪ F )

).

By (10) this implies that E ∪ F ∈M, and M is an algebra.Suppose that E,F ∈M and E ∩ F = ∅. Then

µ∗(E ∪ F ) = µ∗((E ∪ F ) ∩ E

)+ µ∗

((E ∪ F ) ∩ cE

)= µ∗(E) + µ∗(F ) ,

so that µ∗ is finitely addivitive inM. In order to show thatM is a σ-algebra, it suffices to showthat it is closed under countable union of disjoint subsets. Then, let Ej ⊆ M be a sequenceof disjoint sets. Setting F = ∪+∞

j=1Ej , we wish to show that F is µ∗-measurable.To this end, set Fn = ∪nj=1Ej , and let A ⊆ X . Then

µ∗(A ∩ Fn) = µ∗(A ∩ Fn ∩ En) + µ∗(A ∩ Fn ∩ cEn)

= µ∗(A ∩ En) + µ∗(A ∩ Fn−1) ,

as it is easy to check. Arguing by induction we then obtain that

µ∗(A ∩ Fn) =n∑j=1

µ∗(A ∩ Ej) .

It follows that

µ∗(A) = µ∗(A ∩ Fn) + µ∗(A ∩ cFn) ≥n∑j=1

µ∗(A ∩ Ej) + µ∗(A ∩ cF ) ,

for every n ≥ 1. Letting n→ +∞ we obtain

µ∗(A) ≥+∞∑j=1

µ∗(A ∩ Ej) + µ∗(A ∩ cF ) ≥ µ∗(+∞⋃j=1

(A ∩ Ej) + µ∗(A ∩ cF ))

= µ∗(A ∩ F ) + µ∗(A ∩ cF ) = µ∗(A) .

Thus, the above ones are all equalities and F is µ∗-measurable, as we wanted to show. Moreover,taking A = F in the above equalities, we obtain

µ∗(+∞⋃j=1

Ej

)= µ∗(F ) =

+∞∑j=1

µ∗(Ej) ,

that is, µ∗ is countable additive on M.


Finally, we need to show thatM is complete. Suppose that µ∗(N) = 0 and let A ⊆ X be anyset. Then,

µ∗(A) ≤ µ∗(A ∩N) + µ∗(A ∩ cN) = µ∗(A ∩ cN) ≤ µ∗(A) ,

so that N ∈M and M is complete.

The application of Caratheodory’s theorem we have in mind deals with the notion of premea-sure.

Definition 3.4. Given a set X and an algebra A of subsets of X , a set function ρ : A → [0,+∞]is called a premeasure if

(i) ρ(∅) = 0;

(ii) if Aj is a sequence of disjoint sets in A such that ∪+∞j=1Aj ∈ A, then

ρ(+∞⋃j=1

Aj

)=

+∞∑j=1

ρ(Aj) .

Notice that a premeasure satisfies the monotonicity condition: if E,F ∈ A, E ⊆ F , thenρ(E) ≤ ρ(F ). Indeed, writing F = E ∪ (F ∩ cE), the conclusion follows easily.

Lemma 3.5. Let X be a set, A an algebra of subsets of X and ρ : A → [0,+∞] a premeasure,and let µ∗ be given by

µ∗(A) = inf +∞∑j=1

ρ(Ej) : A ⊆+∞⋃j=1

Ej , Ej ∈ E. (11)

Then µ∗ is an outer measure such that the following properties hold true:

(i) µ∗|A = ρ;

(ii) every set E ∈ A is µ∗-measurable.

Proof. By Prop. 3.1 we know that µ∗ is an outer measure. (i) Let E ∈ A. Then E ⊆ ∪+∞j=1Ej ,

where E1 = E and Ej = ∅ for j = 2, 3, . . . . Then

µ∗(E) ≤+∞∑j=1

ρ(Ej) = ρ(E) .

Thus, it suffices to prove the reverse inequality. Let E ∈ A and let Ej be any covering of E

with sets in A, i.e. Ej ∈ A, j = 1, 2, . . . , and E ⊆ ∪+∞j=1Ej . Set An = En \ ∪n−1

j=1Ej . Then theAn’s are dijoint elements of A, as it is easy to check, and their union is E. Therefore, by thedefinition of premeasure,

ρ(E) =

+∞∑n=1

ρ(An) ≤+∞∑n=1

ρ(En) .

Taking the infimum on the right hand side over the coverings of E by sets in A, it follows thatρ(E) ≤ µ∗(E) for E ∈ A. This proves (i).

24 M. M. PELOSO

(ii) Let E ∈ A. In order to show that E is µ∗-measurable, it suffices to prove (10). Let B ⊆ X .Given ε > 0, there exists a collection Aj ⊆ A such that B ⊆ ∪+∞

j=1Aj and∑+∞

j=1 ρ(Aj) ≤µ∗(B) + ε. Then, by the additivity of ρ on A,

µ∗(B) + ε ≥+∞∑j=1

(ρ(Aj ∩ E) + ρ(Aj ∩ cE)

)≥ µ∗(B ∩ E) + µ∗(B ∩ cE) ,

since Aj ∩E and Aj ∩ cE are coverings of B∩E and B∩ cE, resp., by elements of A. Since,ε > 0 was arbitraty, this proves (ii).

Lemma 3.6. Let X ,A, ρ and µ∗ be as in Lemma 3.5. Then

(i) Given any B ⊆ X and ε > 0 there exists a collection Aj ⊆ A be such that B ⊆ ∪+∞j=1Aj

and µ∗(∪+∞j=1 Aj

)≤ µ∗(B) + ε.

Assume further that ρ is σ-finite and let M(A) be the σ-algebra generated by A. The followingproperties hold true.

(ii) A set E ⊆ X is µ∗-measurable if and only if there exists A ∈ M(A) such that E ⊆ Aand µ∗(A \ E) = 0.

(iii) A set E ⊆ X is µ∗-measurable if and only if there exists B ∈ M(A) such that B ⊆ Eand µ∗(E \B) = 0.

Proof. (i) If µ∗(B) we have nothing to prove. Suppose that µ∗(B) is finite. Then, given ε > 0, bydefinition of infimum, there exists Aj ⊆ A such that B ⊆ ∪+∞

j=1Aj and∑+∞

j=1 ρ(Aj) ≤ µ∗(B)+ε.

Using Lemma 3.5 (i) we then have

µ∗(+∞⋃j=1

Aj

)≤

+∞∑j=1

µ∗(Aj) =

+∞∑j=1

ρ(Aj) ≤ µ∗(B) + ε ,

as we wished to prove.(ii) One direction is obvious. If E ⊆ X and there exists B ∈ M(A) such that B ⊆ E and

µ∗(E \B) = 0, then E = B ∪ (E \B). It follows that E is µ∗-measurable since B ∈M(A) andall sets of µ∗-measure 0 are µ∗-measurable.

Conversely, suppose E ⊆ X is µ∗-measurable. Assume first that µ∗(E) < +∞. For each

k = 1, 2, . . . , we apply part (i) to E, with ε = 1/k. We find a collection of sets A(k)j such that

A(k) := ∪+∞j=1A

(k)j ⊇ E and µ∗(A(k)) ≤ µ∗(E) + 1

k . Set

A =

+∞⋂k=1

A(k) .

Then A ⊇ E, so that µ∗(E) ≤ µ∗(A). Moreover, for each k,

µ∗(A) = µ∗( +∞⋂k=1

A(k))≤ µ∗(A(k)) ≤ µ∗(E) +

1

k,

so that µ∗(A) = µ∗(E). Since E is µ∗-measurable,

µ∗(A) = µ∗(A ∩ E) + µ∗(A ∩ cE) = µ∗(E) + µ∗(A \ E) .


Since µ∗(A) = µ∗(E) < +∞, we can substract it from both sides and obtain that µ∗(A\E) = 0.This proves (ii) when µ∗(E) < +∞.

Next, suppose E is µ∗-measurable and µ∗(E) = +∞. Here we use the assumption that ρ isσ-finite, that is that X = ∪+∞

j=1Xj with ρ(Xj) < +∞. We may assume that the Xj ’s are disjoint.

Let Ej = E ∩ Xj . By Lemma 3.5 (ii) we know that the Xj ’s are µ∗-measurable, so is Ej foreach j.

Then, for each j there exists Aj ∈ M(A) such that Ej ⊆ Aj and µ∗(Aj \ Ej) = 0. SettingA = ∪+∞

j=1Aj we have that E ⊆ A and

µ∗(A \ E) = µ∗(+∞⋃j=1

(Aj \ Ej))≤

+∞∑j=1

µ∗((Aj \ Ej)

)= 0 .

This proves (ii).

Finally we prove (iii). We first assume that µ∗(E) < +∞. We apply (ii) ( which is now validfor all µ∗-measurable sets) to F = cE and find A ∈ A, A ⊇ cE and µ∗(A \ cE) = 0. SettingB = cA, we have B ∈M(A), B ⊆ E and since E is µ∗-measurable,

µ∗(B) = µ∗(B ∩ E) + µ∗(B ∩ cE) = µ∗(B) + µ∗(B ∩ cE) .

Now, µ∗(B) ≤ µ∗(E) < +∞, so we can subtract it on both sides of the line of equations above.We then obtain

µ∗(B) + µ∗(B ∩ cE) = 0 .

The proof of the case µ∗(E) is similar to the analogous case in (ii). This completes the proof.

We are finally ready to prove the result about the construction of a complete measure, thatwill be used in the next session to construct the Lebesgue measure on R as a particular instance.

Theorem 3.7. Let X ,A, ρ and µ∗ be as in Lemma 3.5. Further, assume that ρ is σ-finite. LetM be the σ-algebra of the µ∗-measurable sets and let µ = µ∗|M be the complete measure given

by Thm. 3.3.Let M(A) be the σ-algebra generated by A and let ν be any measure on M(A) whose restric-

tion to A coincides with ρ. Then µ is the completion of ν and

M =M(A) ∪N , (12)

where N =N ⊆ X : there exists F ∈M(A), N ⊆ F, ν(F ) = 0

.

Proof. (†) In order to show that (µ,M) is the completion of (ν,M(A)) we need to show thatµ|M(A)

= ν and that (12) holds. We begin with the former one.

Let µ∗ be the outer measure given by (11) in Lemma 3.5. Let (M, µ) be the completemeasure constructed in Thm. 3.3, starting from µ∗. By Lemma 3.5 we know that A ⊆ M.Hence M(A) ⊆M. Now, let E ∈M(A) and let Aj ⊆ A be such that E ⊆ ∪+∞

j=1Aj . Then

ν(E) ≤+∞∑j=1

ν(Aj) =

+∞∑j=1

ρ(Aj)

so that

ν(E) ≤ inf +∞∑j=1

ρ(Aj) : E ⊆+∞⋃j=1

Aj , Aj ∈ A

= µ∗(E) = µ(E) ,

26 M. M. PELOSO

since E in particular is in M. Thus, ν(E) ≤ µ(E) for all E ∈ M(A). Moreover, settingA = ∪+∞

j=1Aj , since ν and µ coincide on A, we have that

ν(A) = limn→+∞

ν( n⋃j=1

Aj

)= lim

n→+∞µ( n⋃j=1

Aj

)= µ(A) .

Conversely, assume first that µ(E) < +∞. Then, given ε > 0, there exists a collectionAj ⊆ A such that E ⊆ ∪+∞

j=1Aj and µ(A) ≤ µ(E) + ε, so that µ(A \ E) < ε. Then, using thefirst part too, we have that

µ(E) ≤ µ(A) = ν(A) = ν(E) + ν(A \ E) ≤ ν(E) + µ(A \ E) ≤ ν(E) + ε .

Since ε > 0 was arbitrary, we have µ(E) ≤ ν(E), hence µ(E) = ν(E), if µ(E) < +∞ andE ∈ M(A). Finally, suppose E ∈ M(A) and µ(E) = +∞. Here we assume that ρ is σ-finite,that is that X = ∪+∞

j=1Aj with ρ(Aj) < +∞. We may assume that the Aj ’s are disjoint. Then

µ(E) =+∞∑j=1

µ(E ∩Aj) =+∞∑j=1

ν(E ∩Aj) ≤ ν(E) .

Hence, ν = µ on M(A).The fact that (12) holds follows from Lemma 3.6 (ii). Indeed, Lemma 3.6 (ii) says that E ⊆ X

is in M if and only if there exists B ∈M(A) such that

E = B ∪ (E \B)

with µ(E \B) = 0, where clearly (E \B) ∈M. This proves (12) and therefore the theorem.

3.2. The Lebesgue measure on R. We now are now ready to introduce the main object ofthis course.

We consider the collection A of finite unions of disjoint left-open/right-closed intervals in R,that is,

A =E ⊆ R : E =

n⋃j=1

Ij , Ij disjoint, Ij = (aj , bj ] or Ij = (aj ,+∞), −∞ ≤ aj < bj < +∞.

Lemma 3.8. Set ρ(∅) = 0, ρ(E) = +∞ if E ∈ A is unbounded, and if the intervals (aj , bj ],j =, 1, 2, . . . , n are disjoint and E = ∪nj=1(aj , bj ],

ρ(E) =

n∑j=1

(bj − aj) .

Then ρ is a premeasure on the algebra A.

Proof. It is clear that E is well defined, that is, if E = ∪nj=1(aj , bj ] = ∪mk=1(ck, dk], then

n∑j=1

(bj − aj) =m∑k=1

(dk − ck) .

It is also easy to see that ρ is finitely additive. In fact, both assertions follow from the factthat an element of A can be written in a unique way as disjoint union of maximal disjointleft-open/right-closed intervals in R.


Thus, it remains to show that if E is countable union of disjoint left-open/right-closed intervalsE = ∪+∞

j=1(aj , bj ], and E ∈ A, then ρ(E) =∑+∞

j=1 ρ((aj , bj ]

)=∑+∞

j=1(bj − aj). Since E ∈ A, it

can be written as finite union of disjoint left-open/right-closed intervals or of the form (a,+∞),say Ij , j = 1, . . . , n. Using the finite additivity of ρ, it suffices to consider the case whenE is itself a left-open/right-closed interval (a, b], or (a,+∞). Suppose then that I = ∪+∞

j=1Ij ,

where Ij = (aj , bj ]. Notice that this includes also the case I = (a,+∞). Since for each n,I =

(∪nj=1 Ij

)∪(∪+∞j=n+1 Ij

)=:(∪nj=1 Ij

)∪ J , with J ∈ A, we have

ρ(I) = ρ( n⋃j=1

Ij

)+ ρ(J) =

n∑j=1

ρ(Ij) + ρ(J) ≥n∑j=1

ρ(Ij)

for all n. Hence,

ρ(I) ≥+∞∑j=1

ρ(Ij) .

Conversely, if∑+∞

j=1 ρ(Ij) = +∞ we have nothing to prove. Hence, assume that∑+∞

j=1 ρ(Ij) <+∞. Notice that this implies that I must be bounded. Let ε > 0 be given. Then, the sets(aj , bj + ε2−j) : j = 1, 2, . . . form an open cover of I. Therefore, there exists a finite

subcollection (aj` , bj` + ε2−j`) : ` = 1, 2, . . . ,m that still covers I. Hence, I ⊆ ∪m`=1(aj` , bj` +ε2−j` ]. Using the monotonicity and the finite subadditivity (that follows from the finite addivity)of ρ, we have

ρ(I) ≤ ρ( m⋃`=1

(aj` , bj` + ε2−j` ])≤

m∑`=1

ρ((aj` , bj` + ε2−j` ]

)≤

m∑`=1

(ρ(Ij`) + ε2−j`

)≤

+∞∑j=1

(ρ(Ij) + ε2−j

)≤

+∞∑j=1

ρ(Ij) + ε ,

for every ε > 0. Since ε > 0 was arbitrary, this proves the inequality ρ(I) ≤∑+∞

j=1 ρ(Ij), hencethe lemma.

Definition 3.9. Let A and ρ be as in Lemma 3.8, µ∗ the outer measure defined in (11) inLemma 3.5. We define the Lebesgue measure space on (R,L,m) where L is the σ-algebra andas the complete measure m = µ constructed in Thm. 3.7. The measure m is called the Lebesguemeasure on R and L the σ-algebra of Lebesgue measurable sets.

Remark 3.10. We collect here some obvious but fundamental properties of the Lebesgue mea-sure in R.

(1) We begin by observing that the σ-algebra of Lebesgue measurable sets L, by Thm. 3.3is the completion of the σ-algebra generated by the algebra A of the finite unions of disjointleft-open/right-closed intervals in R. Then L contains BR, the σ-algebra of Borel sests in R.More precisely, using Thm. 3.7 we have that,

L = BR ∪N ,

28 M. M. PELOSO

where N =N ⊆ R : there exists F ∈ BR, N ⊆ F, m(F ) = 0

.

(2) For each interval I ⊆ R, m(I) equals the length of I (possibly +∞).

(3) Each open set A ⊆ R has positive measure, or possibly = +∞. Indeed, it suffices tonotice that A is at most countable union of disjoint open intervals In, n = 1, 2, . . . , since thenm(A) ≥ m(I) > 0. Then, given x ∈ A, there exists an open interval Ix ⊆ A and containingx. Let I1 be the union of all such intervals. If I1 = A we are done. Otherwise, there exixtsy ∈ A \ I1, and an open interval Iy contained in A and disjoint from I1. Let z ∈ Iy ∩Q. DefineI2 be the union of all open intervals containing z and contained in A \ I1. This process is atmost of countably many steps, and the conclusion follows.

