Real Analysis II, Winter 2018

From the Finnish original “Moderni reaalianalyysi”1 by Ilkka Holopainen

adapted by Tuomas Hytönen

February 22, 2018

1Version dated September 14, 2011 Contents

1 General theory of measure and integration 2 1.1 Measures ...... 2 1.11 Metric outer measures ...... 4 1.20 Regularity of measures, Radon measures ...... 7 1.31 Uniqueness of measures ...... 9 1.36 Extension of measures ...... 11 1.45 Product measure ...... 14 1.52 Fubini’s theorem ...... 16

2 Hausdorff measures 21 2.1 Basic properties of Hausdorff measures ...... 21 2.12 Hausdorff dimension ...... 24 2.17 Hausdorff measures on Rn ...... 25 3 Compactness and convergence of Radon measures 30 3.1 Riesz representation theorem ...... 30 3.13 Weak convergence of measures ...... 35 3.17 Compactness of measures ...... 36

4 On the Hausdorff dimension of fractals 39 4.1 Mass distribution and Frostman’s lemma ...... 39 4.16 Self-similar fractals ...... 43

5 Differentiation of measures 52 5.1 Besicovitch and Vitali covering theorems ...... 52

1 Chapter 1

General theory of measure and integration

1.1 Measures

Let X be a set and P(X) = {A : A ⊂ X} its power set. Definition 1.2. A collection M ⊂ P (X) is a σ-algebra of X if

1. ∅ ∈ M; 2. A ∈ M ⇒ Ac = X \ A ∈ M; S∞ 3. Ai ∈ M, i ∈ N ⇒ i=1 Ai ∈ M. Example 1.3. 1. P(X) is the largest σ-algebra of X;

2. {∅,X} is the smallest σ-algebra of X; 3. Leb(Rn) = the Lebesgue measurable subsets of Rn; 4. If M is a σ-algebra of X and A ⊂ X, then M|A = {B ∩ A : B ∈ M} is a σ-algebra of A. 5. If M is a σ-algebra of X and A ∈ M, then

MA = {B ⊂ X : B ∩ A ∈ M} is a σ-algebra of X. Definition 1.4. If F ⊂ P(X) is any family of subsets of X, then \ σ(F) = {M : M is a σ-algebra of X, F ⊂ M} is the σ-algebra generated by F. It is the smallest σ-algebra that contains F.

2 Example 1.5. A subset I ⊂ Rn is an open n-interval if it has the form

I = {(x1, . . . , xn): aj < xj < bj}, where −∞ ≤ aj < bj ≤ +∞. Then n n σ({I : I is an n-interval}) = σ({A : A ⊂ R is open}) =: Bor(R ) is the σ-algebra of Borel sets of Rn. (Think why there is equality on the left.) n We observe that, in R , all open sets, closed sets, Gδ sets (countable inter- sections of open sets), Fσ sets (countable unions of closed sets), Fσδ sets, Gδσ sets, etc., are Borel sets. Thus e.g. the set Q of rational numbers is a Borel set. Remark 1.6. In every topological space X, one can define the Borel sets Bor(X) = σ({A : A ⊂ X open}. Definition 1.7. Let M be a σ-algebra of X. A mapping µ : M → [0, +∞] is a measure, if

1. µ(∅) = 0, S∞ P∞ 2. µ( i=1 Ai) = i=1 µ(Ai), if Ai ∈ M are disjoint. The triplet (X, M, µ) is called a measure space and the elements of M measur- able sets. Condition (2) above is called countable additivity. The definition implies the monotonicity of the meusure: If A, B ∈ M and A ⊂ B, then µ(A) ≤ µ(B). Remark 1.8. 1. If µ(X) < ∞, the measure µ is finite. 2. If µ(X) = 1, then µ is a probability measure. S∞ 3. If X = i=1 Ai, where µ(Ai) < ∞ for all i, the measure µ is σ-finite. We also say that X is σ-finite with respect to µ. 4. If A ∈ M and µ(A) = 0, then A has (or is of) measure zero. 5. If X is a topological space and Bor(X) ⊂ M (i.e., every Borel set is measurable), then µ is a Borel measure. n n Example 1.9. 1. X = R , M = Leb R , µ = mn = Lebesgue measure. n n n 2. X = R , M = Bor R , µ = mn| Bor R = the restriction of the Lebesgue measure on Borel sets.

3. Let X 6= ∅ be any set. Fix x ∈ X and define for all A ⊂ X ( 1, if x ∈ A; µ(A) = 0, if x∈ / A.

Then µ : P(X) → [0, +∞] is a measure (so called Dirac measure at x). It is often denoted by µ = δx.

3 4. If f : Rn → [0, ∞] is Lebesgue measurable, then µ : Leb(Rn) → [0, ∞] defined by

µ(E) = f(x)dmn(x) ˆE is a measure. (See course “Measure and integral”.)

5. If (X, M, µ) is a measure space and A ∈ M, then µxA : MA → [0, ∞] defined by (µxA)(B) = µ(B ∩ A) is a measure. It is called the restriction of µ on A.

