Analysis I - University of Pittsburghhajlasz/Notatki/Analysis I.pdf · Analysis I Piotr Haj lasz...

Analysis I

Piotr Haj lasz

1 Measure theory

1.1 σ-algebra.

Definition. Let X be a set. A collection M of subsets of X is σ-algebra if M has thefollowing properties

1. X ∈M;

2. A ∈M =⇒ X \ A ∈M;

3. A1, A2, A3, . . . ∈M =⇒⋃∞

i=1Ai ∈M.

The pair (X,M) is called measurable space and elements of M are called measurablesets.

Exercise. Let (X,M) be a measurable space. Show that

1. ∅ ∈M;

2. A,B ∈M =⇒ A \B ∈M;

3. M is closed under finite unions, finite intersections and countable intersections.

Examples.

1. 2X , the family of all subsets of X is a σ-algebra.

2. M = ∅, X is a σ-algebra.

3. If E ⊂ X is a fixed set, then M = ∅, X,E,X \ E is a σ-algebra.

1

Proposition 1 If Mii∈I is a family of σ-algebras, then

M =⋂i∈I

Mi

is a σ-algebra.

A simple proof is left to the reader.

Let R be a family of subsets of X. By σ(R) we will denote the intersection of allσ-algebras that contain R. Note that there is at least one σ-algebra that contains R,namely 2X .

1. σ(R) is a σ-algebra that contains R

2. σ(R) is the smallest σ-algebra that containsR in the sense that if M is a σ-algebrathat contains R, then σ(R) ⊂M.

We say that σ(R) is a σ-algebra generated by R.

Example. If R = E, then σ(R) = ∅, X,E,X \ E.

In the sequel we will need the following result.

Proposition 2 The family M of subsets of a set X is a σ-algebra if and only if itsatisfies the following three properties

1. X ∈M;

2. If A,B ∈M, then A \B ∈M.

3. If A1, A2, A3, . . . ∈M are pairwise disjoint, then⋃∞

i=1Ai ∈M.

We leave the proof as an exercise cf. the proof of Theorem 3(d). 2

1.2 Borel sets.

Let X be a metric space (or more generally a topological space). By B(X) we willdenote the σ-algebra generated by the family of all open sets in X. Elements of B(X)are called Borel sets and B(X) is called σ-algebra of Borel sets.

The following properties are obvious.

1. If U1, U2, U3, . . . ⊂ X are open, then⋂∞

i=1 Ui is a Borel set;

2. All closed sets are Borel;

3. If F1, F2, F3, . . . ⊂ X are closed, then⋃∞

i=1 Fi is a Borel set.

2

1.3 Measure.

Definition. Let (X,M) be a measurable space. A measure (called also positivemeasure) is a function

µ : M→ [0,∞]

such that

1. µ(∅) = 0;

2. µ is countably additive i.e. if A1, A2, A3, . . . ∈M are pairwise disjoint, then

µ

(∞⋃i=1

Ai

)=

∞∑i=1

µ(Ai).

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

If µ(X) <∞, then µ is called finite measure.

If µ(X) = 1, then µ is called probability or probability measure.

If we can write X =⋃∞

i=1Ai, where Ai ∈ M and µ(Ai) < ∞ for all i = 1, 2, 3, . . .,then we say that µ is σ-finite.

Example. If X is an arbitrary set, and µ : 2X → [0,∞] is defined by µ(E) = m ifE is finite and has m elements, µ(E) = ∞ if E is infinite, then µ is a measure. It iscalled counting measure.

Theorem 3 (Elementary properties of measures) Let (X,M, µ) be a measurespace. Then

(a) If the sets A1, A2, . . . , An ∈ M are pairwise disjoint then µ(A1 ∪ . . . ∪ An) =µ(A1) + . . .+ µ(An).

(b) If A,B ∈M, A ⊂ B and µ(B) <∞, then µ(B \ A) = µ(B)− µ(A).

(c) If A,B ∈M, A ⊂ B, then µ(A) ≤ µ(B).

(d) If A1, A2, A3, . . . ∈M, then

µ

(∞⋃i=1

Ai

)≤

∞∑i=1

µ(Ai).

(e) If A1, A2, A3, . . . ∈M, µ(Ai) = 0, for i = 1, 2, 3, . . . then µ(⋃∞

i=1Ai) = 0.

3

(f) If A1, A2, A3, . . . ∈M, A1 ⊂ A2 ⊂ A3 ⊂ . . . then

µ

(∞⋃i=1

Ai

)= lim

i→∞µ(Ai).

(g) If A1, A2, A3, . . . ∈M, A1 ⊃ A2 ⊃ A3 ⊃ . . . and µ(A1) <∞, then

µ

(∞⋂i=1

Ai

)= lim

i→∞µ(Ai).

Proof.

(a) The sets A1, A2, . . . , An, ∅, ∅, ∅, . . . are pairwise disjoint. Hence

µ(A1 ∪ . . . ∪ An) = µ(A1 ∪ . . . ∪ An ∪ ∅ ∪ ∅ ∪ ∅ ∪ . . .)= µ(A1) + . . .+ µ(An) + µ(∅) + µ(∅) + µ(∅) + . . .

= µ(A1) + . . .+ µ(An).

(b) Since B = A ∪ (B \ A) and the sets A, B \ A are disjoint we have

µ(B) = µ(A) + µ(B \ A) (1)

and the claim easily follows.

(c) The claim follows from (1).

(d) We have

A = A1 ∪ A2 ∪ A3 ∪ . . .= A1︸︷︷︸

B1

∪ (A2 \ A1)︸︷︷︸B2

∪ (A3 \ (A1 ∪ A2))︸︷︷︸B3

∪ (A4 \ (A1 ∪ A2 ∪ A3))︸︷︷︸B4

∪ . . .

The sets Bi are pairwise disjoint and Bi ⊂ Ai. Hence

µ(A) =∞∑i=1

µ(Bi) ≤∞∑i=1

µ(Ai).

(e) It easily follows from (d).

(f) Since Ai ⊂ Ai+1, we have

A = A1 ∪ A2 ∪ A3 ∪ . . .= A1︸︷︷︸

B1

∪ (A2 \ A1)︸︷︷︸B2

∪ (A3 \ A2)︸︷︷︸B3

∪ (A4 \ A3)︸︷︷︸B4

∪ . . .

4

The sets Bi are pairwise disjoint and hence

µ(A) =∞∑i=1

µ(Bi) = limi→∞

(µ(B1) + . . .+ µ(Bi)) = limi→∞

µ(B1 ∪ . . . ∪Bi) = limi→∞

µ(Ai).

The limit exist because µ(Ai) ≤ µ(Ai+1) (since Ai ⊂ Ai+1).

(g) It suffice to apply (f) to the sets A1 \ Ai.

The proof is complete. 2

Exercise. Provide a detailed proof of (g). Where do we use the assumption thatµ(A1) < ∞? Show an example that the conclusion of (g) need not be true without theassumption that µ(A1) <∞.

1.4 Outer measure and Caratheodory construction.

It is quite difficult to construct a measure with desired properties, but it is mucheasier to construct so called outer measure which has less restrictive properties. TheCaratheodory theorem shows then how to extract a measure from an outer measure.

Definition. Let X be a set. A function

µ∗ : 2X → [0,∞]

is called outer measure if

1. µ∗(∅) = 0;

2. A ⊂ B =⇒ µ∗(A) ≤ µ∗(B);

3. for all sets A1, A2, A3, . . . ⊂ X

µ∗

(∞⋃i=1

Ai

)≤

∞∑i=1

µ∗(Ai).

We call a set E ⊂ X µ∗-measurable if

∀A ⊂ X µ∗(A) = µ∗(A ∩ E) + µ∗(A \ E). (2)

Condition (2) is called Caratheodory condition. Since the inequality

µ∗(A) ≤ µ∗(A ∩ E) + µ∗(A \ E)

is always satisfied, in order to verify measurability of a set E it suffices to show that

µ∗(A) ≥ µ∗(A ∩ E) + µ∗(A \ E).

5

Proposition 4 All sets with µ∗(E) = 0 are µ∗ measurable.

Proof. If µ∗(E) = 0, then for an arbitrary set A ⊂ X we have

µ∗(A) ≥ µ∗(A \ E) = µ∗(A \ E) + µ∗(A ∩ E)︸︷︷︸0

,

because A ∩ E ⊂ E. 2

Let M∗ be the class of all µ∗-measurable sets.

Theorem 5 (Caratheodory) M∗ is a σ-algebra and

µ∗ : M∗ → [0,∞]

is a measure.

Proof. We will split the proof into several steps.

Step I: X ∈M∗

This is obvious.

Step II: E,F ∈M∗ =⇒ E ∪ F ∈M∗.

Let A ⊂ X. We have

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

Since A ∩ E = A ∩ (E ∪ F ) ∩ E and (A \ E) ∩ F = (A ∩ (E ∪ F )) \ E, we have

µ∗(A) = µ∗(A ∩ (E ∪ F ) ∩ E) + µ∗(A ∩ (E ∪ F ) \ E) + µ∗(A \ (E ∪ F ))

= µ∗(A ∩ (E ∪ F )) + µ∗(A \ (E ∪ F )).

Step III: E,F ∈M∗ =⇒ E \ F ∈M∗.

It follows from the symmetry of the condition (2) that X \E ∈M∗. Since E \F =X \ ((X \ E) ∪ F ) the claim follows.

Step IV: If the sets E1, E2, E3, . . . ∈M∗ are pairwise disjoint, then for every A ⊂ X

µ∗(A ∩∞⋃i=1

Ei) =∞∑i=1

µ∗(A ∩ Ei).

If E,F ∈M∗ are pairwise disjoint, the for every A ⊂ X

µ∗(A∩ (E ∪F )) = µ∗(A∩ (E ∪F )∩E) +µ∗(A∩ (E ∪F )\E) = µ∗(A∩E) +µ∗(A∩F ).

6

This, Step II and the induction argument implies that for every n

µ∗(A ∩n⋃

i=1

Ei) =n∑

i=1

µ∗(A ∩ Ei). (3)

Hence

µ∗(A ∩∞⋃i=1

Ei) ≥n∑

i=1

µ∗(A ∩ Ei)

and thus in the limit

µ∗(A ∩∞⋃i=1

Ei) ≥∞∑i=1

µ∗(A ∩ Ei).

Since the opposite inequality is a property of an outer measure, the claim follows.

Step V: If the sets E1, E2, E3, . . . ∈M∗ are pairwise disjoint, then⋃∞

i=1Ei ∈M∗.

Step II and the induction argument implies that for every n

n⋃i=1

Ei ∈M∗.

Hence (3) gives

µ∗(A) = µ∗

(A ∩

n⋃i=1

Ei

)+ µ∗

(A \

n⋃i=1

Ei

)≥

n∑i=1

µ∗(A ∩ Ei) + µ∗

(A \

∞⋃i=1

Ei

),

and in the limit

µ∗(A) ≥∞∑i=1

µ∗(A ∩ Ei) + µ∗

(A \

∞⋃i=1

Ei

)= µ∗

(A ∩

∞⋃i=1

Ei

)+ µ∗

(A \

∞⋃i=1

Ei

).

Since the opposite inequality is a property of an outer measure the claim follows.

Step VI: Final step. It follows from Steps I, III, V and Proposition 2 that M∗ is aσ-algebra and then Step IV with A = X implies that µ∗ restricted to M∗ is a measure.The proof is complete. 2

We say that a measure µ : M → [0,∞] is complete if every subset of a set ofmeasure zero is measurable (and hence has measure zero). Therefore it follows fromProposition 4 that the measure described by the Caratheodory theorem is complete.

Let (X, d) be a metric space. For E,F ⊂ X we define

dist (E,F ) = infx∈E,y∈F

d(x, y), diamE = supx,y∈E

d(x, y).

Definition. An outer measure µ∗ defined on subsets of a metric space is called metricouter measure if

µ∗(E ∪ F ) = µ∗(E) + µ∗(F ) whenever dist (E,F ) > 0.

7

Theorem 6 If µ∗ is a metric outer measure, then all Borel sets are µ∗-measurable i.e.B(X) ⊂M∗.

Proof. Since M∗ is a σ-algebra, it suffices to show that all open sets belong to M∗. LetG ⊂ X be open. It suffices to show that for every A ⊂ X

µ∗(A) ≥ µ∗(A ∩G) + µ∗(A \G) (4)

(because the opposite inequality is obvious). We can assume that µ∗(A) < ∞ asotherwise (4) is obviously satisfied.

For each positive integer n define

Gn = x ∈ G : dist (x,X \G) >1

n.

Then

dist (Gn, X \G) ≥ 1

n> 0. (5)

Let

Dn = Gn+1 \Gn = x ∈ G :1

n+ 1< dist (x,X \G) ≤ 1

n.

Clearly

G \Gn =∞⋃

i=n

Di (6)

and

dist (Di, Dj) ≥1

i+ 1− 1

j> 0 provided i+ 2 ≤ j.

Since the mutual distances between the sets D1, D3, D5, . . . , D2n−1 are positive we have

µ∗(A∩D1) +µ∗(A∩D3) + . . . µ∗(A∩D2n−1) = µ∗(A∩ (D1∪D3∪ . . .∪D2n−1)) ≤ µ∗(A)

and similarly

µ∗(A ∩D2) + µ∗(A ∩D4) + . . .+ µ∗(A ∩D2n) ≤ µ∗(A).

Hence∞∑i=1

µ∗(A ∩Di) ≤ 2µ∗(A) <∞.

Now (6) yields

µ∗(A ∩ (G \Gn)) ≤∞∑

i=n

µ∗(A ∩Di)

and henceµ∗(A ∩ (G \Gn))→ 0 as n→∞.

8

Inequality (5) gives

µ∗(A ∩Gn) + µ∗(A \G) = µ∗((A ∩Gn) ∪ (A \G)) ≤ µ∗(A)

and thus

µ∗(A ∩G) + µ∗(A \G) ≤ µ∗(A ∩Gn) + µ∗(A ∩ (G \Gn)) + µ∗(A \G)

≤ µ∗(A) + µ∗(A ∩ (G \Gn))

and after passing to the limit

µ∗(A ∩G) + µ∗(A \G) ≤ µ∗(A).


We will use outer measures to prove the following important result.

Theorem 7 Let X be a metric space and µ a measure in B(X). Suppose that X is aunion of countably many open sets of finite measure. Then

µ(E) = infU⊃E

U−open

µ(U) = supC⊂E

C−closed

µ(C). (7)

Before proving the theorem let’s discuss two of its corollaries. The first one shows thatin a situation described by the above theorem in order to prove that two measures areequal it suffices to compare them on the class of open sets.

Corollary 8 If µ is as in Theorem 7 and ν is another measure on B(X) that satisfies

ν(U) = µ(U) for all open sets U ⊂ X

thenν(E) = µ(E) for all E ⊂ B(X).

The second corollary describes an important class of measures satisfying assumptionsof theorem 7.

Definition. Let X be a metric space and µ a measure defined on the σ-algebra ofBorel sets. We say that µ is a Radon measure if µ(K) <∞ for all compact sets and

µ(E) = infU⊃E

U−open

µ(U) = supK⊂E

K−compact

µ(K) for all E ∈ B(X). (8)

Corollary 9 If X is a locally compact and separable metric space1 and µ is a measurein B(X) such that µ(K) <∞ for every compact set K, then X is a union of countablymany open sets of finite measure and µ is a Radon measure.

1Being locally compact means that ever point has a neighborhood whose closure is compact. IRn islocally compact, but also IRn \ 0 is locally compact.

9

Proof. It follows from locally compactness of X that X is a union of a family of opensets with compact closures. Since X is separable, every covering of X by open sets hascountable subcovering and hence we can write

X =∞⋃i=1

Ui, Ui — compact.

This proves the first part of the corollary because µ(Ui) ≤ µ(Ui) <∞. Now (8) followsfrom (7) and Theorem 3(f) because

C =∞⋃

n=1

(C ∩

n⋃i=1

Ui

)︸︷︷︸

compact

.

and henceµ(C) = sup

K⊂CK−compact

µ(K)

for every closed set C. The proof is complete 2

Proof of Theorem 7. For E ⊂ X we set

µ∗(E) = infU⊃E

U−open

µ(U).

It is easy to see that µ∗ is a metric outer measure. Hence µ∗ restricted to the class ofBorel sets is a measure. Clearly

µ(U) = µ∗(U) for all open sets U ⊂ X, (9)

andµ(E) ≤ µ∗(E) for all E ∈ B(X). (10)

We can assume that X is a union of an increasing sequence of open sets with finitemeasure. Indeed, if X is the union of open sets Ui of finite measure, then the setsVn = U1 ∪ U2 ∪ . . . ∪ Un satisfy

X =∞⋃

n=1

Vn, Vn ⊂ Vn+1, µ(Vn) <∞.

Now inequality (10) implies that for all E ∈ B(X)

µ(Vn \ E) ≤ µ∗(Vn \ E) and µ(Vn ∩ E) ≤ µ∗(Vn ∩ E), (11)

because both Vn\E and Vn∩E are Borel. Actually we have equalities in both inequalitiesof (11) because otherwise we would have a sharp inequality

µ(Vn) = µ(Vn \ E) + µ(Vn ∩ E) < µ∗(Vn \ E) + µ∗(Vn ∩ E) = µ∗(Vn),

10

which contradicts (9). Hence in particular µ(Vn ∩ E) = µ∗(Vn ∩ E). Since

µ(Vn ∩ E)→ µ(E) and µ∗(Vn ∩ E)→ µ∗(E)

as n→∞ by Theorem 3(f) we conclude that µ(E) = µ∗(E) and thus

µ(E) = infU⊃E

U−open

µ(U) (12)

by the definition of µ∗. To prove that the measure of a set E ∈ B(X) can be approxi-mated by measures of closed subsets of E observe that Vn \ E has finite measure andhence it follows from (12) that there is an open set Gn such that

Vn \ E ⊂ Gn, µ(Gn \ (Vn \ E)) <ε

2n.

The set G =⋃∞

n=1Gn is open and C = X \G ⊂ E is closed. Now it suffices to observethat

E \ C = E ∩∞⋃

n=1

Gn ⊂∞⋃

n=1

Gn \ (Vn \ E)

and hence µ(E \ C) < ε. The proof is complete. 2

1.5 Hausdorff measure.

Let ωs = πs/2/Γ(1 + s2), s ≥ 0. If s = n is a positive integer, then ωn is volume of the

unit ball in IRn.

Let X be a metric space. For ε > 0 and E ⊂ X we define

Hsε(E) = inf

ωs

2s

∞∑i=1

(diamAi)s

where the infimum is taken over all possible coverings

E ⊂∞⋃i=1

Ai with diamAi < ε.

Since the function ε 7→ Hsε(E) is nonincreasing, the limit

Hs(E) = limε→0Hs

ε(E)

exists. Hs is called Hausdorff measure.

It is easy to see that if s = 0, H0 is the counting measure.

Theorem 10 Hs is a metric outer measure.

11

Proof. ClearlyHs : 2X → [0,∞],

Hs(∅) = 0,

A ⊂ B =⇒ Hs(A) ≤ Hs(B).

In order to show that Hs is an outer measure it remains to show that

Hs

(∞⋃

n=1

En

)≤

∞∑n=1

Hs(En).

We can assume that the right hand side is finite. Hence

∞∑n=1

Hsε(En) <∞ for every ε > 0.

Fix δ > 0. We can find a covering of each set En

En ⊂∞⋃i=1

Ani, diamAni < ε

such that

Hsε(En) ≥ ωs

2s

∞∑i=1

(diamAni)s − δ

2n

Hence∞∑

n=1

Hsε(En) ≥ ωs

ss

∞∑i,n=1

(diamAni)s − δ ≥ Hs

ε

(∞⋃

n=1

En

)− δ,

because Ani∞i,n=1 forms a covering of⋃∞

n=1En. Passing to the limit with δ → 0 yields

∞∑n=1

Hs(En) ≥∞∑

n=1

Hsε(En) ≥ Hs

ε

(∞⋃

n=1

En

).

Now passing to the limit with ε→ 0 yields the result. We are left with the verificationof the metric condition. Let dist (E,F ) > 0. It suffices to show that

Hsε(E ∪ F ) = Hs

ε(E) +Hsε(F ) (13)

for all ε < dist (E,F ). Let

E ∪ F ⊂∞⋃i=1

Ai, diamAi < ε

be a covering of E ∪ F . We can assume that

Ai ∩ (E ∪ F ) 6= ∅ (14)

12

for all i, as otherwise we could remove Ai from the covering. Since diamAi < dist (E,F )it follows from (14) that Ai has a nonempty intersection with exactly one set E or F .Accordingly, the family Aii splits into two disjoint subfamilies

Bjj = all the sets Ai such that Ai ∩ E 6= ∅;

Cjj = all the sets Ai such that Ai ∩ F 6= ∅.Hence

ωs

2s

∞∑i=1

(diamAi)s =

ωs

2s

∑j

(diamBj)s +

ωs

2s

∑j

(diamCj)s

≥ Hsε(E) +Hs

ε(F ).

Since Aii was an arbitrary covering of E ∪ F such that diamAi < ε, we concludeupon taking the infimum that

Hsε(E ∪ F ) ≥ Hs

ε(E) +Hsε(F ).

Since the opposite inequality is obvious (13) follows. 2

Exercise. Show that if

• Hs(E) <∞, then Ht(E) = 0 for all t > s;

• Hs(E) > 0, then Ht(E) =∞ for all 0 < t < s.

Definition. The Hausdorff dimension is defined as follows. If Hs(E) > 0 for alls ≥ 0, then dimH(E) =∞. Otherwise we define

dimH(E) = infs ≥ 0 : Hs(E) = 0.

It follows from the exercise that there is s ∈ [0,∞] such that Ht(E) = 0 for t > sand Ht(E) =∞ for 0 < t < s. Hausdorff dimension of E equals s.

1.6 Lebesgue measure.

For a closed intervalP = [a1, b1]× . . .× [an, bn] ⊂ IRn (15)

we define its volume by|P | = (b1 − a1) · . . . · (bn − an)

and for A ⊂ IRn we define

L∗n(A) = inf∞∑i=1

|Pi|

13

where the infimum is taken over all coverings

A ⊂∞⋃i=1

Pi

by intervals as in (15).

Arguments similar to those in the proof that the Hausdorff measure is an outermetric measure give the following result

Theorem 11 L∗n is a metric outer measure.

Definition. L∗n is called outer Lebesgue measure. L∗n-measurable sets are calledLebesgue measurable. L∗n restricted to Lebesgue measurable sets is called Lebesguemeasure and is denoted by Ln.

Corollary 12 All Borel sets are Lebesgue measurable.

All sets with L∗n(A) = 0 are Lebesgue measurable. Then Ln(A) = 0. A set hasLebesgue measure zero if and only it it can be covered by a sequence of intervals withwhose sum of volumes can be arbitrarily small. Note that this is the same as the setsof measure zero in the setting of Riemann integral.

Theorem 13 If P is a closed interval as in (15), then

Ln(P ) = |P |.

In what follows we will often write |A| to denote the Lebesgue measure of a Lebesguemeasurable set A.

Proof. In order to prove the theorem we have to show that if P ⊂⋃∞

i=1 Pi, then|P | ≤

∑∞i=1 |Pi|.

We will need the following quite obvious special case of this fact.