(4) Each point x has measure 0, so countable sets all have measure 0. In particular, the func-tion χ[0,1]∩Q is integrable and

∫χ[0,1]∩Q dm = 0, and the function χ[0,1]∩(R\Q) is also integrable

and∫χ[0,1]∩(R\Q) dm =

∫ (χ[0,1] − χ[0,1]∩(R\Q)

)dm = 1.

We now see other fundamental properties of the Lebesgue measure.

Proposition 3.11. Let E ∈ L. Then

(i) m(E) = infm(U) : E ⊆ U, U open

,

and also

(ii) m(E) = supm(K) : K ⊆ E, K compact

.

Proof. We preliminary observe that, if E ∈ L, then

m(E) = inf +∞∑j=1

(bj − aj) : E ⊆+∞⋃j=1

(aj , bj). (13)

Call ν(E) the quantity on the right hand side above. By construction we have that

m(E) = inf +∞∑j=1

(bj − aj) : E ⊆+∞⋃j=1

(aj , bj ],

so that clearly m(E) ≤ ν(E) (since in the case of m we take the infimum over a larger numericalset). On the other hand, given ε > 0, let (aj , bj ] be a countable collection of left-open/right-

closed intervals whose union covers E and such that m(E) + ε ≥∑+∞

j=1(bj −aj). Then (aj , bj +

ε2−j) is a collection of open intervals whose union covers E and such that

+∞∑j=1

(bj + ε2−j − aj) =+∞∑j=1

(bj − aj) + ε ≤ m(E) + 2ε .

Since ε > 0 was arbitrary, ν(E) ≤ m(E) and (13) follows.(i) Since, if E ⊆ U we have m(E) ≤ m(U), it is clear that

m(E) ≤ infm(U) : U open, U ⊇ E .From (13) we know that given ε > 0 there exists a collection of open intervals Ij whose unioncovers E and such that

m(E) + ε ≥+∞∑j=1

m(Ij) ≥ m(+∞⋃j=1

Ij

)= m(U) .


This shows that m(E) ≥ infm(U) : U open, U ⊇ E, and proves (i).

(ii) Suppose first that E is bounded. Notice that this implies that E is contained in somebounded interval (a, b], so it has finite measure. If E is closed, then it is compact and theconclusion is obvious. Assume now that E is not closed. Then E \E is not empty, and part (i)gives that given ε > 0 there exists an open set U ⊇ (E \ E) such that m(U) ≤ m(E \ E) + ε.Let K = E \U = E \U . Then K ⊆ E and it is compact, as they are both easy to check. Then,

m(K) = m(E)−m(E ∩ U) = m(E)−[m(U)−m(U \ E)

]≥ m(E)−m(U) +m(E \ E)

≥ m(E)− ε .

Thus (ii) holds in this case. Suppose now E is unbounded and define Ej = E∩(j, j+1]. We applythe argument above and, given ε > 0, there exist compact sets Kj , j ∈ Z such that Kj ⊆ Ej and

m(Ej) ≤ m(Kj) + ε2−|j|. Let Hn = ∪nj=−nKj , so that Hn is compact and Hn ⊆ ∪nj=−nEj ⊆ E.Thus,

m(Hn) =n∑

j=−nm(Kj) ≥

n∑j=−n

(m(Ej)− ε2−|j|

)≥

n∑j=−n

m(Ej)− 2ε .

Since∑n

j=−nm(Ej)→ m(E) as n→ +∞, we have that

m(Hn) ≥ m(E)− 3ε ,

for n large enough. This gives (ii) also for unbounded sets, and we are done.

Example 3.12. Given the sets D = R\Q and D1 = D∩ (0, 1]. Given M, ε > 0 determine H,Kcompacts, H ⊆ D, K ⊆ D1 such that m(H) > M and m(K) > 1− ε.

Notice that the statement about D follows easily from the case of D1 by setting H =∪nj=−n(K + j) and n large enough so that m(H) =

∑nj=−nm(K) > (2n+ 1)(1− ε) > M .

In the case of D1, we observe that the proof of Thm. 3.11 (ii) provides the construction ofsuch a set. Let E = D1, so that E = [0, 1] and E \ E = Q ∩ [0, 1]. Let U be an open setsuch that U ⊇ (Q ∩ [0, 1]) and m(U) ≤ m

((Q ∩ [0, 1])

)+ ε = ε. In fact, given ε > 0, we

may choose U = ∪+∞j=1Iqj , where qj is a denumeration of the rational numbers in [0, 1] and

Iqj = (qj − ε2−(j+1), qj + ε2−(j+1)). Then, we set K = E \ U = E \ U , that is,

K = [0, 1] \+∞⋃j=1

Iqj ,

which is clearly closed, bounded, hence compact, and

m(K) ≥ 1−+∞∑j=1

m(Iqj ) = 1−+∞∑j=1

ε2−j = 1− ε .

Proposition 3.13. Let E ∈ L. Then E + y =x + y : x ∈ E

∈ L for every y ∈ R and

m(E + y) = m(E). If r ∈ R and rE =x+ y : x ∈ E

, then rE ∈ L and m(rE) = |r|m(E).

30 M. M. PELOSO

Proof. Since the collection of left-open/right-closed intervals in R is translation invariant, so isBR. Given y, r ∈ R, we then define my(E) = m(E + y) and mr(E) = m(rE). We consider thecase of my first. It is easy to see that my is a measure on BR. Next, since the premeasure ρ isinvariant by translation, if A ∈ A then

my(A) = m(A+ y) = ρ(A+ y) = ρ(A) ,

so that my |A = ρ. Thus, we can apply the second part of Thm. 3.7 and see that (m,L) is the

completion of (my,BR) and that m = my on BR. In particular, if E ∈ BR and m(E) = 0, thenm(E + y) = my(E) = 0 which implies that the set of Lebesgue measure 0 are preserved bytranslation. Thus, also my is complete and my = m, that is, m is translation invariant.

The same proof shows that mr = rm when r > 0 is positive, since A is preserved by positivedilations, and mr = rρ on A. In this case we apply the second part of Thm. 3.7 to mr, thepremeasure rρ and the complete measure rm. We leave the elementary details to the reader.

To treat the case r < 0, it suffices to consider the case r = −1, that is, to reflection aboutthe origin. In this case we need to start with the algebra A generated by left-closed/right-openintervals instead. Again, we leave the details to the reader.

As an immediate corollary we obtain that the set N constructed in Example 1.1 cannot beLebesgue measurable. Thus, L is strictly contained in P(R).

The next result now follows easily from the previous result. We leave the proof as an exercise.

Proposition 3.14. Let E ⊆ R. Then the following properties are equivalent.

(i) E ∈ L;

(ii) there exists a Gδ-set V and N ∈ L with m(N) = 0 such that E = V \N ;

(iii) there exists an Fσ-set W and N ′ ∈ L with m(N ′) = 0 such that E = W ∪N ′.

We now compare the Lebesgue integral with the Riemann integral. We write R([a, b]

)to

denote the space of Riemann integrable functions on the compact interval [a, b] and we write

R∫ ba f dt to denote the Riemann integral of f ∈ R

([a, b]

). We recall that the definition of

f ∈ R([a, b]

)requires f to be bounded on [a, b].

Proposition 3.15. The following properties hold true.

(i) If f : [a, b]→ R and f ∈ R([a, b]

), then f is Lebesgue integrable on [a, b] and∫

[a,b]f dm = R

∫ b

af(t) dt .

(ii) If f : [a,+∞] → [0,+∞), f ∈ R([a, b]

), for all b > a and there exists the improper

Riemann integral R∫ +∞a f(t) dt, then f is Lebesgue integrable on [a,+∞) and∫

[a,+∞)f dm = R

∫ +∞

af(t) dt .

Analogously, if f : [a, b) → [0,+∞), f ∈ R([a, c]

), for all a < c < b and there exists

the improper Riemann integral R∫ ba f(t) dt, then f is Lebesgue integrable on [a, b) and∫

[a,b)f dm = R

∫ b

af(t) dt .


(iii) If f : [a, b] → R is bounded, then f ∈ R([a, b]

)if and only if the set F of points of

discountinuity of f is such that m(F ) = 0.

Proof. (i) Given a partition P = a = x0 < x1 < · · · < xn = b of [a, b], let

s(f,P) =

n∑i=1

mi(xi − xi−1), S(f,P) =

n∑i=1

Mi(xi − xi−1),

be the inferior and superior sums of f , resp., w.r.t. the partition P, where mi = infx∈(xi−1,xi) f(x)

and Mi = supx∈(xi−1,xi) f(x). Since f ∈ R([a, b]

), given n = 1, 2, . . . , there exist partitions Pn

of [a, b] such that Pn+1 is a refinement of Pn and 0 ≤ S(f,Pn)− s(f,Pn) ≤ 1/n.For each n = 1, 2, . . . fixed, set Ii(xi−1, xi), and define the simple functions

sn =

n∑i=1

miχIi , Sn =

n∑i=1

MiχIi .

Then,0 ≤ s1 ≤ s2 ≤ · · · ≤ f ≤ · · · ≤ S2 ≤ S1 .

Since the sequences sn and Sn are monotone, let L(x) = limn→+∞ sn(x) and U(x) =limn→+∞ Sn(x). Then, 0 ≤ L(x) ≤ f(x) ≤ U(x) in [a, b]. Notice that L,U are bounded andmeasurable. Now, by the MCT and since f ∈ R

([a, b]

),∫

Ldm = limn→+∞

∫sn dm = lim

n→+∞

n∑i=1

mi(xi − xi−1) = limn→+∞

s(f, Pn) = R∫ b

af(t) dt ,

and also,∫U dm = lim

n→+∞

∫Sn dm = lim

n→+∞

n∑i=1

Mi(xi − xi−1) = limn→+∞

S(f, Pn) = R∫ b

af(t) dt .

Therefore, 0 ≤ L ≤ U and∫

[a,b](U − L) dm =∫

[a,b] U dm −∫

[a,b] Ldm = 0. This gives that

U = L = f a.e. and ∫[a,b]

f dm = R∫ b

af(t) dt ,

as we wished to prove.

Next we prove (iii). Let P = ∪+∞n=1Pn. Clearly P is countable, hence of measure 0. For

x ∈ [a, b] \ P , the following are equivalent:

• f is continuous at x;

• U(x) = L(x).

Since x ∈ [a, b] \ P , there exists a sequence of intervals In such that In ⊇ In+1, ∩+∞n=1In = x.

Indeed, given n, x belongs to a unique interval In determined by the partition Pn, Pn+1 is arefinement of Pn. Hence, In ⊇ In+1, and ∩+∞

n=1In is non-empty, but its diameter is 0, sinceeach partition Pn contains intervals of at most length 1/n. Then, L(x) = limn→+∞ infx∈In f(x),U(x) = limn→+∞ supx∈In f(x). It is now easy to convince oneself that the two conditions aboveare equivalent.

Next we observe that the following are equivalent:

(a) f is continuous a.e..;

32 M. M. PELOSO

(b) L = U a.e.;

(c) f ∈ R([a, b]

).

Clearly (a) is equivalent to (b), and notice that in (i) we have proved that (c) implies (b).Suppose (b) does not hold, then 0 ≤

∫Ldm <

∫U dm, that is, U − L > δ > 0 on a set of

positive measure. This implies that difference infP S(f,P) − supP s(f,P) > 0, that is (c) doesnot hold. Hence, (b) implies (c), and we are done.

(ii) We only prove the first part, the proof of the second one being completely analogous.Since f ∈ R

([a, b]

)for all b > a, by (i) we have that∫

[a,n]f dm = R

∫ n

af(t) dt

for all n sufficiently large. Set fn = χ[a,n]f . Then 0 ≤ f1 ≤ f2 ≤ · · · ≤ f and fn → f pointwisein [a,+∞). By the MCT∫

[a,+∞]f dm = lim

n→+∞

∫[a,n]

f dm = limn→+∞

R∫ n

af(t) dt = R

∫ +∞

af(t) dt .

This proves (ii), hence the theorem.

3.3. Examples.Here and in what follows, dx = dm(x) denotes the Lebesgue measure on the real line. We

also “import” the notation∫ ba f(x) dx to denote the Lebesgue integral on the set [a, b], that is,

we write ∫[a,b]

f dm =

∫ b

af(x) dx ,

since, by the privious result the two expressions are equal for f when both integral exist.

(1) Consider the functions fn : [0,+∞) → [0,+∞), fn = 1nχ[n,n+1). Show that fn → 0

uniformly in [0,+∞), but∫ +∞

0 fndx 6→∫ +∞

0 f dx.

On the other hand, show that if fn → f uniformly on a compact interval [a, b], fn ∈ L1([a, b]),

then∫ ba fn dx→

∫ ba f dx.

(2) We evaluate the following limits, if they exist:

limk→+∞

∫ k

0

(1− x

k

)kex/2 dx , lim

k→+∞

∫ k

0

(1 +

x

k

)ke−2x dx .

(3) Let fn : R → [0,+∞) be given by fn(x) = (1 + 4nx2)−1. Setting f =∑+∞

n=1 fn, evaluate∫f dx.We observe that for each n, fn is continuous on R, hence measurable. Since fn ≥ 0, f =∑+∞n=1 fn is well defined, non-negative, and measurable. We may apply Cor. 2.17 (2) to see that

+∞∑n=1

∫fn dx =

∫ +∞∑n=1

fn dx =

∫f dx .


We observe that each fn is integrable and then∫Rf(x) dx =

+∞∑n=1

∫Rfn(x) dx

= 2+∞∑n=1

∫ +∞

0

1

(1 + (2nx)2)dx

=+∞∑n=1

1

2n−1arctan(2nx)

∣∣∣+∞0

= π .

(4) Let fn : (0,+∞)→ [0,+∞), n = 1, 2, . . . be given by

fn(x) =1

xen/x log2(1 + nx).

Show that fn ∈ L1((0,+∞)

)for all n and that limn→+∞

∫ +∞0 fn dx = 0.

(5) Given the functions fn : [0,+∞)→ R,

fn(x) =n4/3x

1 + n5/2x3,

evaluate the limits

(a) limn→+∞

∫ +∞

1fn(x) dx ,

(b) limn→+∞

∫ 1

0fn(x) dx ,

justifying your answers.

It is easy to see that the fn are measurable and non-negative. (a) When x ≥ 1, fn → 0pointwise and

0 ≤ fn(x) ≤ 1

n7/6x2≤ 1

x2∈ L1(1,+∞) .

By the DCT it follows that∫ +∞

1 fn dx→ 0 as n→ +∞. (b) With the change of variables nx = twe obtain that ∫ 1

0fn(x) dx =

∫ +∞

0gn(t) dt ,

where gn(t) = tn2/3+n1/6t

χ(0,n)(t). Then, gn → 0 pointwise on (0,+∞) and |gnt)| ≤ t1+t3

∈L1((0,+∞)

). The conclusion now follows from the DCT.

(5) Let fk, gk : [0,+∞)→ R be given respectively, by

fk(x) =1√kχ(0,k)(x)

1√xe(x− 1

k)2

and gk(x) =1

k2χ[ 1

k,+∞)(x)

1

x3e−(x−k)2 .

Evaluate, if they exist, the limits,

(i) limk→+∞

∫ +∞

0fk(x) dx , and (ii) lim

k→+∞

∫ +∞

0gk(x) dx .

34 M. M. PELOSO

3.4. The Cantor set and its generalizations. In this section we introduce a noticeableexample of a set in [0, 1] that is uncountable and yet it has measure 0, while a variation on thesame theme produces sets in [0, 1] of measures arbitrarily close to 1.

Definition 3.16. Consider the compact interval [0, 1]. Let J1,0 = (1/3, 2/3) and C1 = [0, 1]\J1,0.Then C1 is union of two intervals, namely I1,0 = [0, 1/3] and I1,1 = [2/3, 1], each of length 1/3.Next, we define C2 as the interval C1 taken away the two middle thirds of each I1,j , j = 0, 1.Then C2 is union of 22 intervals, namely I2,0 = [0, 1/9], I2,1 = [2/9, 3/9], I2,2 = [6/9, 7/9], andI2,3 = [8/9, 9/9], each of length 1/32. We now iterate this process. Notice that at each step wedouble the number of intervals. Thus, we can write

Ck =

2k−1⋃j=0

Ik,j , m(Ik,j) = 3−k for all j = 0, . . . , 2k − 1,

and

Ck+1 =

2k−1⋃j=0

(Ik,j \ Jk,j

)=:

2k+1−1⋃j=0

Ik+1,j , with m(Jk,j) = 3−k .

We then construct sets Ck, k = 1, 2, . . . such that

• C1 ⊇ C2 ⊇ · · · ;• each Ck is compact so that ∩+∞

k=1Ck 6= ∅;• each Ck is union of 2k disjoint intervals, each of length 3−k.

These properties are all easy to check. We finally set

C =

+∞⋂k=1

Ck .

The set C is called the Cantor set.

Proposition 3.17. Let C be the Cantor set. Then, the following properties hold true.