Theorem 1.10. Let (X, M, µ) be a measure space and A1,A2,... ∈ M.

∞ [ 1. If A1 ⊂ A2 ⊂ ..., then µ( Ai) = lim µ(Ai). i→∞ i=1

∞ \ 2. If A1 ⊃ A2 ⊃ ... and µ(Ak) < ∞ for some k, then µ( Ai) = lim µ(Ai). i→∞ i=1 Proof. Measure and integral.

1.11 Metric outer measures

Definition 1.12. A mapping µ˜ : P(X) → [0, +∞] is an outer measure (or exterior measure) on X if

1. µ˜(∅) = 0; P∞ S∞ 2. µ˜(A) ≤ i=1 µ˜(Ai), if A ⊂ i=1 Ai ⊂ X. Remark 1.13. 1. The outer measure is thus defined on all subsets of X.

2. Condition (2) implies the monotonicity of the outer measure, i.e., µ˜(A) ≤ µ˜(B) if A ⊂ B ⊂ X. 3. Many books (e.g. Evans–Gariepy, Mattila, . . . ) refer to outer measures simply as measures. 4. Let µ˜ be an outer measure on X and A ⊂ X. Then its restriction to A,

(˜µxA)(B) =µ ˜(B ∩ A), is an outer measure on X. Every outer measure defines a σ-algebra of “measurable” sets through the so called Carathéodory condition:

4 Definition 1.14. Let µ˜ be an outer measure on X. A set E ⊂ X is µ˜- measurable, or just measurable, if

µ˜(A) =µ ˜(A ∩ E) +µ ˜(A \ E) for all A ⊂ X. Theorem 1.15. Let µ˜ be an outer measure on X and

M = Mµ˜ = {E ⊂ X : E is µ˜-measurable}. Then 1. M is a σ-algebram and 2. µ =µ ˜|M is a measure (i.e., µ is countably additive). Proof. Measure and integral. Definition 1.16. An outer measure µ˜ on a topological space X is called a Borel outer measure if every Borel set of X is µ˜-measurable (i.e., the measure determined by µ˜ is a Borel measure). We next investigate, when a given outer measure on a topological space X has this property. Definition 1.17 (Carathéodory). An outer measure µ˜ on a metric space (X, d) is a metric outer measure if

µ˜(A ∪ B) =µ ˜(A) +µ ˜(B) for all A, B ⊂ X such that dist(A, B) = inf{d(a, b): a ∈ A, b ∈ B} > 0. Theorem 1.18. An outer measure µ˜ on a metric space (X, d) is a Borel outer measure, if and only if µ˜ is a metric outer measure. We begin with a lemma: Lemma 1.19. Let µ˜ be a metric outer measure and A ⊂ G ⊂ X, where G is open. If c Ak = {x ∈ A : dist(x, G ) ≥ 1/k}, k ∈ N, then µ˜(A) = limk→∞ µ˜(Ak). S∞ Proof. Since G is open, we have A = k=1 Ak. Let Bk = Ak+1 \ Ak. Then ∞ ∞  [   [  A = A2n ∪ B2k ∪ B2k+1 , k=n k=n so that ∞ ∞ X X µ˜(A) ≤ µ˜(A2n) + µ˜(B2k) + µ˜(B2k+1) =µ ˜(A2n) + (I)n + (II)n. k=n k=n

5 Consider the limit n → ∞. (1) If (I)n, (II)n → 0 as n → ∞, then

µ˜(A) ≤ lim µ˜(A2n) ≤ lim µ˜(A2n+1) ≤ µ˜(A), n→∞ n→∞ and the claim of the lemma is true. X (2) If (I)n 6→ 0 as n → ∞, then µ˜(B2n) = ∞. n n−1 [ On the other hand, A ⊃ A2n ⊃ B2k, where k=1 1 1 dist(B ,B ) ≥ − > 0. 2k 2k+2 2k + 1 2k + 2 Since µ˜ is a metric outer measure, it follows that

n−1 n−1 X  [  µ˜(B2k) =µ ˜ B2k ≤ µ˜(A2n) ≤ µ˜(A2n+1) ≤ µ˜(A). k=1 k=1

As n → ∞, we deduce that µ˜(A) = limk→∞ µ˜(Ak) = ∞, and the claim of the lemma is true again. If (II)n 6→ 0 as n → ∞, the argument is entirely similar. Proof of Theorem 1.18. Let us first assume that µ˜ is a metric outer measure and prove that µ˜ is a Borel outer measure. Since Bor(X) = σ({F : F ⊂ X closed}) and Mµ˜ is a σ-algebra, it suffices to prove that every closed set F ⊂ X is µ˜-measurable. Let E ⊂ X be an arbitrary set, for which we check the Carathéodory condi- tion µ˜(E) =µ ˜(E ∩ F ) +µ ˜(E \ F ). We apply Lemma 1.19 with

A = E \ F ⊂ X \ F = G.