Lemma 14 If P ⊂⋃k

i=1 Pi is a finite covering of a closed interval P by closed intervals

Pi, then |P | ≤∑k

i=1 |Pi|.

Proof. This lemma follows from the theory of Riemann integral.2 Indeed, volume ofthe interval equals the integral of the characteristic function. Hence inequality χP ≤∑k

i=1 χPiimplies

|P | =∫

IRn

χP ≤k∑

i=1

∫IRn

χPi=

k∑i=1

|Pi|.

2This lemma can be proved in a more elementary way without using the Riemann integral, but theproof would be longer.

14


Now we can complete the proof of the theorem. Observe that each interval Pi iscontained in a slightly bigger open interval P ε

i such that

|P εi | = |Pi|+

ε

2i,

where P εi is the closure of P ε

i . Now from the covering P ⊂⋃∞

i=1 Pεi we can select a

finite subcovering

P ⊂k⋃

j=1

P εij

(compactness) and hence the lemma yields

|P | ≤k∑

j=1

|P εij| ≤

∞∑i=1

|P εi | =

∞∑i=1

|Pi|+ ε.

Since ε > 0 was arbitrary the theorem follows. 2

Proposition 15

• If P = (a1, b1)× . . . (an, bn) is a bounded interval, then

Ln(P ) = (b1 − a1) · . . . · (bn − an).

• Ln(∂P ) = 0.

• If P is an unbounded interval in IRn and has nonempty interior (i.e. each sidehas positive length), then Ln(P ) =∞.

Proof. The first claim follows from the observation that P can be approximated frominside and from outside by closed intervals that differs arbitrarily small in volume. Thesecond claim follows from the first one Ln(∂P ) = Ln(P )− Ln(intP ) = 0, and the latsclaim follows from the fact that P contains closed intervals of arbitrarily large measure.2

IRk ⊂ IRn, k < n is a subset generated by the first k coordinates.

Corollary 16 If k < n, then Ln(IRk) = 0.

The next result shows that we can compute Lebesgue measure of an open set byrepresenting it as a union of cubes. We call a cube dyadic if its sidelength equals 2k forsome k ∈ ZZ.

15

Theorem 17 An arbitrary open set in IRn is a union of closed dyadic cubes withpairwise disjoint interiors. Hence Lebesgue measure of the open set equals the sum ofthe measures of these cubes.

Proof. Instead of a formal proof we will show a picture which explains how to representany open set as a union of cubes with sidelength 2−k, k = 0, 1, 2, 3, . . .

The proof is complete 2

Compare the following result with Theorem 7.

Theorem 18 For an arbitrary set E ⊂ IRn

L∗n(E) = inf Ln(U),

where the infimum is taken over all open sets U such that E ⊂ U .

Proof. L∗n = inf∑

i |Pi| and each interval Pi is contained in an open set (interval) ofslightly bigger measure. 2

Every Borel set B is Lebesgue measurable. Every set E with L∗(E) = 0 is Lebesguemeasurable. Hence sets of the form B ∪ E are Lebesgue measurable. These are allLebesgue measurable sets. More precisely we have the following characterization.

16

Definition. Let X be a metric space. By a Gδ set we mean a set of the formA =

⋂∞i=1Gi, where the sets Gi ⊂ X are open and by Fσ we mean a set of the form

B =⋃∞

i=1 Fi, where the sets Fi ⊂ X are closed. Clearly all Gδ and Fσ sets are Borel.

Theorem 19 Let A ⊂ IRn. Then the following conditions are equivalent

1. A is Lebesgue measurable;

2. For every ε > 0 there is an open set G such that A ⊂ G and L∗n(G \ A) < ε;

3. There is a Gδ set H such that A ⊂ H and L∗n(H \ A) = 0;

4. For every ε > 0 there is a closed set F such that F ⊂ A and L∗n(A \ F ) < ε;

5. There is a Fσ set M such that M ⊂ A and L∗n(A \M) = 0;

6. For every ε > 0 there is an open set G and a closed set F such that F ⊂ A ⊂ Gand Ln(G \ F ) < ε.

Proof. (1)⇒(2) Every measurable set can be represented as a union of sets with finitemeasure A =

⋃∞i=1Ai, Ln(Ai) <∞. It follows from Theorem 18 that for every i there

is an open set Gi such that Ai ⊂ Gi and Ln(Gi \ Ai) < ε/2i. Hence A ⊂ G =⋃∞

i=1Gi,Ln(G \ A) < ε.

(2)⇒(3) We define H =⋂∞

i=1 Ui, where Ui are open sets such that A ⊂ Ui, L∗n(Ui\A) <1/i.

(3)⇒(1) This implication is obvious.

(1)⇔(4)⇔(5) this equivalence follows from the equivalence of the conditions (1), (2)and (3) applied to the set IRn \ A.

(1)⇒(6) If A is Lebesgue measurable then the existence of the sets F and G followsfrom the conditions (2) and (4).

(6)⇒(1) Take closed and open sets Fi, Gi such that Fi ⊂ A ⊂ Gi, Ln(Gi \ Fi) < 1/i.Then the set H =

⋂∞i=1Gi is Gδ and L∗n(H \ A) = 0. 2

The Lebesgue measure has an important property of being invariant under trans-lations i.e. if A is a translation of a Lebesgue measurable set B, then A is Lebesguemeasurable and Ln(A) = Ln(B). This property follows immediately from the definitionof the Lebesgue measure. We will show not that this property implies that there is aset which is not Lebesgue measurable.

Theorem 20 (Vitali) There is a set E ⊂ IRn which is not Lebesgue measurable.3

3The reader will easily check that the same proof applies to any translation invariant measure µsuch that the measure of the unit cube if positive and finite; see also Theorem 21.

17

Proof. We will prove the theorem for n = 1 only, but the same argument works foran arbitrary n. The idea of the proof is to find a countable family of pairwise disjointand isometric (by translation)sets whose union E contains (0, 1) and is contained in(−1, 2). If the sets in the family are isometric to a set V , then V cannot be Lebesguemeasurable, because measurability of V would imply that a number between 1 and3 (measure of E) be equal to infinite sum of equal numbers (equal to measure of V )which is impossible. It should not be surprising that the construction of the set V hasto involve axiom of choice.

For x, y ∈ (0, 1) we write x ∼ y if x − y is a rational number. Clearly ∼ is anequivalence relation and hence (0, 1) is the union of a family F of pairwise disjoint sets

[x] = y ∈ (0, 1) : x ∼ y.

It follows from the axiom of choice that there is a set V ⊂ (0, 1) which contains exactlyone element from each set in the family F . Let

E =⋃

a∈Q∩(−1,1)

Va, where Va = V + a = x+ a : x ∈ V .

It is easy to see that

• Va ∩ Vb = ∅ if a, b ∈ Q, a 6= b;

• (0, 1) ⊂ E ⊂ (−1, 2).

Suppose that V is measurable. Then each Va is measurable, L1(V ) = L1(Va) and E ismeasurable. If L1(V ) > 0, then L1(E) =∞ which is impossible and if L1(V ) = 0, thenL1(E) = 0 which is also impossible. That proves that the set V cannot be measurable.2

The Lebesgue measure is translation invariant and Ln([0, 1]n) = 1. It turns out thatthe above two properties uniquely determine Lebesgue measure. This proves that theLebesgue measure is in a sense the only natural way of measuring length, area, volumeof general sets in one, two, three,. . . dimensional Euclidean spaces.

Theorem 21 If µ is a measure on B(IRn) such that µ(a+E) = µ(E) for all a ∈ IRn,E ∈ B(IRn) and µ([0, 1]n) = 1, then µ(E) = Ln(E) on B(IRn).

Proof. According to Corollary 8 it suffices to prove that µ(U) = Ln(U) for all opensets U .

Consider an open cube Qk = (0, 2−k)n, where k is a positive integer. There are 2kn

pairwise disjoint open cubes contained in the unit cube [0, 1]n, each being a translationof Qk. Since the measure is invariant under translations we conclude that

µ(Qk) ≤ 2−kn.

18

It is easy to see that we can cover the boundary ∂Q by translations of Qk in a way thatthe sum of µ-measures of the cubes covering ∂Q goes to zero as k → ∞. This provesthat µ(∂Q) = 0.

Now [0, 1]n can be covered by 2kn cubes each being a translation of the closed cube

Qk. Since the measure is invariant under translations we conclude that

µ(Qk) ≥ 2−kn

and henceµ(Qk) = µ(Qk) = 2−kn = Ln(Qk). (16)

As we know (Theorem 17) an arbitrary open set U is a union of cubes with pairwisedisjoint interiors and sidelength 2−k (k depends on the cube). Cubes are not disjoint,but meet along the boundary which has µ-measure zero. Hence the µ(U) equals thesum of the µ-measures of the cubes which according to (16) equals Ln(U). The proofis complete. 2

Theorem 22 Hn = Ln on B(IRn).

Sketch of the proof. Observe that the Hausdorff measure Hn in IRn has the propertythat

Hn(E + a) = Hn(E) for all E ∈ B(IRn).

Therefore in order to prove that Hn = Ln on B(IRn) it suffices to prove that

Hn([0, 1]n) = 1

This innocently looking result is however very difficult and we will not prove it here. 2

Exercise. Using Theorem 22 prove that Hn = L∗n on the class of all subsets of IRn.

If s < n, then Hs is the s-dimensional measure in IRn. Not that s is not requiredto be an integer. If k < n is an integer and M ⊂ IRn is a k-dimensional manifold,then one can prove that Hk coincides on M with the natural k-dimensional Lebesguemeasure on M .4

Examples. If C ⊂ [0, 1] is the standard ternary Cantor set, then one can prove thatdimH(C) = log 2/ log 3 = 0.6309 . . . Moreover Hs(C) = 1 for s = log 2/ log 3.

The Van Koch curve is defined as the limiting curve for the following construction

4We have already seen how in the course of Advanced Calculus to measure subsets of M in thesetting of the Riemann integral. This construction can easily be generalized to the Lebesgue measureon M . We will discuss it with more details later.

19

The Van Koch curve, denoted by K is homeomorphic to the circle S1, but dimH K =log 4/ log 3 = 1, 261 . . .

Proposition 23 A set E ⊂ IRn has Lebesgue measure zero if and only if for every ε >there is a family of balls B(xi, ri)∞i=1 such that

E ⊂∞⋃i=1

B(xi, ri),∞∑i=1

rni < ε.

Proof. ⇒ Each bass B(xi, ri) is contained in a cube with sidelength 2ri and hence Ecan be covered by a sequence of cubes whose sum of volumes is

∑∞i=1(2ri)

n < 2nε.

⇐ If a set E ⊂ IRn has Lebesgue measure zero, then for every ε > 0 there is an openset U such that E ⊂ U and Ln(U) < ε. The set U can be represented as union of cubeswith pairwise disjoint interiors (Theorem 17). each cube Q of sidelength ` is containedin a ball of radius r =

√n` and hence rn = nn/2|Q|. Therefore E can be covered by a

sequence of balls B(xi, ri)∞i=1 such that∑∞

i=1 rni < nn/2ε. 2

Definition. We say that a function between metric spaces f : (X, d) → (Y, ρ) isLipschitz continuous if there is a constant L > 0 such that

ρ(f(x), f(y)) ≤ Ld(x, y) for all x, y ∈ X.

In this case we say that f is L-Lipschitz continuous and call L Lipschitz constant if f .

Proposition 24 If f : IRn → IRn is Lipschitz continuous and E ⊂ IRn has Lebesguemeasure zero, then f(E) has Lebesgue measure zero too.

Proof. It follows from Proposition 23 and the fact that the L-Lipschitz image of a ballof radius r is contained in a ball of radius Lr. 2

Exercise. Let f : IRn → IRn be a homeomorphism. Prove that f(A) is Borel if andonly if A is Borel.

Although the theory of Lebesgue measure seems to work very well something veryimportant is missing: we haven’t proved so far that if Q is a unit cube whose sides arenot parallel to coordinate axes, then its measure equals 1. Indeed we considered onlyintervals with sides parallel to coordinate directions. Fortunately this important factis contained in a result that we will prove now.

20

Theorem 25 If L : IRn → IRn is a nondegenerate linear transformation with thematrix A (detA 6= 0), then L(E) ⊂ IRn is a Borel (Lebesgue measurable) set if andonly if E ⊂ IRn is Borel (Lebesgue measurable) set. Moreover

Ln(L(E)) = | detA|Ln(E). (17)

Proof. Since L is a homeomorphism it preserves the class of Borel sets (see Exercise).Since both L and L−1 are Lipschitz continuous L preserves the class of sets of measurezero (Proposition 24). Therefore L preserves the class of Lebesgue measurable sets(because each Lebesgue measurable set is the union of a Borel set and a set of measurezero). It follows from Proposition 24 that the formula (17) is satisfied for sets of measurezero. Therefore it remains to prove (24) for Borel sets.

Define a new measure µ(E) = Ln(L(E)). Clearly µ is a measure on B(IRn) that istranslation invariant. Let µ([0, 1]n) = a > 0. Hence the measure a−1µ satisfies all theassumptions of Theorem 21 and therefore a−1µ(E) = Ln(E), so we have

Ln(L(E)) = aLn(E) for all E ∈ B(IRn).

It remains to prove that a = | detA|. Clearly a = a(A) : GL (n)→ (0,∞) is a functiondefined on the class of invertible matrices. We have

a(A1A2) = a(A1)a(A2). (18)

Indeed, if A1, A2 are matrices representing linear transformations L1 and L2, then

a(A1A2)Ln(E) = Ln(L1(L2(E)) = a(L1)Ln(L2(E)) = a(L1)a(L2)Ln(E).

Alsoa(snI) = sn, s > 0. (19)

Here I is the identity matrix. This property is quite obvious and we leave a formalproof to the reader. Now it remains to prove the following fact.

Lemma 26 Let a : GL (n)→ (0,∞) be a function with properties (18) and (19). Thena(A) = | detA|.

Proof. Let Ai(s) be the diagonal matrix with −s on the ith place and s on all otherplaces on the diagonal. Since Ai(s)

2 = s2I we have a(Ai(s))2 = a(Ai(s)

2) = s2n andhance a(Ai(s)) = |sn| = | detAi(s)|.

For k 6= l let Bkl(s) = [aij] be the matrix such that akl = s, aii = 1, i = 1, 2, . . . andall other entries equal zero. Multiplication by the matrix Bkl(s) from the right (left) isequivalent to adding kth column (`th row) multiplied by s to `th column (kth row).

It is well known from linear algebra (and easy to prove) that applying such opera-tions to any nonsingular matrix A it can be transformed to a matrix of the form tI or

21

An(t). Since multiplication by Bkl(s) does not change determinant, t = | detA|1/n. Itremains to prove that a(Bkl(s)) = 1. Since Bkl(−s) = Ak(1)Bkl(s)Ak(1) we havethat a(Bkl(−s)) = a(Bkl(s)). On the other hand Bkl(s)Bkl(−s) = I and hencea(Bkl(s))

2 = 1, so a(Bkl(s)) = 1. The proof of the lemma and hence the proof ofthe theorem is complete. 2

2 Integration

2.1 Measurable functions.

Definition. Let (X,M) be a measurable space and Y a metric space. We say that afunction f : X → Y is measurable if

f−1(U) ∈M for every open set U ⊂ Y .

If E ∈M is a measurable set, then we say that f : E → Y is measurable if f−1(U) ∈M

for every open U ⊂ Y .

Clearly, f : E → Y is measurable if it can be extended to a measurable functionf : X → Y , but there is also another point of view.

ME = A ⊂ E : A ∈M = B ∩ E : B ∈M

is a σ-algebra and measurability of f : E → Y is equivalent to measurability of f withrespect to ME.

Theorem 27 Let (X,M) be a measurable space, f : X → Y a measurable functionand g : Y → Z a continuous function. Then the function g f : X → Z is measurable.

Proof. For every open set U ⊂ Z we have

(g f)−1(U) = f−1(g−1(U)︸︷︷︸open in Y

) ∈M.


Theorem 28 Let (X,M) be a measurable space and Y a metric space. If the functionsu, v : X → IR are measurable and Φ : IR2 → Y is continuous, then the function

h(x) = Φ(u(x), v(x)) : X → Y

is measurable.

22

Proof. Put f(x) = (u(x), v(x)), f : X → IR2. According to Theorem 27 it suffices toprove that f is measurable. Let R = (a, b)× (c, d) be an open rectangle. Then

f−1(R) = f−1((a, b)× (c, d)) = u−1((a, b)) ∩ u−1((c, d)) ∈M.

Every open set U ⊂ IR2 is a countable union of open rectangles

U =∞⋃i=1

Ri

and hence

f−1(U) =∞⋃i=1

f−1(Ri) ∈M.


This theorem has many applications.

(a) If f = u+ iv : X → C, where u, v : X → IR are measurable, then f is measurable.

Indeed, we take Φ(u, v) = u+ iv.

(b) If f = u + iv : X → C is measurable, then the functions u, v, |f | : X → IR aremeasurable.

Indeed, we take Φ1(u+ iv) = u, Φ2(u+ iv) = v, Φ3(u+ iv) = |u+ iv|.

(c) If f, g : X → C are measurable, then the functions f+g, fg : X → C are measurable.

Indeed, we take Φ1 : C × C → C, Φ1(z1, z2) = z1 + z2 and Φ2 : C × C → C,Φ2(z1, z2) = z1z2.

(d) A set E ⊂ X is measurable if and only if the characteristic function if E

χE(x) =

1 if x ∈ E,0 if x 6∈ E,

is measurable.

Indeed, this follows from the fact that for any open set U ⊂ IR

(χE)−1(U) =

∅ if 0 6∈ U , 1 6∈ U ,E if 0 6∈ U , 1 ∈ U ,

X \ E if 0 ∈ U , 1 6∈ U ,X if 0 ∈ U , 1 ∈ U ,

23

(e) If f : X → C is a measurable function, then there is a measurable function α : X →C such that |α| = 1 and f = α|f |.

The set E = x : f(x) = 0 is measurable because

E =∞⋂i=1

f−1(B(0,

1

i)︸︷︷︸

disc of radius 1/i

).

Hence the function h(x) = f(x)+χE(x) is measurable and h(x) 6= 0 for all x. Thereforeh is a measurable function as a function h : X → C \ 0. Since ϕ : C \ 0 → C,ϕ(z) = z/|z| is continuous we conclude that

α(x) = ϕ(f(x) + χE(x))

is measurable. If x ∈ E, then α(x) = 1 and if x 6∈ E, then α(x) = f(x)/|f(x)|.Therefore f(x) = α(x)|f(x)| for all x ∈ X.

Definition. Let X, Y be two metric spaces. We say that f : X → Y is a Borelmapping if it is measurable with respect to the σ-algebra of Borel sets in X i.e.

f−1(U) ∈ B(X) for every open set U ⊂ Y .

Clearly every continuous mapping is Borel.

Theorem 29 Let (X,M) be a measurable space and Y a metric space. Let also f :X → Y be a measurable mapping.

(a) If E ⊂ Y is a Borel set, then f−1(E) ∈M.

(b) If Z is another metric space and g : Y → Z is a Borel mapping, then gf : X → Zis measurable.

Proof. (a) Let R = A ⊂ Y | f−1(A) ∈ M. Clearly R contains all open sets. If weprove that R is a σ-algebra, we will have that B(Y ) ⊂ R which is the claim.

• Y ∈ R

Indeed, f−1(Y ) = X ∈M.

• If A ∈ R, then Y \ A ∈ R.

Indeed, f−1(Y \ A) = X \ f−1(A) ∈M.

24

• If A1, A2, . . . ∈ R, then⋃∞

i=1Ai ∈ R.

Indeed, f−1(⋃∞

i=1Ai) =⋃∞

i=1 f−1(Ai) ∈ R. The above three properties imply that R

is a σ-algebra.

(b) If U ⊂ Z is open, then g−1(U) ⊂ Y is Borel and hence (gf)−1(U) = f−1(g−1(U)) ∈M by (a). 2

Definition. Let (X,M) be a measurable space. Let IR = IR ∪ −∞,∞. We saythat a function f : X → IR is measurable if f−1(U) is measurable for every open setU ⊂ IR and if the sets f−1(+∞), f−1(−∞) are measurable. Similarly for E ∈ M wedefine measurable functions f : E → IR.

Exercise. Let (X,M) be a measurable space. Prove that a function f : X → IR ismeasurable if and only if f−1((a,∞]) ∈M for all a ∈ IR.

Let an be a sequence in IR. For k = 1, 2, . . . put

bk = supak, ak+1, ak+2, . . . (20)

andβ = infb1, b2, . . . . (21)

We call β upper limit of an and write

β = lim supn→∞

an

Clearly bk is a decreasing sequence, so β = limk→∞ bk. It is easy to see that thereis a subsequence ani

of an such that limi→∞ ani= β. Moreover β is the largest

possible value for limits of all subsequences of an.

The lower limit is defined by

lim infn→∞

an = − lim supn→∞

(−an). (22)

Equivalently the lower limit can be defined by interchanging sup and inf in (20) and(21). The lower limit equals to the lowest possible value for limits of subsequences ofan.

For a sequence of functions fn : X → IR we define new functionssupn fn, lim supn fn : X → IR by(

supnfn

)(x) = sup

nfn(x) for every x ∈ X,

(lim sup

nfn

)(x) = lim sup

nfn(x) for every x ∈ X.

25

Similarly we can define functions infn fn and lim infn→∞ fn.

If f(x) = limn→∞ fn(x) for every x ∈ X, then we say that f is a pointwise limit offn.

Theorem 30 Suppose that fn : X → IR is a sequence of measurable functions. Thenthe functions supn fn, infn fn, lim supn→∞ fn, lim infn→∞ fn are measurable.

Proof. Since (sup

nfn

)−1

((a,∞]) =∞⋃

n=1

f−1n ((a,∞]) ∈M,

measurability of supn fn follows from the exercise. By a similar argument we provethat the function infn fn is measurable. The two facts imply measurability of lim sup fn

becauselim sup

n→∞fn = inf

k≥1sup

i≥kfi.

Now measurability of lim infn fn is an obvious consequence of the equality

lim infn→∞

fn = − lim supn→∞

(−fn).


Corollary 31 The limit of every pointwise convergent sequence of functions fn : X →C (or fn : X → IR) is measurable.

Corollary 32 If f, g : X → IR are measurable functions, then the functions maxf, g,minf, g are measurable. In particular the fuctions

f+ = maxf, 0, f− = −minf, 0

are measurable.

The functions f+ and f− are called positive part and negative part of f . Note thatf = f+ − f−, |f | = f+ + f−.

Corollary 31 has the following generalization to the case of metric space valuedmappings.

Theorem 33 Let fn : X → Y be a sequence of measurable mappings with values ina separable metric space. If fn(x) → f(x) for every x ∈ X, then f : X → Y ismeasurable.