(i) C is compact, nowhere dense and totally disconnected;

(ii) C has the cardinality of continuum;

(iii) C ∈ L and m(C) = 0.

We recall that a set is said to be nowhere dense if its closure has empty interior and a set istotally disconnected if its connected components are single points.

Proof. It is clear that C is compact and non-empty, as intersection of incapsulated compact sets.For any given ε > 0 and an interval I ⊆ [0, 1], let k be such that 2−k < ε. Since C ⊆ Ck foreach k, and Ck is union of disjoint compact intervals of length 3−k, I cannot be contained inCk, hence in C. Thus, C does not contain any interval. This implies that C is nowhere denseand totally disconnected. This proves (i).

To show (ii), it is easy to construct an injective correspondence between the set of sequencesof 0 and 1 and C. Given the sets I1,0 and I1,1 we associate the values 0 and 1, resp. SplittingI1,0 into I2,0 and I2,1 we associate to each of them the terms 0, 0 and 0, 1, resp., and so on.At each step we have an interval Ik,j to which it is associated the finite sequence of length kof 0’s and 1’s, when we split it into two intervals, we associate to them the sequences of lengthk+1 by adding a 0 to the sequence corresponding to interval on the left, and a 1 to the sequence


corresponding to interval on the right. In this way we associate to any given sequence of 0’s and1’s the a sequence of incapsulated compact intervals Ik`,j`, which has non-empty intersectionand thus contains at least one element of C. This proves (ii).

(iii) Notice that C = limk→+∞m(Ck). The set Ck+1 is obtained by removing 2k intervals

each of length 3−(k+1) from Ck. Then,

m(C) = limk→+∞

m(Ck+1) = limk→+∞

(m(Ck)−

2k−1∑j=0

m(Jk,j

))= lim

k→+∞

(m(Ck)−

2k

3k+1

)= lim

k→+∞

(m(Ck−1)− 2k−1

3k− 2k

3k+1

)= lim

k→+∞

(1− 1

3− · · · − 2k−1

3k− 2k

3k+1

)= lim

k→+∞1− 1

2

k+1∑j=1

2j

3j

= 1− 1

2·

23

1− 23

= 0 .

This proves the proposition.

Using a similar procedure, for any ε > 0 we can construct a nowhere dense, totally discon-nected subset of [0, 1], of measure greater than 1− ε.

Proposition 3.18. Let ε > 0. Then, there exists α, 0 < α ≤ 1/3 and a set C(α) ⊆ [0, 1], which

is compact, nowhere dense, and totally disconnected set, such that m(C(α)) ≥ 1− ε.

Proof. We imitate the construction of the Cantor set. Given [0, 1], we remove the central interval

of length α, that is we define C(α)1 = [0, 1] \

(12 −

α2 ,

12 + α

2

). From the resulting two intervals, we

remove the central intervals, each of length α2, Then, at the step k + 1, we remove from C(α)k

the central intervals of length αk+1, and we set C(α) = ∩+∞k=1C

(α)k . Notice that, if α = 1

3 , then

C(α) = C, the Cantor set.The topological properties of C(α) are proved as in the case of C. Finally, observing that the

set C(α)k+1 is obtained by removing 2k intervals each of length αk+1 from C

(α)k , we have

m(C(α)) = limk→+∞

m(C(α)k+1) = lim

k→+∞

(m(C

(α)k )− 2kαk+1

)= lim

k→+∞1− 1

2

k+1∑j=1

(2α)j

= 1− 1

2· 2α

1− 2α

=1− 3α

1− 2α.

Hence, it suffices to select α so that 1−3α1−2α > 1− ε.

Corollary 3.19. There exist open sets U ⊆ R such that m(U \ U) > 0.

36 M. M. PELOSO

Proof. Notice that, since U is open, U \ U = ∂U , the boundary of U .

We observe that for any 0 < α ≤ 1/3, C(α) is a perfect set, that is, it coincides with its derived

set, the set of its accumulation points. Since C(α) is closed, it suffices to notice that each pointof C(α) is of accumulation for C(α) itself. Thus, it follows that C(α) = ∂C(α).

It suffices to consider U = cC(α), with 0 < α < 1/3, so that m(C(α)) > 0. Now, ∂U =

∂( cU) = ∂C(α) = C(α), so that m(∂U) > 0.

3.5. Integrals depending on a parameter. The following result is quite often very useful.We state it for a generic measure space (X ,M, µ), although we will essentially use it in the caseof the Lebesgue measure (R,L,m).

Theorem 3.20. Let (X ,M, µ) be a measure space, f : X × [a, b] → R (or C), such thatf(·, t) ∈ L1(µ) for every t ∈ [a, b] and let

F (t) =

∫Xf(x, t) dµ(x) .

(1) Suppose that there exists g ∈ L1(µ) such that |f(x, t)| ≤ g(x) for all t ∈ [a, b] and supposethat limt→t0 f(x, t) = f(x, t0). Then

limt→t0

F (t) = F (t0) .

In particular, if f(x, ·) is continuous for every x ∈ X , then F is continuous.(2) Suppose that ∂tf exists and that there exists g ∈ L1(µ) such that |∂tf(x, t)| ≤ g(x) for all

(x, t) ∈ X × [a, b]. Then F is differentiable and

F ′(t) =

∫X∂tf(x, t) dµ(x) .

We remark that in the applications this result is often used in its local form, that is, in thengbh of a given point t0 of the parameter. It is in fact sufficient just take [a, b] to be a sufficientlysmall ngbh of t0.

Proof. (1) Let tn ⊆ [a, b] be any sequence converging to t0 and set fn(x) = f(x, tn). Thenfn → f(x) := f(x, t0) |fn| ≤ g ∈ L1(µ), and we can apply the DCT to fn and obtain

limn→+∞

F (tn) = limn→+∞

∫Xfn(x) dµ =

∫Xf(x, t0) dµ = F (x, t0) .

Since the sequence tn → t0 was arbitrary, the conclusion follows.(2) Again, let tn ⊆ [a, b] be any sequence converging to t0. Set

hn(x) =f(x, tn)− f(x, t0)

tn − t0.

Then, limn→+∞ hn(x) = ∂tf(x, t0). Hence, ∂tf(·, t0) is measurable. Moreover, using the meanvalue theorem we see that

|hn(x)| ≤ supt∈[a,b]

∣∣∂tf(x, t)∣∣ ≤ g(x) .


Therefore, we can apply the DCT to hn, using again the fact that the sequence tn → t0 wasarbitrary, and obtain that

F ′(t0) = limn→+∞

F (tn)− F (t0)

tn − t0= lim

n→+∞

∫X

f(x, tn)− f(x, t0)

tn − t0dµ(x) = lim

n→+∞

∫Xhn(x) dµ

=

∫X∂tf(x, t0) dµ ,


Example 3.21. The Gamma function. For z ∈ C, Re z > 0 we set

Γ(z) =

∫ +∞

0tz−1e−t dt .

(We recall that tz−1 = e(z−1) log t.)Then:

(1) for all a ∈ C, Re z > 0, zΓ(z) = Γ(z + 1);

(2) Γ(1) = 1;

(3) for all non-negative integer n, Γ(n+ 1) = n!;

(4) Γ ∈ C∞((0,+∞)

).

Furthermore, identifying z = x+ iy ∈ C with (x, y) ∈ R2, the function Γ is C∞ in (x, y) : x >0.4 We prove the above statements in the case z real, that is, z = x > 0. The case of thecomplex z is proven in exactly the same way, however we do not need in the present class.

We begin by observing that, for every fixed x > 0, tx−1e−t ∈ L1((0,+infty)

). For, tx−1e−t ≤

tx−1 for 0 < t < 1, while tx−1e−t is easily see to be integrable on (1,+∞). Then, integrating byparts we see that

Γ(x) =

∫ +∞

0tx−1e−t dt = lim

ε→0+, b→+∞

∫ b

εtx−1e−t dt

= limε→0+, b→+∞

1

x

(txe−t

∣∣∣bε

)+

1

x

∫ b

εtxe−t dt

= 0 +1

x

∫ +∞

0txe−t dt

=Γ(x+ 1)

x.

This shows (1), while (2) it is obvious. To show (3), assume by induction that Γ(n) = (n− 1)!.By (1) we now have that Γ(n+ 1) = nΓ(n) = n(n− 1)! = n!. It easily follows from Thm. 3.20(2), and we leave the details as an exercise.

4In fact, Γ is differentiatble in the complex sense, that is, Γ is holomorphic. We will not stress this, howeverfundamental, aspect at the present time.

38 M. M. PELOSO

3.6. More on L1(m). In this section we prove some more results about the Lebesgue spaceL1(m), that by Thm. 2.31, we recall, is a Banach space. We recall, if X is a topological space,that the support of a continuous function g is the closure of the set x ∈ X : g(x) 6= 0. Wedenote by Cc(X ) the space of continuous functions with compact support in X .

We begin with an elementary lemma, which holds true on any σ-finite measure space.

Lemma 3.22. Let (X ,M, µ) be σ-finite measure space and let f ∈ L1(µ). For any ε > 0 thereexists a set A such that

∫cA |f | dµ < ε and f is bounded on A.

Proof. For sake of simplicity, we prove only in the case of (R,L,m). We invite the reader toprove it in the general case.

DefineBn =

x : |x| ≤ n and |f(x)| ≤ n

,

so that B1 ⊆ B2 ⊆ · · · , and ∪+∞n=1Bn = R. Define the sequence fn = χBnf . Then, fn → f

pointwise, |fn| ≤ |f | ∈ L1(m), so that fn → f in L1(m) as n → +∞, by the DTC. It sufficesto choose nε so that

∫cBnε|f | dm = ‖f − fnε‖L1(m) < ε and set A = Bnε . The conclusion now

follows.

Proposition 3.23. (Absolute continuity of the measure) Let f ∈ L1(R). For any ε > 0there exists δ > 0 such that if m(F ) < δ, then∫

F|f | dm < ε .

Proof. Given ε > 0, let A be a set such that∫

cA |f | dm < ε/2 and f is bounded on A, sayM = supx∈R |f(x)|. Now, chose δ = ε/(2M). Then, if F is a measurable set with m(F ) < δ wehave ∫

F|f(x)| dx ≤

∫cA∩F

|f(x)| dx+

∫A∩F|f(x)| dx

≤∫

cA|f(x)| dx+

∫A∩F|f(x)| dx

<ε

2+Mm(F ) <

ε

2+Mδ

= ε .

We now recall

Lemma 3.24. (Urysohn’s Lemma)5 Let X be a locally compact Hausdorff space, K compactand U open, K ⊆ U . Then there exists g continuous such that g = 1 on K and g = 0 on cU .

In particular, if U is taken such that U is compact, then the function g is continuous and hascompact support.

Theorem 3.25. (Lusin’s Theorem) (1) Let f be a measurable function on R. Suppose thatthere exists A ⊆ R such that m(A) < +∞ and f = 0 on cA. Then, for any ε > 0, there existsg ∈ Cc(R) such that

m(x ∈ R : f(x) 6= g(x)

)< ε.

If f is bounded, g can be chosen so that |g| ≤ |f .(2) The space Cc(R) is dense in L1(m).

5See e.g. [Rudin, Real and Complex Analysis, 3rd Ed., McGraw–Hill Editor]


Proof. Assume first that f is bounded, say |f | ≤ 1, and that A is compact. Then, by Thm. 2.9(2), there exists a sequence of simple functions ϕn such that |ϕ1| ≤ |ϕ2| ≤ · · · ≤ |f |, ϕn → fpointwise and uniformly on every set on which f is bounded – hence on all of R in this case. Setψ1 = ϕ1 and ψn = ϕn − ϕn−1 for n ≥ 2. Then,

∑nk=1 ψk = ϕn so that

∑+∞n=1 ψn = f , pointwise

and uniformly since f is bounded. Observe that, with the notation of Thm. 2.9, assuming freal-valued for simplicity,

ψn = ϕn − ϕn−1 =(s+n − s+

n−1

)−(s−n − s−n−1

)We claim that ψn = 2−nχTn , for a certain set Tn. Indeed, assume for a moment that 0 ≤ f ≤ 1,and recall the construction of sn:

sn(x) =2n−1∑k=0

k2−nχEn,k(x) ,

where

En,k = f−1((k2−n, (k + 1)2−n]

),

and similarly for sn−1. (Notice that 0 ≤ k ≤ 2n − 1 only, since f ≤ 1.) The function sn+1 wasconstructed by splitting the sets En,k = En+1,2k∪En+1,2k+1 as in (3), we have that sn+1 = sn on

En+1,2k, while sn+1 − sn = 2−(n+1) on En+1,2k+1, for each k = 0, . . . , 2n+1 − 1. This establishes

the claim for s±n − s±n−1, hence for ψn.

Next, fix an open set U such that A ⊆ U and U is compact. Given ε > 0 there exist setsKn, Un, where Kn is compact, Un is open, and such that

Kn ⊆ Tn ⊆ Un ⊆ U and m(Un \Kn) < ε2−n .

By Urysohn’s Lemma there exist continuous functions hn such that 0 ≤ hn ≤ 1, hn = 1 on Kn,and hn = 0 on cUn. Define

g(x) =

+∞∑n=1

2−nhn(x) .

Then, the series converges uniformly, g is continuous, and it is 0 outside U , which is is bounded,so that supp(g) is compact. Moreover, observe that hn = 1 on Kn, so that 2−nhn = ψn on Kn,and that hn = 0 = ψn on cUn. Hence,

x : g(x) 6= f(x) ⊆+∞⋃n=1

Un \Kn

and

m(x : g(x) 6= f(x)

)≤

+∞∑n=1

m(Un \Kn) < ε .

This proves (1) when |f | ≤ 1 and A is compact.Now, let A be any set such that m(A) < +∞. Then, given any ε > 0, there exist U,K, with

K compact, U open K ⊆ A ⊆ U , and such that m(U \K) < ε/2. Then, by the argument above,we can find a countinuous function with support in K such that

m(x ∈ K : g(x) 6= f(x)

)< ε/2 .

40 M. M. PELOSO

Since also f vanishes outside U

m(x : g(x) 6= f(x)

)= m

(x ∈ U : g(x) 6= f(x)

)≤ m

(x ∈ K : g(x) 6= f(x)

)+m(U \K)

< ε .

Thus, the conclusion holds true when f is bounded.

Finally, we remove the assumption that f is bounded. For each n, let Bn = x : |f(x)| ≤ n.Since f is real valued, ∪+∞

n=1Bn = R, so that cB1 ⊇ cB2 ⊇ · · · , and ∩+∞n=1

cBn = ∅. Thus,m( cBn) → 0 as n → +∞. Then, given ε > 0 we can find an integer n such that m( cBn) ≤ε/2 and then g continuous with compact support contained in Bn such that m

(x : g(x) 6=

(χBnf)(x))< ε/2. Therefore,

m(x : g(x) 6= f(x)

)≤ m

(x ∈ Bn : g(x) 6= f(x)

)+m( cBn)

< ε .

This proves (1).

In order to prove (2), given f ∈ L1(m), given any ε > 0, by Lemma 3.22 there exists A ⊆ Rsuch that m(A) < +∞ and

∫cA |f | dm < ε/2 and f is bounded on A, say |f | ≤M on A. Part (1)

now gives that there exists g continuous and with support in A and such that m(A1) < ε/4M ,where A1 = x ∈ A : g(x) 6= f(x), and |g| ≤ |f |. Then

‖f − g‖L1(m)

∫A1

|f − g| dm∫A\A1

|f − g| dm+

∫cA|f − g| dm

≤ ε/2 + ε/2 .

The proof is now complete.

4. Product measure spaces and the Lebesgue integral in Rn

Our current goal is to define the Lebesgue measure on the higher-dimensional eucledean spaceRn, and to reduce the computations to integrals in lower dimensions. In order to do this, wefirst present the theory of integration on product measure spaces and the fundamental theoremsof Tonelli and Fubini on the equality of iterated integrals.

4.1. Product measure spaces.

Definition 4.1. Let (X ,M) and (Y,N ) be measurable spaces. A subset of X × Y of the formA×B where A ∈M and B ∈ N is called a measurable rectangle. We define the product σ-algebrain X × Y as the σ-algebra generated by the collection of measurable rectangles and we denoteit by M×N .

Next, given two measure spaces (X ,M, µ) and (Y,N , ν) we wish to construct a measure onM×N such that the measure of any measurable rectangle A×B equals µ(A)ν(B).

Throughout this section, we denote by A the collection of finite unions of disjoint measueablerectangles in X × Y.

Lemma 4.2. The following properties hold:


(i) A is an algebra;

(ii) if M(A) denotes the σ-algebra generated by A, then M(A) =M×N .

Proof. (†) In order to prove that A is an algebra we need to show that it is closed under thecomplement and finite unions of sets in A. Given two measurable rectangles A1×B1 and A2×B2,we preliminary observe that (A1×B1)∩ (A2×B2) = (A1 ∩A2)× (B1 ∩B2) is also a measurablerectangle.