When Ak ⊂ A is as in the lemma, we have

dist(Ak,F ∩ E) ≥ dist(Ak,F ) ≥ 1/k > 0, lim µ˜(Ak) =µ ˜(A) =µ ˜(E \ F ). k→∞ By the assumption that µ˜ is a metric outer measure and observing that Ak ⊂ A ⊂ E, we have

µ˜(Ak) +µ ˜(F ∩ E) =µ ˜(Ak ∪ (F ∩ E)) ≤ µ˜(E).

When k → ∞, the left-hand side tends to µ˜(E \ F ) +µ ˜(F ∩ E), and we get the Carathéodory condition. (Recall that the other inequality in the equality is automatic for any sets.) Thus the arbitrary closed F ăis µ˜-measurable, and hence µ˜ is a Borel outer measure. The other direction is left as an exercise.

6 1.20 Regularity of measures, Radon measures

Among all outer measures, particularly useful are those that have many mea- surable sets. These have a name: (We denote by µ the measure determined by the outer measure µ˜.) Definition 1.21. An outer measure µ˜ on X is called regular, if for every A ⊂ X, there is a µ˜-measurable set E such that A ⊂ E and µ(E) =µ ˜(A).(E is a measurable cover of A.) Definition 1.22. Let X be a topological space.

1. An outer measure µ˜ on X is called Borel regular, if µ is a Borel measure and for every A ⊂ X, there is a Borel set B ∈ Bor(X) such that A ⊂ B and µ(B) =µ ˜(A). 2. If (X, M, µ) is a measure space such that Bor(X) ⊂ M, the measure µ is called Borel regular if for every A ∈ M there is B ∈ Bor(X) such that A ⊂ B and µ(A) = µ(B). Lemma 1.23. Let µ˜ be a Borel regular outer measure on X and A ⊂ X be µ˜-measurable. If µ(A) < ∞ or A ∈ Bor(X), then µ˜xA is Borel regular. Proof. Exercise.

Theorem 1.24. Let µ˜ be a Borel regular outer measure on a metric space X, let A ⊂ X be µ˜-measurable and ε > 0. 1. If µ(A) < ∞, there is a closed set C ⊂ A such that µ(A \ C) < ε. S∞ 2. If there are open sets V1,V2,... such that A ⊂ i=1 Vi and µ(Vi) < ∞ for all i, then there exists an open set V ⊃ A such that µ(V \ A) < ε.

Proof. Considering the Borel regular outer measure µ˜xA in place of µ˜ (see Lemma 1.23), we may assume without loss of generality that µ˜(X) < ∞. Let

D = {A ⊂ X : ∀ε > 0 ∃ closed C ⊂ A and open V ⊃ A such that µ(V \C) < ε}.

We check that D is a σ-algebra. It is easy to see that ∅ ∈ D and A ∈ D implies c S∞ A ∈ D. Let then A1,A2,... ∈ D, let A = i=1 Ai and ε > 0. There are closed i sets Ci and open sets Vi such that Ci ⊂ Ai ⊂ Vi and µ(Vi \ Ci) < ε/2 . Then S∞ V = i=1 Vi ⊃ A is open and, by Theorem 1.10,

n n ∞ ∞ ∞ [ [ [  [  X lim µ( V \ Ci) = µ(V \ Ci) ≤ µ (Vi \ Ci) ≤ µ(Vi \ Ci) < ε. n→∞ i=1 i=1 i=1 i=1 i=1 Sn Sn and thus µ(V \ i=1 Ci) < ε for some n, where C := i=1 Ci ⊂ A is closed. Thus C ⊂ A ⊂ V and µ(V \ C) < ε, and thus A ∈ D. Hence D is indeed a σ-algebra.

7 We check that D contains all closed sets. So let C be closed and

Vi := {x ∈ X : dist(x, C) < 1/i}. T Then each Vi is open, V1 ⊃ V2 ⊃ ... and i Vi = C. Hence limi→∞ µ(Vi) = µ(C) and limi→∞ µ(Vi \ C) = 0. (Here we use the assumption that µ(X) < ∞.) This implies that C ∈ D. Thus D is a σ-algebra that contains all closed sets and hence all Borel sets. In particular, the claims of the theorem are true in the case of a Borel set A. Let us then consider an arbitrary µ˜-measurable A with µ(A) < ∞. Since µ˜ is Borel regular, there exists a Borel set B ⊃ A such that µ(A) = µ(B) and thus µ(B \ A) = 0. For the same reason, there is a Borel set D ⊃ B \ A such that µ(D) = µ(B \ A) = 0. Then E = B \ D ⊂ A is a Borel set and µ(A \ E) ≤ µ(B \ E) = µ(D) = 0. By the already proved result for Borel sets, there is a closed C ⊂ E ⊂ A such that µ(E \ C) < ε and hence µ(A \ C) ≤ µ(A \ E) + µ(E \ C) < 0 + ε, so the first part of the theorem holds. For part (2), we apply part (1) to the sets Vi \A to find closed sets Ci ⊂ Vi \A −i S∞ such that µ((Vi \ A) \ Ci) < ε2 . Then V = i=1(Vi \ Ci) is open, A ⊂ V , and µ(V \ A) < ε. Remark 1.25. The Borel regularity of µ˜ was not used in the proof in the case when A is a Borel set. Thus the above theorem also holds for arbitrary Borel outer measures µ˜ and Borel sets A. In the sequel, a central role is played by Radon measures that we define next. We recall that a topological space X is locally compact if every x ∈ X has a neighbourhood whose closure is compact. A topological space is a Hausdorff space if its distinct points have disjoint neighbourhoods. Definition 1.26. Let X be a locally compact Hausdorff space. A measure µ is called a Radon measure if it is a Borel measure and 1. µ(K) < ∞ for all compact K ⊂ X; 2. µ(V ) = sup{µ(K): K ⊂ V compact} for all open V ⊂ X; 3. µ(B) = inf{µ(V ): B ⊂ V,V ⊂ X open} for all Borel sets B ∈ Bor(X). Remark 1.27. 1. In general, a Borel regular measure (on a locally compact Hausdorff space) need not be a Radon measure. (Exercise.)