26

Proof. Let yi∞i=1 be a countable dense set. Then the functions di : Y → IR, di(y) =d(yi, y) are continuous. Hence di fn : X → IR are measurable. Since di fn → di f ,the functions di f : X → IR are measurable too by Corollary 31. In particular thesets

x : d(yi, f(x)) ≥ d(yi, F ) = (di f)−1([d(yi, F ),∞))

are measurable. To prove measurability of f it suffices to prove measurability of f−1(F )for every closed set F and it follows from the equality

f−1(F ) =∞⋂i=1

x : d(yi, f(x)) ≥ d(yi, F ) . (23)

To prove this equality observe that if x ∈ f−1(F ), then f(x) ∈ F and henced(yi, f(x)) ≥ d(yi, F ) for every i, so x belongs to the right hand side of (23). Onthe other hand if x 6∈ f−1(F ), then f(x) 6∈ F and hence there is i such thatd(yi, f(x)) < d(yi, F ) (because yi∞i=1 is a dense subset of Y and so we can find yi

arbitrarily close to f(x)), therefore x cannot belong to the right hand side of (23). 2

Definition. Let (X,M) be a measurable space. A measurable function s : X → C iscalled simple function if it has only a finite number of values. Simple functions can berepresented in the form

s =n∑

i=1

αiχAi,

where αi 6= αj for i 6= j and Ai are disjoint measurable sets.

Theorem 34 If f : X → [0,∞] is measurable, then there is a sequence of simplefunctions sn on X such that

1. 0 ≤ s1 ≤ s2 ≤ . . . ≤ f ;

2. limn→∞ sn(x) = f(x) for every x ∈ X.

If in addition f is bounded, then sn converges to f uniformly.

Proof. It is easy to see that the sequence

sn(x) =

n if f(x) ≥ n,k2k if k

2n ≤ f(x) < k+12n ≤ n,

has all properties that we need. 2

27

2.2 Integral

In the integration theory we will assume that 0 · ∞ = 0.

Definition. If s : X → [0,∞) is a simple function of the form

s =n∑

i=1

αiχAi,

where αi 6= αj for i 6= j and Ai are pairwise disjoint measurable sets, then for anyE ∈M we define ∫

E

s dµ =n∑

i=1

αiµ(Ai ∩ E) .

According to our assumption 0 · ∞ = 0, so 0 · µ(Ai ∩ E) = 0, even if µ(Ai ∩ E) =∞.

If f : X → [0,∞] is measurable, and E ∈ M, we define the Lebesgue integral of fover E by ∫

E

f dµ = sup

∫E

s dµ,

where the supremum is over all simple functions s such that 0 ≤ s ≤ f .

(a) If 0 ≤ f ≤ g, then∫

Ef dµ ≤

∫Eg dµ.

(b) If A ⊂ B and f ≥ 0, then∫

Af dµ ≤

∫Bf dµ.

(c) If f ≥ 0 and c is a constant 0 ≤ c <∞, then∫

Ef dµ = c

∫Ef dµ.

(d) If f(x) = 0 for all x ∈ E, then∫

Ef dµ = 0, even if µ(E) =∞.

(e) If µ(E) = 0, then∫

Ef dµ = 0, even if f(x) =∞ for all x ∈ E.

(f) If f ≥ 0, then∫

Ef dµ =

∫XχEf dµ.

In the proof of the next two theorems we will need the following lemma.

Lemma 35 Let s and t be nonnegative measurable simple functions on X. Then thefunction

ϕ(E) =

∫E

s dµ, E ∈M

defines a measure in M. Moreover∫X

(s+ t) dµ =

∫X

s dµ+

∫X

t dµ . (24)

28

Proof. Since ϕ(∅) = 0, it remains to prove that ϕ is countably additive. If E1, E2, . . .are disjoint measurable sets, then

ϕ

(∞⋃

k=1

Ek

)=

n∑i=1

αiµ

(∞⋃

k=1

Ek ∩ Ai

)=

n∑i=1

αi

∞∑k=1

µ(Ek ∩ Ai)

=∞∑

k=1

n∑i=1

αiµ(Ek ∩ Ai) =∞∑

k=1

ϕ(Ek) .

This completes the proof of the first part. Let now

s =k∑

i=1

αiχAi, t =

m∑j=1

βjχBj

be a representation of the simple functions as in the definition, then for Eij = Ai ∩Bj

we have ∫Eij

s+ t dµ = (αi + βj)µ(Eij) =

∫Eij

s dµ+

∫Eij

t dµ .

Since X is a union of disjoint sets Eij we conclude (24) from the first part of the lemma.2

The following theorem is one of the most important results of the theory of inte-gration.

Theorem 36 (Lebesgue monotone convergence theorem) Let fn be a se-quence of measurable functions on X such that:

1. 0 ≤ f1 ≤ f2 ≤ . . . ≤ ∞ for every x ∈ X;

2. limn→∞ fn(x) = f(x) for every x ∈ X.

Then f is measurable and

limn→∞

∫X

fn dµ =

∫X

f dµ

Proof. Measurability of f follows from Corollary 31. The sequence∫

Xfn dµ is increasing

and hence it has a limit α = limn→∞∫

Xfn dµ. Since fn ≤ f , we have

∫Xfn dµ ≤

∫Xf dµ

and hence α ≤∫

Xf dµ . Let 0 ≤ s ≤ f be a simple function. Fix a constant 0 < c < 1

and defineEn = x : fn(x) ≥ cs(x), n = 1, 2, 3, . . .

It is easy to see that E1 ⊂ E2 ⊂ . . . and⋃∞

n=1En = X (because fn(x)→ f(x) for everyx ∈ X.) Hence ∫

X

fn dµ ≥∫

En

fn dµ ≥ c

∫En

s dµ . (25)

29

Since ϕ(E) =∫

Es dµ is a measure and E1 ⊂ E2 ⊂ . . . we conclude (Theorem 3(f)) that∫

En

s dµ = ϕ(En)→ ϕ

(∞⋃

n=1

En

)= ϕ(X) =

∫X

s dµ .

Thus after passing to the limit in (25) we obtain α ≥ c∫

Es dµ and hence∫

X

f dµ ≥ α ≥∫

X

s dµ ,

because 0 < c < 1 was arbitrary. Taking supremum over all simple functions 0 ≤ s ≤ fyields ∫

X

f dµ ≥ α ≥∫

X

f dµ ,

i.e.∫

Xf dµ = α and the theorem follows. 2

Theorem 37 If fn : X → [0,∞] is measurable, for n = 1, 2, 3, . . . and

f(x) =∞∑

n=1

fn(x) for all x ∈ X,

then f is measurable and ∫X

f dµ =∞∑i=1

∫X

fn dµ .

Proof. The function f is a limit of measurable functions, so it is measurable. Let siand ti be increasing sequences of simple nonnegative function such that si → f1,ti → f2 (Theorem 34). Then si + ti → f1 + f2 and hence (24) combined with themonotone convergence theorem gives∫

X

f1 + f2 dµ =

∫X

f1 dµ+

∫X

f2 dµ .

Applying an induction argument we obtain∫X

(f1 + f2 + . . .+ fN) dµ =N∑

n=1

∫X

fn dµ .

Since gN = f1 + . . .+fN is an increasing sequence of measurable functions and gN → fthe theorem follows from the monotone convergence theorem. 2

Theorem 38 (Fatou’s lemma) If fn : X → [0,∞] is measurable, for n = 1, 2, 3, . . .,then ∫

X

(lim infn→∞

fn

)dµ ≤ lim inf

n→∞

∫X

fn dµ .

30

Exercise. Construct a sequence of measurable functions fn : IR→ [0,∞] such that theinequality in Fatou’s lemma is sharp.

Proof of the theorem. Let gn = inffn, fn+1, fn+2, . . .. Then gn ≤ fn and hence∫X

gn dµ ≤∫

X

fn dµ . (26)

Since 0 ≤ g1 ≤ g2 ≤ . . ., all the functions gn are measurable (Theorem 30) andgn → lim infk→∞ fn, inequality (26) combined with the monotone convergence theoremyields the result. 2

Theorem 39 Let f : X → [0,∞] be measurable. Then

ϕ(E) =

∫E

f dµ for E ∈M

defines a measure in M. Moreover∫X

g dϕ =

∫X

gf dµ (27)

for every measurable function g : X → [0,∞].

Proof. Clearly ϕ(∅) = 0. To prove that ϕ is a measure it remains to prove countableadditivity. Let E1, E2, . . . ∈M be disjoint sets and E =

⋃∞n=1En. Then obviously

χEf =∞∑

n=1

χEnf

and

ϕ(E) =

∫X

χEf dµ =

∫X

∞∑n=1

χEnf dµ

=∞∑

n=1

∫X

χEnf dµ =∞∑

n=1

ϕ(En) ,

by Theorem 37. To prove (27) observe that it is true if g = χE, E ∈M. Hence it is alsotrue for simple functions and the general case follows from the monotone convergencetheorem combined with Theorem 34. 2

Definition. Let (X,M, µ) be a measure space. The space L1(µ) consists of allmeasurable functions f : X → C for which ‖f‖1 =

∫X|f | dµ < ∞. Elements of L1(µ)

are called Lebesgue integrable functions or summable functions.

31

If f = u + iv, then the functions u+, u−, v+, v− are measurable and Lebesgueintegrable as bounded by |f |. For a set E ∈M we define∫

E

f dµ =

∫E

u+ dµ−∫

E

u− dµ+ i

(∫E

v+ dµ−∫

E

v− dµ

).

If f : X → IR is measurable and each of the integrals∫

Ef+ dµ,

∫Ef− dµ is finite, then

we also define ∫E

f dµ =

∫E

f+ dµ−∫

E

f− dµ .

Theorem 40 Suppose that f, g ∈ L1(µ). If α, β ∈ C, then αf + βg ∈ L1(µ) and∫X

(αf + βg) dµ = α

∫X

f dµ+ β

∫X

g dµ .

We leave the proof as a simple exercise.

This theorem says that L1(µ) is a linear space and f 7→∫

Xf dµ is a linear functional

in L1(µ).

Similarly we define L1(µ, IRn) as the class of all measurable mappings f =(f1, f2, . . . , fn) : X → IRn such that

∫X|f | dµ <∞ and then we set∫

X

f dµ =

(∫X

f1 dµ, . . . ,

∫X

fn dµ

).

Theorem 41 If f ∈ L1(µ, IRn), then∣∣∣∣∫X

f dµ

∣∣∣∣ ≤ ∫X

|f | dµ .

Proof. Let α = (α1, . . . , αn) ∈ IRn be a unit vector such that |∫

Xf dµ| = 〈α,

∫Xf dµ〉.

Then ∣∣∣∣∫X

f dµ

∣∣∣∣ = 〈α,∫

X

f dµ〉 =n∑

i=1

αi

∫X

fi dµ

=

∫X

n∑i=1

αifi dµ =

∫X

〈α, f〉 dµ ≤∫

X

|f | dµ

by Schwarz inequality. 2

Theorem 42 (Lebesgue’s dominated convergence theorem) Suppose fn is asequence of complex valued functions on X such that the limit f(x) = limn→∞ fn(x)exists for every x ∈ X. If there is a function g ∈ L1(µ) such that

|fn(x)| ≤ g(x) for every x ∈ X and all n = 1, 2, . . .,

32

then f ∈ L1(µ),

limn→∞

∫X

|f − fn| dµ = 0

and

limn→∞

∫X

fn dµ =

∫X

f dµ.

Proof. Clearly f is measurable and |f | ≤ g. Hence f ∈ L1(µ). It is easy to see that2g − |f − fn| ≥ 0 and hence Fatou’s lemma yields∫

X

2g dµ ≤ lim infn→∞

∫X

(2g − |f − fn|) dµ

=

∫X

2g + lim infn→∞

(−∫

X

|f − fn| dµ)

=

∫X

2g − lim supn→∞

∫X

|f − fn| dµ .

The integral∫

X2g dµ is finite and hence

lim supn→∞

∫X

|fn − f | dµ ≤ 0 .

This immediately implies

limn→∞

∫X

|fn − f | dµ = 0 .

Now ∣∣∣∣∫X

fn dµ−∫

X

f dµ

∣∣∣∣ ≤ ∫X

|fn − f | dµ→ 0 ,

so

limn→∞

∫X

fn dµ =

∫X

f dµ .


2.3 Almost everywhere

Definition. If there is a set N ∈M of measure zero, µ(N) = 0 such that a propertyP (x) is satisfied for all x ∈ X \ N , then we say that the property P (x) is satisfiedalmost everywhere (a.e.)

For example fn → f a.e. means that there is a set N ∈ M, µ(N) = 0 such thatfn(x)→ f(x) for all x ∈ X \N .

33

If f = g a.e., then we write f ∼ g. It is easy to see that ∼ is an equivalence relation(f ∼ g, g ∼ h⇒ f ∼ h follows from the fact that the union of two sets of measure zerois a set of measure zero). Note that if f ∼ g, then∫

E

f dµ =

∫E

g dµ for every X ∈M.

Therefore from the point of view of the integration theory, functions which are equala.e. cannot be distinguished. Moreover if

∫Ef dµ =

∫Eg dµ for all E ∈M, then f = g

a.e. This follows from part (b) of Theorem 43 below applied to f − g. Thus twofunctions f and g cannot be distinguished in the integration theory if and only if theyare equal a.e. Therefore it is natural to identify such functions.

Theorem 43

(a) Suppose f : X → [0,∞] is measurable, E ∈M and∫

Ef dµ = 0, then f = 0 a.e.

in E.

(b) Suppose f ∈ L1(µ) and∫

Ef dµ = 0 for every E ∈M. Then f = 0 a.e. in X.

Proof. (a) Suppose An = x ∈ E : f(x) ≥ 1/n. Then

1

nµ(An) ≤

∫An

f dµ ≤∫

E

f dµ = 0.

Hence µ(An) = 0 and therefore the set x ∈ E : f(x) > 0 =⋃∞

n=1An has measurezero.

(b) Let = u+ iv. Define E = x ∈ X : u(x) ≥ 0. Then

0 = real part of

∫E

f dµ =

∫E

u dµ =

∫E

u+ dµ,

and hence u+ = 0 a.e. by (a). Similarly u−, v+, v− = 0 a.e. Accordingly, f = 0 a.e. 2

Recall that a measure µ is complete if every subset of a set of measure zero ismeasurable (and hence has measure zero). All the measures obtained through theCaratheodory construction are complete (Proposition 4), e.g. the Hausforff measureand the Lebesgue measure are complete. However, we can assume that any measureis complete, because we can add all subsets of sets of measure zero to the σ-algebra.More precisely one can prove.

Theorem 44 Let (X,M, µ) be a measure space, let M∗ be the collection of all E ⊂ Xfor which there exist sets A,B ∈M such that A ⊂ E ⊂ B and µ(B \A) = 0 and defineµ(E) = µ(A) in this situation. Then M∗ is a σ-algebra and µ is a complete measureon M∗.

34

For a proof, see Rudin page 28.

Therefore, whenever needed, we can assume that the measure is complete.

Remark. If µ is a measure on B(IRn) and there is an uncountable set E ∈ B(IRn)such that µ(E) = 0, then µ cannot be complete, because one can prove that thecardinality of all Borel sets in B(IRn) is the same as the cardinality of real numbersand the cardinality of all subsets of E is the same as the cardinality of all subsets of realnumbers which is bigger than the cardinality of real numbers. Hence in order to makethe measure µ complete we have to enlarge B(IRn) to a bigger σ-algebra by includingalso sets which are not Borel. This is the case of the Lebesgue measure. The σ-algebraof Lebesgue measurable sets is larger, than the σ-algebra of Borel sets.

Definition. Suppose now that µ is a complete measure. If a function with values ina metric space Y is defined only at almost all points of X, i.e.

f : X \N → Y, µ(N) = 0 ,

then we say that f is measurable if there is y0 ∈ Y such that

f(x) =

f(x) if x ∈ X \N ,y0 if x ∈ N ,

is measurable. In such a situation we say that f : X → Y is a measurable functiondefined a.e.

Note that it follows from the assumption that µ is complete that f can be defined

in N in an arbitrary way and still the resulting function ˜f is measurable and ˜f ∼ f .

In particular functions in L1(µ) need to be defined a.e. only. L1(µ) is a linear space.Since we identify functions that are equal a.e. we identify L1(µ) with the quotient spaceL1(µ)/ ∼ (check that L1(µ)/ ∼ is a linear space). Therefore f ∈ L1(µ) satisfies f = 0 inL1(µ) if and only if f = 0 a.e., if and only if ‖f‖1 =

∫X|f | dµ = 0 (see Theorem 43(a)).

In all the previously discussed convergence theorems (Theorem 36, Theorem 37,Theorem 38, Theorem 42) we can replace the requirement of everywhere convergenceby the a.e. convergence. For example the Lebesgue dominated convergence theoremcan be formulated as follows.

Theorem 45 (Lebesgue dominated convergence theorem) Suppose that fn isa sequence of complex valued functino on X defined a.e. If the limit f(x) =limn→∞ fn(x) exists a.e. and if there is a function g ∈ L1(µ) such that

|fn(x)| ≤ g(x) a.e., for all n = 1, 2, 3, . . .

then f ∈ L1(µ),

limn→∞

∫X

|f − fn| dµ = 0

35

and

limn→∞

∫X

fn dµ =

∫X

f dµ .

Not surprisingly, this theorem is an easy consequence of the previous version of theLebesgue dominated convergence theorem. We leave details to the reader.

Recall that if Y is a metric space, then a mapping f : IRn → Y is Lebesguemeasurable if f−1(U) is a Lebesgue measurable set for every open U ⊂ Y . Since theσ-algebra of Lebesgue measurable sets is larger, than that of Borel sets, the mappingf need not be Borel. I turns out, however, that f equals a.e. to a Borel mapping, atleast when Y is separable.

Theorem 46 If Y is a separable metric space and f : IRn → Y is Lebesgue measurable,then there is a Borel mapping g : IRn → Y such that f = g a.e.

Proof. Since Y is separable, there is a countable family of open sets Ui∞i=1 in Y suchthat every open set in Y is a union of open sets from the family. Indeed, it suffices totake the family of balls with rational length radii and centers in a countable dense setin Y .

For i = 1, 2, . . . let Mi ⊂ IRn be a Fσ set such that

Mi ⊂ f−1(Ui), Ln(f−1(Ui) \Mi) = 0 ,

see Theorem 19. Clearly the set E =⋃∞

i=1(f−1(Ui) \Mi) has Lebesgue measure zero

and hence there is a Gδ set H such that E ⊂ H, Ln(H) = 0 (Theorem 19). Fix y0 ∈ Yand define

g(x) =

f(x) if x 6∈ H,y0 if x ∈ H.

Clearly f = g a.e. and it remains to prove that the mapping g is Borel. It is easy tosee that5 if y0 6∈ Ui, then g−1(Ui) = Mi \ H and if y0 ∈ Ui, then g−1(Ui) = Mi ∪ H.In each case g−1(Ui) is a Borel set. Since every open set U ⊂ Y can be represented asU =

⋃∞j=1 Uij , the set

g−1(U) =∞⋃

j=1

g−1(Uij )

is Borel. 2

5A good picture is useful here.

36

2.4 Theorems of Lusin and Egorov

Theorem 47 (Lusin) Let X be a metric space and µ a measure in B(X) such thatX is a union of countably many open sets of finite measure. If f : X → Y is a Borelmapping with values in a separable metric space, then for every ε > 0 there is a closedset F ⊂ X such that µ(X \ F ) < ε and f |F is continuous.

Proof. Let Ui∞i=1 be a countable family of open sets such that every open set in Y isa union of open sets from the family. It follows from Theorem 7 that for every i thereis an open set Vi ⊂ X such that6

f−1(Ui) ⊂ Vi, µ(Vi \ f−1(Ui)) < ε2−i−1 .

Let

E =∞⋃i=1

Vi \ f−1(Ui).

Clearly µ(E) < ε/2. We will prove that the mapping g = f |X\E is continuous. Observethat7

g−1(Ui) = Vi ∩ (X \ E) . (28)

Indeed, g−1(Ui) ⊂ Vi ∩ (X \ E), but on the other hand

Vi ∩ (X \ E) ⊂ Vi ∩ (X \ (Vi \ f−1(Ui))) = f−1(Ui) ,

soVi ∩ (X \ E) ⊂ f−1(Ui) ∩ (X \ E) = g−1(Ui) .

In order to prove that g is continuous it suffices to prove that for every open set U ⊂ Ythere is an open set V ⊂ X such that8 g−1(U) = V ∩ (X \E). Let U =

⋃∞j=1 Uij . Then

(28) gives g−1(U) = (⋃∞

j=1 Vij )∩ (X \E) which proves continuity of g in X \E. The setE need not be closed, but it follows from Theorem 7 that there is an open set G ⊂ Xsuch that E ⊂ G and µ(G \ E) < ε/2. Now f |F is continuous, where F = X \G, F isclosed and µ(X \ F ) = µ(G) = µ(E) + µ(G \ E) < ε. 2

As a corollary we will prove the following characterization of Lebesgue measurablefunctions also known as the Lusin theorem.

Let E ⊂ IRn be a Lebesgue measurable set. Recall that f : E → IR is Lebesguemeasurable if f−1(U) is Lebesgue measurable for every open U ⊂ IR (f−1(U) is a subsetof E).

6Actually to prove this fact we need first to split f−1(Ui) into sets of finite measure and then applyTheorem 7 to each such set.

7It is easy to see if you make an appropriate picture, but a formal proof is enclosed below.8Because the class of open sets in the metric space X \ E is exactly the class of sets of the form

V ∩ (X \ E), where V ⊂ X is open.

37

Theorem 48 (Lusin) Let E ⊂ IRn be Lebesgue measurable. Then a function f : E →IR is Lebesgue measurable if and only if for every ε > 0 there is a closed set F ⊂ Esuch that f |F is continuous and Ln(E \ F ) < ε.

Proof. Since f equals a.e. to a Borel function necessity follows from Theorem 47. Toprove sufficiency it suffices to show that for every a ∈ IR the set Ea = x : f(x) ≥ a isLebesgue measurable (this is a variant of the exercise from page 25.) Let F be a closedset such that f |F is continuous and Ln(E \ F ) < ε. Then the set F ′ = x : (f |F )(x) ≥a = Ea ∩ F is closed and L∗n(Ea \ F ′) ≤ Ln(E \ F ) < ε, hence measurability of Ea

follows from Theorem 19. 2

In the next theorem µ is a measure on an arbitrary set X (not necessarily a metricspace), but it is crucial that µ(X) <∞.

Theorem 49 (Egorov) Let µ be a finite measure on a set X. Let fn : X → Y be asequence of measurable functions with values in a separable metric space. If fn → fa.e., then f is measurable and for every ε > 0 there is a measurable set E ⊂ X suchthat µ(X \ E) < ε and fn ⇒ f uniformly on E.

Proof. Measurability of f follows from Theorem 33. Let d be the metric in Y . Define

Ci,j =∞⋃

n=j

x : d(fn(x), f(x)) ≥ 2−i .

We will prove that this set is measurable. First observe that the mapping F = (fn, f) :X → Y ×Y is measurable. Indeed, if Ui∞i=1 is a collection of open sets in Y such thatevery open set in Y is a countable union of open sets from the family, it easily followsthat every open set in Y × Y is a countable union of open sets of the form Ui × Uj.Therefore to prove measurability of F it suffices to observe that each set

F−1(Ui × Uj) = f−1n (Ui) ∩ f−1(Uj)

is measurable. Since the function d : Y × Y → IR is continuous, the composed mapd F : X → IR is measurable and hence

x : d(fn(x), f(x)) ≥ 2−i = F−1(y1, y2) ∈ Y × Y : d(y1, y2) ≥ 2−i= F−1(d−1([2−i,∞)) = (d F )−1([2−i,∞))

is measurable. That proves measurability of the sets Ci,j. Now observe that Ci,j is adecreasing sequence of sets with respect to j, µ(Ci,1) ≤ µ(X) <∞ and hence

µ(Ci,j)→ µ(∞⋂

k=1

Ci,k) = 0 as j →∞ (29)

38

by Theorem 3(g). Indeed, this intersection has measure zero because if x ∈⋂∞

k=1Ci,k,then there is a subsequence fnj

(x) such that d(fnj(x), f(x)) ≥ 2−i for all j, which can be

true only for x in a set of measure zero. Now it follows from (29) that for every i thereis an integer J(i) such that µ(Ci,J(i)) < ε2−i. It is easy to see that µ(

⋃∞i=1Ci,J(i)) < ε

and that fn converges uniformly to f on E = X \⋃∞

i=1Ci,J(i), because for every id(fn(x), f(x)) < 2−i for all n ≥ J(i) and all x ∈ E. 2

Again if X is a metric space and µ is a finite measure of B(X) we can assume thatthe set E in the above theorem is closed, see Theorem 7.