Now, given a measurable rectangle A×B we have thatc(A×B) = (X × cB) ∪ ( cA×B) ,

that is, c(A×B) is finite union of disjoint measurable rectangles, hence in A. Next, let A1×B1

and A2 × B2 be measurable rectangles A not necessarily disjoint (otherwise we have nothingto prove). By the first part, A 3 c(A2 × B2) = ∪nj=1(Ej × Fj), where Ej × Fj are disjointmeasurable rectangles. Then,

(A1 ×B1) \ (A2 ×B2) = (A1 ×B1)⋂(

(E1 × F1) ∪ · · · ∪ (En × Fn))

=(

(A1 ∩ E1)× (B1 ∩ F1))∪ · · · ∪

((A1 ∩ En)× (B1 ∩ Fn)

),

hence a finite union of disjoint measurable rectangles, which belongs to A. Therefore,

(A1 ×B1) ∪ (A2 ×B2) =(

(A1 ×B1) ∩ c(A2 ×B2))∪ (A2 ×B2) ∈ A .

By induction, we obtain that

(A1 ×B1) ∪ · · · ∪ (An ×Bn) ∈ A ,that is, A is closed under finite unions. Finally, let A×B ∈ A, that is, A×B = ∪nj=1(Ej ×Fj),where Ej × Fj are disjoint measurable rectangles. Then,

c(A×B) = c(

(E1 × F1) ∪ · · · ∪ (En × Fn))

= c(E1 × F1) ∩ · · · ∩ c(En × Fn)

=((X × cF1) ∪ ( cE1 × F1)

)∩ · · · ∩

((X × cFn) ∪ ( cEn × Fn)

)=

n⋃j=1

(Aj ×Bj) ∩ (A′j ×B′j) : Aj , A

′j = X or cE`, Bj , B

′j = F`

.

Hence, c(A×B) is union of intersections of disjoint measurable rectangles, hence it is union ofdisjoint measurable rectangles. This shows that A is an algebra, i.e. (i).

In order to prove (ii), observe that M×N contains the measurable rectangles, hence theirfinite unions. Thus, M×N contains A and therefore the σ-algebra generated by it, that is,M(A). Conversely,M(A) contains all the measurable rectangles, hence the σ-algebra generatedby them, that is, M×N . This completes the proof.

Now, let (X ,M, µ) and (Y,N , ν) be measure spaces and let A be the algebra of finite unionsof disjoint measurable rectangles. If E ⊆ X ×Y and E = ∪nj=1Aj ×Bj is an element of A (that

is, Aj ×Bj are disjoint measurable rectangles, j = 1, . . . , n), we set

ρ(E) =n∑j=1

µ(Aj)ν(Bj) . (14)

42 M. M. PELOSO

Lemma 4.3. With the hypotheses as above, ρ is a premeasure on A.

Proof. (†) First of all, we need to show that ρ is well defined, that is, for any E ∈ A, the value ofρ(E) is independent of the decomposition of E as finite unions of disjoint measurable rectangles.Observe that, if A×B is a measurable rectangle and A×B = ∪mj=1A

′j ×B′j , where A′j ×B′j are

disjoint measurable rectangles, we have

χA(x)χB(y) = χA×B(x, y) =

m∑j=1

χA′j×B′j (x, y) =

m∑j=1

χA′j (x)χB′j (y) .

We now integrate w.r.t. x on both sides of the equalities above and see that

µ(A)χB(y) =

∫χA(x)χB(y) dµ(x) =

m∑j=1

∫χA′j (x)χB′j (y) dµ(x) =

m∑j=1

µ(A′j)χB′j (y) .

We proceed by integrating w.r.t. y and obtain

ρ(A×B) = µ(A)ν(B) =

∫µ(A)χB(y) dν(y) =

∫ m∑j=1

µ(A′j)χB′j (y) dν(y)

=

m∑j=1

µ(A′j)ν(B′j) . (15)

This shows that ρ(A×B) =∑m

j=1 µ(A′j)ν(B′j) for any decomposition of A×B as finite unionsof disjoint measurable rectangles. This easily implies that ρ is well defined. Now, ρ is finitelyadditive by construction.

We need to show ρ is countably additive, that is, if Aj × Bj is a collection of of disjointmeasurable rectangles whose union is in A, then

ρ(+∞⋃j=1

(Aj ×Bj))

=+∞∑j=1

ρ(Aj ×Bj) . (16)

By the finite additivity of ρ, we first show that it suffices to consider the case in which ∪+∞j=1(Aj×

Bj) is a single measurable rectangle. The argument is analogous to the one in the proof of Lemma

3.8, but we repeat it for sake of completeness. If A×B = ∪nk=1(A(k) ×B(k)), with A(k) ×B(k))

disjoint measurable rectangles, it is possible to find subcollections (A(k)j ×B

(k)j ), k = 1, 2, . . . , n,

of (Aj ×Bj) such that, for each k = 1, 2, . . . , n,

A(k) ×B(k) =

+∞⋃j=1

(A(k)j ×B

(k)j ) .

If we know that for each k, ρ(A(k) × B(k)

)=∑+∞

j=1 ρ(A

(k)j × B

(k)j

), the conclusion then follows

from the finite additivity of ρ.

Thus, let us show (16) when ∪+∞j=1(Aj × Bj) = (A × B). We proceed as in the argument

leading to(15). We have that χA(x)χB(y) =∑+∞

j=1 χAj (x)χBj (y). Integrating first w.r.t. x wehave

µ(A)χB(y) =

∫χA(x)χB(y) dµ(x) =

+∞∑j=1

µ(Aj)χBj (y) .


Next we integrate w.r.t. y and obtain

ρ(+∞⋃j=1

(Aj ×Bj))

= ρ(A×B) = µ(A)ν(B)

=

∫µ(A)χB(y) dν(y) =

∫ +∞∑j=1

µ(Aj)χBj (y) dν(y)

=m∑j=1

µ(Aj)ν(Bj) .

This shows that ρ is a premeasure and we are done.

Definition 4.4. Let (X ,M, µ), (Y,N , ν) be measure spaces, M×N the product σ-algebra.We define an outer measure (µ× ν)∗ on P(X × Y) as

(µ× ν)∗(E) = inf +∞∑j=1

ρ(Aj) : E ⊆+∞⋃j=1

Aj , Aj ∈ A for all j.

Lemma 3.5 shows that indeed (µ×ν)∗ is an outer measure, that A is contained in the σ-algebraof (µ × ν)∗-measurable sets, and (µ × ν)∗ restricted to A coincides with ρ. As a consequence,we also obtain that M×N is contained in the σ-algebra of (µ× ν)∗-measurable sets.

Furthermore, if we also assume that (X ,M, µ) and (Y,N , ν) are σ-finite, Thm. 3.7 showsthat there exists a measure defined on the σ-algebra of (µ× ν)∗-measurable sets.

Thus, we define the product measure µ× ν on the product σ-algebra M×N(µ× ν)(E) = (µ× ν)∗(E) , E ∈M×N .

Finally, µ× ν is the unique measure on M×N such that (µ× ν)|A = ρ.

We observe that the σ-algebra of (µ× ν)∗-measurable sets equals (M×N ) union the subsetsof sets of (µ× ν)∗-measure 0,

4.2. Integration on product measure spaces. Having constructed the product measurespace (X × Y,M×N , µ × ν) we now have all the results about the integration theory in anabstract measure space at our disposal. However, we wish to relate the calculus of integrals onX × Y to the integrals on X and on Y. This is the goal of the current section.

Thus, let the measure spaces (X ,M, µ) and (Y,N , ν) be given. Given a set E ∈ M×N wedefine its x-section as

Ex =y ∈ Y : (x, y) ∈ E

and its y-section as

Ey =x ∈ X : (x, y) ∈ E

.

If f is a function on X × Y we define the x-section as the function fx(y) = f(x, y) and they-section as fy(x) = f(x, y). We now have

Proposition 4.5. With the notation above, the following hold true:

(i) if E ∈M×N then, for every x ∈ X , Ex ∈ N and, for every y ∈ Y, Ey ∈M;

(ii) if f is a function on X × Y that is M × N -measurable, for every x ∈ X , fx is N -measurable and, for every y ∈ Y, fy is M-measurable.

44 M. M. PELOSO

Proof. This is easy. Let

R =E ∈M×N : Ex ∈ N and y ∈ Y for all x ∈ X , y ∈ Y

.

Then, R contains all measurable rectangles A×B, since

(A×B)x =

B if x ∈ A∅ if x 6∈ A

and (A×B)y =

A if y ∈ B∅ if y 6∈ B .

(17)

Next, notice that R is a σ-algebra, since(+∞⋃j=1

Ej

)x

=+∞⋃j=1

Ejx and(+∞⋃j=1

Ej

)y=

+∞⋃j=1

Ejy ,

and also (cE)x

= c(Ex)

and(cE)y

= c(Ey).

Therefore, R contains the σ-algebra generated by the measurable rectangles, hence the σ-algebragenerated by A, which equals M×N . Hence, R =M×N , and this shows (i).

In order to show (ii), it suffices notice that (f−1(U))x = (fx)−1(U) and (f−1(U))y = (fy)−1(U).

Next, we need the following notion.

Definition 4.6. A collection C of subsets of a set X is called a monotone class if it is closedwith respect to countable increasing unions and countable decreasing intersections.

Clearly, a σ-algebra is a monotone class, and in particular P(X ) is a monotone class. It iseasy to see that intersections of monotone classes is still a monotone class. Hence, given anyE ⊆ P(X ) there exists the smallest monotone class containing E ; namely the intersection of allmonotone classes containing E , family that is not empty since it contains P(X ). We denote byC(E) such monotone class and we call it the monotone class generated by E .

Theorem 4.7. (Monotone Class Lemma) Let A be an algebra of subsets of X . Then,M(A) = C(A), that is, the monotone class generated by A coincides with the σ-algebra generatedby A.

Proof. (†) See [F], Monotone Class Lemma 2.35.

We use the Monotone Class Lemma to prove the following result, which is fundamental forthe proof of the Tonelli–Fubini theorem that will follow.

Proposition 4.8. Let (X ,M, µ), (Y,N , ν) be σ-finite measure spaces. Let E ∈M×N . Then,for every x ∈ X , the function y 7→ µ(Ey) is N -measurable and for every y ∈ Y, the functionx 7→ ν(Ex) is M-measurable. Moreover,

(µ× ν)(E) =

∫Xν(Ex) dµ(x) =

∫Yµ(Ey) dν(y) .

Proof. We first assume that µ and ν are finite measures, that is, µ(X ), ν(Y) < +∞. Let C bethe collection of all sets E ∈ M×N for which the conclusions in the statement are true. Weclaim that all measurable rectangles A×B belong to C. Indeed, by (17) it follows that

µ((A×B)y

)=

µ(A) if y ∈ B0 if y 6∈ B

= µ(A)χB(y) ,


so that the function y 7→ µ((A×B)y

)is ν-measurable. Analogously,

ν((A×B)x

)=

ν(B) if x ∈ A0 if x 6∈ A

= ν(B)χA(x) ,

so that the function x 7→ ν((A×B)x

)is µ-measurable. Moreover,

(µ× ν)(A×B) = µ(A)ν(B) =

∫Yµ(A)χB(y) dν(y) =

∫Yµ((A×B)y

)dν(y) ,

and also

(µ× ν)(A×B) = µ(A)ν(B) =

∫Xν(B)χA(x) dµ(x) =

∫Xν((A×B)x

)dµ(x) .

This proves the claim. By finite addivity, also finite unions of disjoint measurable rectanglesalso verify the conclusions in the statement, so that C contains the algebra A of finite unions ofdisjoint measurable rectangles.

If we show that C is a monotone class, then C would contain the monotone class generated bythe algebra A, that, by the Monotone Class Lemma, coincides with M×N . This would implythe result, under the assumption that both µ and ν are finite measures. In order to show thatC is a monotone class, we need to show that it closed under countable increasing unions andcountable decreasing intersections of sets. Suppose then that En ⊆ C and E1 ⊆ E2 ⊆ · · · .Set E = ∪+∞

n=1En. Then µ(En

y) is an increasing sequence of non-negative, N -measurable

functions converging to µ(Ey). Thus, µ(Ey) is also N -measurable, and by the MCT we havethat ∫

Yµ(Ey) dν = lim

n→+∞

∫Yµ(En

y)dν = lim

n→+∞(µ× ν)(En) = (µ× ν)(E) .

By switching the roles of µ and ν and arguing in the same way, we also obtain that∫Xν(Ex) dµ = (µ× ν)(E) .

Thus, C is closed under countable increasing unions.On the other hand, suppose that En ⊆ C and E1 ⊇ E2 ⊇ · · · . Set E = ∩+∞

n=1En. Herewe are going to use the assumption that µ and ν are finite measures. We have that µ

(En

y)

is a decreasing sequence of non-negative, N -measurable functions converging to µ(Ey). Thus,µ(Ey) is also N -measurable. Moreover,

0 ≤ µ(En

y)≤ µ(X ) ∈ L1(ν) ,

since µ(X ), ν(Y) < +∞. Thus, we can apply the DCT and obtain∫Yµ(Ey) dν = lim

n→+∞

∫Yµ(En

y)dν = lim

n→+∞(µ× ν)(En) = (µ× ν)(E) .

Again, by switching the roles of µ and ν and arguing in the same way, we have∫Xν(Ex) dµ = (µ× ν)(E) .

Thus, C is closed under countable decreasing intersections and C is a monotone class. Thisproves the assertion in the case µ, ν are finite measures.

46 M. M. PELOSO

Suppose now that (X ,M, µ), (Y,N , ν) are σ-finite measure spaces. Let Xj ⊆ M, Yj ⊆ Nbe increasing sequences of sets such that ∪+∞

j=1Xj = X , ∪+∞j=1Yj = Y. Let E be any set inM×N .

The previous argument applies to the set Ej = E∩(Xj×Yj) to give that for every j, the functionsy 7→ µ((Ej)

y) and x 7→ ν((Ej)x) are N -measurable and M-measurable, resp., for every y ∈ Yj ,x ∈ Xj , resp. Therefore, the functions y 7→ µ((Ej)

y)χYj (y) and x 7→ ν((Ej)x)χXj (x) areN -measurable and M-measurable, resp., for every y ∈ Y, x ∈ X , resp. Moreover,

(µ× ν)(Ej) =

∫Xj

ν((Ej)x) dµ(x) =

∫Xν((Ej)x)χXj (x) dµ(x)

=

∫Yj

µ((Ej)y) dν(y) =

∫Yµ((Ej)

y)χYj (y) dν(y) .

Finally, we apply the MCT to the increasing sequences ν((Ej)x)χXj and µ((Ej)y)χYj, that

converge to ν(Ex) and mu(Ey), resp., and obtain that

(µ× ν)(E) = limj→+∞

(µ× ν)(Ej)

= limj→+∞

∫Xν((Ej)x)χXj (x) dµ =

∫Xν(Ex) dµ

= limj→+∞

∫Yµ((Ej)

y)χYj (y) dν =

∫Yµ(Ey) dν .

This completes the proof.

Theorem 4.9. (The Tonelli–Fubini Theorem.) Let (X ,M, µ), (Y,N , ν) be σ-finite measurespaces.(1) (Tonelli) Let f be a non-negativeM×N -measurable function on X ×Y. Then, the functionsg(y) =

∫X f

y(x) dµ(x) and h(x) =∫Y fx(y) dν(y) are N -measurable and M-measurable, resp.,

and ∫X×Y

f(x, y) d(µ× ν)(x, y) =

∫Yg(y) dν(y) =

∫Xh(x) dµ(x) . (18)

(2) (Fubini) Let f ∈ L1(X ×Y, (µ×ν)(x, y)

). Then, the function g(y) =

∫X f

y(x) dµ(x) belgongs

to L1(Y, ν) for a.e. x and the function h(x) =∫Y fx(y) dν(y) belgongs to L1(X , µ) for a.e. y

and equalities (18) hold.

Notice that (18) can also be written in terms of iterated integrals∫X×Y

f(x, y) d(µ× ν)(x, y) =

∫Y

(∫Xf(x, y) dµ(x)

)dν(y) =

∫X

(∫Yf(x, y) dν(y)

)dµ(x) .

It is custumary to use the notations∫X×Y

f(x, y) d(µ× ν)(x, y) =

∫X

∫Yf(x, y) dν(y) dµ(x) =

∫∫f dµdν .

Proof. (1) Observe that Tonelli’s theorem reduces to Prop. 4.8 in the case of characteristicfunctions. By finite addivity of the integrals and measures, the result follows for non-negativesimple functions. Let f ≥ 0 be (M×N )-measurable. By Thm. 2.9 (1) there exists an increasing


sequence sn of non-negative simple functions, such that sn → f pointwise. Then, we can applythe MCT to obtain∫

X×Yf(x, y) d(µ× ν)(x, y) = lim

n→+∞

∫X×Y

sn(x, y) d(µ× ν)(x, y)

= limn→+∞

∫Y

(∫Xsn(x, y) dµ(x)

)dν(y) =

∫Y

(∫Xf(x, y) dµ(x)

)dν(y) , (19)

and also ∫X×Y

f(x, y) d(µ× ν)(x, y) = limn→+∞

∫X×Y

sn(x, y) d(µ× ν)(x, y)

= limn→+∞

∫X

(∫Ysn(x, y) dν(y)

)dµ(x) =

∫X

(∫Yf(x, y) dν(y)

)dµ(x) . (20)

This proves (1).