2. On the other hand, a Radon measure need not be Borel regular: Let A ⊂ R ∗ be non-Lebesgue-measurable, µ˜ = m xA and µ =µ ˜|{E ⊂ R : E is µ˜-measurable}. Then µ is a Radon measure but it is not Borel regular. (Exercise.)

8 In some cases, Radon measures can be easily characterised.

Theorem 1.28. Let µ be a Borel measure on Rn. Then µ is a Radon measure if and only if µ is locally finite, i.e.

n ∀ x ∈ R ∃rx > 0 : µ((B(x, r)) < ∞ for 0 < r < rx. Proof. By definition, Radon measures are locally finite, so it remains to prove the converse. Let µ be a locally finite Borel measure on Rn. If K ⊂ Rn is compact, for every x ∈ K we choose a ball centred at x with finite measure. Since K is compact, it can be covered by finitely many such balls.Thus the measure of K is finite, which proves (1) of Definition 1.26. We will prove both (2) and (3) of Definition 1.26 for all Borel sets A ⊂ Rn. Consider the Borel sets Ai = A ∩ B(0, i) of finite measure. By Theorem 1.24 and Remark 1.25, there are closed sets Ci ⊂ Ai such that µ(Ai \ Ci) < 1/i. Since Ci ⊂ Ai ⊂ B(0, i), the set Ci is also bounded and thus compact. (Here we use the fact that we are in Rn.) Now

µ(A) ≥ µ(Ai) ≥ µ(Ci) > µ(Ai) − 1/i → µ(A) by Theorem 1.10, and thus (2) of Definition 1.26 holds. S∞ Since A ⊂ i=1 B(0, i), where Vi = B(0, i) are open sets with µ(Vi) < ∞, Theorem 1.24 (and Remark 1.25) prove the existence of an open V ⊃ A such that µ(V \ A) < ε and hence µ(V ) < µ(A) + ε, which proves (3) of Definition 1.26.

Corollary 1.29. Let µ˜ be a locally finite metric outer measure on Rn. Then its restriction µ to µ˜-measurable subsets is a Radon measure. Remark 1.30. Theorem 1.28 is valid under more general assumptions, e.g. with Rn replaced by a locally compact metric space whose topology has a countable basis.

1.31 Uniqueness of measures

We next investigate conditions for a collection F ⊂ P(X) so that two measures µ, ν : σ(F)) → [0, +∞] coincide, as soon as they coincide on F. Definition 1.32. 1. A nonempty collection F ⊂ P(X) is a π-system if A ∩ B ∈ F whenever A, B ∈ F. 2. A collection D ⊂ P(X) is a d-system (or Dynkin class) if

(a) X ∈ D; (b) if A, B ∈ D and A ⊂ B, then B \ A ∈ D; S∞ (c) if A1,A2,... ∈ D and A1 ⊂ A2 ⊂ ..., then k=1 Ak ∈ D.

9 3. The d-system generated by a collection A ⊂ P(X) is the smallest d-system that contains A, namely \ d(A) = {D : D ⊃ A is a d-system}.