Exercise. Show an example that if µ(X) = ∞, then the claim of Egorov’s theoremmay be false.

2.5 Convergence in measure

Definition. We say that a sequence of measurable functions fn : X → IR convergesin measure to a measurable function f : X → IR if for every ε > 0

limn→∞

µ(x : |fn(x)− f(x)| ≥ ε) = 0.

We denote convergence in measure by fnµ→ f . As usual we assume that the functions

fn and f are defined a.e.

Theorem 50 (Lebesgue) If µ(X) < ∞ and a sequence of measurable functions fn :

X → IR converges to f : X → IR a.e., then it converges in measure, fnµ→ f .

Proof. Given ε > 0, letEi = x : |fi(x)− f(x)| ≥ ε .

The sequence of sets⋃∞

i=nEi is a decreasing sequence of sets of finite measure andhence

µ

(∞⋃

i=n

Ei

)→ µ

(∞⋂

n=1

∞⋃i=n

Ei

)= 0.

The last set has measure zero, because if x ∈⋃∞

i=nEi for every n, then x belongs toinfinitely many Ei’s so there is a subsequence fni

such that |fni(x) − f(x)| ≥ ε which

is true only on a set of measure zero. Hence

µ(x : |fn(x)− f(x)| ≥ ε) = µ(En) ≤ µ(∞⋃

i=n

Ei)→ 0,

which proves convergence in measure. 2

39

Theorem 51 (Riesz) If fnµ→ f , then there is a subsequence fni

such that fni→ f

a.e.

Remark. In this theorem measure of the space can be infinite, µ(X) =∞.

Proof. For every i there is ni such that

µ(x : |fni(x)− f(x)| ≥ 1

i) ≤ 1

2i.

We can assume that n1 < n2 < n3 . . . Let

Fk = X \∞⋃

i=k

x : |fni(x)− f(x)| ≥ 1

i .

Clearly µ(X \ Fk) ≤∑∞

i=k 2−i = 2−k+1 and hence µ(X \⋃∞

k=1 Fk) = 0. We willprove that fni

→ f pointwise on⋃∞

k=1 Fk. To this end it suffices to show that fni

converges to f pointwise on every set Fk, but we will actually show a stronger factthat fni

⇒ f uniformly on Fk. Indeed, for all x ∈ Fk, x does not belong to the set⋃∞i=kx : |fni

(x)− f(x)| ≥ 1/i, so for every j ≥ k and all x ∈ Fk

|fnj(x)− f(x)| ≤ 1

j

which proves uniform convergence of fnjto f on Fk 2

Remark. We actually proved not only convergence a.e. but also uniform convergenceon subsets of X whose complement has arbitrary small measure. Note that Lebesgue’stheorem combined with this stronger conclusion of Riesz’ theorem implies Egorov’stheorem for sequences of real valued functions.

Exercise. Generalize the Lebesgue and the Riesz theorems to the case of sequences ofmeasurable mappings with values into separable metric spaces.

2.6 Absolute continuity of the integral

The following result is known as the absolute continuity of the integral.

Theorem 52 Let f ∈ L1(µ). Then for every ε > 0 there is δ > 0 such that∫

E|f | dµ <

ε whenever µ(E) < δ.

Proof. For if not, there would exist ε0 > 0 and sets En ∈ M such that µ(En) ≤ 2−n

and∫

En|f | dµ ≥ ε0. The sequence of sets Ak =

⋃∞n=k En is decreasing and

µ

(∞⋂

k=1

Ak

)= 0 , (30)

40

because µ(⋂∞

k=1Ak) ≤ µ(An) ≤∑∞

i=n µ(Ei) ≤ 2−n+1 for every n. Since ϕ(E) =∫E|f | dµ is a measure on M (Theorem 39) and ϕ(A1) =

∫A1|f | dµ ≤

∫X|f | dµ < ∞,

Theorem 3(g) implies that

ϕ(Ak)→ ϕ

(∞⋂

k=1

Ak

)=

∫⋂∞

k=1 Ak

|f | dµ = 0

by (30). This is, however, a clear contradiction, because ϕ(Ak) ≥ ϕ(Ek) =∫

Ek|f | dµ ≥

ε0. 2

2.7 Riesz representation theorem

Definition. If f : X → IR is a continuous function on a metric space X, thensupp f = x ∈ X : f(x) 6= 0 is called the support of f and Cc(X) = Cc(X, IR) denotesthe linear space of continuous functions on X with compact support. We say that alinear functional I : Cc(X)→ IR is positive if

I(f) ≥ 0 whenever f ≥ 0.

Let X be a locally compact metric space and µ a measure on B(X) such thatµ(X) < ∞ for every compact set K ⊂ X, i.e. µ is a Radon measure, see Corollary 9.Then

I(f) =

∫X

f dµ

is a positive linear functional on Cc(X). Surprisingly, every positive linear functionalon Cc(X) can be represented as an integral with respect to a Radon measure.

Theorem 53 (Riesz representation theorem) Let X be a locally compact metricspace9 and let I : Cc(X) → IR be a positive linear functional. Then there is a uniqueRadon measure µ such that

I(f) =

∫X

f dµ . (31)

Proof. For an open set G ⊂ X denote by Γ(G) the class of continuous functions0 ≤ f ≤ 1 with compact support in G.

Lemma 54 Let G1, G2, . . . , Gk ⊂ X be open sets. If f ∈ Γ(G1 ∪ . . . ∪Gk), then thereexist functions fi ∈ Γ(Gi), i = 1, 2, . . . such that f = f1 + . . .+ fk.

9This theorem is true for a more general class of locally compact Hausdorff topological spaces.

41

Proof. Since X is locally compact, there is a compact set E such that

supp f ⊂ intE, E ⊂ G1 ∪ . . . ∪Gk .

The sets Hi = Gi ∩ E form an open covering of the compact space E. If we can findcontinuous nonnegative functions ϕ1, . . . , ϕk in E such that

ϕ = ϕ1 + . . .+ ϕk > 0 in E, suppϕi ⊂ Hi,

then the functions

fi(x) =

f(x)ϕi(x)/ϕ(x) if x ∈ E,

0 if x 6∈ Ewill have desired properties. To construct functions ϕi observe that there is an opencovering of E by sets Di, i = 1, 2, . . . , k such that Di ⊂ Hi and then we define ϕi(x) =dist (X,E \Di). 2

Let now I : Cc(X)→ IR be a positive linear functional. Obviously

I(0) = 0 and I(f) ≤ I(g) for f ≤ g

(because I(g − f) ≥ 0).

Exercise. Prove that if ψ : [0,∞)→ [0,∞) is an increasing function such that ψ(s+t) = ψ(s) + ψ(t), then ψ(t) = tψ(1).

It follows from the exercise that

I(tf) = tI(f) for all t ≥ 0 and f ∈ Cc(X). (32)

Now for E ∈ B(X) we define

µ(E) = infE⊂G−open

λ(G) , (33)

whereλ(G) = sup

f∈Γ(G)

I(f) .

In order to prove that µ is a measure on B(X) it suffices to prove that for any opensets G1, G2, . . .

λ(∞⋃

n=1

Gn) ≤∞∑

n=1

λ(Gn) (34)

andλ(G1 ∪G2) = λ(G1) + λ(G2) provided G1 ∩G2 = ∅. (35)

Indeed, the two properties easily imply that the function µ∗ defined on the class of allsubsets of X by (33) is a metric outer measure.

42

Let Gn∞n=1 be a sequence of open sets and L < λ(⋃∞

n=1Gn). Then there is a

function f ∈ Γ(⋃∞

n=1Gn) such that I(f) > L. Since supp f is compact, f ∈ Γ(⋃N

n=1Gn)for some finite N . According to the lemma f = f1 + . . .+ fN , fi ∈ Γ(Gi), so

L < I(f) =N∑

n=1

I(fn) ≤∞∑

n=1

λ(Gn)

and (34) easily follows.

If f ∈ Γ(G1), g ∈ Γ(G2), G1 ∩ G2 = ∅, then f + g ∈ Γ(G1 ∪ G2). Indeed, thecondition G1 ∩G2 = ∅ implies that f + g ≤ 1. Hence

I(f) + I(g) = I(f + g) ≤ λ(G1 ∪G2)

and (35) follows upon taking the supremum over f and g.

We proved that µ is a measure in B(X). The measure µ is finite on compact sets.Indeed, if K ⊂ G ⊂ G and G is compact, then there is a function h ∈ Cc(X) such thath = 1 on G and therefore

µ(K) ≤ supf∈Γ(G)

I(f) ≤ I(h) <∞ .

Clearlyµ(G) = sup

f∈Γ(G)

I(f) for every open G ⊂ X

andµ(E) = inf

E⊂G−openµ(G) for every E ∈ B(X).

Let g ∈ Cc(X), g ≥ 0. Since Ig(f) = I(fg) is a positive functional it follows from whatwe have already proved that

µg(G) = supf∈Γ(G)

I(fg) for every open G ⊂ X

andµg(E) = inf

E⊂G−openµg(G) for every E ∈ B(X)

is a measure in B(X) that is finite on compact sets.

Lemma 55 I(g) = µg(X).

Proof. For f ∈ Γ(X), fg ≤ g and hence I(fg) ≤ I(g). On the other hand if f ∈ Γ(X),f = 1 on supp g, then fg = g and hence I(fg) = I(g). Therefore

µg(X) = supf∈Γ(X)

I(fg) = I(g) .

2

43

Lemma 56 (infE g)µ(E) ≤ µg(E) ≤ (supE g)µ(E) for every E ∈ B(X).

Proof. If supp f ⊂ G, then it follows from (32) that

(infGg)I(f) ≤ I(fg) ≤ (sup

Gg)I(f) .

Taking the supremum over f ∈ Γ(G) yields

(infGg)µ(G) ≤ µg(G) ≤ (sup

Gg)µ(G)

and then the lemma follows after taking the infimum over all open sets G such thatE ⊂ G. This is because continuity of g implies

infE⊂G−open

(infGg) = inf

Eg and inf

E⊂G−open(sup

Gg) = sup

Eg .

2

Lemma 57 µg(E) =∫

Eg dµ for all E ∈ B(X).

Proof. The measure µ is finite on compact sets and hence every Borel set is a union ofsets of finite measure (Corollary 9). Therefore it suffices to prove the equality for setsE of finite measure µ(E) < ∞. Given ε > 0 let s =

∑nk=1 akχEk

be a simple functionsuch that s ≤ g ≤ s + ε (Theorem 34). Here the sets Ek are pairwise disjoint and10⋃n

k=1Ek = X.

Then ∫E

s dµ =n∑

k=1

akµ(Ek ∩ E) ≤n∑

k=1

( infEk∩E

g)µ(Ek ∩ E)

≤n∑

k=1

µg(Ek ∩ E) = µg(E)

and ∫E

(s+ ε) dµ =n∑

k=1

(ak + ε)µ(Ek ∩ E) ≥n∑

k=1

( supEk∩E

g)µ(Ek ∩ E)

≥n∑

k=1

µg(Ek ∩ E) = µg(E) .

We proved that∫E

(g − ε) dµ ≤∫

E

s dµ ≤ µg(E) ≤∫

E

(s+ ε) dµ ≤∫

E

(g + ε) dµ .

10i.e. we include the set where ak = 0. For example if s = χE , then we write s = 1 · χE + 0 · χX\E .

44

Since ε > 0 was arbitrary and µ(E) <∞, the lemma easily follows. 2

Comparing Lemma 55 and Lemma 57 we obtain that I(g) =∫

Xg dµ which proves

(31) for g ≥ 0. The general case follows from the fact that every function in Cc(X) isa difference of two nonnegative functions.

We are left with the proof of uniqueness of the measure. Suppose that

I(f) =

∫X

f dµ =

∫X

f dν for all f ∈ Cc(X),

where µ and ν are measures on X that are finite on compact sets. According toCorollary 9 and Corollary 8 it suffices to prove that the measures µ and ν agree on theclass of open sets.

Let G be an open set and K ⊂ G a compact set. Let f ∈ Γ(G) be such that f = 1on K. Then

µ(G) ≥∫

G

f dµ = I(f) =

∫G

f dν ≥ ν(K) .

Taking supremum over compact sets K ⊂ G and applying Corollary 9 we obtain thatµ(G) ≥ ν(G). Similarly we prove the opposite inequality. The proof of the theorem iscomplete. 2

3 Integration on product spaces

This should be a word for word copy of Chapter 8 from Rudin’s book. There is no needto rewrite the whole chapter, so I will not include anything here.

4 Vitali covering lemma, doubling and Hausdorff

measures

4.1 Doubling measures

If σ > 0 and B is a ball in a metric space, then by σB we will denote a ball concentricwith B and with the radius σ times that of B.

Definition. Let X be a metric space and µ a measure on B(X). We say that µ is adoubling measure if there is a constant Cd > 0 such that for every ball B, 0 < µ(B) <∞and µ(2B) ≤ Cdµ(B).

The Lebesgue measure is an example of a doubling measure with Cd = 2n. It turnsout that general doubling measures inherit many properties of the Lebesgue measure.

45

Definition. We say that a metric space X is a doubling-metric space if there is aconstant M > 0 such that every ball B can be covered by no more than M balls of halfthe radius.

Examples. IRn is metric-doubling. If X is a metric-doubling space, then every subsetof X is metric-doubling. In particular every subset of IRn is metric-doubling. If X isan infinite discrete space, then X is not metric-doubling.

Proposition 58 If a metric space is complete and metric-doubling, then it is locallycompact.

Proof. It suffices to prove that any closed ball is compact. It easily follows from themetric-doubling condition that each closed ball is totally bounded. Since it is complete(as a closed subset of a complete metric space) it is compact. One can also prove thistheorem more directly by mimicking the proof that the interval [0, 1] is compact. 2

Proposition 59 If µ is a doubling measure on a metric space X, then X is metric-doubling.

Proof. We will prove that X satisfies the metric-doubling condition with M = C4d . Let

xii∈I ⊂ B(x, r) be a maximal r/2-separated set, i.e.

∀i,j∈I d(xi, xj) ≥r

2

∀y∈B(x,r) ∃i∈I d(xi, y) <r

2.

From the second condition we have B(x, r) ⊂⋃

i∈I B(xi, r/2) and it suffices to showthat the number of elements in I is bounded by C4

d , i.e. #I ≤ C4d . The balls B(xi, r/4)

are pairwise disjoint and B(xi, r/4) ⊂ B(x, 2r) ⊂ B(xi, 4r). We have

µ(B(x, 2r)) ≥∑i∈I

µ(B(xi, r/4)) ≥ C−4d

∑i∈I

µ(B(xi, 4r)) ≥ C−4d µ(B(x, 2r))(#I)

and hence #I ≤ C4d . 2

Theorem 60 (Volberg-Konyagin) If X is a complete metric-doubling metric space,then there is doubling measure on X.

This theorem is very difficult and we will not prove it. Note that according to the pre-vious proposition metric-doubling condition is necessary for the existence of a doublingmeasure, so Volberg-Konyagin’s theorem is very sharp.

The Euclidean space is metric doubling and hence any subset of the Euclidean spaceis metric-doubling. Therefore we have

46

Corollary 61 If X ⊂ IRn is closed, then there is a doubling measure on X.

Proposition 62 Let µ be a doubling measure on a metric space X. Then there is aconstant C > 0 such that

µ(B(x, r))

µ(B(x0, r0))≥ C

(r

r0

)s

whenever x ∈ B(x0, r0), r ≤ r0 and s = logCd/ log 2.

Remark. For the Lebesgue measure in IRn, Cd = 2n, logCd/ log 2 = n and henceLn(B(x, r))/L(B(0, 1)) ≥ Crn which is the sharp lower bound estimate for theLebesgue measure of the ball. That shows that the exponent s in the above lemmacannot be any smaller.

Proof. Let x ∈ B(x0, r0), r ≤ r0. Let k be the largest integer such that 2k−1r ≤ 2r0.Hence 2kr > 2r0 and thus B(x0, r0) ⊂ B(x, 2kr). We have

µ(B(x0, r0)) ≤ µ(B(x, 2kr)) ≤ Ckdµ(B(x, r)),

µ(B(x, r))

µ(B(x0, r0)≥ C−k

d . (36)

Since 2k−1r ≤ 2r0, r/r0 ≤ 4−12−k and hence(r

r0

)s

=

(r

r0

)log Cd/ log 2

≤ (4−1)log Cd/ log 2︸︷︷︸A

(2−k)log Cd/ log 2 = AC−kd .

This estimate together with (36) yields the claim. 2

4.2 Covering lemmas

Theorem 63 (5r-covering lemma) Let B be a family of balls in a metric space suchthat supdiamB : B ∈ B < ∞. Then there is a subfamily of pairwise disjoint ballsB′ ⊂ B such that ⋃

B∈B

B ⊂⋃

B∈B′5B .

If the metric space is separable, then the family B′ is countable and we can arrange itas a sequence B′ = Bi∞i=1, so

⋃B∈B

B ⊂∞⋃i=1

5Bi .

47

Remark. Here B can be either a family of open balls or closed balls. In both casesproof is the same.

Proof. Let supdiamB : B ∈ B = R < ∞. Divide the family B according to thediameter of the balls

Fj = B ∈ B :R

2j< diamB ≤ R

2j−1 .

Clearly B =⋃∞

j=1Fj. Define B1 ⊂ F1 to be the maximal family of pairwise disjointballs. Suppose the families B1, . . . ,Bj−1 are already defined. Then we define Bj to bethe maximal family of pairwise disjoint balls in

Fj ∩ B : B ∩B′ = ∅ for all B′ ∈j−1⋃i=1

Bi .

Next we define B′ =⋃∞

j=1 Bj. Observe that every ball B ∈ Fj intersects with a ball in⋃ji=1 Bj. Suppose that B ∩B1 6= ∅, B1 ∈

⋃ji=1 Bi. Then

diamB ≤ R

2j−1= 2 · R

2j≤ 2 diamB1

and hence B ⊂ 5B1. The proof is complete. 2

Remark. We have actually proved more: every ball B ∈ B is contained in 5B1 forsome ball B1 ∈ B′ such that B1∩B 6= ∅. This fact will be used in the proof of the nextresult.

Definition. We say that a family F of closed balls covers a set A in the Vitali senseif for every a ∈ A there is a ball B ∈ F of arbitrarily small radius such that a ∈ B. Inother words

∀a∈A infdiamB : a ∈ B ∈ F = 0.

Theorem 64 (Vitali covering theorem) Let µ be a doubling measure on X andA ⊂ X a measurable set. Let F be a family of closed balls that covers A in the Vitalisense. Then there is a subfamily G ⊂ F of pairwise disjoint balls such that

µ(A \

⋃B∈G

B)

= 0 . (37)

Proof. Let F ′ ⊂ F be a subfamily of all balls with diameters less than 1. Observe thatthe family F ′ also covers A in the Vitali sense. According to the 5r-covering lemmathere is a countable subfamily G ⊂ F ′ of pairwise disjoint balls such that

A ⊂⋃

B∈F ′B ⊂

⋃B∈G

5B .

48

We will prove that the family G satisfies (37). To this end it suffices to prove that forany ball Q of radius bigger than 1

µ(Q ∩ (A \

⋃B∈G

B))

= 0 .

Observe thatQ ∩

(A \

⋃B∈G

B)

= Q ∩(A \

⋃B∈G

B⊂10Q

B).

Indeed, if B is not contained in 10Q, then11 B ∩ (Q ∩ A) = ∅. The family of ballsB such that B ∈ G and B ⊂ 10Q is countable and hence we can name the balls byB1, B2, B3, . . . We have

∞∑i=1

µ(5Bi) ≤ C

∞∑i=1

µ(Bi) ≤ Cµ(10Q) <∞ .

The first inequality follows from the doubling condition and the second inequality fromthe fact that the balls Bi are pairwise disjoint and contained in 10Q. The aboveinequality implies that

µ( ∞⋃

i=N

5Bi

)≤

∞∑i=N

µ(5Bi)→ 0 as N →∞

and it remains to prove that

Q ∩(A \

⋃B∈G

B⊂10Q

B)⊂ Q ∩

(A \

N⋃i=1

Bi

)⊂

∞⋃i=N

5Bi . (38)

The first inclusion is obvious. To prove the second one let a ∈ Q ∩ (A \⋃N

i=1Bi). The

set⋃N

i=1Bi is closed and a 6∈⋃N

i=1Bi, so there is a ball Ba of sufficiently small radiusr < 1 such that a ∈ Ba ∈ F ′ and

Ba ∩N⋃

i=1

Bi = ∅ . (39)

On the other hand it follows from the remark stated after the proof of the 5r-coveringlemma that there is B ∈ G such that Ba ∩ B 6= ∅, Ba ⊂ 5B. Since diamB < 1 anda ∈ Q, it follows that B ⊂ 10Q and hence B = Bj for some j. Now (39) implies thatj > N and hence

a ∈ 5Bj ⊂∞⋃

i=N

5Bi

which proves (38). 2

11Because diam B < 1 and the radius of Q is bigger than 1.

49

Corollary 65 For an open set U ⊂ IRn and every ε > 0 there is a sequence of pairwisedisjoint closed balls B1, B2, B3, . . . contained in U and of radii less than ε such that

Ln

(U \

∞⋃i=1

Bi

)= 0 .

Exercise. Show a direct proof of the corollary without using the Vitali covering lemma.

4.3 Doubling measures and the Hausdorff measure

We will snow now how to use the covering lemmas to prove important results aboutHausdorff measures.

Theorem 66 For any Borel set A ⊂ IRn and any 0 < δ ≤ ∞

Ln(A) = Hnδ (A) = Hn(A) .

Proof. Step 1. If Ln(Z) = 0, then Hnδ (Z) = 0.

Indeed, for every ε > 0, Z can be covered by cubes Qi, each of diameter less thanδ such that

∑i |Qi| < ε. Since12 (diamQi)

n = C(n)|Qi| we have that

Hnδ (Z) ≤ ωn

2n

∑i

(diamQi)n =

ωn

2nC(n)

∑i

|Qi| < εωn

2nC(n)

and hence Hnδ (Z) = 0.

Step 2. Hnδ (A) ≤ Ln(A).

Let A ⊂⋃∞

i=1Qi, where the cubes Qi have pairwise disjoint interiors and

∞∑i=1

|Qi| ≤ Ln(A) + ε .

Fix i. It follows from Corollary 65 that there are pairwise disjoint balls Bj ⊂ Qi,j = 1, 2, 3, . . . such that

Ln(Qi \∞⋃

j=1

Bj) = 0 .

12By C(n) we denote a constant which depends on n only.