In order to prove Fubini’s theorem, we proceed in an analogous fashion. For f ∈ L1(µ × ν),we Thm. 2.9 (2) to find a sequnce ϕn of simple functions, such that 0 ≤ |ϕ1| ≤ |ϕn| ≤ · · · |f |,ϕn → f pointwise. Then, we argue as in (19) and (20), using the DCT, instead of the MCT.This completes the proof of Tonelli–Fubini’s theorem.

Example 4.10. The hypotheses in the Tonelli–Fubini theorem cannot be relaxed, as the fol-lowing examples show.

(1) Let f : (0, 1)× (−1, 1)→ R be given by f(x, y) = y/x. Then, clearly∫ 1

0

(∫ 1

−1

y

xdy)dx =

∫ 1

00 dx = 0 ,

while, ∫ 1

−1

(∫ 1

0

y

xdx)dy =

∫ 1

−1y(∫ 1

0

1

xdx)dy =

∫ 1

−1g(y) dy

where g(y) = +∞ if 0 < y < 1 and g(y) = −∞ if −1 < y < 0. This last integral does not existand the iterated integrals are not equal.

(2) Let T1 =

(x, y) : 0 < x < 1, 0 < y < x

, T2 =

(x, y) : 0 < y < 1, 0 < x < y

, and define

f : (0, 1)× (0, 1)→ R be given f(x, y) = 1x2χT1(x, y)− 1

y2χT2(x, y). Then,∫ 1

0

(∫ 1

0f(x, y) dy

)dx =

∫ 1

0

(∫ x

0

1

x2dy −

∫ 1

x

1

y2dy)dx

=

∫ 1

0

(1

x+(1

y

∣∣∣1x

))dy =

∫ 1

0

(1

x+ 1− 1

x

)dy

= 1 .

48 M. M. PELOSO

On the other hand,∫ 1

0

(∫ 1

0f(x, y) dx

)dy =

∫ 1

0

(−∫ y

0

1

y2dx+

∫ 1

y

1

x2dx)dy

=

∫ 1

0

(− 1

y+(− 1

x

∣∣∣1y

))dy =

∫ 1

0

(− 1

y− 1 +

1

y

)dy

= −1 .

Thus, again the iterated integrals are not equal.

Remark 4.11. The measure (µ × ν) is not complete, even if (µ,M) and(ν,N ) are completemeasures. As an example consider (X , µ,M) = (cY, ν,N ) = (R,m,L), the Lebesgue measureon R. Let A be any set of measure 0 and B the example of a non-measurable set constructed inExample 1.1. Then (A×B) 6∈ L × L, but it is contained in a set of (m×m)-measure 0.

This is not a significant drawback, since we can always complete the measure (µ×ν). However,the Tonelli–Fubini theorem takes a slightly more complicated form in the case of completemeasures.

Theorem 4.12. (The Tonelli–Fubini Theorem for complete measures) Let (X ,M, µ),(Y,N , ν) be complete σ-finite measure spaces. Let (X ×Y, T , τ) be the completion of the productmeasure space (X × Y,M×N , µ× ν) and f be T -measurable.

(1-Tonelli) If f ≥ 0 then fx is N -measurable for µ-a.e. x ∈ X and fy is M-measurable forν-a.e. y ∈ Y and∫

X×Yf(x, y) dτ(x, y) =

∫Y

(∫Xf(x, y) dµ(x)

)dν(y) =

∫X

(∫Yf(x, y) dν(y)

)dµ(x) . (21)

(2-Fubini) If f ∈ L1(τ) then fx is N -measurable for µ-a.e. x ∈ X and fy is M-measurablefor ν-a.e. y ∈ Y. Moreover fx ∈ L1(ν) for µ-a.e. x ∈ X , fy ∈ L1(µ) for ν-a.e. y ∈ Y and (21)holds.

Proof. (†) We sketch the proof, leaving the details to the reader.Let f be T -measurable. Since (T , τ) is the completion of (M×N , µ × ν), by Prop. 2.12 it

follows that there exists an (M×N )-measurable function g such that g = f τ -a.e. Let h = f−g.We can apply Tonelli–Fubini theorem to g in either case f ∈ L+(τ) or f ∈ L1(τ) and obtainthat ∫

X×Yg(x, y) dτ(x, y) =

∫X×Y

g(x, y) d(µ× ν)(x, y)

=

∫Y

(∫Xg(x, y) dµ(x)

)dν(y) =

∫X

(∫Yg(x, y) dν(y)

)dµ(x) .

Hence, it suffices to show that if h is T -measurable and h = 0 τ -a.e., then hx, hy are in L1(ν)and in L1(µ), hence measurable, resp., and∫

Yhx dν =

∫Xhy dµ = 0 (22)


for µ-a.e. x, and ν-a.e. y, resp. Indeed, using the observation at the end of Section 2, (22)would imply that∫X×Y

(f(x, y)− g(x, y)

)dτ(x, y) =

∫X×Y

h(x, y) dτ(x, y) =

∫X×Y

h(x, y) d(µ× ν)(x, y)

=

∫Y

(∫Xh(x, y) dµ(x)

)dν(y) =

∫X

(∫Yh(x, y) dν(y)

)dµ(x)

= 0 .

Thus, we prove (22). Since h = 0 τ -a.e., there exists E ∈ M×N such that (µ × ν)(E) = 0and h = 0 on cE. We then have∫

X×Yh(x, y) dτ =

∫X×Y

h(x, y) d(µ× ν) =

∫Eh(x, y) d(µ× ν) .

Since (µ× ν)(E) = 0, we have

0 =

∫Yµ(Ey) dν =

∫Xν(Ex) dµ

which imply that

µ(Ey) = 0 ν − a.e. and ν(Ex) = 0 µ− a.e. . (23)

Then, there exists a sequence of simple functions τ -measurable ϕn such that 0 ≤ |ϕ1| ≤|ϕ2| ≤ · · · ≤ |h| and ϕn → h. Notice that ϕnx → hx and ϕn

y → hy, for all x, y.Observe also that, clearly, ϕn = 0 on cE, so that ϕnx = 0 on c(Ex) and ϕn

y = 0 on c(Ey),for all x, y. Thus, by (23), ϕnx = χEx µ-a.e. and ϕn

y = χEy ν-a.e. Since χEx and χEy are Nand M-measurable, resp., and µ and ν are complete, by Prop. 4.8 it follows that also ϕnx andϕn

y are N and M-measurarble, resp. This easily implies that hx ∈ L1(ν) and hy ∈ L1(µ) forµ-a.e. x and ν-a.e. y and thus (22), and hence the theorem, follow.

4.3. The Lebesgue integral in Rn. Clearly, the results of the previous section on productmeasure spaces can be extended to the case of cartesian products of any finite number of measurespaces. We assume the validity of all these results, without formally state any of them.

We define the Lebesgue measure in mn on Rn as the completion of m× · · · ×m (n copies ofm) on BR × · · · × BR, or, equivalently, on L × · · · × L. We denote by Ln the σ-algebra domainof mn, and call these sets Lebesgue measurable sets in Rn.

In this section we prove some properties of mn that extend similar properties valid in the casen = 1. If R = R1 × · · · ×Rn ⊆ Rn is a rectangle, we call Rj its sides, j = 1, . . . , n.

Theorem 4.13. Let E ∈ Ln. Then, the following properties hold true.

(i) mn(E) = infmn(U) : E ⊆ U, U open

= sup

mn(K) : K ⊆ E, K compact

;

(ii) E = A1 \N1, where A1 is a Gδ-set and mn(N1) = 0, and also E = A2 ∪N2, where A2

is an Fσ-set and mn(N2) = 0;

(iii) if mn(E) < +∞, then for every ε > 0 there exists a finite collection of rectangles Rj,j = 1, . . . , N , whose sides are intervals and such that mn

(E 4∪Nj=1Rj

)< ε.

50 M. M. PELOSO

Proof. (†) (i) By construction of the product measure, given E ∈ Ln and ε > 0, there exists acollection Rj of rectangles such that E ⊆ ∪+∞

j=1Rj and∑+∞

j=1 mn(Rj) ≤ mn(E) + ε. In order

to prove the first equality, notice that we may assume that mn(E) < +∞, since otherwise wehave nothing to prove. Then,

∑+∞j=1 mn(Rj) is a convergent series, so that there exists M > 0

such that mn(Rj) ≤M for all j.

For each j fixed, Rj = R(1)j × · · · × R

(n)j , with R

(k)j ∈ L, k = 1, . . . , n. By Prop. 3.11, for

each j fixed, we can find open sets U(k)j such that R

(n)j ⊆ U

(k)j and m

(U

(k)j

)< m

(R

(k)j

)+

2−jε/(NMn−1), where N is a sufficiently large fixed positive integer. Then

R ⊆ U :=+∞⋃j=1

(U

(1)j × · · · × U

(n)j

).

Notice that the Uj are open rectangles. Hence U is open, and

mn(U) ≤+∞∑j=1

mn(Uj) =+∞∑j=1

n∏k=1

m(U

(k)j

)≤

+∞∑j=1

n∏k=1

(m(R

(k)j

)+ 2−jε/(NMn−1)

)

≤+∞∑j=1

(mn(Rj) + ε2−j

)≤ mn(E) + 2ε .

Since ε > 0 was arbitrary, the first equality in (i) is proved. In order to prove the second equality,we proceed exactly as in the proof of (ii) in Prop. 3.11, replacing the intervals (j, j+ 1] with thecubes Qj = (0, 1]n + j, where j ∈ Zn.

The proofs of (ii), which exactly the same as the ones in Prop. 3.14, is again left as an exercise.(iii) Let mn(E) < +∞, and let ε > 0 be given. We can find open rectangles Uj such that∑+∞j=1 mn(Uj) ≤ mn(E) + ε/2. Hence, mn(Uj) < +∞ for all j. For each j, the sides of Uj are

countable unions of disjoint open intervals, say U(`)j = ∪+∞

k`=1I(`)j,k`

, ` = 1, . . . , n. Thus, for each j

fixed,

Uj =( +∞⋃k1=1

I(1)j,k1

)× · · · ×

( +∞⋃kn=1

I(n)j,kn

)=

+∞⋃k1,...,kn=1

I(1)j,k1× · · · × I(n)

j,kn.

Hence, for each j, we can find a finite subunion Vj of the I(1)j,k1× · · · × I(n)

j,kn’s such that

mn(Uj) ≤ mn(Vj) + 2−jε/2 .

Notice that the sides of the Vj ’s are finite unions of intervals. Then, if N is large enough so that∑+∞j=N+1mn(Uj) < ε/2 we have

mn

(E \

N⋃j=1

Vj

)≤ mn

( N⋃j=1

(Uj \ Vj

)∪

+∞⋃j=N+1

Uj

)

≤ mn

( N⋃j=1

(Uj \ Vj

))+mn

( +∞⋃j=N+1

Uj

)< ε.


On the other hand,

mn

( N⋃j=1

Vj \ E)≤ mn

(+∞⋃j=1

Uj \ E)≤

+∞∑j=1

mn(Uj)−mn(E) < ε .

This proves (iii), and we are done.

Next, we prove that mn is also translation invariant.

Proposition 4.14. Let E ∈ Ln, and f ∈ L+(mn) or ∈ L1(mn). For a ∈ R define E + a =x+ a x ∈ E

∈ Ln and fa = f(·+ a). Then, mn(E + a) = mn(E) and

∫fa dmn =

∫f dmn.

Proof. (†) By composition of translations along each each axis, it suffices to consider the caseof translations along one axis, that is, a = (0, . . . , aj , 0, . . . , 0). Then, if E = E1 × · · · × En isa rectangle, then mn(E + a) = m(E1) · · ·m(Ej + aj) · · ·m(En) = mn(E), by the 1-dimensionalresult Prop. 3.13. Then, the result is true for the elements of algebra A of finite unions ofdisjoints measurable rectangles. By the construction of the product measure, the invariancenow follows. Notice also that in particular the class of sets of measure zero is invariant undertranslations.

Next, let f be Lebesgue measurable and fa = f(· + a). Let B be a Borel set in R (or in C,if f is complex-valued). Then, f−1(B) is a Lebesgue measurable set, hence f−1(B) = E ∪ N ,where E is a Borel set in Rn and N is contained in a Borel set of measure 0. Then, denoting byτa the traslation by a ∈ Rn, so that f(·+ a) = f τa,

(fa)−1(B) = τ−1

a

(f−1(B)

)= τ−1

a

(E ∪N

)= (E − a) + (N − a) ,

that is, (fa)−1(B) is union of a Borel set and of a set of measure 0. Hence, it is Lebesgue

measurable, and fa is measurable.Finally, the identity

∫fa dmn =

∫f dmn holds true when f = χE , and E is a measurable

set. Hence, it holds true for simple functions, and by the MCT and the DCT it holds true forf ∈ L+(mn) and f ∈ L1(mn), resp.

We now wish to show how the Lebsegue measure behaves under change of variables. Webegin with a linear transformation. Let T be an invertible linear transformation and we writeT ∈ GL(n,R). This means that T : Rn → Rn is a linear bijection, and this is the case if andonly if detT 6= 0. We denote by 〈·, ·〉 the inner product in Rn. Then, we identify T with thematrix

(Tjk)j,k=1,...,n

, where Tjk = 〈Tej , ek〉, and e1, . . . , en is the canonical basis of Rn.

It is well known that every T ∈ GL(n,R) can be decomposed as composition of elementarytransformations. Namely,

• translations in each direction by a scalar multiple of another component of the variable:T (x1, . . . , xj , . . . , xn) = (x1, . . . , xj + cxk, . . . , xn), with j = 1, . . . , n and c ∈ R;

• dilations in each direction:T (x1, . . . , xj , . . . , xn) = (x1, . . . , cxj , . . . , xn), with c ∈ R \ 0, j = 1, . . . , n;

• interchanges of coordinates:T (x1, . . . , xj , . . . , xk, . . . , xn) = (x1, . . . , xk, . . . , xj , . . . , xn), 1 ≤ j < k ≤ n.

Theorem 4.15. (Linear change of variables formula) Let T ∈ GL(n,R). Then the follow-ing properties hold.

52 M. M. PELOSO

(1) If f is Ln-measurable, then f T is also Ln-measurable. If f ∈ L+(mn) or f ∈ L1(mn)then ∫

f(x) dmn = |detT |∫

(f T )(x) dmn .

(2) If E ∈ Ln, then T (E) ∈ Ln and mn

(T (E)

)= |detT |mn(E).

Proof. If f is Borel measurable, then f T is Borel measurable, since T is continuous. Moreover,since f (T S) = (f T ) S, and det(T S) = (detT )(detS), it suffices to prove the resultfor the elementary matrices as above. We may apply the Tonelli–Fubini theorem, according towhether f ∈ L+(mn) or f ∈ L1(mn). We can integrate in the variable xj first and notice that∫

Rf(x1, . . . , xj + cxk, . . . , xn) dxj =

∫Rf(x1, . . . , xj , . . . , xn) dxj

and also that ∫Rf(x1, . . . , cxj , . . . , xn) dxj = |c|

∫Rf(x1, . . . , xj , . . . , xn) dxj ,

using the 1-dimensional results. Analogously, integrating in the variables (xj , xk) first∫R2

f(x1, . . . , xk, . . . , xj , . . . , xn) dxjdxk =

∫R2

f(x1, . . . , xj , . . . , xk, . . . , xn) dxjdxk .

This proves (1) for f Borel measurable. If E is a Borel set, then (1) applied to χE gives (2) inthis case. In particular, T and T−1 preserve the Borel sets of measure 0. This easily imply thatevery T ∈ GL(n,R) preserves the Lebesgue sets of measure 0.

Finally, we are in the position to prove (1) for a general Lebesgue measurable funciton f .Indeed, there exists g which is Borel measurable such that g = f a.e., that is, f = g on a Borelset E and mn( cE) = 0. Then, f T = g T on the Borel set T−1(E), mn

(c(T−1E)

)) = 0, i.e.

f T = g T a.e., and we have∫(f T )(x) dx =

∫(g T )(x) dx = | detT |

∫g(x) dx = | detT |

∫f(x) dx .

This proves the theorem.

We extend the previous result to C1-diffeomorphisms of Rn. We recall that a map Φ : Rn →Rn is called a C1-diffeomorphisms if Φ is C1, is injective, and its Jacobian matrix (Jac Φ)(x)is invertible at every point x ∈ Rn. We recall also that the Jacobian matrix is the matrixof the partial derivatives of the components of Φ, that is, denoting by ∇f the gradient of ascalar-valued function f ,

Φ =

Φ1...

Φn

and Jac Φ =

∇Φ1...∇Φn

=

∂x1Φ1 · · · ∂xnΦ1... · · ·

...∂x1Φn · · · ∂xnΦn

.

We also recall that, if Φ is a C1-diffeormophism, then Φ−1 is well defined, and the inversefunction theorem gives that also Φ−1 is a C1-diffeormophism and it holds that

Jac(Φ−1

)(Φ(x)) =

((Jac Φ)(x)

)−1,

for all x ∈ Rn.We now have the general change of variables formula.


Theorem 4.16. (Change of variables formula for C1-diffeormophisms) Let Ω be an openset in Rn and let Φ : Ω→ Rn be a C1-diffeormophism. Then the following properties hold.