Theorem 1.33 (Dynkin’s lemma). Let A ⊂ P(X) be a π-system. If D ⊃ A is a d-system, then D ⊃ σ(A). Proof. It is enough to consider the case that D = d(A), and then it is enough to check that D is a σ-algebra. Namely, D ⊃ A by assumption, and σ(A) is the smallest σ-algebra that contain A. If A ∈ D, then A ⊂ X ∈ D, and hence Ac = X \ A ∈ D. Thus ∅ = Xc ∈ D. So it only remains to check that D is closed under countable unions. For this, the main point is to check that D is closed under intersections: A, B ∈ D implies A ∩ B ∈ D. Assume this for a moment. Since D is also closed under complements (as we checked), it is then closed under unions (using A ∪ B = c c c 0 Sk (A ∩ B ) ). Finally, if A1,A2,... ∈ D, then Ak := j=1 Aj ∈ D are increasing, S∞ S∞ 0 and k=1 Ak = k=1 Ak ∈ D since D is a d-system. So D is closed under arbitrary countable unions, as required for a σ-algebra. So it only remains to check the closedness of D under intersections. For this purpose, we make an auxiliary definition: If F ⊂ D is any subcollection, then F 0 := {A ∈ D : A ∩ F ∈ D ∀F ∈ F}. It follows directly from the definition that for any F, G ⊂ D: (∗) G ⊂ F 0 ⇔ G ∩ F ∈ D ∀F ∈ F, ∀G ∈ G ⇔ F ⊂ G0. In particular, we see that A ⊂ A0 (since A is a π-system) and we need to prove that D ⊂ D0. For this purpose, we prove that for any F ⊂ D, the collection F 0 is a d-system. Assume this for a moment. Since D is the smallest d-system that contain A, it implies that (∗∗) A ⊂ F 0 ⇒ D ⊂ F 0 and allows us to conclude that

(∗∗) (∗) (∗∗) A ⊂ A0 ⇒ D ⊂ A0 ⇒ A ⊂ D0 ⇒ D ⊂ D0, which is what we wanted to prove, since the last inclusion means that A∩B ∈ D for all A, B ∈ D. It remains to check that F 0 is a d-system. For this, let F ∈ F be arbitrary. Then X ∩ F = F ∈ D, and thus X ∈ F 0. If A, B ∈ F 0 and A ⊂ B, then (B \ A) ∩ F = (B ∩ F ) \ (A ∩ F ) ∈ D since both A ∩ F,B ∩ F ∈ D and A ∩ F ⊂ B ∩ F , and thus B \ A ∈ F 0. If 0 A1,A2,... ∈ F is an increasing sequence, then A1 ∩F,A2 ∩F ∈ D is an increas-  S∞  S∞ S∞ 0 ing sequence, and k=1 Ak ∩ F = k=1(Ak ∩ F ) ∈ D so that k=1 Ak ∈ F . Thus F 0 satisfies all three properties of a d-system, and we are done.

10 Theorem 1.34. Let F be a π-system, M = σ(F), and µ, ν : M → [0, 1] be probability measures such that µ(A) = ν(A) for all A ∈ F. Then µ ≡ ν. Proof. We denote D = {A ∈ M : µ(A) = ν(A)} and prove that D = M. It is enough to prove that D is a d-system, since D contains the π-system F by assumption, and hence M = σ(F) ⊂ D ⊂ M.

1. Since µ(X) = 1 = ν(X), we have X ∈ D. 2. If A, B ∈ D and A ⊂ B, then also B \ A ∈ D, since

µ(B \ A) = µ(B) − µ(A) = ν(B) − ν(A) = ν(B \ A).

S∞ 3. If A1,A2,... ∈ D and A1 ⊂ A2 ⊂ ..., then also k=1 Ak ∈ D, since

∞ ∞  [   [  µ Ak = lim µ(Ak) = lim ν(Ak) = ν Ak . k→∞ k→∞ k=1 k=1

Thus D is a d-system, and the theorem is proved.

Corollary 1.35. If µ, ν : Bor(Rn) → [0, +∞) are finite measures and µ(I) = ν(I) for every n-interval I, then µ ≡ ν.

n S∞ n Proof. Since R = j=1 Ij for the n-intervals Ij = (−j, j) , the assumption implies that µ(Rn) = ν(Rn) = c, say. If c = 0, then µ ≡ 0 ≡ ν, and the result is clear. Otherwise, c−1µ, c−1ν : Bor(Rn) → [0, 1] are probability measures that coincide on all n-intervals, which clearly form a π-system F. Since σ(F) = Bor(Rn), the result follows from Theorem 1.34.

1.36 Extension of measures

Consider a collection F ⊂ P(X) and a mapping µ : F → [0, +∞] that is σ- additive on F, i.e.,

∞ ∞ ∞  [  X [ (1.37) µ Ak = µ(Ak) if Ak ∈ F are disjoint and Ak ∈ F. k=1 k=1 k=1

We investigate whether µ can be extended to a measure on σ(F). In view of the previous section, we expect that F should be a π-system. In general this is not enough, but by assuming a bit more, we get a positive result. Definition 1.38. 1. A family A ⊂ P(X) is an algebra if

(a) ∅ ∈ A; (b) A, B ∈ A implies A ∩ B ∈ A; (c) A ∈ A implies X \ A ∈ A.

11 2. A family A ⊂ P(X) is a semialgebra if it satisfies (1a) and (1b) above, but (1c) is replaced by n [ X \ A = Ak k=1

for some n ∈ N and disjoint Ak ∈ A, whenever A ∈ A. G Let us introduce the notation Ai for the union of disjoint sets Ai. i Lemma 1.39. Let P be a semialgebra. Then

n ¯ n G o P = Ai : A : i ∈ P, i ∈ N i=1 is an algebra (called the algebra generated by P).