50

Now applying Step 1 w have

Hnδ (Qi) = Hn

δ (∞⋃i=1

Bi) ≤∞∑i=1

Hnδ (Bi) ≤

∞∑i=1

ωnrni

=∞∑i=1

Ln(Bi) = Ln(∞⋃i=1

Bi) = Ln(Qi) .

Therefore

Hnδ (A) ≤ Hn

δ (∞⋃i=1

Qi) ≤∞∑i=1

Hnδ (Qi) ≤

∞∑i=1

Ln(Qi) ≤ Ln(A) + ε .

Since ε > 0 was arbitrary, the claim follows.

Step 3. Ln(A) ≤ Hnδ (A).

We will need the following fact.

Lemma 67 (Isodiametric inequality) For any Lebesgue measurable set A ⊂ IRn

Ln(A) ≤ ωn

(diamA

2

)n

We will prove this lemma in Section 6. Observe that if A is a ball, then we have theequality. Therefore the lemma says that among all sets of fixed diameter, ball has thelargest volume.

Now we can complete the proof of Step 3. Let A ⊂⋃∞

i=1Ci, diamCi < δ. Then

Ln(A) ≤∞∑i=1

Ln(Ci) ≤∞∑i=1

ωn

(diamCi

2

)n

=ωn

2n

∞∑i=1

(diamCi)n .

Taking the infimum over all such coverings we obtain

Ln(A) ≤ Hnδ (A) .

Step 4. General case. We proved in steps 2 and 3 that Ln(A) = Hnδ (A) for all δ.

HenceLn(A) = lim

δ→0Hn

δ (A) = Hn(A) .


Now we will compare doubling measures with Hausdorff measures in metric spaces.Recall that in a given metric space there is 0 ≤ s ≤ ∞ such that

Ht(X) =∞ for t < s and Ht(X) = 0 for t > s.

This unique number s is called Hausdorff dimension ofX and is denoted by s = dimH X.

51

Theorem 68 Let µ be a doubling measure on a metric space X, µ(2B) ≤ Cdµ(B) forevery ball B. Let s = logCd/ log 2. Then

dimH X ≤ s .

Proof. It suffices to prove that Hs(B) < ∞ for every ball B. Indeed, then Ht(B) = 0for t > s and hence Ht(X) = 0 for t > s. This, however, implies that dimH X ≤ s.

Let B = B(x0, R). According to Proposition 62 there is a constant C > 0 such thatfor every x ∈ B and r < R, µ(B(x, r)) ≥ Crs.

Fix 0 < δ < R. Let xi∞i=1 ⊂ B be a maximal δ/2-separated set, i.e. d(xi, xj) ≥ δ/2for i 6= j and for every x ∈ B there is i ∈ I such that d(x, xi) < δ/2. Therefore the ballsB(xi, δ/4) are disjoint, contained in B(x, 2R) and B ⊂

⋃i∈I B(xi, r/2). We estimate

the number of elements in I.

µ(B(x, 2R)) ≥∑i∈I

µ(B(xi, δ/4)) ≥∑i∈I

C(δ/4)s = (#I)C(δ/4)s .

Hence#I ≤ µ(B(x, 2R))C−14s︸︷︷︸

does not depend on δ

δ−s = C ′δ−s .

Now B ⊂⋃

i∈I B(xi, δ/2) is a covering of the ball by sets of diameters less than orequal to δ, and hence

Hsδ(B) ≤ ωs

2s

∑i∈I

(diamB(xi, δ/2))s ≤ ωs

2sδs(#I) ≤ ωs

2sδsC ′δ−s =

ωs

2sC ′

HenceHs(B) = lim

δ→0Hs

δ(B) ≤ ωs

2sC ′ <∞

Definition. We say that a Borel measure µ on a metric space X is s-regular, if thereis a constant C ≥ 1 such that

C−1rs ≤ µ(B(x, r)) ≤ Crs

for all x ∈ X and all 0 < r < diamX.

Clearly, every s-regular measure is doubling.

Theorem 69 Let µ be an s-regular measure on X. Then there are constants C1, C2 > 0such that

C1µ(E) ≤ Hs(E) ≤ C2µ(E)

for every Borel set E ⊂ X. In particular Hs is s-regular.

52

Proof. First we will prove that there is a constant C1 such that C1µ(E) ≤ Hs(E) forevery Borel set E. Obviously we can assume that Hs(E) < ∞. Since every boundedset is contained in a ball of radius diamA, then µ(A) ≤ C(diamA)s by the s-regularityof the measure. For every ε > 0 there is a covering E ⊂

⋃∞i=1Ei such that

ωs

2s

∞∑i=1

(diamEi)s ≤ Hs(E) + ε .

We have

µ(E) ≤ µ( ∞⋃

i=1

Ei

)≤

∞∑i=1

µ(Ei) ≤ C∞∑i=1

(diamEi)s

= C2s

ωs

ωs

2s

∞∑i=1

(diamEi)s ≤ C

2s

ωs

(Hs(E) + ε) .

Since ε > 0 was arbitrary we conclude that C1µ(E) ≤ Hs(E), where C1 = C−1ωs/2s.

The proof of the opposite inequality is more difficult.

Lemma 70 If µ is an s-regular measure on a metric space X, then there is a constantM > 0 such that for every ball B of radius R and any r < R, the ball B can be coveredby no more than M(R/r)s balls of radius r.

Proof. Let xii∈I ⊂ B be a maximal r-separated set, i.e. d(xi, xj) ≥ r for i 6= jand for every x ∈ B there is i ∈ I such that x ∈ B(xi, r). Hence B ⊂

⋃i∈I B(xi, r)

and B(xi, r/2) ∩ B(xj, r/2) = ∅ for i 6= j. Since B(xi, r/2) ⊂ 2B and µ(B(xi, r/2)) ≥C−1(r/2)s we have

C(2R)s ≥ µ(2B) ≥ (#I)C−1(r/2)s .

Hence #I ≤ C24s(R/r)s and hence the lemma follows with M = C24s. 2

The lemma implies that Hsδ(B) ≤ C ′µ(B) where B is a ball of radius R. Indeed,

we can cover B by M(R/r)s balls of radius r and we can take r so small that 2r < δ.Hence

Hsδ(B) ≤ ωs

2sM

(R

r

)s

(2r)s = ωsMRs ≤ C ′µ(B) .

Since the constant C ′ is independent of δ we conclude that Hs(B) ≤ C ′µ(B). We willnow prove a stronger inequality which says that for an arbitrary open set U ⊂ X

Hs(U) ≤ C ′′µ(U) .

This will imply that Hs(E) ≤ C ′′µ(E) for any Borel set E ⊂ X. Indeed, both measuresµ andHs have the property that X is a union of a countably family of open sets of finite

53

measure. Therefore both measures have the property that the measure of any Borelset equals infimum of measures of open sets that contain given Borel set (Theorem 7).

It follows from the proof of Lemma 70 that X is separable, so according to the5r-covering lemma we can find disjoint balls Bi ⊂ U such that U ⊂

⋃∞i=1 5Bi. Hence

Hs(U) ≤∞∑i=1

Hs(5Bi) ≤ C ′∞∑i=1

µ(5Bi) ≤ C ′′∞∑i=1

µ(Bi) ≤ C ′′µ(U).


5 Differentiation

5.1 Lebesgue differentiation theorem

Definition. We say that f ∈ L1loc(µ), if for every x ∈ X there is r > 0 such that∫

B(x,r)|f | dµ <∞.

The integral average of f over a measurable set E of finite measure will be denoteby

fE = µ(E)−1

∫E

f dµ =

∫E

f dµ .

Theorem 71 (Lebesgue differentiation theorem) If µ is a doubling measure ona metric space X and f ∈ L1

loc(µ), then

limr→0

∫B(x,r)

f dµ = f(x) for µ-a.e. x ∈ X . (40)

Proof. We can assume that f ≥ 0. Since X is separable, it is a countable union ofballs such that f is integrable on each such ball. Therefore, it suffices to prove that if∫

Bf dµ <∞, then (40) is satisfied for µ-a.e. x ∈ B.

If for a measurable set A ⊂ B we have

lim infr→0

∫B(x,r)

f dµ ≤ t for all x ∈ A ,

then ∫A

f dµ ≤ tµ(A) .

Indeed, given ε > 0, let U be an open set such that A ⊂ U , µ(U) ≤ µ(A) + ε. Forevery x ∈ A there is arbitrarily small r > 0 such that B(x, r) ⊂ U and∫

B(x,r)

f dµ ≤ t+ ε .

54

Such closed balls form a covering of U in the Vitali sense and hence we can selectpairwise disjoint balls B1, B2, B3, . . . from the covering such that

µ(U \∞⋃i=1

Bi) = 0,

∫Bi

f dµ ≤ t+ ε .

We have ∫A

f dµ ≤∫

U

f dµ =∞∑i=1

∫Bi

f dµ ≤ (t+ ε)∞∑i=1

µ(Bi)

= (t+ ε)µ(U) ≤ (t+ ε)(µ(A) + ε) .

Since ε > 0 was arbitrary, we conclude∫A

f dµ ≤ tµ(A)

Similarly, if A ⊂ B is a measurable set such that

lim supr→0

∫B(x,r)

f dµ ≥ t for all x ∈ A ,

then ∫A

f dµ ≥ tµ(A) . (41)

Let

As,t = x ∈ B : lim infr→0

∫B(x,r)

f dµ ≤ s < t ≤ lim supr→0

∫B(x,r)

f dµ .

Then

tµ(As,t) ≤∫

As,t

f dµ ≤ sµ(As,t)

and hence µ(As,t) = 0. This easily implies that

lim infr→0

∫B(x,r)

f dµ = lim supr→0

∫B(x,r)

f dµ for µ-a.e. x ∈ B ,

i.e. the limit limr→0

∫B(x,r)

f dµ exists for µ-a.e. x ∈ B. Denote this limit by

g(x) = limr→0

∫B(x,r)

f dµ

whenever the limit exists. We have to prove that f(x) = g(x) for µ-a.e. x ∈ B.

Fix a measurable set F ⊂ B and ε > 0. Let

An = x ∈ F : (1 + ε)n ≤ g(x) < (1 + ε)n+1, n ∈ ZZ .

55

Observe that

lim supr→0

∫B(x,r)

f dµ = g(x) ≥ (1 + ε)n for all x ∈ An ,

and hence∫

Anf dµ ≥ (1 + ε)nµ(An) by (41). We have∫

F

g dµ =∞∑

n=−∞

∫An

g dµ ≤∞∑

n=−∞

(1 + ε)n+1µ(An)

≤ (1 + ε)∞∑

n=−∞

∫An

f dµ = (1 + ε)

∫F

f dµ .

Similarly we prove that ∫F

g dµ ≥ (1 + ε)−1

∫F

f dµ .

Since ε > 0 was arbitrary we conclude that for every measurable set F ⊂ B∫F

f dµ =

∫F

g dµ

and hence f = g µ-a.e. 2

Among all representatives of f in the class of functions that are equal f a.e. weselect the following one

f(x) := lim supr→0

∫B(x,r)

f dµ . (42)

It follows from the Lebesgue differentiation theorem that the right hand side of (42)which is defined everywhere equals f a.e.

Definition. Let f ∈ L1loc(µ). We say that x ∈ X is a Lebesgue point of f if

limr→0

∫B(x,r)

|f(y)− f(x)| dµ(y) = 0 ,

where f(x) is defined by (42).

Theorem 72 Let µ be a doubling measure and f ∈ L1loc(µ). Then µ-a.e. point x ∈ X

is a Lebesgue point of f .

Proof. For every c ∈ IR

limr→0

∫B(x,r)

|f(y)− c| dµ(y) = |f(x)− c| µ-a.e. (43)

56

LetE = x ∈ X : (43) holds for all c ∈ Q .

Since Q is countable, µ(X \ E) = 0. Let x ∈ E, be such that |f(x)| < ∞, and letQ 3 cn → f(x). Then it follows from (43) that

limr→0

∫B(x,r)

|f(y)− f(x)| dµ = |f(x)− f(x)| = 0 .

2

Definition. We say that a family F of measurable sets in X is regular at x ∈ X ifthere is a constant C > 0 such that for every S ∈ F there is a ball B(x, rS) such that

S ⊂ B(x, rS), µ(B(x, rS)) ≤ Cµ(S) ,

and for every ε > 0 there is a set S ∈ F with µ(S) < ε.

Theorem 73 Let µ be a doubling measure and f ∈ L1loc(µ). If x ∈ X is a Lebesgue

point of f and F is a regular family at x ∈ X, then

limS∈F

µ(S)→0

∫S

f dµ = f(x) .

Proof. For S ∈ F let rs be defined as above. Observe that if µ(S) → 0, then rs → 0by Proposition 62. We have∣∣∣∣∫

S

f dµ− f(x)

∣∣∣∣ ≤ ∫S

|f(y)− f(x)| dµ(y) ≤ µ(S)−1

∫B(x,rS)

|f(y)− f(x)| dµ(y)

=µ(B(x, rS))

µ(S)

∫B(x,rS)

|f(y)− f(x)| dµ(y) ≤ C

∫B(x,rS)

|f(y)− f(x)| dµ(y)→ 0

as µ(S)→ 0. 2

Corollary 74 If f ∈ L1loc(IR

n) and x ∈ IRn is a Lebesgue point of f , then for anysequence of cubes Qi such that x ∈ Qi, diamQi → 0 we have

limi→∞

∫Qi

f dµ = f(x) .

Corollary 75 Let F (x) =∫ x

af(t) dt, where f ∈ L1(a, b). Then F ′(x) = f(x) for a.e.

x ∈ (a, b).

Proof. We have

F (x+ h)− F (x)

h=

∫ x+h

x

f(t) dt→ f(x) as h→ 0,

whenever x is a Lebesgue point of f . 2

57

Theorem 76 If A ⊂ IRn is a measurable set, then for almost all x ∈ A

limr→0

|B(x, r) ∩ A||B(x, r)|

= 1

and for almost all x ∈ IRn \ A

limr→0

|B(x, r) ∩ A||B(x, r)|

= 0 .

Proof. Applying the Lebesgue differentiation theorem to f = χA ∈ L1loc(IR

n) we havethat for a.e. x ∈ A

|B(x, r) ∩ A||B(x, r)|

=

∫B(x,r)

χA dLn → χA(x) = 1

and for a.e. x ∈ IRn \ A|B(x, r) ∩ A||B(x, r)|

→ χA(x) = 0 .

2

Definition. Let A ⊂ IRn be a measurable set. We say that x ∈ IRn is a density pointof A if

limr→0

|A ∩B(x, r)||B(x, r)|

= 1 .

Thus the above theorem says that almost every point of a measurable set A ⊂ IRn is adensity point.

6 Brunn-Minkowski inequality

In this section we will prove the Brunn-Minkowski inequality, isodiametric inequalityand the isoperimetric inequality.

Theorem 77 (Brunn-Minkowski inequality) If A,B ⊂ IRn are compact sets, then

Ln(A+B)1/n ≥ Ln(A)1/n + Ln(B)1/n .

where A+B = a+ b : a ∈ A, b ∈ B.

Proof. Observe that if we translate the sets, then the set A + B will also translate,so the inequality will remain the same. In particular, we can locate the origin of thecoordinate system at any point.

We divide the proof into several steps.

58

Step 1. A and B are rectangular boxes with sides parallel to the coordinate axes.

Denote the edges of A and B by a1,. . . ,an and b1,. . . ,bn respectively. Then A + Bis also a rectangular box with edges a1 + b1,. . . ,an + bn. Hence the Brunn-Minkowskiinequality takes the form

n∏i=1

(ai + bi)1/n ≥

n∏i=1

a1/ni +

n∏i=1

b1/ni

which is equivalent to

7 Lp spaces

Definition. Let (X,µ) be a measure space. For 0 < p <∞, Lp(µ) denotes the classof all complex valued measurable functions such that

‖f‖p =

(∫X

|f |p dµ)1/p

<∞

and we define Lp(µ) = Lp(µ)/ ∼, where f ∼ g if f = g a.e.

For p =∞ we define L∞ to be the class of essentially bounded measurable functions,i.e. such that there is M > 0 with

|f(x)| ≤M µ-a.e. (44)

It is easy to see that there is a smallest number M satisfying (44). We denote thisnumber by ‖f‖∞. Therefore

|f(x)| ≤ ‖f‖∞ µ-a.e.

and for any ε > 0, µ(x : |f(x)| > ‖f‖∞ − ε) > 0. Finally we set L∞(µ) = L(µ)/ ∼.

Example. Let µ be a counting measure on IN = 1, 2, 3, . . ., i.e. for A ⊂ IN,µ(A) = #A, the number of elements in the set A if A is finite and µ(A) = ∞ is A isinfinite. Clearly if f : IN→ C, then∫

IN

f dµ =∞∑

n=1

f(n).

The functions f : IN → C can be identified with complex sequences and it is easy tosee that

Lp(µ) = x = (xn)∞n=1 : ‖x‖p =( ∞∑

n=1

|xn|p)1/p

<∞ ,

andL∞(µ) = x = (xn)∞n=1 : ‖x‖∞ = sup

n∈IN|xn| <∞ .

In this particular situation we denote Lp(µ) by `p for all 0 < p ≤ ∞.

59

Theorem 78 (Jensen’s inequality) Let (X,µ) be a measure space such that µ(X) =1. If f ∈ L1(µ) satisfies a < f(x) < b for a.e. x and ϕ is a convex function on (a, b),then

ϕ

(∫X

f dµ

)≤∫

X

(ϕ f) dµ .

Remark. We do not exclude the cases a = −∞ and b =∞.

Recall that ϕ is convex on (a, b) if

ϕ((1− λ)x+ λy) ≤ (1− λ)ϕ(x) + λϕ(y) for all x, y ∈ (a, b) and 0 ≤ λ ≤ 1.

It is easy to see that this condition is equivalent to

ϕ(t)− ϕ(s)

t− s≤ ϕ(u)− ϕ(t)

u− tfor all s, t, u such that a < s < t < u < b (45)

Lemma 79 Every convex function on (a, b) is continuous.

60

Proof. Fix a < s < t < b and take x, y such that a < s < x < y < t < b. Denote byS,X, Y, T the corresponding points on the graph of ϕ.

The point Y belongs to the angle with the vertex X formed by the half-lines−→SX

and−−→XT . This easily implies that if y is decreasing to x, then Y → X, which proves the

right hand side continuity of ϕ. The proof of the left hand side continuity is similar. 2

Proof of Jensen’s inequality. For t =∫

Xf dµ we have a < t < b. Fix this value of t and

define

β = supa<s<t

ϕ(t)− ϕ(s)

t− s.

It follow from (45) that

β ≤ ϕ(u)− ϕ(t)

u− tfor all t < u < b

and henceϕ(u) ≥ ϕ(t) + β(u− t) for all t < u < b. (46)

On the other handϕ(t)− ϕ(s)

t− s≤ β for all a < s < t

and henceϕ(s) ≥ ϕ(t) + β(s− t) for all a < s < t (47)

The inequalities (46) and (47) imply that

ϕ(s) ≥ ϕ(t) + β(s− t) for all a < s < b.

In particularϕ(f(x))− ϕ(t)− β(f(x)− t) ≥ 0 a.e.

61

Integrating with respect to µ yields∫X

ϕ(f(x))− ϕ(∫

X

f dµ)

︸︷︷︸constant

−β(f(x)−

∫X

f dµ︸︷︷︸constant

)dµ ≥ 0 .

Since µ(X) = 1 we conclude∫X

(ϕ f) dµ− ϕ(∫

X

f dµ)− β

(∫X

f dµ−∫

X

f dµ︸︷︷︸0

)≥ 0

and hence ∫X

(ϕ f) dµ ≥ ϕ(∫

X

f dµ).

2

The numbers 1 < p, q < ∞ such that 1/p + 1/q = 1 or p = 1 and q = ∞ arecalled Holder conjugate or just conjugate exponents. Observe that if 1 < p <∞, thenq = p/(p− 1).

Theorem 80 (Holder’s inequality) Let (X,µ) be a measure space and 1 < p, q <∞be conjugate exponents. Then for any two complex valued measurable functions f andg we have ∫

X

|fg| dµ ≤ ‖f‖p‖g‖q .

Proof. We will need the following lemma.

Lemma 81 (Young’s inequality) If a, b > 0, then

ab ≤ ap

p+bq

q.

Proof. We can assume that a, b > 0, otherwise the inequality is obvious. Then a = es/p,b = et/q for some s, t ∈ IR and the inequality is equivalent to

es/p+t/q ≤ es/p+ et/q .

Now the lemma follows from the convexity of ex. 2

We can assume that 0 < ‖f‖p < ∞ and 0 < ‖g‖q < ∞, otherwise the Holderinequality is obvious. The lemma yields

|f(x)g(x)|‖f‖p‖g‖q

≤ 1

p

(|f(x)|‖f‖p

)p

+1

q

(|g(x)|‖g‖q

)q

.

62

Integrating this inequality yields∫X

|fg| dµ ≤ ‖f‖p‖g‖q( 1

p‖f‖pp

∫X

|f |p dµ+1

q‖g‖qq

∫X

|g|q dµ︸︷︷︸=1/p+1/q=1

)

and the theorem follows. 2

Theorem 82 (Minkowski’s inequality) Let (X,µ) be a measure space and 1 ≤ p ≤∞. Then for any two complex valued measurable functions f and g we have

‖f + g‖p ≤ ‖f‖p + ‖g‖p .

Proof. If p = 1, the inequality is obvious. If p =∞, then

|f(x) + g(x)| ≤ |f(x)|+ |g(x)| ≤ ‖f‖∞ + ‖g‖∞ a.e.

Hence ‖f + g‖∞ ≤ ‖f‖∞ + ‖g‖∞. If 1 < p < ∞, then we need to use the Holderinequality.∫

X

|f + g|p dµ ≤∫

X

|f ||f + g|p−1 dµ+

∫X

|g||f + g|p−1 dµ

≤(∫

X

|f |p dµ)1/p(∫

X

|f + g|(p−1)q dµ

)1/q

+

(∫X

|g|p dµ)1/p(∫

X

|f + g|(p−1)q dµ

)1/q

≤ (‖f‖p + ‖g‖p)

(∫X

|f + g|p dµ)1/q

and hence (∫X

|f + g|p dµ)1−1/q

≤ ‖f‖p + ‖g‖p.

Since 1− 1/q = 1/p the theorem follows. 2

Example. In the case of space `p, the Holder and the Minkowski inequality read asfollows

∞∑n=1

|xnyn| ≤

(∞∑

n=1

|xn|p)1/p( ∞∑

n=1

|yn|q)1/q

,

(∞∑

n=1

|xn + yn|p)1/p

≤

(∞∑

n=1

|xn|p)1/p

+

(∞∑

n=1

|yn|p)1/p

.

Definition. Let K denote IR or C. The normed space is a pair (X, ‖ · ‖), where X isa linear space over K and

‖ · ‖ : X → [0,∞)

is a function such that:

63

(a) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for all x, y ∈ X;

(b) ‖αx‖+ |α|‖x‖ for all x ∈ X and α ∈ K;

(c) ‖x‖ = 0 iff x = 0.

The function ‖ · ‖ is called a norm.

Since ‖x− y‖ = ‖(x− z) + (z − y)‖ ≤ ‖x− z‖+ ‖z − y‖ we see that

d(x, y) = ‖x− y‖

defines a metric in X. Normed spaces are always regarded as metric spaces with respectto this metric.