(1) If f is Ln-measurable on Φ(Ω), then f Φ is Ln-measurable on Ω. If f ∈ L+(mn) orf ∈ L1(mn) then ∫

Φ(Ω)f(x) dx =

∫Ω

(f Φ)(x) | det(Jac Φ(x))| dx .

(2) If E ∈ Ln, then Φ(E) ∈ Ln and

mn

(Φ(E)

)=

∫E| det(Jac Φ(x))| dx .

Proof. (†) See [F], Theorem 2.47.

We conclude this section with the extension of a classical result that relates the area underthe graph of a (positive) function and the integral of the function itself.

Proposition 4.17. Let f : Ω ⊆ Rn → R be measurable, where Ω is a measurable set. Let Gfdenote the region between the graph of f and its domain:

Gf =

(x, y) ∈ Rn × R : 0 ≤ y ≤ f(x).

Then Gf is measurable and m(Gf ) =∫

Ω f dmn.

Proof. (†) We leave the proof as an exercise. In order to prove measurability, notice that themap (x, y) 7→ f(x)− y is the composition of (x, y) 7→ (f(x), y) and (z, y) 7→ z − y.

4.4. Polar coordinates in Rn. We wish to extend the familiar formulas for integration inpolar coordinates in R2 and R3, that were defined using the Riemann integral in one and twovariables, and that now we recall.

With the polar coordinates Φ : R2 \ (0, 0) → (0,+∞) × [0, 2π), Φ(x, y) = (r cos θ, r sin θ),we had ∫

R2

f(x, y) dxdy =

∫ +∞

0

∫ 2π

0f(r cos θ, r sin θ) dθ rdr .

In the case of R3, we used what are called spherical coordinates Φ : R3 \ (0, 0, 0) → (0,+∞)×[0, π]× [0, 2π), Φ(x, y, z) = (r sinφ cos θ, r sinφ sin θ, r cosφ), and had the formula∫

R3

f(x, y, z) dxdydz =

∫ +∞

0

∫ π

0

∫ 2π

0f(r sinφ cos θ, r sinφ sin θ, r cosφ) dθ r2 sinφdφdr .

Again, these formulas were given using the theory of the Riemann integral. We wish to putthem in the context of abstract integration, with respect to well-defined measures.

We observe that, in both cases, we can define the length of an arc on unit circle S1 andthe surface area of a “rectangle” on the unit sphere S2 as follows. The area of the sectorΩθ0 =

(x, y) = (r cos θ, r sin θ) : 0 < r < R, α− θ0

2 < θ < α+ θ02

of angle θ0 is m(Ωθ0) = 1

2R2θ0.

This equation can be used to define the length of an arc Aθ0 , of angle θ0, by solving for θ0: setting

σ(Aθ0) =m(Ωθ0)

12R

2= θ0 .

54 M. M. PELOSO

Analogous reasoning can be made in the case of the “rectangle” on the unit sphere S2. LetΩ(φ1,φ2,θ0) =

(x, y, z) = (r sinφ cos θ, r sinφ sin θ, r cosφ) : 0 < r < R, φ1 < φ < φ2, α − θ0

2 <

θ < α+ θ02

. An elementary calculation shows that

m(Ω(φ1,φ2,θ0)) = θ0

(cosφ1 − cosφ2

).

Again, we can define the surface area of the corresponding rectangle R(φ1,φ2,θ0) on the unitsphere, as

σ(R(φ1,φ2,θ0)) =m(Ω(φ1,φ2,θ0))

13r

3.

Let us go back to the case of Rn. We denote by Sn−1 the unit sphere in Rn: Sn−1 =x ∈

Rn : |x| = 1

. For x ∈ Rn \ 0 we write

x = rx′, where r = |x|, x′ = x

|x|.

These are called the polar coordinates in Rn. The corresponding map Φ : Rn \ 0 → (0,+∞)×Sn−1 is then Φ(x) = (r, x′). It is a continuous bijection, and its inverse is also continuous, beinggiven by Φ−1(r, x′) = rx′. We denote by (mn)∗ the measure on (0,+∞)× Sn−1 induced by mn

by the formula

(mn)∗(A) = mn

(Φ−1(A)

),

where A is a Borel set in (0,+∞)× Sn−1. We denote by ρ = ρn the measure on (0,+∞) givenby, for E ⊆ (0,+∞),

ρ(E) =

∫Ern−1 dr .

Definition 4.18. Let E be a Borel set in Sn−1. For a > 0 define

Ea = Φ−1((0, a]× E

)=x ∈ Rn : x = rx′, 0 < r ≤ a, x′ ∈ E

,

and set

σ(E) = n ·mn(E1) .

It is clear that σ is a measure on the σ-algebra of Borel sets in Sn−1, since the map E 7→ E1

takes Borel sets in Sn−1 to Borel sets in Rn and commutes with unions, intersections andcomplements.

We wish to show that the measures (mn)∗ and (ρ × σ) coincides on the Borel subsets of(0,+∞)× Sn−1.

Theorem 4.19. With the above definition, σ is the unique measure on the unit sphere Sn−1 inRn such that (ρ× σ) = (mn)∗. Morevoer, if f ∈ L+(mn) or f ∈ L1(mn), we have∫

f(x) dmn(x) =

∫ +∞

0

∫Sn−1

f(rx′) dσ(x′) rn−1dr . (24)

Proof. We have just observed that σ is indeed a measure on the σ-algebra of Borel sets in Sn−1.Moreover, writing Ta(x) = ax for a > 0 and x ∈ Rn,

mn(Ea) = mn(Ta(E1)) = anmn(E1) =an

nσ(E) ,


using Thm. 4.15 (2) and the definition of σ. Therefore, for 0 < a < b,

(mn)∗((a, b]× E

)= mn

(Eb \ Ea

)= mn(Eb)−mn(Ea)

=bn

nσ(E)− an

nσ(E) = σ(E)

bn − an

n= σ(E)

∫ b

arn−1 dr

= (ρ× σ)((a, b]× E

).

Therefore, the measures (mn)∗ and (ρ × σ) coincides on the sets of the form (a, b] × E, whereE is a Borel in Sn−1. It is easy to see that the σ-algebra generated by such sets is indeed theσ-algebra of Borel sets of (0,+∞)× Sn−1, so that (mn)∗ = (ρ× σ).

Formula (24) holds true when f = χE1 , since∫χE1(x) dmn(x) = mn(E1) =

σ(E)

n= σ(E)

∫ 1

0rn−1 dr

=

∫ 1

0

∫Sn−1

χE(x′) dσ(x′) rn−1dr

=

∫ +∞

0

∫Sn−1

χE1(rx′) dσ(x′) rn−1dr .

Then, (24) holds true for simple functions and passing to the limit, using the MCT if f ∈ L+(mn)or the DCT if f ∈ L1(mn), we obtain the result.

Recall that a function f on Rn is said to be radial if it depends on |x|, that is, if there existsg defined on (0,+∞) such that f(x) = g(|x|), for all x.

Corollary 4.20. (1) If f ∈ L+(mn) or if f ∈ L1(mn) is radial, and f(x) = g(|x|), for all x,then ∫

f(x) dmn(x) = ωn

∫ +∞

0g(|x|) rn−1dr

where ωn = σ(Sn−1) denotes the surface area of Sn−1.(2) If Br = x : |x| < r denotes the ball centered at the origin and radius r > 0, and

fa(x) = |x|−a, we have

(i) fa ∈ L1(Br) if a < n, while if g(x) ≥ c|x|−n, then g 6∈ L1(Br);

(ii) fa ∈ L1( cBr) if a > n, while if g(x) ≥ c|x|−n, then g 6∈ L1( cBr).

Proof. The part (1) follows at once from the previous result, while part (2) follows from (1).

We conclude this part with the evaluation of some fundamental explicit integrals.

Proposition 4.21. We have:

(i) for a > 0,

∫e−a|x|

2dx =

(πa

)n/2;

(ii) σ(Sn−1) =2πn/2

Γ(n/2);

(iii) mn(B1) =πn/2

Γ(n2 + 1).

56 M. M. PELOSO

Proof. (i) Let In =∫Rn e

−a|x|2 dx. We observe that, by Cor. 4.20 (1),

I2 = 2π

∫ +∞

0e−r

2r dr = −

(πa

)e−ar

2∣∣∣+∞0

=π

a.

Next, we observe that, since e−a|x|2

=∏nk=1 e

−ax2k , using Tonelli’s theorem we have

In =

∫R· · ·∫R

( n∏k=1

e−ax2k

)dx1 · · · dxn = (I1)n .

In particular, I1 =√I2 =

(πa

)1/2and In =

(πa

)n/2, as we wished to show.

(ii) We now observe that

πn/2 =

∫e−|x|

2dx = σ(Sn−1)

∫ +∞

0e−r

2rn−1 dr

=σ(Sn−1)

2

∫ +∞

0e−ss

n2−1 ds

=σ(Sn−1)

2Γ(n/2) .

This gives (ii).Finally, we know that mn(B1) = σ(Sn−1)/n, so that

mn(B1) =πn/2

n2 Γ(n2

) =πn/2

Γ(n2 + 1

) ,by the properties of the Gamma function.

5. Hausdorff measaures

5.1. A quick review of submanifolds in Rn.

Definition 5.1. A subset M of Rn is called a C1-submanifold, of dimension k, 1 ≤ k ≤ n− 1,if for every x ∈M there exist a ngbh U of x, an open set V ⊆ Rk and ϕ : V → U such that

• ϕ is injective;

• ϕ(V ) = M ∩ U ;

• the rank of the Jacobian of ϕ is maximal at every point x ∈ V , that is, rank(

Jacϕ(x))

=k for all x ∈ V .

The map ϕ, or the pair (ϕ, V ), is called a local parametrization for M ∩ U , or local chart.We also assume that M can be covered by at most finitely many of such sets M ∩ U .

Remark 5.2. Using the implict function theorem, it is easy to see that this definition is equiv-alent to requiring the following: For each x ∈ M there exist a nbgh U of x and a C1-functionF : U → Rn−k such that

• U ∩M =x ∈ Rn : F (x) = 0

;

• the rank of the Jacobian of F is maximal at every point x ∈ U , that is, rank(

Jac(F )(x))

=n− k for all x ∈ U .


Indeed, we may apply the implict function theorem to F at the point x where F (x) = 0,separating the variable x as x = (x′, x′′), with x′ ∈ Rk and x′′ ∈ Rn−k in such a way thatdet Jacx′′(F )(x′, x′′) 6= 0 in a ngbh U of x. Then, there exist a nbgh V ′ ⊆ Rk of x′, a ngbh V ′′

of x′′ in Rn−k and a unique C1 function ϕ : V ′ → V ′′ such that F (x′, ϕ(x′)) = 0. Then, the setM ∩ U =

(x′, x′′) ∈ U : F (x′, x′′) = 0

= ϕ(V ′), and ϕ satisfies the hypothes as in Def. 5.1.

Remark 5.3. We quickly review the familiar cases when n = 2, 3, and k = 1, 2, that is, theconcepts of surfaces and surface integrals in R3, in analogy with the theory of (piecewise) C1

curves and line integrals.Let A be a connected open set in R2 and D a set such that A ⊆ D ⊆ A, so that, in particular

D = A. Let ϕ : D → R3, injective, ϕ ∈ C1(D) and such that rank(

Jac(ϕ))

= 2 on D. Setting

Σ = ϕ(D), the pair (Σ, ϕ) is called a surface in R3 and ϕ its parametrization. Notice that weallow ϕ also to be defined on a portion or the whole of the boundary of A. We will in particularuse this observation in the case of the sphere in R3.

We observe that rank(

Jac(ϕ))

= 2 if and only if the matrix(ϕu ϕv

)has rank 2, if and only

if ϕu ∧ ϕv 6= 0, where we denote by (u, v) the coordinates in D ⊆ R2,

ϕu =

∂uϕ1

∂uϕ2

∂uϕ3

and ϕv =

∂vϕ1

∂vϕ2

∂vϕ3

.

Then, the surface area of Σ is defined as

Area(Σ) =

∫D‖ϕu ∧ ϕv‖ dudv ,

and the surface integral of a continuous function defined on M as∫Σf dσ =

∫Df(ϕ(u, v)) ‖ϕu ∧ ϕv‖ dudv .

The reader should check that the unit sphere in R3, parametrized by the spherical coordinatesϕ : [0, π]× [0, 2π), ϕ(φ, θ) = (sinφ cos θ, sinφ sin θ, cosφ), is an example of surface in R3.

5.2. Hausdorff measures. Although we will work in Rn, we begin with a few definitions andfacts that have their natural setting in the class of metric spaces.

Definition 5.4. Let (X , d) be a metric space, with distance d. Let µ∗ be an outer measure onX . We say that µ∗ is a metric outer measure if for any A,B ⊆ X with d(A,B) > 0 we have that

µ∗(A ∪B) = µ∗(A) + µ∗(B) .

We recall that d(A,B) = infd(x, y) : x ∈ A, y ∈ B. It is immediate to check thatd(A,B) > 0 implies that A ∩ B = ∅, while d(A,B) = d(A,B) so that d(A,B) = 0 doesnot imply that A ∩B = ∅.Proposition 5.5. Let µ∗ be a metric outer measure in a metric space (X , d). Then, the Borelsets are µ∗-measurable.

Proof. (†) In order to check that the Borel σ-algebra BX is contained in the σ-algebra of µ∗-measurable sets, it suffices to check that every closed F is µ∗-measurable. In turn, by (10), itsuffices to chech that

µ∗(A) ≥ µ∗(A ∩ F ) + µ∗(A ∩ cF ) for all A with µ∗(A) < +∞ .

58 M. M. PELOSO

For the complete proof, we refer the reader to [F] Prop. 11.16.

We now introduce the concept of Hausdorff measure. For α ≥ 0 we set

γα =πα/2

2αΓ(α2 + 1

) . (25)

Recall that in Prop. 4.21 (iii) we computed the Lebesgue volume of the unit ball in Rn. Notice

that we have that γn = mn(B1)2n .

Definition 5.6. Let (X , d) be a second-countable6 metric space, δ > 0 and α ≥ 0. For A ⊆ Xset

H(δ)α (A) = γα inf

+∞∑j=1

(diamBj

)α: A ⊆

+∞⋃j=1

Bj , diamBj ≤ δ. (26)

Notice that the the quantity H(δ)α (A) is increasing as δ decreases, since we are taking the infimum

over a smaller set of coverings. Thus, we define

Hα(A) = limδ→0+

H(δ)α (A) . (27)

We call Hα(A) the α-dimensional Hausdorff outer measure of A.

Remarks 5.7. A few remarks are in order.(1) In fact, it is not necessary to restrict the definition of Hausdorff measure to second-

countable metric spaces. This assuption guarantees that every open cover has a countablesubcover. On a general metric space, we need to allow the open covers also to be uncountable.

(2) We will soon show, see Prop. 5.8 that Hα is indeed a metric outer measure, and by Prop.5.4 it will then follow that Hα|BX is a measure; see Cor. 5.9.

(3) In the definition (26) the Bj ’s are generic subsets of X . It is equivalent to require that

the Bj ’s are closed, since diamBj = diamBj , or open, since one replaces Bj by an open set ofdiameter ≤ diamBj + ε2−j . In the case of X = Rn, it is possible to show that it is equivalent totake the Bj ’s to be open, or closed, (euclidean) balls. However, this fact is not elementary andwe refer to [KP], Cor. 4.3.9.

Proposition 5.8. Hα is a metric outer measure.

Proof. Prop. 3.1 gives that H(δ)α is an outer measure for each δ > 0. Then, clearly Hα(∅) =

limδ→0+ H(δ)α (∅) = 0, if A ⊆ B are subsets in X , then,

Hα(A) = limδ→0+

H(δ)α (A) ≤ lim

δ→0+H(δ)α (B) = Hα(B) .

Finally, if A = ∪+∞j=1Aj , then, using the MCT we have

Hα(A) = limδ→0+

H(δ)α (A) ≤ lim

δ→0+

+∞∑j=1

H(δ)α (Aj) =

+∞∑j=1

limδ→0+

H(δ)α (Aj) =

+∞∑j=1

Hα(Aj) .

6We recall that a topological space is called second-countable if its topology has a countable base. In particular,every cover by open sets as a countable subcollection that is still a cover, see Theorem 16.9 in S. Willard, GeneralTopology, (1970) Addison-Wesley Publishing.


Now we show that Hα is a metric outer measure. Let A,B ⊆ X , d(A,B) > 0. Let δ besuch that 0 < δ < d(A,B), and let Ej be a covering of A ∪ B with diamEj < δ for each j.Then, each Ej can intersect only one among A and B, that is, for every j, if Ej ∩ A 6= ∅, thenEj ∩ B = ∅, and viceversa, if Ej ∩ B 6= ∅, then Ej ∩ A = ∅. Then, we split the covering Ejinto two disjoint collections E(A)

j and E(B)j such that A ⊆ ∪+∞

j=1E(A)j and B ⊆ ∪+∞

j=1E(B)j .

We then have,

H(δ)α (A) +H(δ)

α (B) ≤+∞∑j=1

H(δ)α

(E

(A)j

)+

+∞∑j=1

H(δ)α

(E

(A)j

)=

+∞∑j=1

H(δ)α (Ej) .