Proof. Clearly ∅ ∈ P¯. If n m G G A = Ai,B = Bj ∈ P¯,Ai,Bj ∈ P, i=1 j=1 then G A ∩ B = Ai ∩ Bj, i,j where Ai ∩ Bj ∈ P and thus A ∩ B ∈ P¯. Moreover, since Ai ∈ P, we have X \ Ai ∈ P¯ by the semialgebra property, and thus

n \ X \ A = (X \ Ai) ∈ P¯ i=1 by the closedness of P¯ under intersections, which we just proved. Definition 1.40. A measure on an algebra A is a σ-additive mapping µ : A → [0, ∞] such that µ(∅) = 0. Lemma 1.41. Let P be a semialgebra and µ : P¯ → [0, ∞] be σ-additive with µ(∅) = 0. Then there is a unique measure µ¯ : P¯ → [0, ∞] that agrees with µ on P. Proof. For n G A = Ai ∈ P˜,Ai ∈ P, i=1 we set

n X (1.42) µ¯(A) = µ(Ai). i=1

12 Fn Fm 1. µ¯ is well defined: If A = i=1 Ai = j=1 Bj, then

m n G G Ai = Ai ∩ A = Ai ∩ Bj,Bj = A ∩ Bj = Ai ∩ Bj, j=1 i=1

where Ai ∩ Bj ∈ P, and hence, by the σ-additivity of µ on P, X X X X X X µ(Ai) = µ(Ai ∩ Bj) = µ(Ai ∩ Bj) = µ(Bj). i i j j i j

¯ ¯ 2. µ¯ is a measure on P: Clearly µ¯(∅) = 0. For σ-additivity, consider Ai ∈ P F∞ ¯ with A = i=1 Ai ∈ P. Thus

n n G Gi A = Bk (Bk ∈ P),Ai = Aij (Aij ∈ P), k=1 j=1

and hence G G Bk = A∩Bk = Aij∩Bk,Aij = Aij∩A = Aij∩Bk,Aij∩Bk ∈ P. i,j k

By the definition of µ¯, and the σ-additivity of µ on P, X X X X X µ¯(A) = µ(Bk) = µ(Aij ∩ Bk) = µ(Aij) = µ¯(Ai). k k i,j i,j i

3. Uniqueness: If µ is a measure on P¯ with

(1.43) µ¯(A) = µ(A) ∀A ∈ P,

it is immediate from σ-additivity that µ¯ must be given by (1.42) on all Fn ¯ A = i=1 Ai ∈ P.

Theorem 1.44 (Carathéodory–Hahn extension theorem). Let A ⊂ P(X) be an algebra and µ : A → [0, ∞] a measure on A. 1. There is a measure ν : σ(A) → [0, ∞] such that ν|A = µ. S∞ 2. If µ is σ-finite with respect to A (i.e., X = i=1 Ai, where Ai ∈ A and µ(Ai) < ∞), then ν is unique. Proof. We define ν˜ : P(X) → [0, ∞] by

∞ ∞ n X [ o ν˜(E) = inf µ(Ai): E ⊂ Ai,Ai ∈ A for all i . i=1 i=1

13 1. This ν˜ is an outer measure and ν˜(A) = µ(A) for all A ∈ A. (Exercise.) 2. Every A ∈ A is ν˜-measurable. (Exercise.) 3. If M is the family of all ν˜-measurable sets, then ν˜|M is a measure by Theorem 1.15. Since A ⊂ M and M is a σ-algebra, we have σ(A) ⊂ M. Then ν =ν ˜|σ(A) is the required measure.

4. To prove uniqueness, consider two measures νi : σ(A) → [0, ∞] (i = 1, 2) such that ν1(A) = ν2(A) = µ(A) for all A ∈ A. If A ∈ A satisfies −1 0 < µ(A) < ∞, then B 7→ µ(A) νi(A ∩ B) are two probability measures on σ(A) that agree on the π-system A, and hence ν1(A ∩ B) = ν2(A ∩ B) for all B ∈ σ(A) by Theorem 1.34. If µ(A) = 0, the same conclusion is obvious, since 0 ≤ νi(A ∩ B) ≤ νi(A) = µ(A) = 0 for both i = 1, 2 in this case. By σ-finiteness, we can then choose sets A1 ⊂ A2 ⊂ ... with S∞ A1,A2,... ∈ A and X = i=1. From the continuity of measure it then follows that

ν1(B) = lim ν1(B ∩ Ak) = lim ν2(B ∩ Ak) = ν2(B) k→∞ k→∞

for all B ∈ σ(A), and thus ν1 ≡ ν2.

1.45 Product measure

We apply the Carathéodory–Hahn extension theorem for the construction of the product measure. Let (X, M, µ) and (Y, N , ν) be measure spaces and

S = {A × B : A ∈ M,B ∈ N } ⊂ P(X × Y ) the collection of so called measurable rectangles.

Lemma 1.46. S is a semialgebra.