Definition. We say that a normed space (X, ‖ · ‖) is a Banach space if it is completewith respect to the metric d(x, y) = ‖x− y‖.

It follows from the Minkowski inequality that (Lp(µ), ‖ · ‖p) is a normed space forall 1 ≤ p ≤ ∞.

Theorem 83 Lp(µ) is a Banach space for all 1 ≤ p ≤ ∞.

Proof. We will prove this theorem for 1 ≤ p < ∞ and leave the case p = ∞ as anexercise. Let fn be a Cauchy sequence in Lp(µ). Let fni

be a subsequence suchthat

‖fni+1− fni

‖p < 2−i, i = 1, 2, 3, . . .

Let

gk =k∑

i=1

|fni+1− fni

|, g =∞∑i=1

|fni+1− fni

| .

It follows from the Minkowski inequality that ‖gk‖p < 1, and then Fatou’s lemmaapplied to the sequence gp

k yields∫X

gp dµ =

∫X

(lim infk→∞

gpk) dµ ≤ lim inf

k→∞

∫X

gpk dµ ≤ 1 ,

so ‖g‖p ≤ 1. In particular g(x) <∞ a.e. Thus the series

fn1(x) +∞∑i=1

(fni+1(x)− fni

(x))

converges absolutely for a.e. x. Since

fn1(x) +k∑

i=1

(fni+1(x)− fni

(x)) = fnk+1(x)

64

we conclude that the sequence fnk(x) converges for a.e. x. Denote its limit by f(x),

i.e.limk→∞

fnk(x) = f(x) a.e.

It suffices to prove that f ∈ Lp and ‖f − fn‖p → 0 as n→∞.

For every ε > 0 there is N such that ‖fn − fm‖p < ε for all n,m ≥ N . ApplyingFatou’s lemma to |fni

− fm|p with m ≥ N fixed and i = 1, 2, 3, . . . we have∫X

|f − fm|p dµ ≤ lim infi→∞

∫X

|fni− fm|p dµ ≤ εp .

Hence f − fm ∈ Lp and thus f ∈ Lp. Moreover ‖f − fm‖p → 0 as m→∞. 2

Exercise. Prove the above theorem for p =∞.

The following fact follows from the proof of the above theorem.

Corollary 84 If fn → f in Lp(µ), then there is a subsequence fniwhich converges to

f a.e.

Lemma 85 Let S be the class of complex, measurable, simple functions s on X suchthat

µ(x : s(x) 6= 0) <∞ .

If 1 ≤ p <∞, then S is dense in Lp(µ).

Proof. Clearly S ⊂ Lp(µ). If f ∈ Lp(µ) and f ≥ 0, let sn be a sequence of simplefunctions such that 0 ≤ sn ≤ f , sn → f pointwise. Since sn ∈ Lp it easily follows thatsn ∈ S. Now inequality 0 ≤ |f − sn|p ≤ fp and the dominated convergence theoremimplies that sn → f in Lp. In the general case we write f = (u+ − u−) + i(v+ − v−)and apply the above argument to each of the functions u+, u−, v+, v− separately. 2

Theorem 86 Let X be a locally compact metric space and µ a Radon measure on X.Then the class of compactly supported continuous functions Cc(X) is dense in Lp(µ)for all 1 ≤ p <∞.

Proof. According to the lemma it suffices to prove that the characteristic function ofa set of finite measure can be approximated in Lp by compactly supported continuousfunctions. Let E be a measurable set of finite measure. Given ε > 0 let K ⊂ E be acompact set such that µ(E \K) < (ε/2)p. Let U be an open set such that K ⊂ U , Uis compact and µ(U \ K) < (ε/2)p. Finally let ϕ ∈ Cc(X) be such that suppϕ ⊂ U ,0 ≤ ϕ ≤ 1, ϕ(x) = 1 for x ∈ K. We have

‖χE − ϕ‖p =

(∫X\K|χE − ϕ|p dµ

)1/p

≤(∫

X\K|χE|p dµ

)1/p

+

(∫X\K|ϕ|p dµ

)1/p

≤ µ(E \K)1/p + µ(U \K)1/p < ε .

65

2

7.1 Convolution

Let us start with an observation that since the Lebesgue measure in invariant undertranslations and under the mapping x 7→ −x, for any f ∈ L1(IRn) and any y ∈ IRn wehave ∫

IRn

u(x) dx =

∫IRn

u(x+ y) dx =

∫IRn

u(−x) dx .

Also since for every measurable set E and any t > 0 the set tE = tx : x ∈ Ehas measure Ln(tE) = tnLn(E) we easily conclude that the Lebesgue integral has thefollowing scaling property ∫

IRn

u(x/t) dx = tn∫

IRn

u(x) dx .

Observe that in the case of the Riemann integral the above equalities are direct con-sequences of the change of variables formula. We will prove a corresponding changeof variables formula for the Lebesgue integral later, but as for the proof of the aboveequalities we do not have to refer to the general change of variables formula as theyfollow directly from the properties of the Lebesgue measure mentioned above.

Definition. For measurable functions f and g on IRn we define the convolution by

(f ∗ g)(x) =

∫IRn

f(x− y)g(y) dy

and it is a question under what conditions the convolution is well defined.

If f ∈ L1loc(IR

n) and g is bounded, measurable and vanishes outside a bounded set,then the function y 7→ f(x− y)g(y) is integrable, so (f ∗ g)(x) is well defined and finitefor every x ∈ IRn. If f, g ∈ L1(IRn), then it can happen that for a given x the functiony 7→ f(x − y)g(y) is not integrable and hence (f ∗ g)(x) is not defined. However as apowerful application of the Fubini theorem we can prove the following surprising result.

Theorem 87 If f, g ∈ L1(IRn), then for a.e. x ∈ IRn the function y 7→ f(x − y)g(y)is integrable and hence (f ∗ g)(x) exists. Moreover f ∗ g ∈ L1(IRn) and

‖f ∗ g‖1 ≤ ‖f‖1‖g‖1 .

Proof. The function |f(x − y)g(y)| as a function of a variable (x, y) ∈ IR2n is measur-

66

able.13 Hence Fubini’s theorem yields∫IRn

∫IRn

|f(x− y)g(y)| dx dy =

∫IRn

(∫IRn

|f(x− y)||g(y)| dx)dy

=

∫IRn

(∫IRn

|f(x− y)| dx)|g(y)| dy = ‖f‖1‖g‖1 .

Therefore

‖f‖1‖g‖1 =

∫IRn

(∫IRn

|f(x− y)g(y)| dy)dx ≥

∫IRn

∣∣∣∣∫IRn

f(x− y)g(y) dy

∣∣∣∣ dx = ‖f∗g‖1 .

2

Theorem 88 If f, g, h ∈ L1(IRn) and α, β ∈ C, then

1. f ∗ g = g ∗ f ;

2. f ∗ (g ∗ h) = (f ∗ g) ∗ h;

3. f ∗ (αg + βh) = αf ∗ g + βf ∗ h.

Remark. This theorem says that (L1, ∗) is a commutative algebra.

Proof. (1) We have

(f∗g)(x) =

∫IRn

f(x−y)g(y) dy =

∫IRn

f(−y)g(y+x) dy =

∫IRn

f(y)g(x−y) dy = (g∗f)(x) .

(2) We have

f ∗ (g ∗ h)(x) =

∫IRn

f(x− y)(g ∗ h)(y) dy =

∫IRn

f(x− y)

(∫IRn

g(y − z)h(z) dz

)dy

=

∫IRn

h(z)

(∫IRn

f(x− y)g(y − z) dy

)dz =

∫IRn

h(z)

(∫IRn

f(−y)g(y − z + x) dy

)dz

=

∫IRn

h(z)

(∫IRn

f(y)g((x− z)− y) dy

)dz =

∫IRn

h(z)(f ∗ g)(x− z) dz = (f ∗ g) ∗ h(x) .

(3) It is easy and left to the reader. 2

Theorem 89 If f ∈ L1loc(IR

n) and g is continuous with compact support, then

1. f ∗ g is continuous on IRn.

13Why?

67

2. If f ∈ C1(IRn), then f ∗ g ∈ C1(IRn) and

∂

∂xi

(f ∗ g)(x) =( ∂f∂xi

∗ g)(x) .

3. If g ∈ C1(IRn), then f ∗ g ∈ C1(IRn) and

∂

∂xi

(f ∗ g)(x) =(f ∗ ∂g

∂xi

)(x) .

4. If f ∈ Cr(IRn) or g ∈ Cr(IRn), then f ∗ g ∈ Cr(IRn).

Proof. (1) Since g is bounded and has compact support, (f ∗ g)(x) is well defined andfinite for all x ∈ IRn. Fix x ∈ IRn. The function y 7→ f(y)g(x − y) vanishes outside asufficiently large ball B (because g as compact support). Let xn → x. Then there is aball B (perhaps larger than the one above), so that all the functions

y 7→ f(y)g(xn − y)

vanish outside B. Hence

|f(y)g(xn − y)| ≤ ‖g‖∞|f(y)|χB(y) ∈ L1(IRn)

and the dominated convergence theorem yields

(f ∗ g)(xn) =

∫IRn

f(y)g(xn − y) dy →∫

IRn

f(y)g(x− y) dy = (f ∗ g)(x)

which proves continuity of f ∗ g.

(2) The function ∂f/∂xi is bounded on bounded subsets of IRn (because it is continu-ous). Fix x ∈ IRn. For any r > 0 there is a constant M > 0 such that∣∣∣∣ ∂f∂xi

∣∣∣∣ ≤M for all z ∈ B(x, r + 1).

Let tk → 0, |tk| < 1. We have

f(x+ tkei − y)− f(x− y)

tkg(y)→ ∂f

∂xi

(x− y)g(y) .

The functions on the left hand side are bounded by M‖g‖∞ for all y with |y| < r (bythe mean value theorem). Take r so large that supp g ⊂ B(0, r). Then

(f ∗ g)(x+ tkei)− (f ∗ g)(x)

tk

=

∫B(0,r)

f(x+ tkei − y)− f(x− y)

tkg(y) dy →

∫B(0,r)

∂f

∂xi

(x− y)g(y) dy =( ∂f∂xi

∗ g)(x) .

68

We could pass to the limit under the sign of the integral, because the functions wereuniformly bounded.

(3) The proof is similar to that in the case (2); details are left to the reader.

(4) This result follows from (2) and (3) by induction. 2

Let ϕ ∈ C∞0 (B(0, 1)) be such a function that ϕ ≥ 0 and

∫IRn ϕ(x) dx = 1. A function

with the given properties can be constructed explicitly. Indeed,

ϕ(x) =

exp

(1

|x|2−1

)if |x| < 1,

0 if |x| ≥ 1.

belongs to C∞0 (IRn) and we define

ϕ(x) =ϕ(x)∫

IRn ϕ(y) dy.

For ε > 0 we setϕε(x) = ε−nϕ(x/ε) .

Then suppϕε ⊂ B(0, ε), ϕε ≥ 0 and∫

IRn ϕε(x) dx = 1. For f ∈ L1loc(IR

n) we put

fε = f ∗ ϕε.

It follows from Theorem 89 that fε ∈ C∞.

Theorem 90 Let f be a function defined on IRn.

1. If f is continuous, then fε ⇒ f uniformly on compact sets as ε→ 0.

2. If f ∈ Cr, then Dαfε ⇒ Dαf uniformly on compact sets for all |α| ≤ r.

Proof. (1) Uniform continuity of f on compact sets implies that for any compact set K

sup|y|<ε

supx∈K|f(x)− f(x− y)| → 0 as ε→ 0.

Since∫

IRn ϕε(y) dy = 1, we have∫

IRn f(x)ϕε(y) dy = f(x) and hence for any compactset K

supx∈K|f(x)− fε(x)| = sup

x∈K

∣∣∣∣∫IRn

(f(x)− f(x− y))ϕε(y) dy

∣∣∣∣≤ sup

x∈K

∫B(0,ε)

|f(x)− f(x− y)|ϕε(y) dy → 0 as ε→ 0 .

(2) It follows from Theorem 89 and the induction argument that Dα(f∗ϕε) = (Dαf)∗ϕε

for all |α| ≤ r and hence the theorem follows from the part (1). 2

69

Lemma 91 If f ∈ Lp, 1 ≤ p <∞, then fε ∈ Lp(IRn) and ‖fε‖p ≤ ‖f‖p.

Proof. If p = 1, then

‖fε‖1 = ‖f ∗ ϕε‖1 ≤ ‖f‖1‖ϕε‖1 = ‖f‖1

by Theorem 87. Let then 1 < p <∞. Holder’s inequality yields

|fε(x)|p =

∣∣∣∣∫IRn

f(x− y)ϕε(y) dy

∣∣∣∣p ≤ (∫IRn

|f(x− y)|ϕε(y)1/pϕε(y)1/q dy

)p

≤(∫

IRn

|f(x− y)|pϕε(y) dy

)(∫IRn

ϕε(y) dy

)p/q

= |f |p ∗ ϕε(x) .

Hence

‖fε‖pp ≤∫

IRn

|f |p ∗ ϕε(x) dx = ‖ |f |p ∗ ϕε‖1 ≤ ‖ |f |p‖1‖ϕε‖1 = ‖f‖pp .

2

The following result proves density of the class of smooth functions in Lp(IRn).

Theorem 92 If f ∈ Lp(IRn) and 1 ≤ p <∞, then fε ∈ Lp(IRn) and ‖f − fε‖p → 0 asε→ 0.

Proof. We will need the following result.

Lemma 93 If f ∈ Lp(IRn), 1 ≤ p <∞, then

limy→0

∫IRn

|f(x+ y)− f(x)|p dx = 0 .

Proof. For y ∈ IRn let τy : Lp(IRn)→ Lp(IRn) be the translation operator defined by

(τyf)(x) = f(x+ y) for f ∈ Lp(IRn).

The lemma says that τy is continuous, i.e. ‖τyf − f‖p → 0 as y → 0. Given ε > 0let g be a compactly supported continuous function such that ‖f − g‖p < ε/3, seeTheorem 86. Then

‖τyf−f‖p ≤ ‖τyf−τyg‖p+‖τyg−g‖p+‖f−g‖p = 2‖f−g‖p+‖τyg−g‖p < 2ε/3+‖τyg−g‖p .

Since τyg ⇒ g converges uniformly, there is δ > 0 such that ‖τyg−g‖p < ε/3 for |y| < δand hence

‖τyf − f‖p < ε for |y| < δ,

70

which proves the lemma. 2

Now we can complete the proof of the theorem. Assume first that 1 < p < ∞.Holder’s inequality and Fubini’s theorem yield

‖f − fε‖p =

∫IRn

∣∣∣∣∫IRn

(f(x)− f(x− y))ϕε(y) dy

∣∣∣∣p dx≤

∫IRn

(∫IRn

|f(x)− f(x− y)|ϕε(y)1/pϕε(y)1/q dy)p

dx

≤∫

IRn

(∫IRn

|f(x)− f(x− y)|pϕε(y) dy( ∫

IRn

ϕε(y) dy︸︷︷︸1

)p/q)dx

=

∫IRn

∫B(0,ε)

|f(x)− f(x− y)|pϕε(y) dy dx

=

∫B(0,ε)

(∫IRn

|f(x)− f(x− y)|p dx)ϕε(y) dy → 0 as ε→ 0.

by Lemma 93. If p = 1, then the above argument simplifies, because we do not haveto use Holder’s inequality. 2

Corollary 94 C∞0 is dense in Lp(IRn) for all 1 ≤ p <∞.

Proof. Since continuous functions with compact support are dense in Lp it suffices toprove that every compactly supported continuous function can be approximated in Lp

by C∞0 . This, however, immediately follows from Theorem 92, because if f vanishes

outside a bounded set, then fε has compact support. 2

8 Functions of bounded variation

Definition. We say that a function f : [a, b] → IR has bounded variation if itsvariation defined by

b∨a

f = supn−1∑i=0

|f(xi+1)− f(xi)| <∞

is finite, where the supremum is taken over all n and all partitions a = x0 < x1 < . . . <xn = b.

Clearly monotonic functions have bounded variation

b∨a

f = |f(b)− f(a)| .

71

If a function f : [a, b] → IR has bounded variation, then the function x 7→∨x

a f isnondecreasing on [a, b]. Also the function x 7→

∨xa f − f(x) is nondecreasing, because

f(y) − f(x) ≤∨y

x f for y > x. Therefore every function of bounded variation is adifference of two nondecreasing functions.

f(x) =x∨a

f −( x∨

a

f − f(x)). (48)

We proved

Theorem 95 (Jordan) A function f : [a, b] → IR has bounded variation if and onlyif it is a difference of two nondecreasing functions.

Exercise. Prove that if f : [a, b] → IR is a continuous function of bounded variation,then the function x 7→

∨xa f is continuous.

The exercise and the formula (48) implies

Theorem 96 A continuous function has bounded variation if and only if it is a differ-ence of two continuous nondecreasing functions.

Geometrically bounded variation is related to the length of a curve. Recall that ifϕ = (ϕ1, . . . , ϕk) : [a, b]→ IRk is a continuous curve, then its length is defined by

72

`(ϕ) = supn−1∑i=0

‖ϕ(xi+1)− ϕ(xi)‖

= supn−1∑i=1

√(ϕ1(xi+1)− ϕ1(xi))2 + . . .+ (ϕk(xi+1)− ϕk(xi))2

where the supremum is taken over all n and all partitions a = x0 < x1 < . . . < xn = b.

Definition. We say that a curve ϕ : [a, b] → IRk is rectifiable if it has finite length`(ϕ) <∞.

The following result easily follows from the definition of length and the definitionof variation.

Theorem 97 A continuous curve ϕ : [a, b] → IRk is rectifiable if and only if thecoordinate functions ϕi : [a, b]→ IR are of bounded variation for i = 1, 2, . . . , k.

The following result is an easy exercise.

Theorem 98 If functions f, g : [a, b]→ IR have bounded variation, then the functionsf ± g and fg have bounded variation. If in addition g ≥ c > 0, then the function f/ghas bounded variation.

Exercise. Prove the above theorem.

Proposition 99 Let f be a monotonic function on (a, b). Then the set of points of(a, b) at which f is discontinuous is at most countable.

Proof. Suppose f is nondecreasing. If f is discontinuous at x ∈ (a, b), then

limt→x−

f(t) < limt→x+

f(t)

and hence there is a rational number r(x) such that

limt→x−

f(t) < r(x) < limx→t+

f(t) .

It is easy to see that if f is discontinuous at x1 and at x2, x1 6= x2, then r(x1) 6= r(x2).Therefore there is a one-to-one correspondence between the set of points where f isdiscontinuous and the following subset of Q

r(x) : f is discontinuous at x

which clearly is at most countable. 2

73

Corollary 100 If f : [a, b]→ IR has bounded variation, then the set of points at whichf is discontinuous is at most countable.

We will prove now a deep and important result.

Theorem 101 Functions of bounded variation are differentiable a.e.

Proof. It suffices to prove that nondecreasing functions are differentiable a.e.

Lemma 102 If f : [a, b]→ IR is nondecreasing, then

lim suph→0+

f(x+ h)− f(x)

h<∞ a.e. (49)

Proof. Let E be the set of points where the upper limit (49) equals∞. Fix an arbitraryK > 0. For every x ∈ E there is an arbitrarily small interval [x, x+ h] such that

f(x+ h)− f(x)

h> K ,

i.e.h < K−1(f(x+ h)− f(x)) . (50)

Such intervals cover the set E in the Vitali sense. Accordingly, we can select disjointintervals I1, I2, . . . such that |E \

⋃∞k=1 Ik| = 0, so |E| ≤

∑∞k=1 |Ik| and hence

|E| ≤∞∑

k=1

|Ik| < K−1(f(b)− f(a))

by (50) and the fact that f is nondecreasing. Since the above inequality holds for anyK > 0 it follows that |E| = 0. 2

Definition. The Dini derivatives are defined as follows

D+f(x) = lim suph→0+

f(x+ h)− f(x)

h, D+f(x) = lim inf

h→0+

f(x+ h)− f(x)

h,

D−f(x) = lim suph→0−

f(x+ h)− f(x)

h, D−f(x) = lim inf

h→0−

f(x+ h)− f(x)

h.

The lemma says that D+f <∞ a.e. Similarly we prove that all other Dini deriva-tives D+f,D

−f,D−f are finite a.e.

We need to prove that

D+f = D+f = D−f = D−f a.e. (51)

74

To this end it suffices to prove that

D−f ≥ D+f a.e. and D+f ≥ D−f a.e.

because then (51) will follow from the inequality

D+f ≥ D+f ≥ D−f ≥ D−f ≥ D+f a.e.

We will prove only that D−f ≥ D+f a.e. as the prof of the inequality D+f ≥ D−fa.e. is very similar.

We need to prove that the set D−f < D+f has measure zero. We have

D−f < D+f =⋃

0<u<vu,v∈Q

D−f < u < v < D+f︸︷︷︸E(u,v)

.

This is a countable union of sets E(u, v) and thus it suffices to prove that each setE(u, v) has measure zero. Let U be an arbitrary open set such that E(u, v) ⊂ U . Forevery x ∈ E(u, v) there is an arbitrarily small h > 0 such that [x − h, x] ⊂ U and(f(x− h)− f(x))/(−h) < u, i.e.14

x∨x−h

f < hu . (52)

Such intervals cover the set E(u, v) in the Vitali sense, so we can select pairwise disjointintervals I1, I2, . . . such that

|E(u, v) \∞⋃

k=1

Ik| = 0 .

We have∞∑

k=1

∨Ik

f ≤ u∞∑

k=1

|Ik| ≤ u|U | .

The first inequality follows from (52) and the second inequality from the fact that theintervals I1, I2, . . . are disjoint and contained in U .

Denote15 G = E(u, v)∩⋃∞

k=1 int Ik. Clearly |E(u, v)\G| = 0. For every x ∈ G thereis an arbitrarily short interval [x, x+ h] ⊂

⋃k int Ik such that (f(x+ h)− f(x))/h > v

and hencex+h∨

x

f > hv .

14We take [x− h, x] because D−f < u is a condition for the limit as h→ 0−.15int Ik denotes the interior of Ik.

75

These intervals cover the set G in the Vitali sense, so we can select pairwise disjointintervals J1, J2, . . . ⊂

⋃k Ik such that |G \

⋃k Jk| = 0 and

∞∑k=1

∨Jk

f > v

∞∑k=1

|Jk| = v

∣∣∣∣∣∞⋃

k=1

Jk

∣∣∣∣∣ .We have

v|E(u, v)| = v|G| ≤ v

∣∣∣∣∣∞⋃

k=1

Jk

∣∣∣∣∣ <∞∑

k=1

∨Jk

f ≤∞∑

k=1

∨Ik

f ≤ u|U | .

Since the measure of U can be arbitrarily close to the measure of E(u, v) we obtain

v|E(u, v)| ≤ v|E(u, v)|

and hence |E(u, v)| = 0, because v > u. The proof is complete. 2

Theorem 103 If a function f : [a, b] → IR has bounded variation, then f ′ ∈ L1(a, b).If in addition f is nondecreasing, then∫ b

a

f ′(x) dx ≤ f(b)− f(a) .

Example. The Cantor Staircase Function. This is a continuous nondecreasing functionf : [0, 1]→ [0, 1] such that f(0) = 0, f(1) = 1 and f is constant in each of the intervalsin the complement of the Cantor set. Since the Cantor set has measure zero we concludethat f ′(0) = 0 a.e.