This inequality holds for all coverings Ej of A ∪B, with diameters ≤ δ. Thus, by taking theinfimum over such coverings on the right hand side, we see that

H(δ)α (A) +H(δ)

α (B) ≤ H(δ)α

(A ∪B

),

and then we pass to the limit as δ → 0+ to obtain

Hα(A) +Hα(B) ≤ Hα

(A ∪B

),


Corollary 5.9. If (X , d) is a second-countable metric space, then (X , Hα,BX ) is a measurespace.

Proof. The previous proposition shows that Hα is a outer measure, which is also metric, in thesense of Def. 5.4. Prop. 5.5 gives that Borel sets BX in X are Hα-measurable, so Hα|BX is ameasure, by Caratheodory’s Theorem 3.3; hence the conclusion.

The proposition that follows will imply that we can define the Hausdorff dimension of any setA ⊆ X .

Proposition 5.10. Let A ⊆ X . Then, the following properties hold true:

(i) if Hα(A) < +∞ for some α, then Hβ(A) = 0 for all β > α;

(ii) if Hα(A) > 0 for some α, then Hγ(A) = +∞ for all γ < α;

(iii) supγ : Hγ(A) = +∞

= inf

β : Hβ(A) = 0

:= α.

Such value α is called the Hausdorff dimension of A.

Proof. (i) Suppose Hα(A) < +∞. Let ε0 > 0 be fixed. For any δ > 0, there exists a covering

Bj of A with diamBj ≤ δ such that γα∑+∞

j=1

(diamBj

)α ≤ H(δ)α (A) + ε0. Since H

(δ)α (A) ≤

Hα(A) for every δ > 0, for such covering of A we have

γα

+∞∑j=1

(diamBj

)α ≤ Hα(A) + ε0 .

Now, if β > α we have

+∞∑j=1

(diamBj

)β ≤ δβ−α +∞∑j=1

(diamBj

)α ≤ 1

γαδβ−α

(Hα(A) + ε0

),

so thatH

(δ)β (A) ≤

γβγαδβ−α

(Hα(A) + ε0

).

60 M. M. PELOSO

Letting δ → 0+ we see that Hβ(A) = 0. This proves (i).In order to prove (ii), we simply notice that, seeking a contradiction, if Hγ(A) < +∞ for

some γ < α, then part (i) implies that Hα(A) = 0, a contradiction. Then (ii) follows.Finally, given A ⊆ X , parts (i) and (ii) show that the value α is well defined, and we are

done.

5.3. Hausdorff measures in Rn. We now restrict ourselves to the case of Rn, endowed with thestandard euclidean distance. From now on therefore, Hα will denote the α-Hausdorff measureon the σ-algebra of Borel sets in Rn. Our goal is to show that when α = k = 0, 1, . . . , n, thenthe Hausdorff measure Hk provides the right notion of measure on a k-dimensional submanifoldin Rn.

Theorem 5.11. On the σ-algebra of Borel sets in Rn, Hn = mn.

Proof. (†) We prove that there exist constants Cn ≤ C ′n such that

Hn ≤ Cnmn ≤ C ′nHn . (28)

To show that it is possible to choose Cn = C ′n = γn requires a couple of results that fall beyondthe scope of this course. For the proof we refer to [L] Teorema 2.1.

We now give a few examples of the Hausdorff dimesion of some given sets.

Example 5.12. (1) Let A = p be the set consisting a single point in Rn. Then H0(p) = 1,and therefore the Hausdorff dimension of p is 0.

Indeed, for any covering Bj of p by sets of diameter ≤ δ, since γ0 = 1,

γ0

∑j

(diamBj

)0= cardinality(Bj) ≥ 1 .

Thus, H(δ)0 (p) ≥ 1 for all δ > 0, which implies H0(p) ≥ 1. Conversely, a single ball B(p, δ/2)

covers p so that, H(δ)0 (p) ≤ 1 and therefore also H0(p) ≤ 1.

(2) Using the fact that H0 is a measure, it follows at once from (1) that if A ⊆ Rn is finite orcountable, then H0(A) = cardinality(A).

(3) Is A ⊆ Rn is a Borel measurable set such that mn(A) > 0, then, its Hausdorf dimensionequals n. The conclusion follows at once from Prop. 5.10.

(4) Let C be the Cantor set in R. Then, the Hausdorff dimension of C is log 2/ log 3. For aproof, see [L].

We now wish to define an intrisic notion of measure on a k-dimensional submanifold M ofRn. We begin with a lemma.

Lemma 5.13. Let k be an integer, 1 ≤ k ≤ n and let T : Rk → Rn be an injective lineartransformation and T ∗ its adjoint. Then, setting J (T ) =

√det(T ∗T ), for any Borel set A ⊆ Rk,

Hk

(T (A)

)= J (T )Hk(A) .

Proof. We begin by observing that T ∗T : Rk → Rk is positive semi-definitive, since 〈T ∗Tv, v〉 =〈Tv, Tv〉 = ‖Tv‖2, where 〈·, ·〉 denotes the inner product in Rk and ‖ · ‖ the norm in Rk.

If k = n, then T ∈ GL(n,R), and det(T ∗T ) = (detT )2, so that J (T ) = | detT |. Hence, theresult follows from Thm.’s 4.15 and 5.11.


Assume now k < n. Then, there exists a rotation R such that

R(T (Rk)

)=x = (x1, . . . , xn) ∈ Rn : xk+1 = · · · = xn = 0

= Rk × 0 .

Set S = RT . Then, identifying Rk × 0 with Rk itself, S becomes a linear transformation ofRk into itself. Using the result for transformation in the same dimension k, we have that

Hk

(S(A)

)= J (S)Hk(A) .

Moreover, S∗S = (RT )∗RT = T ∗R∗RT = T ∗T , so that J (S) = J (T ) and, using the fact thatHk is rotational invariant, we have

Hk

(T (A)

)= Hk

(RT (A)

)= Hk

(S(A)

)= J (S)Hk(A) = J (T )Hk(A) .

This completes the proof.

We now generalize the previous result to C1-injections, as in Def. 5.1.

Theorem 5.14. Let V be an open set in Rk, k ≥ 1, ϕ : V → Rn a C1, injective map suchthat Jacϕ(x) has rank k at every point x ∈ V . Let A ⊆ V be any Borel set, then ϕ(A) is Borelmeasurable in Rn and

Hk

(ϕ(A)

)=

∫AJ(

Jacϕ(x))dHk(x) . (29)

Moreover, if f is a Borel measurable function on ϕ(V ) = M such that, either f ∈ L+(M,Hk)or f ∈ L1(M,Hk), then∫

Mf(y) dHk(y) =

∫Vf(ϕ(x))J

(Jacϕ(x)

)dHk(x) . (30)

Notice that the integrals on the right hand sides of 29 and (30) are evaluated on Borel sets ofRk, so the measures dHk equal dmk in both cases.

Proof. (†) See [F], Lemma 11.22 and Thm. 11.25.

Example 5.15. Let γ be a simple regular curve in Rn, that for the moment we take withoutits end points. This means that we are given an injective, C1 map, ϕ : (a, b) → Rn, such that|ϕ′(t)| 6= 0, for all t ∈ (a, b). Then, the length of γ was defined as

`(γ) =

∫ b

a|γ′(t)| dt ,

and the line integral of a continuous function defined on γ as∫γf ds =

∫ b

af(γ(t))|γ′(t)| dt .

We set M = ϕ((a, b)

)and apply Thm. 5.14. Observe that Jacϕ =

ϕ′1...ϕ′n

so that

J(

Jacϕ(t))2

=(ϕ′1(t) · · · ϕ′n(t)

)ϕ′1(t)...

ϕ′n(t)

= |ϕ′(t)|2 .

62 M. M. PELOSO

Therefore, by (29) we have that

H1(M) = H1

(ϕ((a, b)

))=

∫(a,b)J(

Jacϕ(t))dH1(t) =

∫ b

a|ϕ′(t)| dt = `(M) .

Similarly, if f is a continuous function non-negative on M , then it is Borel measurable andby (30) we have∫

Mf(y) dH1(y) =

∫(a,b)

f(ϕ(t))J(

Jacϕ(t))dH1(t) =

∫ b

af(ϕ(t))|ϕ′(t)| dt .

Therefore, the integral w.r.t. the 1-dimensional Hausdorff measure generalizes both the notionof length of a regular curve and the one of line integral of continuous function.

It is also easy to extend the above argument to the case of a piecewise C1, simple curve, andincluding the endpoints. We have seen in Example 5.12 (1) that H0 coincides with the countingmeasure in Rn, so that H0(p1, . . . , pN) = N , for any finite collection of points. This alsoimplies that Hα(p1, . . . , pN) = 0 for any α > 0. Then, if M = ϕ

((a, b)

)∪ ϕ(a), ϕ(b), then

H1(M) = H1

(ϕ((a, b)

))+H1

(ϕ(a)

)+H1

(ϕ(b)

)= H1

(ϕ((a, b)

)),

and then the argument proceeds as before. Analogously for the integral of f :∫Mf(y) dH1(y) =

∫(a,b)

f(ϕ(t))J(

Jacϕ(t))dH1(t) + 0 =

∫ b

af(ϕ(t))|ϕ′(t)| dt .

Example 5.16. Let M be a surface in R3 and assume that it is parametrized by a single C1,injective map, ϕ : V → R3, where V is an open set in R2, and Jacϕ(y) has rank 2 for all y ∈ V .Then, adopting the notation in Example 5.3, we have that

Area(M) =

∫V‖ϕu ∧ ϕv‖ dudv ,

and the surface integral of a continuous function defined on M as∫Mf dσ =

∫Vf(ϕ(u, v)) ‖ϕu ∧ ϕv‖ dudv .

Again, we have want to apply Thm. 5.14. Observe that Jacϕ =

∇ϕ1

∇ϕ2

∇ϕ3

=(ϕu ϕv

)so that

J(

Jacϕ(t))2

=

(tϕutϕv

)(ϕu ϕv

)=

(|ϕu|2 ϕu · ϕvϕu · ϕv |ϕv|2

).

Then, it is elementary to check the equality

J(

Jacϕ(t))

= ‖ϕu ∧ ϕv‖ .


H2(M) = H2

(ϕ(V )

)=

∫VJ(

Jacϕ(y))dH2(y) =

∫V‖ϕu ∧ ϕv‖ dudv = Area(M) .


Similarly, if f is a continuous function non-negative on M , then it is Borel measurable andby (30) we have∫

Mf(y) dH2(y) =

∫Vf(ϕ(u, v))J

(Jacϕ(u, v)

)dH2(u, v) =

∫Vf(ϕ(u, v))‖ϕu ∧ ϕv‖ dudv .

Therefore, the integral w.r.t. the 2-dimensional Hausdorff measure generalizes both the notionof area of a regular surface and the one of surface integral of a continuous function.

Example 5.17. Let M be the graph of a C1 function f : V ⊆ Rk → R, that is, M =y ∈

Rk+1 : y = (u, f(u)), u ∈ V

. Then, ϕ : V → Rk+1 given by ϕ(u) = (u, f(u)) is clearly injective,

C1 and its Jacobian has maximal rank 1. Then,

(Surface area)(M) =

∫V

√1 + |∇f(u)|2 du ,

and the surface integral of a continuous function defined on M as∫Mf dσ =

∫Vf(ϕ(u))

√1 + |∇f(u)|2 du .

Again, we have want to apply Thm. 5.14. Observe that Jacϕ =(1 ∇f

)so that

J(

Jacϕ(t))2

= 1 + |∇f(u)|2 ,as it is elementary to check.


Hk(M) = Hk

(ϕ(V )

)=

∫VJ(

Jacϕ(y))dH2(y) =

∫V

√1 + |∇f(u)|2 du = (Surface area)(M) .

Similarly, if g is a continuous function non-negative on M , then it is Borel measurable andby (30) we have∫

Mg(y) dH2(y) =

∫Vg(ϕ(u))J

(Jacϕ(u)

)dHk(u) =

∫Vg(ϕ(u))

√1 + |∇f(u)|2 du .

Again, the integral w.r.t. the k-dimensional Hausdorff measure generalizes both the notion ofsurface area of a regular surface and the one of surface integral of a continuous function.

64 M. M. PELOSO

6. The fundamental theorem of calculus and the theorems of Gauss and Green

In this final part we generilized the fundamental theorem of calculus and the notion andtechnique of integration by parts to the case of domains in higher dimensions.

One of the main issues that appear while passing from 1 to higher dimensions is the shapeof open, and in particular, of connected open sets. It is well known that on the real line theconnected open sets are just the open intervals, and that every open set can be written as atmost countable union of disjoint open intervals. The situation is considerably more complicatedin higher dimensions, and we begin by addressing this issue.

6.1. The fundamental theorem of calculus and Gauss theorem. In order to simplifyour presentation, but also because in practice it suffices to consider this case, we restrict ourattention to connected open sets.

Definition 6.1. A bounded, connected open set Ω ⊆ Rn, n ≥ 2, is called a Ck-regular domainif:

(i) Ω = Ω;

(ii) the boundary ∂Ω of Ω is a Ck, (n− 1)-dimensional submanifold in Rn.

Notice that condition (i) rules out domains such as

A =

(x, y) ∈ R2 : x2 + y2 < 4\

(x, y) ∈ R2 : x2 + y2 = 1, y ≥ 0.

It is easy to see that A is bounded, open, connected, and that A = B(0, 2), so that A = B(0, 2)which is strictly larger than A.

Our first goal is to show that Ck-regular domains can be analytically described in a fairlysimply way.

Lemma 6.2. Let Ω ⊆ Rn be a bounded set given by

Ω =x ∈ Rn : ψ(x) < 0

,

where ψ : Rn → R is a Ck-function such that and ∇ψ(x) 6= 0 if ψ(x) = 0. Then Ck-regulardomain.

Such a function ψ is called a defining function for Ω. As an example we consider the case ofthe ball of radius 1 in Rn, B = B(0, 1), when, clearly

B =x ∈ Rn : 1− |x|2 < 0

that is, ψ(x) = 1− |x|2 is a defining function for B.

Proof. (†) Clearly, Ω is open and it is bounded by assumption. It is it is connected as continuousimage of a connected set.

We claim that∂Ω = ∂Ω =

x ∈ Rn : ψ(x) = 0

. (31)

If is clear that both ∂Ω, ∂Ω ⊆x ∈ Rn : ψ(x) = 0

. Thus, it suffices to show that if ψ(x0) = 0,

then x0 belongs to both ∂Ω and ∂Ω. By the mean value property, for each x ∈ Rn, there existsξx ∈ sx0,x, the segment joining x to x0, such that

ψ(x)− ψ(x0) = ∇ψ(ξx) · (x− x0) . (32)


As customary, we denote by ν(x0) the unit vector ∇ψ(x0)/|∇ψ(x0)| (which is normal to thelevel curves of ψ), and observe that is well defined since ∇ψ(x0) 6= 0. If we take x on the straightline passing through x0 with direction ν(x0), that is, x is of the form x = x0 + tν(x0), witht ∈ R. Plugging into (32) we obtain,

ψ(x) = t〈∇ψ(ξx), ν(x0)〉 .Next, since 〈∇ψ(ξx), ν(x0)〉 → 〈∇ψ(x0), ν(x0)〉 = |∇ψ(x0)| > 0 as x→ x0, it follows that thereexists δ > 0 such that if x ∈ B(x0, δ) (that is, |x−x0| < δ), then ψ(x) > 0 if t > 0 and ψ(x) < 0if t < 0. These conditions can be expressed as

ψ(x0 + tν(x0)

)> 0 if 0 < t < δ ,

andψ(x0 + tν(x0)

)< 0 if − δ < t < 0 .

Therefore,x0 + tν(x0) ∈ Ω and x0 + tν(x0) 6∈ Ω if − δ < t < 0 .

This clearly implies that x0 ∈ ∂Ω ∩ ∂Ω, and proves the claim (31).Now, using (31), we see that condition (i) in Def. 6.1 follows since

Ω ⊆ Ω = Ω \ ∂Ω ⊆ x : ψ(x) ≤ 0 \ ∂Ω

= x : ψ(x) ≤ 0 \ x : ψ(x) = 0 = x : ψ(x) < 0 = Ω .

Finally, using (31) again, we have that ∂Ω = x : ψ(x) = 0 and the conclusion follows fromRemark 5.2.

We now wish to show that every Ck-regular domains can be described as in Lemma 6.2, thusproving the converse implication. We need a few preliminary notation and definitions.

Definition 6.3. Given x ∈ Rn and k ∈ 1, . . . , n, we denote by xk the element of Rn−1 obtainedfrom x = (x1, . . . , xn) by removing the coordinate xk. Thus, e.g. x = (x1, x1) ∈ R× Rn−1, andalso x = (xn, xn) ∈ Rn−1 × R.

We say that a domain D ⊆ Rn is normal w.r.t. to xk, or also k-normal, if there exist an openset V ⊆ Rn−1, b > 0 and a function g : V → (0, b) (or g : V → (−b, 0), resp.) such that

D =

(xk, xk) ∈ V × (0, b) : 0 < xk < g(xk)

orD =

(xk, xk) ∈ V × (−b, 0) : g(xk) < xk < 0

,

resp.