Proof. Clearly ∅ = ∅×∅ ∈ S. If A×B,A0 ×B0 ∈ S, then (A×B)∩(A0 ×B0) = (A ∩ A0) × (B ∩ B0) ∈ S. Moreover,

(A × B)c = (Ac × B) t (A × Bc) t (Ac × Bc), where each term belongs to S, has the required form. We define λ : S → [0, ∞] by

(1.47) λ(A × B) = µ(A)ν(B).

Clearly λ(∅) = 0. We next prove that λ is σ-additive on S.

14 Lemma 1.48. If Ai × Bi ∈ S are disjoint sets such that ∞ G Ai × Bi = A × B ∈ S, i=1 then ∞ X λ(Ai × Bi) = λ(A × B). i=1 Proof. From the assumed equality of sets, we get G B = Bi ∀x ∈ A,

i:x∈Ai and hence X ν(B) = ν(Bi) ∀x ∈ A.

i:x∈Ai This can be further written as ∞ X χA(x)ν(B) = χAi (x)ν(Bi). i=1 Integrating the positive series with respect to µ, we deduce ∞ ∞ X X µ(A)ν(B) = χ (x)ν(B)dµ(x) = χ (x)ν(B )dµ(x) = µ(A )ν(B ), ˆ A ˆ Ai i i i X i=1 X i=1 which was the claim, recalling the definition of λ. Theorem 1.49 (Existence of product measure). Let (X, M, µ) and (Y, N , ν) be σ-finite measure spaces, and S ⊂ P(X × Y ) the family of measurable rectangles. Then the measurable space (X ×Y, σ(S)) has a unique measure, denoted by µ×ν and called the product measure of µ and ν, such that (µ × ν)(A × B) = λ(A × B) = µ(A)ν(B) ∀A ∈ M,B ∈ N .

Proof. Since λ is σ-additive on the semialgebra S and λ(∅) = 0, Lemma 1.41 guarantees the existence of a unique measure λ¯ on the generated algebra S¯ such that λ¯(A × B) = λ(A × B) for all A × B ∈ S. The σ-finiteness of µ and ν immediately implies that λ¯ is σ-finite with respect to S¯. Hence the theorem follows from the Carathéodory–Hahn extension theorem. We recall from Real Analysis I that a measure space (X, M, µ) is complete if all subsets of sets of measure zero are measurable (i.e., if A ⊂ B ∈ M and µ(B) = 0, then A ∈ M), and that a measure can always be completed: Theorem 1.50. Let (X, M, µ) be a measure space. We define M¯ = {A ∪ F : A ∈ M,F ⊂ E for some E ∈ M, µ(E) = 0} and µ¯ : M¯ → [0, ∞] by µ¯(A ∪ F ) = µ(A), when A and F are as above. Then

15 1. M¯ is a σ-algebra of X; 2. µ¯ is a complete measure; 3. µ =µ ¯|M. Here µ¯ is called the completion of µ, and (X, M¯ , µ¯) the completion of (X, M, µ). Remark 1.51. 1. If µ˜ is an outer measure on X, its associated measure space (X, Mµ˜, µ) is always complete. 2. If µ˜ is a Borel regular metric outer measure on X, its associated measure space (X, Mµ˜, µ) is the completion of (X, Bor(X), µ). 3. Even if (X, M, µ) and (Y, N , ν) are complete σ-finite measure spaces, the product measure space (X × Y, σ(S), µ × ν) need not be complete. For example, if ∅ 6= A ∈ M and µ(A) = 0 and B ∈ P(Y ) \N , then A × B ⊂ A × Y , where µ(A × Y ) = 0, but A × B/∈ σ(S) (see Lemma 1.53). n k In particular, the product measure space (R × R , σ(S), mn × mk) of Lebesgue measures mn and mk is not complete, and hence mn × mk 6= mn+k. Instead, mn+k is the completion of mn × mk.

1.52 Fubini’s theorem

Let (X, M, µ) and (Y, N , ν) be σ-finite measure spaces, and µ × ν the product measure constructed above. We study conditions under which f : X × Y → R˙ satisfies   f(x, y)d(µ × ν)(x, y) = f(x, y)dν(y) dµ(x) ˆX×Y ˆX ˆY   = f(x, y)dµ(x) dν(y). ˆY ˆX If E ⊂ X × Y , we denote

y Ex = {y ∈ Y :(x, y) ∈ E},E = {x ∈ X :(x, y) ∈ E}.

y Lemma 1.53. If E ∈ σ(S), then Ex ∈ N and E ∈ M for all x ∈ X, y ∈ Y .

Proof. Let C = {E ∈ σ(S): Ex ∈ N ∀x ∈ X}. If E = A × B ∈ S, then Ex = B ∈ N if x ∈ A, and Ex = ∅ ∈ N if x∈ / A, and thus S ⊂ C. It suffices to prove that C is a σ-algebra, since then σ(S) ⊂ C, and Ex ∈ N for allăx ∈ X holds for all E ∈ σ(S) as claimed. (The result concerning the sets Ey is completely analogous.) The main point is that the operation E 7→ Ex commutes with the standard set operations, namely

c c  [  [ (E )x = (Ex) , Ei = (Ei)x. x i i

16 c c Hence, if E ∈ C, then Ex ∈ N by definition, hence (Ex) ∈ N , i.e., (E )x ∈ N ? and thus Ec ∈ C. The proof concerning unions is similar, and ∅ ∈ C is obvious; thus C is a σ-algebra as required.