Therefore the function f has bounded variation (as a nondecreasing function) and

0 =

∫ 1

0

f ′(x) dx < f(1)− f(0) = 1 .

76

Actually Sierpinski provided a more surprising example. He constructed a strictlyincreasing function with the derivative equal to zero a.e.

Proof of the theorem. We can assume that f is nondecreasing. Let aε > a and bε < bbe points of differentiability of the function F (x) =

∫ x

af(t) dt such that |a − aε| < ε,

|b− bε| < ε and F ′(aε) = f(aε), F′(bε) = f(bε). Since

limh→0+

f(x+ h)− f(x)

h= f ′(x) a.e.

Fatou’s lemma yields∫ bε

aε

f ′(x) dx ≤ lim infh→0+

∫ bε

aε

f(x+ h)− f(x)

hdx = lim inf

h→0

(∫ bε+h

bε

f −∫ aε+h

aε

f

)= f(bε)− f(aε) ≤ f(b)− f(a) .

Passing to the limit with ε→ 0 yields the result. 2

8.1 Absolutely continuous functions

Definition. We say that a function f : [a, b]→ IR is absolutely continuous if for everyε > 0 there is δ > 0 such that if (x1, x1 + h1), . . . , (xk, xk + hk) are pairwise disjointintervals in [a, b] of total length less than δ, i.e.

∑ki=1 hi < δ, then

k∑i=1

|f(xi + hi)− f(xi)| < ε .

Theorem 104 Absolutely continuous functions have bounded variation.

We leave the proof as an exercise.

Theorem 105 If f, g : [a, b] → IR are absolutely continuous, then also the functionsf ± g and fg are absolutely continuous. If, in addition, g ≥ c > 0 on [a, b], then f/gis absolutely continuous.


The following result provides a complete characterization of absolutely continuousfunctions as the functions for which the fundamental theorem of calculus holds.

Theorem 106 If f ∈ L1([a, b]), then the function F (x) =∫ x

af(t) dt is absolutely

continuous. On the other hand if F is absolutely continuous, then its derivative isintegrable F ′ ∈ L1([a, b]) and

F (x) = F (a) +

∫ x

a

F ′(t) dt for all x ∈ [a, b].

77

Proof. Absolute continuity of the function F (x) =∫ x

af(t) dt follows from the absolute

continuity of the integral, Theorem 52.

If F is absolutely continuous, then it has bounded variation and hence F ′ ∈ L1([a, b])by Theorem 103. This implies that the function

x 7→∫ x

a

F ′(t) dt

is absolutely continuous. Hence the function

ϕ(x) = F (x)− F (a)−∫ x

a

F ′(t) dt

is absolutely continuous, ϕ(a) = 0, and16 ϕ′(x) = 0 for a.e. x. It suffices to prove thatϕ(x) = 0 for all x. This will immediately follow from the next lemma.

Lemma 107 If ϕ : [a, b] → IR is absolutely continuous and ϕ′(x) = 0 a.e., then ϕ isconstant.

Proof. Fix β ∈ (a, b) and ε > 0. For almost every x ∈ (a, β) there is an arbitrary smallinterval [x, x+ h] ⊂ (a, β) such that∣∣∣∣ϕ(x+ h)− ϕ(x)

h

∣∣∣∣ < ε .

Such intervals form a Vitali covering of the set of points in (a, β) where the derivative ofϕ equals 0 (almost all points in (a, β)). Hence we can select pairwise disjoint intervalsfrom this covering [x1, x1 + h1], [x2, x2 + h2], . . . ⊂ (a, β) that cover almost all points of(a, β).

For a given ε > 0 we choose δ > 0 as in the condition for the absolute continuityof ϕ. For N sufficiently large, the set (a, β) \

⋃Ni=1[xi, xi + hi] has measure less than

δ. This set is a union of N + 1 open intervals. Denote these intervals by (ξj, ξ′j),

j = 1, 2, . . . , N + 1. We have

|ϕ(β)− ϕ(a)| ≤N∑

i=1

|ϕ(xi + hi)− ϕ(xi)|+N+1∑j=1

|ϕ(ξ′j)− ϕ(ξj)| ≤ ε(β − a) + ε .

Since this inequality holds for any ε > 0 we conclude that ϕ(β) = ϕ(a), so the functionϕ is constant. This proves the lemma and also completes the proof of the theorem. 2

Theorem 108 (Integration by parts) If the functions f, g : [a, b] → IR are abso-lutely continuous, then ∫ b

a

fg′ = fg|ba −∫ b

a

f ′g .

16Because ddt

∫ x

aF ′(t) dt = F ′(x) a.e. by the Lebesgue differentiation theorem.

78

Proof. Clearly fg′ = (fg)′ − f ′g a.e. and absolute continuity of fg implies∫ b

a

fg′ =

∫ b

a

(fg)′ −∫ b

a

f ′g = fg|ba −∫ b

a

f ′g .

2

Theorem 109 Every function of bounded variation is a sum of an absolutely contin-uous function and a singular function, i.e. such a function that its derivative equals 0a.e.

Proof. The theorem follows immediately from the equality

f(x) =(f(x)−

∫ x

a

f ′(t) dt)

+

∫ x

a

f ′(t) dt .

2

Example. Every Lipschitz function f : [a, b]→ IR is absolutely continuous and henceit is differentiable a.e. and satisfies the claim of the fundamental theorem of calculus.

Definition. For a function f : [a, b]→ IR, the Banach indicatrix is defined by

N(f, y) = #f−1(y) for y ∈ IR,

i.e. N(f, y) is the number of points in the preimage of y (it can be equal to 0 or to ∞for some y).

Theorem 110 (Banach) For a continuous function f : [a, b] → IR its indicatrix ismeasurable and

b∨a

f =

∫IR

N(f, y) dy . (53)

Remark. We do not assume that the function f has bounded variation, so if thefunction f has unbounded variation, then the theorem says that the integral of thefunction on the right hand side of (53) equals ∞.

Proof. Under construction.

Theorem 111 If f : [a, b]→ IR is absolutely continuous, then

b∨a

f =

∫ b

a

|f ′(t)| dt .

79

Proof. Under construction.

The above two theorems immediately imply the following result.

Corollary 112 If f : [a, b]→ IR is absolutely continuous, then∫ b

a

|f ′(t)| dt =

∫IR

N(f, y) dy .

Theorem 113 (Rademacher) If f : Ω→ IR is Lipschitz continuous, where Ω ⊂ IRn

is open, then f is differentiable a.e., i.e.

∇f(x) =

⟨∂f

∂x1

(x), . . . ,∂f

∂xn

(x)

⟩exists a.e. and

limy→x

f(y)− f(x)−∇f(x) · (y − x)

‖y − x‖= 0 for a.e. x ∈ Ω .

Proof. Let ν ∈ Sn−1 and

Dνf(x) =d

dtf(x+ tν)|t=0

be the directional derivative whenever it exists. Since Dνf(x) exists precisely when theBorel-measurable functions

lim supt→0

f(x+ tν)− f(x)

tand lim inf

t→0

f(x+ tν)− f(x)

t

coincide, the set Aν on which Dνf fails to exist is Borel-measurable. The functiont 7→ f(x+tν) is absolutely continuous and hence differentiable a.e. Thus the intersectionof the set Aν with any line parallel to ν has one dimensional measure zero and henceFubini’s theorem implies that Aν has measure zero. Accordingly, for every ν ∈ Sn−1

Dνf(x) exists for a.e. x ∈ Ω

Take any ϕ ∈ C∞0 (Ω) and note that for each sufficiently small h > 0∫

Ω

f(x+ hν)− f(x)

hϕ(x) dx = −

∫Ω

ϕ(x− hν)− ϕ(x)

−hf(x) dx .

Indeed, invariance of the integral with respect to translations yields∫Ω

f(x+ hν)ϕ(x) dx =

∫Ω

f(x)ϕ(x− hν) dx

80

and h has to be so small that x+hν ∈ Ω for all x ∈ suppϕ. The dominated convergencetheorem yields ∫

Ω

Dνf(x)ϕ(x) dx = −∫

Ω

f(x)Dνϕ(x) dx .

This is true for any ν ∈ Sn−1 and in particular we have∫Ω

∂f

∂xi

(x)ϕ(x) dx = −∫

Ω

f(x)∂ϕ

∂xi

(x) dx for i = 1, 2, . . . , n.

Now ∫Ω

Dνf(x)ϕ(x) dx = −∫

Ω

f(x)Dνϕ(x) dx = −∫

Ω

f(x) (∇ϕ(x) · ν) dx

= −n∑

i=1

∫Ω

f(x)∂ϕ

∂xi

(x) νi dx =n∑

i=1

∫Ω

ϕ(x)∂f

∂xi

(x) νi dx

=

∫Ω

ϕ(x)(∇f(x) · ν) dx .

Lemma 114 If g ∈ L1loc(Ω) and

∫Ωg(x)ϕ(x) dx = 0 for every ϕ ∈ C∞

0 (Ω), then g = 0a.e.

We leave the proof of this lemma as an exercise. The lemma implies that

Dνf(x) = ∇f(x) · ν a.e.

Now let ν1, ν2, . . . be a countable dense subset of Sn−1 and let

Ak = x ∈ Ω : ∇f(x), Dνkf(x) exists and Dνk

f(x) = ∇f(x) · νk .

Let A =⋂∞

k=1Ak. Clearly |Ω \ A| = 0 and

Dνkf(x) = ∇f(x) · νk for all x ∈ A and all k = 1, 2, . . .

We will prove that f is differentiable at each point of the set A. To this end, for anyx ∈ A, ν ∈ Sn−1 and h > 0 define

Q(x, ν, h) =f(x+ hν)− f(x)

h−∇f(x) · ν

and it suffices to prove that if x ∈ A, then for every ε > 0 there is δ > 0 such that

|Q(x, ν, h)| < ε whenever 0 < h < δ and ν ∈ Sn−1.

Let the function f be L-Lipschitz, i.e. |f(x) − f(y)| ≤ L|x − y| for all x, y ∈ Ω. Thisimplies that |∂f/∂xi| ≤ L a.e. and thus |∇f(x)| ≤

√nL a.e. Hence for any x ∈ A,

ν, ν ′ ∈ Sn−1 and h > 0

|Q(x, ν, h)−Q(x, ν ′, h)| ≤ (√n+ 1)L|ν − ν ′| .

81

Given ε > 0 let p be so large that for each ν ∈ Sn−1

|ν − νk| <ε

2(√n+ 1)L

for some k = 1, 2, . . . , p.

Sincelim

h→0+Q(x, νi, h) = 0 for all x ∈ A and all i = 1, 2, . . .

we see that for a given x ∈ A there is δ > 0 such that

|Q(x, νi, h)| < ε/2 whenever 0 < h < δ and i = 1, 2, . . . , p .

Now

|Q(x, ν, h)| ≤ |Q(x, νk, h)|+ |Q(x, νk, h)−Q(x, ν, h)| < ε/2 + (√n+ 1)L|νk − ν| < ε

whenever ν ∈ Sn−1 and 0 < h < δ and the theorem follows. 2

Theorem 115 (McShane) If f : A → IR is an L-Lipschitz function defined on asubset A ⊂ X of a metric space X, then there is an L-Lipschitz function f : X → IRsuch that f |A = f .

Proof. For x ∈ X we define

f(x) = infy∈A

(f(y) + Ld(x, y))

and it a a routine exercise to check that f is L-Lipschitz and satisfies f |A = f . 2

Theorem 116 (Stepanov) Let Ω ⊂ IRn be open. Then a measurable function f :Ω→ IR is differentiable a.e. in Ω if and only if

lim supy→x

|f(f)− f(x)|‖y − x‖

<∞ for a.e. x ∈ Ω. (54)

The implication ⇒ is obvious. To prove the implication ⇐ first we show that (54)implies that f is Lipschitz continuous on a big set A in the sense that the measure ofΩ \ A is small. Then we extend f |A to a Lipschitz function on IRn using McShane’stheorem. Now f is differentiable a.e. by Rademacher theorem. In particular it isdifferentiable at a.e. point of A and it remains to prove that f is differentiable at everydensity point of A at which f is differentiable. We leave details as an exercise. 2

82

9 Signed measures

Definition. Let M be a σ-algebra. A function µ : M → [−∞,∞] is called signedmeasure if

1. µ(∅) = 0;

2. µ attains at most one of the values ±∞;

3. µ is countably additive.

Examples.

1. µ = µ1 − µ2, where µ1 and µ2 are (positive) measures and one of them is finite.

2.

µ(E) =

∫E

f dν, f ∈ L1(ν) .

Definition. A measurable set E ∈M is called positive if µ(A) > 0 for every measur-able subset A ⊂ E.

Example. If E = E1 ∪ E2, E1 ∩ E2 = ∅, µ(E1) = 7, µ(E2) = −3, then µ(E) = 4 > 0,but E is not positive.

Lemma 117 If µ is a signed measure, then every measurable set such that 0 < µ(E) <∞ contains a positive set A of positive measure.

Proof. If E is positive, then we take A = E. Otherwise there is B ⊂ E, µ(B) < 0. Letn1 be the smallest positive integer such that there is B1 ⊂ E with

µ(B1) ≤ −1

n1

.

If A1 = E \ B1 is positive, then we take A = A1 (observe that µ(A1) > 0 as otherwiseµ(E) < 0). If A1 is not positive, then let n2 be the smallest positive integer such thatthere is B2 ⊂ E \B1 with

µ(B2) ≤ −1

n2

.

If A2 = E\(B1∪B2) is positive, then we take A = A2 (observe that µ(A2) > 0). If A2 isnot positive, then we define A3 as above and so on. If Am = E \

⋃mj=1Bj is positive for

some m, then we take A = Am (µ(Am) > 0). Otherwise we obtain an infinite sequenceBj and we set

A = E \∞⋃

j=1

Bj .

83

We will prove that A is positive and µ(A) > 0. We have

IR 3 µ(A) = µ(E)−∞∑

j=1

µ(Bj)

and hence

0 < µ(E) = µ(A) +∞∑

j=1

µ(Bj) ≤ µ(A)−∞∑

j=1

1

nj

∞∑j=1

1

nj

< µ(A) . (55)

Note that µ(A) < ∞ as otherwise µ(E) = µ(A) + µ(E \ A) = ∞ + µ(E \ A) = ∞.Hence (55) yields µ(A) > 0 and

∑∞j=1 1/nj < ∞. In particular nj → ∞ ad j → ∞.

It remains to prove that A is positive. If C ⊂ A = E \⋃∞

j=1Bj is measurable, the forevery m, C ⊂ E \

⋃mj=1Bj and the definition of nm+1 yields

µ(C) > − 1

nm+1 − 1→ 0 as m→∞

and hence µ(C) ≥ 0. This proves positivity of A. 2

Theorem 118 (Hahn’s decomposition theorem) If µ is a signed measure, thenthere exist two disjoint measurable sets X+ and X− such that

X = X+ ∪X−, X+ ∩X− = ∅ ,

µ is positive on X+ and µ is negative on X− (i.e. −µ is positive on X−).

Therefore every signed measure is on the form

µ = µ+ − µ−

where µ+ and µ− are positive measures defined by µ+(E) = µ(E ∩ X+), µ−(E) =−µ(E ∩X−). Moreover the measures µ+ and µ− are concentrated on disjoint sets X+

and X−.

Exercise. Prove that the decomposition X = X+∪X− is unique up to sets of µ-measurezero, i.e. if X = X+ ∪ X− is another decomposition, then

µ(X+ \ X+) = µ(X+ \X+) = µ(X− \ X−) = µ(X− \X−) = 0 .

Example. Let µ(E) =∫

Ef dν, where f ∈ L1(ν). Then µ is a signed measure and

µ(E) =

∫E

f+ dν −∫

E

f− dν .

84

Hence we can take

X+ = x : f(x) ≥ 0, X− = x : f(x) < 0 ,

but we also can take

X+ = x : f(x) > 0, X− = x : f(x) ≤ 0 .

In general X+ 6= X+, X− 6= X−, however, the sets differ by the set where f equals 0and this set has measure zero

µ(x : f(x) = 0) =

∫f=0

f dν = 0 .

Proof of Hahn’s theorem. Suppose µ does not attain value +∞. Let

M = supµ(A) : A ∈M, A is positive .

Then there is a sequence of positive sets

A1 ⊂ A2 ⊂ A3 ⊂ . . . ; µ(Ai)→M

and if A =⋃∞

i=1Ai, then A is positive with µ(A) = M <∞. It remains to prove thatX \A is negative (then we take X+ = A, X− = X \A). For if not, there is E ⊂ X \Awith 0 < µ(E) <∞. Hence Lemma 117 implies that there is a positive subset C ⊂ Eof positive measure. Now A ∪ C is positive and

µ(A ∪ C) = µ(A) + µ(C) > M

which is an obvious contradiction. 2

As an application we will prove the Radon-Nikodym-Lebesgue theorem and thenwe will classify all continuous linear functionals on Lp(µ) for 1 ≤ p <∞.

Definition. Let µ be a (positive) measure in a σ-algebra M. If ν is another (positive)measure in M such that

E ∈M, µ(E) = 0 ⇒ ν(E) = 0 ,

then we say that ν is absolutely continuous with respect to µ and we write

ν µ .

Example. If f ≥ 0 is measurable, and a measure ν is defined by

ν(E) =

∫E

f dµ ,

85

then ν µ.

Definition. Let λ be a measure in M. We say that λ is concentrated on A ∈M if

λ(E) = λ(A ∩ E) for all E ∈M .

Equivalently λ is concentrated on A ∈M if λ(X \ A) = 0.

Definition. Let λ1, λ2 be two measures in M. If there are sets

A,B ∈M, A ∩B = ∅

such thatλ1 is concentrated on A ,

λ2 is concentrated on B ,

then we say that the measures λ1 and λ2 are mutually singular and we write

λ1 ⊥ λ2 .

Lemma 119 Let λ1, λ2 be two measures in M. If there is a set E ∈ M such thatλ1(E) = 0 and λ2 is concentrated on E, then λ1 ⊥ λ2.

Proof. It suffices to take A = X \ E and B = E. 2

Examples.

1. For a ∈ IRn the Dirac mass is a Borel measure defined by δa(E) = 1 if a ∈ E andδa(E) = 0 if a 6∈ E. The lemma yields Ln ⊥ δa.

2. If M is a k-dimensional submanifold in IRn, k < n, and λ(E) = Hk(E ∩M) forE ⊂ IRn, then Ln ⊥ λ.

3. if C is the standard ternary Cantor set and λ(E) = Hlog 2/ log 3(E∩C) for E ⊂ IR,then λ ⊥ L1.

Theorem 120 (Radon-Nikodym-Lebesgue) Let µ and ν be two (positive), σ-finitemeasures on M. Then

(a) There is a unique pair of measures λa, λs on M such that

λ = λa + λs, λa µ, λs ⊥ µ .

(b) There is a M-measurable function h ≥ 0 on X such that

λa(E) =

∫E

h dµ for all E ∈M.

If in addition λ is finite (i.e. λ(X) <∞), then the measures λa, λs are finite andh ∈ L1(µ).

86

Remark. Part (a) is due to Lebesgue and part (b) due to Radon and Nikodym.

In particular if λ µ, then

λ(E) =

∫E

h dµ (56)

for some h ≥ 0 and all E ∈ M which is a striking characterization of all absolutelycontinuous measures. The function h is called Radon-Nikodym derivative and is oftendenoted by h = dλ/dµ. With this notation (56) can be rewritten as

λ(E) =

∫E

dλ

dµdµ for all E ∈M .

Proposition 121 If λ is a finite measure, absolutely continuous with respect to theLebesgue measure, λ Ln, then

dλ

dLn

= limr→0+

λ(B(x, r))

|B(x, r)|for a.e. x ∈ IRn .

Proof. It follows from the Radon-Nikodym theorem that dλ/dLn ∈ L1(IRn) and

λ(B(x, r))

|B(x, r)|=

∫B(x,r)

dλ

dLn

dLn →dλ

dLn

(x) as r → 0+

for a.e. x ∈ IRn by the Lebesgue differentiation theorem. 2

Proof of the Radon-Nikodym-Lebesgue theorem. First we will prove the second part ofthe theorem which is due to Radon and Nikodym. We can state it as follows.

Theorem 122 (Radon-Nikodym) If λ and µ are σ-finite measures on a σ-algebraM and λ µ, then there is a nonnegative measurable function h : X → [0,∞] suchthat

λ(E) =

∫E

h dµ for all E ∈M .

If in addition λ is finite, then h ∈ L1(µ).

Proof. We will prove the theorem under the additional assumption that µ(X) < ∞and λ(X) < ∞. The general case of σ-finite measures easily follows from this specialcase and the details are left to the reader. Let

Φ = ϕ : X → [0,∞] : ϕ is measurable and

∫E

ϕdµ ≤ λ(E) for all E ∈M .

Clearly Φ 6= ∅, because ϕ ≡ 0 ∈ Φ. If ϕ1, ϕ2 ∈ Φ, then maxϕ1, ϕ2 ∈ Φ. Indeed,∫E

maxϕ1, ϕ2 dµ =

∫E∩ϕ1≥ϕ2

ϕ1 dµ+

∫E∩ϕ1<ϕ2

ϕ2 dµ

≤ λ(E ∩ ϕ1 ≥ ϕ2) + λ(E ∩ ϕ1 < ϕ2) = λ(E) .

87

Define

M = supϕ∈Φ

∫X

ϕdµ ≤ λ(X) <∞ .

There is a sequence ϕn ∈ Φ such that∫X

ϕn dµ→M as n→∞ .

Let hn = maxϕ1, ϕ2, . . . , ϕn ∈ Φ. Then hn is an increasing sequence of functionspointwise convergent to a function h. Since

∫Ehn dµ ≤ λ(E) for all E ∈ M it follows

from the monotone convergence theorem that∫

Eh dµ ≤ λ(E) for all E ∈ M. Hence

h ∈ Φ. Moreover ∫X

h dµ = limn→∞

∫X

hn dµ = M .

We will show that ∫E

h dµ = λ(E) for all E ∈M .

Clearly

η(E) = λ(E)−∫

E

h dµ

defines a positive measure in M and we need to show that η ≡ 0. For if not there wouldexist a set A ∈M such that η(A) > 0. This implies that λ(A) > 0 and hence µ(A) > 0(because λ µ). Thus there is ε > 0 such that η(A)− εµ(A) > 0. Define ξ = η − εµ.Then ξ is a signed measure and we can assume that A is positive with respect to ξ byLemma 117. Hence if E ∈M, ξ(A ∩ E) ≥ 0 and thus

0 ≤ ξ(A ∩ E) = η(A ∩ E)− εµ(A ∩ E) = λ(A ∩ E)−∫

A∩E

h dµ− εµ(A ∩ E) ,∫A∩E

h dµ+ εµ(A ∩ E) ≤ λ(A ∩ E) ,

but also ∫E\A

h dµ ≤ λ(E \ A)

because h ∈ Φ. Adding up the two inequalities yields∫E

(h+ εχA) dµ =

∫E

h dµ+ εµ(A ∩ E) ≤ λ(E) .

This implies that h+ εχA ∈ Φ. Since∫X

(h+ εχA) dµ = M + εµ(A) > M ,

we get a contradiction. The proof is complete. 2

Now we can prove the first part of the theorem which is due to Lebesgue.