In other words, D is k-normal if it is the portion of space lying between the domain V in thehyperplane xk = 0 and the graph of the function g, that may lie above or below the hyperplanexk = 0.

Next, we recall the definition of partition of unity. Below we denote by supp g the support ofa continuous function g.

Definition 6.4. Given a compact subset K of Rn and a finite open cover Bj of K, j =1, . . . ,m, we say that a collection pj of functions, j = 1, . . . ,m, is a C∞ partition of unitysubordinated to the covering pj if

(i) for each j, pj ∈ C∞(Rn) and supp pj ⊆ Bj ;

66 M. M. PELOSO

(ii) 0 ≤ pj ≤ 1, j = 1, . . . ,m, and∑m

j=1 pj(x) = 1 for all x ∈ K.

Theorem 6.5. Given a compact set K and a finite open cover Bj of K, there exists a partitionof unity subordinated to the cover Bj.Proof. (†) This is a standard result and it appears in many analysis and geometry books. For aproof we refer to [F] or [L].

Before our first main result of this section, we need a further lemma.

Lemma 6.6. Let Ω be a Ck-regular domain, k ≥ 1. Let A be open, A ⊃ Ω and h : A→ R be aC1-function, whose support is contained in Ω. Then, for every j = 1, . . . , n,∫

Ω∂xjh(x) dx = 0 .

Proof. We use Fubini’s theorem and the 1-dimensional result. Observe that supp(∂xjh) ⊆supph, and moreover supp(∂xjh) ⊆ Ω implies that also supp(∂xjh) ∩ ∂Ω = ∅ Moreover, if

R > 0 is large enough so that Ω ⊆ [−R,R]n, which is possible since Ω is bounded, then h and∇h equal 0 on ⊆ [−R,R]n \ Ω. Therefore,∫

Ω∂xjh(x) dx =

∫[−R,R]n

∂xjh(x) dx =

∫[−R,R]n−1

(∫ R

−R∂xjh(xj , xj) dxj

)dxj

=

∫[−R,R]n−1

(h(xj , R)− h(xj ,−R)

)dxj

= 0 ,

since h(xj , R) = h(xj ,−R) = 0.

Lemma 6.7. Let D be a k-normal domain, say

D =

(xk, xk) ∈ V × (0, b) : 0 < xk < g(xk),

where g : V → (0, b). Let f : D → R be a C1-function, whose support is contained in the set(xk, xk) ∈ V × (0, b)

. Then, for every j = 1, . . . , n,∫

Ω∂xjf dHn =

∫G(g)

fνj dHn−1 , (33)

where G(g) =x ∈ Rn : x = (xk, xk), xk = g(xk)

.

Notice that the measure dHn on the left hand side is just the Lebesgue measure, while themeasure dHn−1 on the right hand side is the induced surface measure, and the integration isrestricted to the portion of the boundary of D where the function f does not vanish.

Proof. We need to distinguish two cases: k = j and k 6= j. We begin with the former one, k = j.We have ∫

Ω∂xjf dHn =

∫V

(∫ g(xk)

0∂xjf(xj , xj) dxj

)dxj

=

∫Vf(xj , g(xj))− f(xj , 0) dxj

=

∫Vf(xj , g(xj))dxj , (34)


since we are taking j = k and the support of f is contained in

(xk, xk) ∈ V × (0, b)

, so thatf(xk, 0) = 0.

Consider now the integral on the right hand side of (33). Notice that D =

(xk, xk) ∈V × (0, b) : ψ(x) := xk − g(xk) < 0

, so that the normal at the points of G(g) is

ν =∇ψ|∇ψ|

. (35)

Again since j = k, we have

νj =1√

1 + |∇g(xj)|2x ∈ G(g) .

Since G(g) is the the graph of the function g, by Example 5.17 the integral over G(g) w.r.t.

dHn−1 is given by integration against the density√

1 + |∇g(xj)|2. Precisely,∫G(g)

fνj dHn−1 =

∫Vf(xj , g(xj))

1√1 + |∇g(xj)|2

√1 + |∇g(xj)|2 dxj

=

∫Vf(xj , g(xj)) dxj .

Substituting this equality into the right hand side of (34) we obtain the result in the case j = k.

Next, suppose j 6= k. Using the fundamental theorem of calculus in one variable we have,∫Ω∂xjf dHn =

∫V

(∫ g(xk)

0∂xjf(xk, xk) dxk

)dxk

=

∫V

∂

∂xj

(∫ g(xk)

0f(xk, xk) dxk

)dxk −

∫Vf(xk, xk)∂xjg(xk) dxk .

We now use Lemma 6.6 applied to the function

h(xk) =

∫ g(xk)0 f(xk, xk) dxk xk ∈ V

0 xk 6∈ V .

Then, h is a C1 function, with support contained in the regular domain V ⊆ Rn−1, so thatLemma 6.6 gives that ∫

V

∂

∂xj

(∫ g(xk)

0f(xk, xk) dxk

)dxk = 0 . (36)

Therefore, ∫Ω∂xjf dHn = −

∫Vf(xk, xk)∂xjg(xk) dxk .

Again, consider the integral on the right hand side of (33). Since ν is given by equation (35)and j 6= k we have,

νj = −∂xjg√

1 + |∇g|2.

68 M. M. PELOSO

Hence,

∫G(g)

fνj dHn−1 =

∫Vf(xj , g(xk))

−∂xjg(xk)√1 + |∇g(xk)|2

√1 + |∇g(xk)|2 dxk

= −∫Vf(xk, g(xk))∂xjg(xk) dxj .

Substituting this equality into the right hand side of (36) we obtain the result in the case j 6= k.This completes the proof.

We now come to one of the main results of this part.

Theorem 6.8. Let Ω ⊆ Rn be a C1-regular domain, A ⊇ Ω an open set, f ∈ C1(A). Then, forevery j ∈ 1, . . . , n we have

∫Ω∂xjf dHn =

∫∂Ωfνj dHn−1 . (37)

Remark 6.9. We remark that the above identity (37) is often written as

∫Ω∂xjf(x) dx =

∫∂Ωf(x′)νj(x

′) dσ(x′) ,

thus writing the Lebesgue measure on the left hand side with its standard notation dx andindicating the induced surface measure on the right hand side as dσ.

Proof. By the implicit function theorem, see also Remark 5.2, for every x0 ∈ ∂Ω, there exists angbh Ux0 of x0 such that Ω ∩ Ux0 is a k-normal domain, for some k ∈ 1, . . . , n, and ∂Ω ∩ Ux0can be expressed as the graph of a Ck-function. We can use the compactness of Ω we can selecta finite such cover B`, ` = 1, . . . , N , and B` ∩Ω is a k-normal domain, for some k. Moreover,there exists an open set B0 such that

B0 ⊇ Ω \N⋃`=1

B` , and B0 ⊆ Ω .

Then, B` : ` = 0, 1, . . . , N is an open cover of Ω, and by Prop. 6.5 there exists a C∞ partitionof unity subordinated to B`, ` = 0, . . . , N ; call it p`, ` = 0, . . . , N . In particular we have

supp p` ⊆ B`, ` = 0, . . . , N, and

N∑`=0

p`(x) = 1 on Ω .


Therefore, using the fact that p0 has compact support in Ω together with Lemma 6.6, and thenLemma 6.7, we have∫

Ω∂xjf(x) dx =

∫Ω

∂

∂xj

( N∑`=0

f(x)p`(x))dx =

N∑`=0

∫Ω

∂

∂xj

(f(x)p`(x)

)dx

=

N∑`=1

∫Ω

∂

∂xj

(f(x)p`(x)

)dx =

N∑`=1

∫Ω∩B`

∂

∂xj

(f(x)p`(x)

)dx

=

N∑`=1

∫∂Ω∩B`

f(x′)p`(x′)νj(x

′) dHn−1(x′) =

N∑`=1

∫∂Ωf(x′)p`(x

′)νj(x′) dHn−1(x′)

=

∫∂Ωf(x′)νj(x

′) dHn−1(x′) ,

since supp p` ⊆ B` and∑N

`=1 p`(x′) = 1 on ∂Ω. This completes the proof.

Corollary 6.10. (Integration by parts in higher dimension) Let Ω ⊆ Rn be a C1-regulardomain, A ⊇ Ω an open set, f, g ∈ C1(A). Then, for every j ∈ 1, . . . , n we have∫

Ωg∂xjf dHn =

∫∂Ωgfνj dHn−1 −

∫Ωf∂xjg dHn . (38)

Proof. The proof is now immediate. By the previous theorem we have∫Ωg∂xjf dHn +

∫Ωf∂xjg dHn =

∫Ω

∂

∂xj

(fg)dHn

=

∫∂Ωgfνj dHn−1 ,

and the conclusion follows.

Recall that if F = (F1, . . . , Fn) is a C1 vector field on an open set A ⊆ Rn, then the divergenceof F is defined as

div F =n∑j=1

∂Fj∂xj

.

Corollary 6.11. (Divergence Theorem, or Gauss Theorem) Let Ω ⊆ Rn be a C1-regulardomain, A ⊇ Ω an open set, F ∈ C1(A,Rn) a C1 vector field on A. Then,∫

Ωdiv F dHn =

∫∂Ω〈F, ν〉 dHn−1 . (39)

Notice that the divergence theorem says that the outward flux of a vector field through aclosed surface equals the integral of the divergence over the region inside the surface w.r.t. thevolume measure.

70 M. M. PELOSO

Proof. This is also immediate consequence of Thm. 6.8:∫Ω

div F dHn =

n∑j=1

∫Ω

∂Fj∂xj

dHn

=

n∑j=1

∫∂ΩFjνxj dHn−1

=

∫∂Ω〈F, ν〉 dHn−1 .

6.2. Extesion to domains with lower regularities. (†)7

As we wish to extend the previous result to more general domain, such the cube for istance,whose boundary is piecewise regular. We begin by making this idea into a precise definition.

Definition 6.12. We say that a bounded domain Ω ⊆ Rn, with n ≥ 2, is a piecewise Ck-regulardomain if

(i) Ω = Ω;

(ii) there exists an integer N such the boundary ∂Ω = Σ1 ∪ · · ·ΣN ∪ S such that(ii-1) for each j = 1, . . . , N , Σj is a relatively open subset of ∂Ω;

(ii-2) for each j = 1, . . . , N there exists a Ck, (n − 1)-dimensional submanifold Mj such

that Σj ⊆;

(ii-3) the subset S is compac and it is contained in an (n− 2)-submanifold;

(ii-4) Σj ∩ Σk ⊆ S, for j 6= k.

In this case, we call Σ1 ∪ · · ·ΣN the regular part of the boundary, and we denote it by ∂regΩ.

Example 6.13. The most typical example of such domain is a cube, or rectangle in Rn. Let uslimit ourselves to the the cases of n = 2, 3.

(1) Let R = (a, b)× (c, d) be an open rectangle in R2. Then the ∂R = Σ1 ∪ . . .Σ4 ∪ S, whereΣ1, . . . ,Σ4 are the four open sides, say Σ1 = (x, c) : a < x < b, Σ2 = (b, y) : c < y < d,etc., while S is the union of the four vertices S = (a, c), (b, c), (b, d), (a, d). It is elementary tocheck that R is a piecewise C1-regular domain.

(2) The same reasoning applies to the case of a rectangle in R3,

R = (a, b)× (c, d)× (α, β) =

(x, y, z) ∈ R3 : a < x < b, c < y < d, α < z < β.

Now

∂R = Σ1 ∪ · · ·Σ8 ∪ Swhere for j = 1, . . . , 8, Σj is one of the open faces, and S is the union of the open edge, that are1-dimensional manifolds, and of the vertices, that have Hausdorff dimension 0.

With the above definitions, it is easy to extend the results of the previous section to the caseof piecewise Ck-regular domains. We observe that, if Ω is a piecewise Ck-regular domain, thenon the regular part of its boundary ∂regΩ, the outward unit normal vector ν is well defined.Hence we have the following results.

7This part is only required for the written part of the exam.


Theorem 6.14. Let Ω ⊆ Rn be a piecewise C1-regular domain, A ⊇ Ω an open set, f ∈ C1(A).Then, for every j ∈ 1, . . . , n we have∫

Ω∂xjf dHn =

∫∂regΩ

fνj dHn−1 . (40)

Corollary 6.15. (Integration by parts in higher dimension) Let Ω ⊆ Rn be a piecewiseC1-regular domain, A ⊇ Ω an open set, f, g ∈ C1(A). Then, for every j ∈ 1, . . . , n we have∫

Ωg∂xjf dHn =

∫∂regΩ

gfνj dHn−1 −∫

Ωf∂xjg dHn . (41)

Corollary 6.16. (Divergence Theorem, or Gauss Theorem) Let Ω ⊆ Rn be a piecewiseC1-regular domain, A ⊇ Ω an open set, F ∈ C1(A,Rn) a C1 vector field on A. Then,∫

Ωdiv F dHn =

∫∂regΩ〈F, ν〉 dHn−1 . (42)

6.3. Green’s Theorem and identities. We begin with the classical Green’s formula in R2.

Theorem 6.17. (Green’s Theorem) Let Ω ⊆ R2 be a C1-regular domain, A ⊇ Ω an openset, ω = fdx+ gdy a 1-form in R2 with coefficients in C1(A). Then,∫

∂Ωfdx+ gdy =

∫Ω

(∂g∂x− ∂f

∂y

)dH2 , (43)

where ∂Ω is oriented counter-clockwise.

Proof. We recall that, if ∂Ω is parametrized by a (piecewise) C1 curve γ, parametrized asγ : [a, b]→ R2 then ∫

∂Ωfdx+ gdy =

∫ b

af(γ(t)

)γ′1(t) + g

(γ(t)

)γ′2(t) dt .

Observe now that, defining the vector field F = (g,−f), the right hand side above can be writtenas ∫ b

a

(f(γ(t)

) γ′1(t)

|γ′(t)|+ g(γ(t)

) γ′2(t)

|γ′(t)|

)|γ′(t)| dt

=

∫ b

a

(g(γ(t)

) γ′2(t)

|γ′(t)|+((−f

(γ(t)

))−γ′1(t)

|γ′(t)|+)|γ′(t)| dt

=

∫∂Ω〈F, ν〉 dH1 ,

since, given the orientation of ∂Ω, the outward normal ν equals( γ′2(t)|γ′(t)| ,

−γ′1(t)|γ′(t)|

). Thus, applying

the Divergence Theorem to the term on the right hand side we see that∫∂Ω〈F, ν〉 dH1 =

∫Ω

div F dH2 =

∫Ω

∂g

∂x− ∂f

∂ydH2 ;

hence the conclusion.

72 M. M. PELOSO

Definition 6.18. (†) The Laplace operator, or laplacian, in Rn is the second order partialdifferential operator

∆ =n∑j=1

∂2

∂x2j

.

Notice that, if f is twice differentiable in Rn then ∆f = div∇f .

Theorem 6.19. (†) (Green’s identities) Let Ω ⊆ R2 be a C1-regular domain, A ⊇ Ω an openset, ω = fdx+ gdy a 1-form in R2 with coefficients in C1(A). Then, denoting by ∂u

∂ν := ν · ∇uthe normal derivative of u on ∂Ω,∫

Ωv∆u+ 〈∇v,∇u〉 dHn =

∫∂Ωv∂u

∂νdHn−1 ; (44)∫

Ω

(u∆v − v∆u

)dHn =

∫∂Ω

(u∂v

∂ν− v∂u

∂ν

)dHn−1 ; (45)∫

Ω

(u∆u+ |∇u|2

)dHn =

∫∂Ωu∂u

∂νdHn−1 . (46)

Proof. All identities follow easily from the divergence theorem. We observe, that

div(v∇u) =

n∑j=1

∂

∂xj

(v∂u

∂xj

)= 〈∇v,∇u〉+ v∆u ,

and

〈v∇u, ν〉 = v∂u

∂ν,

so that (44) follows at once.Now (45) follows from (44) by taking a difference, and (45) follows from (44) by taking

v = u.


References

[FMS] N. Fusco, P. Marcellini, C. Sbordone, Analisi Matematica 2, Liguori Editore.[F] G. B. Folland, Real Analysis: Modern Techniques and Their Applications, 2th Edition,, 1999 Wilely-

Interscience Press.[KP] S. G. Krantz, H. R. Parks, Geometric Integration Theory, Springer-Verlag 2008.[L] E. Lanconelli, Lezioni di Analisi Matematica 2, Seconda Parte, 1997 Pitagora Editrice Bologna,[PS] C. Pagani, S. Salsa, Analisi Matematica 2, Sec. Ed., Zanichelli Editore.[R] W. Rudin, Real and Complex Analysis, 3rd Ed., McGraw-Hill.[SS] E. M. Stein, R. Shakarchi, Real Analysis, Princeton Lectures in Analysis II, 2003 Princeton Univ. Press.

Dipartimento di Matematica, Universita degli Studi di Milano, Via C. Saldini 50, 20133 Milano,Italy

E-mail address: [email protected]

URL: http://wwww.mat.unimi.it/~peloso/

AN INTRODUCTION TO MEASURE THEORY AND THE … · AN INTRODUCTION TO MEASURE THEORY AND THE LEBESGUE INTEGRAL MARCO M. PELOSO Contents 1. First elements of measure theory 1 1.1. Measure

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