88

Theorem 123 (Lebesgue) If λ and µ are σ-finite measures, then there exist uniquemeasures λa and λs such that

λ = λa + λs, λa µ, λs ⊥ µ .

Proof. We leave the proof of the uniqueness as an exercise and so we are left with theproof of the existence of λa and λs. Let ν = µ + λ. Then µ ν and λ ν. By theRadon-Nikodym theorem

µ(E) =

∫E

f dν, λ(E) =

∫E

g dν

for some measurable functions f, g : X → [0,∞] and all E ∈M. Define

λ0(E) = λ(E ∩ f = 0), λ1(E) = λ(E ∩ f > 0) .

Since λ = λ0 + λ1 it suffices to prove that

λ0 ⊥ µ and λ1 µ .

The measure λ0 is concentrated on the set f = 0. Since µ(f = 0) = 0, we concludethat λ0 ⊥ µ. To prove that λ1 µ let µ(E) = 0. Then f = 0, ν-a.e. in the set E, i.e.ν(E ∩ f > 0) = 0 and thus

λ1(E) =

∫E∩f>0

g dν = 0 .

2

We will apply now the Radon-Nikodym theorem to classify all continuous linearfunctionals on Lp(µ) for 1 ≤ p < ∞. Let us start first with some general facts aboutcontinuous linear mappings between normed spaces.

Theorem 124 Let L : X → Y be a linear mapping between normed spaces. Then thefollowing conditinos are equivalent

(a) L is continuous;

(b) L is continuous at 0;

(c) L is bounded, i.e. there is C > 0 such that

‖Lx‖ ≤ C‖x‖ for all x ∈ X .

89

Remark. Formally we should use different symbols to denote norms in spaces X andY . Since it will always be clear in which space we take the norm it is not dangerous touse the same symbol ‖ · ‖ to denote apparently different norms in X and Y .

Proof. The implication (a)⇒(b) is obvious.

(b)⇒(c). Suppose L is continuous at 0 but not bounded. Then there is a sequencexn ∈ X such that

‖Lxn‖ ≥ n‖xn‖ .

Hence0←

∥∥∥L xn

n‖xn‖

∥∥∥ ≥ 1

which is a contradiction17.

(c)⇒(a). Let xn → x. Then

‖Lx− Lxn‖ = ‖L(x− xn)‖ ≤ C‖x− xn‖ → 0

and hence Lxn → Lx which proves continuity of L. 2

In particular if X is a normed space over IR (or C), then a linear functional

Λ : X → IR (or C)

is continuous if and only if it is bounded, i.e. there is C > 0 such that

|Λx| ≤ C‖x‖ for all x ∈ X. (57)

The number‖Λ‖ = sup

‖x‖≤1

|Λx|

is called the norm of Λ.

Lemma 125 The norm Λ is the smallest number C for which the inequality (57) issatisfied.

Proof. Suppose |Λx| ≤ C‖x‖ for all x ∈ X. If ‖x‖ ≤ 1, then |Λx| ≤ C‖x‖ ≤ C andhence

‖Λ‖ = sup‖x‖≤1

|Λx| ≤ C .

Thus the number C is (57) cannot be smaller than ‖Λ‖. It remains to prove that

|Λx| ≤ ‖Λ‖ ‖x‖ for all x ∈ X .

17Convergence to 0 follows from the continuity of L at 0.

90

If x = 0, then the inequality is obvious. If x 6= 0, then

|Λx| =∣∣∣∣Λ x

‖x‖

∣∣∣∣ ‖x‖ ≤ ‖Λ‖‖x‖ .The last inequality follows from the fact that ‖x/‖x‖‖ = 1. 2

Example. If 1/p+ 1/q = 1, 1 < p, q <∞ are Holder conjugate, and g ∈ Lq, then

Λgf =

∫X

fg dµ

defines a bounded linear functional on Lp(µ) and ‖Λg‖ ≤ ‖g‖q. Indeed, Holder’sinequality yields

|Λgf | =∣∣∣∣∫

X

fg dµ

∣∣∣∣ ≤ ‖g‖q‖f‖p .Similarly if g ∈ L∞(µ), then

Λgf =

∫X

fg dµ

defines a bounded linear functional on L1(µ) and ‖Λg‖ ≤ ‖g‖∞.

It turns out that under reasonable assumptions every functional on Lp(µ), 1 ≤ p <∞ is on the form Λg for some g ∈ Lq (or g ∈ L∞ if p = 1).

Theorem 126 Suppose 1 ≤ p < ∞ and µ is a σ-finite measure. Assume that Λ is abounded linear functional on Lp(µ). Then there is a unique g ∈ Lq(µ), 1/p + 1/q = 1such that

Λf =

∫X

fg dµ for all f ∈ Lp(µ) .

Moreover‖Λ‖ = ‖g‖q .

Proof. Uniqueness. Suppose

Λf =

∫X

fg1 dµ =

∫X

fg2 dµ

for some g1, g2 ∈ Lq(µ) and all f ∈ Lp(µ). We need to show that g1 = g2 a.e. ForE ∈M with µ(E) <∞ we have∫

E

g1 dµ = Λ(χE) =

∫E

g2 dµ .

Hence ∫E

(g1 − g2) dµ = 0

91

for every E ∈M with µ(E) <∞. This easily implies that g1 − g2 = 0 a.e. and henceg1 = g2 µ-a.e.

We will prove the theorem under the additional assumption that µ(X) < ∞. Thegeneral case of σ-finite measure can be reduced to the case of finite measure and weleave details to the reader. Define

λ(E) = Λ(χE) .

It is obvious that λ is finitely additive

λ(A ∪B) = λ(A) + λ(B) for A,B ∈M, A ∩B = ∅ .

Indeed,

λ(A ∪B) = Λ(χA∪B) = Λ(χA + χB) = Λ(χA) + Λ(χB) = λ(A) + λ(B) .

We will use this fact to prove that actually λ is conuntably additive. Let E =⋃∞

i=1Ei,where Ei ∈M and Ei ∩ Ej = ∅ for i 6= j. Let Ak = E1 ∪ . . . ∪ Ek. We have

‖χE − χAk‖p = µ(E \ Ak)1/p → 0 as k →∞ ,

because the sets E \Ak form a decreasing sequence of sets with empty intersection andµ(E \ A1) ≤ µ(X) <∞. Continuity of the functional λ implies

ΛχAk→ ΛχE

and hence finite additivity of λ yields

k∑i=1

λ(Ei) = λ(Ak)→ λ(E) as k →∞ .

Accordingly∞∑i=1

λ(Ei) = λ(E)

which is countable additivity of λ. Note that λ is not necessarily positive, so

λ : M→ IR

is a signed measure. By the Hahn decomposition theorem λ = λ+ − λ−, where λ+ andλ− are positive finite measures concentrated on disjoint sets. Clearly µ(E) = 0 impliesthat λ(E) = 0 and also λ+(E) = λ−(E) = 0. Hence λ+, λ− µ. The Radon-Nikodymtheorem yields the existence of 0 ≤ g± ∈ L1(µ) such that

λ±(E) =

∫E

g± dµ .

92

Now

λ(E) =

∫E

g dµ where g = g+ − g− ∈ L1(µ).

We can write the last statement as

Λ(χE) =

∫X

χEg dµ .

Since simple functions are linear combinations of characteristic functions

Λf =

∫X

fg dµ (58)

whenever f is a simple function. Now every f ∈ L∞(µ) is a uniform limit of a sequenceof simple functions and hence (58) holds for all f ∈ L∞(µ). It remains to prove thefollowing three statements

(a) g ∈ Lq(µ) (we know that g ∈ L1(µ));

(b) ‖Λ‖ = ‖g‖q (we know that ‖Λ‖ ≤ ‖g‖q by Holder’s inequality);

(c) Λf =∫

Xfg dµ for all f ∈ Lp(µ) (we know it for f ∈ L∞(µ)) .

We will complete the proof in the case 1 < p <∞. The case p = 1 is easier and left tothe reader. Let

α(x) =

g(x)/|g(x)| if g(x) 6= 0,

0 if g(x) = 0,

andEn = x : |g(x)| ≤ n .

Definef = |g|q/pαχEn .

Clearly f ∈ L∞(µ) (because |f | ≤ nq/p). Moreover

|f |p (a)= |g|qχEn

(b)= fg .

Hence∫En

|g|q dµ by (b)=

∫X

fg dµf∈L∞

= Λf ≤ ‖Λ‖‖f‖pby (a)

= ‖Λ‖(∫

En

|g|q dµ)1/p

.

This yields (∫En

|g|q dµ)1−1/p

≤ ‖Λ‖ .

93

The left hand side converges to ‖g‖q as n → ∞ and hence ‖g‖q ≤ ‖Λ‖. Since wealready know the opposite inequality ‖Λ‖ ≤ ‖g‖q we conclude

‖Λ‖ = ‖g‖q .

This yields the claims (a) and (b). Now (c) follows from the continuity of the functional.Indeed, there is a sequence18 L∞ 3 fk → f in Lp and hence∣∣∣∣Λf − ∫

X

fg dµ

∣∣∣∣ ≤ |Λf − Λfk|+∣∣∣∣Λfk −

∫X

fg dµ

∣∣∣∣ = |Λf − Λfk|+∣∣∣∣∫

X

(fk − f)g dµ

∣∣∣∣≤ |Λf − Λfk|+ ‖fk − f‖p‖g‖q → 0 as k →∞ .

which proves (c). 2

Definition. Let X be a locally compact metric space. The class of continuous func-tions vanishing at infinity C0(X) consists of all continuous functions u : X → IR suchthat for every ε > 0 there is a compact set K ⊂ X such that |u(x)| < ε for all x ∈ X\K.

Exercise. Prove that u ∈ C0(IRn) if and only if u is continuous on IRn and

lim|x|→∞ u(x) = 0.

Example. The set of positive integers IN is a locally compact metric space withrespect to the metric |x− y|. Functions on IN can be identified with infinite sequences.Indeed, a sequence (xi)

∞i=1 corresponds to a function f(i) = xi. Every function on IN

is continuous and compact sets in IN are exactly finite subsets. It easily follows thatC0(IN) consists of sequences such that |xi| → 0 as i → ∞. This space is denoted byc0 = C0(IN). Thus

c0 = (xi)∞i=1 : xi ∈ IR, |xi| → 0 as i→∞ .

It is easy to see that c0 is a Banach space with respect to the norm ‖x‖∞ =supi=1,2,... |xi|, where x = (xi)

∞i=1. Therefore c0 is a closed subspace in `∞.

Recall that Cc(X) consists of all continuous functions on X that have compactsupport. Clearly Cc(X) ⊂ C0(X).

Theorem 127 Let X be a locally compact metric space. Then C0(X) is a Banachspace with the norm

‖u‖∞ = supx∈X|u(x)| .

Moreover Cc(X) is a dense subset of C0(X).


Observe that if µ is a measure on X, then C0(X) is a closed subspace of L∞(µ).Hence C0(X) is the closure of Cc(X) in the metric of L∞(µ). If µ is a Radon measure on

18Because simple functions are dense in Lp.

94

a locally compact metric space, then the closure of Cc(X) in the metric of Lp(µ) equalsLp(µ) for all 1 ≤ p <∞ (Theorem 86). However the closure of Cc(X) in the metric ofL∞(µ) equals C0(X) L∞(µ). We managed to characterize bounded linear functionalon Lp(µ) for all 1 ≤ p < ∞ (Theorem 126). It is not so easy to characterize boundedlinear functionals on L∞(µ). However Theorem 128 below provides a characterizationof bounded linear functionals on C0(X).

Recall that if µ is a signed measure, then Hahn’s decomposition theorem provides aunique representation µ = µ+−µ−, where µ+ and µ− are positive measures concentratedon disjoint sets. We define the measure |µ| by |µ| = µ++µ− and call the number |µ|(X)the total variation of µ.

If a signed measure µ is given by

µ(E) =

∫E

f dλ for some f ∈ L1(λ) ,

then

|µ|(E) =

∫E

|f | dλ .

If µ is a signed measure and f ∈ L1(|µ|), then we define∫X

fdµ =

∫X

f dµ+ −∫

X

f dµ− .

Theorem 128 (Riesz Representation Theorem) Let X be a locally compact met-ric space and let Λ be a bounded linear functional on C0(X). Then there is a uniqueBorel signed measure µ of finite total variation such that

Λf =

∫X

f dµ for all f ∈ C0(X) . (59)

Moreover‖Λ‖ = |µ|(X) .

In the proof we will need the following abstract and general result.

Lemma 129 Let X0 ⊂ X be a dense linear subspace in a normed space X. If Λ :X0 → IR is a bounded linear functional (with respect to the norm in X), then there isa unique functional Λ : X → IR (called extension of Λ) such that

Λx = Λx for all x ∈ X0 and ‖Λ‖ = ‖Λ‖ .

Remarks. Here‖Λ‖ = sup

‖x‖≤1x∈X

|Λx| and ‖Λ‖ = sup‖x‖≤1x∈X0

|Λx| .

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Usually the extension is denoted by the same symbol Λ rather than Λ.

Example. Let X be a locally compact metric space and λ a Radon measure on X. IfΛ : Cc(X)→ IR is a linear functional such that

|Λf | ≤M‖f‖L1(λ) for all f ∈ Cc(X) ,

then there is unique bounded linear functional Λ : L1(λ) → IR such that Λf = Λf forf ∈ Cc(X) and

|Λf | ≤M‖f‖L1(λ) for all f ∈ L1(λ) .

Indeed, Cc(X) is a linear and dense subspace of L1(λ) by Theorem 86.

Proof of the lemma. Let X0 3 xn → x ∈ X. Then we define

Λx = limn→∞

Λxn (60)

and we have to prove that:

(a) the limit exists;

(b) the limit does not depend on the particular choice of a sequence xn ∈ X0 thatconverges to x;

(c) Λx = Λx for x ∈ X0;

(d) ‖Λ‖ = ‖Λ‖.

If xn → x, xn ∈ X0, then

|Λxn − Λxm| = |Λ(xn − xm)| ≤ ‖Λ‖‖xn − xm‖ → 0 as n,m→∞ ,

which shows that Λxn is a Cauchy sequence of real numbers and hence it is convergent.This proves (a). If xn → x, yn → x, xn, yn ∈ X0, then

|Λxn − Λyn| = |Λ(xn − yn)| ≤ ‖Λ‖‖xn − yn‖ → 0 as n→∞

which proves (b). If x ∈ X0, then taking xn = x we have xn → x and hence Λx =limn→∞ Λx = Λx which proves (c). Clearly

‖Λ‖ = sup‖x‖≤1x∈X0

|Λx| ≤ sup‖x‖≤1x∈X

|Λx| = ‖Λ‖ ,

but we actually have equality because if x ∈ X, then there is a sequence xn ∈ X0,xn → x and hence

|Λx| = limn→∞

|Λxn| ≤ ‖Λ‖ limn→∞

‖xn‖ = ‖Λ‖‖x‖ .

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which proves that ‖Λ‖ ≤ ‖Λ‖ (see Lemma 125). The above two inequalities prove (d).2

Remark. In the proof we used the fact that every Cauchy sequence in IR is convergent,i.e. we used the fact that IR is Banach space. Therefore the same proof gives a moregeneral result about extensions of bounded linear mappings L : X0 → Y defined on adense linear subspace X0 ⊂ X of a noremd space X into a Banach space Y . The readershould find an appropriate statement.

Proof of Theorem 128. First we prove uniqueness of the measure µ. To this end we needto prove that if µ is a signed Borel measure of finite total variation such that

∫Xf dµ = 0

for all f ∈ C0(X), then µ ≡ 0. Let µ = µ+ − µ− be the Hahn decomposition withmeasures µ+ and µ− concentrated on disjoint sets X+ and X− respectively. If

h(x) =

1 if x ∈ X+,−1 if x ∈ X−,

then h2 = 1 and dµ = hd|µ|. Hence

|µ|(X) =

∫X

d|µ| =∫

X

h2 d|µ| . (61)

Let fn ∈ Cc(X) be a sequence that converges to h in L1(|µ|) (Theorem 86). Note that

0 =

∫X

fn dµ =

∫X

fnh d|µ| . (62)

Now (61) and (62) yield

|µ|(X) =

∫X

(h2 − fnh) d|µ| =∫

X

(h− fn)h d|µ| ≤∫

X

|h− fn| d|µ| → 0 as n→∞ .

Hence |µ|(X) = 0 and thus µ = 0.

Now let Λ be a bounded linear functional on C0(X). We can assume without lostof generality that19 ‖Λ‖ = 1. We would like to apply the Riesz representation theoremfor positive functionals on Cc(X), but unfortunately the functional Λ is not necessarilypositive. So we want to construct a positive functional Φ such that

|Λf | ≤ Φ(|f |) ≤ ‖f‖∞ for all f ∈ Cc(X) . (63)

Once the functional Φ is constructed we can finish the proof of the theorem as follows.By the Riesz representation theorem there is a (positive) Radon measure λ such that

Φ(f) =

∫X

f dλ for all f ∈ Cc(X) .

19Otherwise we multiply Λ by ‖Λ‖−1.

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Observe thatλ(X) = supΦ(f) : 0 ≤ f ≤ 1, f ∈ Cc(X) . (64)

Indeed, for 0 ≤ f ≤ 1, f ∈ Cc(X) we have∫

Xf dλ ≤ λ(X). On the other hand for

every compact set K ⊂ X there is 0 ≤ f ≤ 1, f ∈ Cc(X) such that f = 1 on K andhence

∫Xf dλ ≥ λ(K). Now (64) follows from the fact that

λ(X) = supK⊂X

K−compact

λ(K) .

Observe that (64) impliesλ(X) ≤ 1. (65)

It follows from (63) that

|Λ(f)| ≤ Φ(|f |) =

∫X

|f | dλ = ‖f‖L1(λ) .

Thus Λ is a bounded linear functional defined on a linear subspace20 Cc(X) ⊂ L1(λ).Hence the functional can be uniquely extended to a bounded linear functional Λ onL1(λ) such that Λf = Λf for f ∈ Cc(X) and

|Λf | ≤ ‖f‖L1(λ) for all f ∈ L1(λ) .

According to Theorem 126 there is a unique function g ∈ L∞(λ), ‖g‖∞ ≤ 1 such that

Λf =

∫X

fg dλ for all f ∈ L1(λ).

If f ∈ C0(X) and fn ∈ Cc(X), fn → f in the norm of C0(X), then

Λf = limn→∞

∫X

fng dλ = Λfn → Λf

and hence Λf = Λf for all f ∈ C0(X). Thus

Λf =

∫X

fg dλ for all f ∈ C0(X) .

Accordingly, we have the representation (59) with dµ = gdλ. Since ‖Λ‖ = 1 andd|µ| = |g| dλ it follows that

|µ|(X) =

∫X

|g| dλ ≥ sup|Λf | : f ∈ C0(X), ‖f‖∞ ≤ 1 = ‖Λ‖ = 1 .

On the other hand ‖g‖∞ ≤ 1 and λ(X) ≤ 1 so

|µ|(X) =

∫X

|g| dλ ≤ 1 .

20Bounded with respect to the norm of L1(λ).

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Thus|µ|(X) = 1 = ‖Λ‖

and the proof is complete. Therefore we are left with the construction of the positivefunctional Φ satisfying (63).

Denote by C+c (X) the class of nonnegative functions in Cc(X) and for f ∈ C+

c (X)define

Φ(f) = sup|Λh| : h ∈ Cc(X), |h| ≤ f .

If f ∈ Cc(X), then f = f+ − f−, f+, f− ∈ C+c (X) and we define

Φ(f) = Φ(f+)− Φ(f−) .

Clearly Φ(f) ≥ 0 for f ∈ C+c (X), so Φ is positive and Φ satisfies (63). Therefore we

only need to show that Φ is linear. Obviously Φ(cf) = cΦ(f) and hence we are leftwith the proof that

Φ(f + g) = Φ(f) + Φ(g) for all f, g ∈ Cc(X). (66)

Observe that it suffices to verify (66) for f, g ∈ C+c (X). Indeed, for f, g ∈ Cc(X) we

will have then21

(f + g)+ − (f + g)− = f+ − f− + g+ − g−

and hence(f + g)+ + f− + g− = (f + g)− + f+ + g+ .

Now, the assumed linearity of Φ on Cc(X)+ yields

Φ((f + g)+) + Φ(f−) + Φ(g−) = Φ((f + g)−) + Φ(f+) + Φ(g+)

Φ((f + g)+)− Φ((f + g)−)︸︷︷︸Φ(f+g)

= Φ(f+)− Φ(f−)︸︷︷︸Φ(f)

+ Φ(g+)− Φ(g−)︸︷︷︸Φ(g)

.

Thus we are left with the proof of (66) for f, g ∈ Cc(X)+. Fix f, g ∈ C+c (X). Given

ε > 0, there exist h1, h2 ∈ Cc(X) such that |h1| ≤ f , |h2| ≤ g and

Φ(f) ≤ |Λh1|+ ε/2, Φ(g) ≤ |Λh2|+ ε/2 .

Let c1 = ±1, c2 = ±1 be such that |Λh1| = c1Λh1, |Λh2| = c2Λh2. We have

Φ(f) + Φ(g) ≤ |Λh1|+ |Λh2|+ ε = Λ(c1h1 + c2h2) + ε

≤ Φ(|h1|+ |h2|) + ε ≤ Φ(f + g) + ε

and henceΦ(f) + Φ(g) ≤ Φ(f + g) .

21Because both sides are equal f + g.

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Now we need to prove the opposite inequality. Let h ∈ Cc(X) satisfy |h| ≤ f + g, letV = x : f(x) + g(x) > 0, and define

h1(x) =

f(x)h(x)

f(x)+g(x)if x ∈ V ,

0 if x 6∈ V ,h2(x) =

g(x)h(x)

f(x)+g(x)if x ∈ V ,

0 if x 6∈ V .

It is easy to see that the functions h1 and h2 are continuous, h = h1 +h2, and |h1| ≤ f ,|h2| ≤ g. We have

|Λh| = |Λh1 + Λh2| ≤ |Λh1|+ |Λh2| ≤ Φ(f) + Φ(g)

and after taking the supremum over all functions h we have

Φ(f + g) ≤ Φ(f) + Φ(g)

which completes the proof. 2

10 Maximal function

Definition. For a function u ∈ L1loc(IR

n) we define the Hardy-Littlewood maximalfunction by

Mu(x) = supr>0

∫B(x,r)

|u(y)| dy .

Clearly Mu : IRn → [0,∞] is a measurable function.

Theorem 130 (Mardy-Littlewood maximal theorem) If u ∈ Lp(IRn), 1 ≤ p ≤∞, then Mu <∞ a.e. Moreover

(a) If u ∈ L1(IRn), then for every t > 0

|x : Mu(x) > t| ≤ C(n)

t

∫IRn

|u| . (67)

(b) If u ∈ Lp(IRn), 1 < p ≤ ∞, then Mu ∈ Lp(IRn) and

‖Mu‖p ≤ C(n, p)‖u‖p . (68)

Remarks. Let u = χ[0,1]. Then for x ≥ 1

Mu(x) ≥ 1

2x

∫ 2x

0

χ[0,1](y) dy =1

2x6∈ L1 .

Hence the inequality (68) does not hold for p = 1. If g ∈ L1(IRn), then

t|x : |g(x) > t| =∫

IRn

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