If f : X × Y → R˙ , we denote

y fx(y) = f (x) = f(x, y).

Lemma 1.54. If f : X × Y → R˙ is µ × ν-measurable, then ˙ • fx : Y → R is ν-measurable for every x ∈ X, • f y : X → R˙ is µ-measurable for every y ∈ Y .

Proof. The main point is that {fx ∈ U} = {f ∈ U}x for any set U. If U = (t, ∞], t ∈ R˙ , then {f ∈ U} ∈ σ(S) by the assumed measurability of f, thus {f ∈ U}x ∈ N by the previous lemma, hence {fx ∈ U} by the observation just y made, and thus fx is measurable. The case of f is analogous.

y Lemma 1.55. If E ∈ σ(S), then x 7→ ν(Ex) is µ-measurable, y 7→ µ(E ) is ν-measurable, and

y ν(Ex)dµ(x) = µ(E )dµ(y) = (µ × ν)(E). ˆX ˆY Proof. We consider the family n o F = E ∈ σ(S): x 7→ ν(Ex) is µ-measurable, ν(Ex)dµ(x) = (µ × ν)(E) . ˆX We need to prove that F = σ(S). (The claim concerning Ey is analogous.) If E = A × B ∈ S, then ( B, x ∈ A, Ex = ν(Ex) = χA(x)ν(B), ν(Ex)dµ(x) = µ(A)ν(B), ∅, x∈ / A, ˆX so that S ⊂ F. Since S is clearly a π-system, it is thus enough to check that F ⊃ S is a d-system, since Dynkin’s lemma then implies that F ⊃ σ(S). It is immediate that X × Y ∈ S ⊂ F. S∞ Let then E1,E2,... ∈ F satsify E1 ⊂ E2 ⊂ ... and E = i=1 Ei. Then S∞ Eix ∈ N , E1x ⊂ E2x ⊂ ... and Ex = i=1 Eix so that x 7→ ν(Ex) = limk→∞ ν(Ekx) is µ-measurable as an increasing limit of µ-measurable func- tions, and monotone convergence theorem and continuity of measure imply that

ν(Ex)dµ(x) = lim ν(Ekx)dµ(x) = lim µ × ν(Ek) = µ × ν(E). ˆX k→∞ ˆX k→∞ S∞ Thus E = i=1 Ei ∈ F.

17 It remains to check that F \ E ∈ F whenever E,F ∈ F and E ⊂ F . We first do this under the simplifying additional assumption that µ(X), ν(Y ) < ∞. Then x 7→ ν((F \ E)x) = ν(Fx \ Ex) = ν(Fx) − ν(Ex) is µ-measurable as a difference of two such functions. (The assumption ν(Y ) < ∞, makes sure that we don’t encounter any undefined ∞ − ∞ here.) These functions are also pointiweise bounded by ν(Y ) < ∞, and hence integrable over the finite measure space (X, M, µ). Then

ν((F \ E)x)dµ(x) = ν(Fx)dµ(x) − ν(Ex)dµ(x) ˆX ˆX ˆX = µ × ν(F ) − µ × ν(E) = µ × ν(F \ E).

Thus the result is proved in the case of finite measures. In general, let µ(A), ν(B) < ∞, and apply this special case to the finite measures µxA and νxB in place of µ and ν. This gives

νxB(Ex)d(µxA)(x) = (µxA) × (νxB)(E) ∀E ∈ σ(S). ˆX

Both (µxA)×(νxB) and (µ×ν)x(A×B) are measures on σ(S) that coincide on the π-system S (which contains X × Y ), and hence they are equal. Moreover,

X fd(µxA) = A fdµ is clear when f is a characteristic function and follows for ´simple functions´ by linearity and for positive measurable functions by monotone convergence. Hence the previous identity can be rewritten as

χA(x)ν(B ∩ Ex)dµ(x) = (µ × ν)((A × B) ∩ E) ˆX for all E ∈ σ(S) and A, B respectively of finite µ and ν measure. S∞ S∞ We finally use σ-finiteness to write X = i=1 Xi, Y = i=1 Yi, with in- creasing sequences Xi and Yi such that µ(Xi), ν(Yi) < ∞. We use the previous identity with A = Xi, B = Yi. Then χXi (x) → 1 and ν(Yi ∩ Ex) → ν(Ex) as increasing limits, and hence monotone convergence implies that

ν(Ex)dµ(x) = lim χXi (x)ν(Yi ∩ Ex)dµ(x) ˆX ˆX i→∞

= lim χXi (x)ν(Yi ∩ Ex)dµ(x) i→∞ ˆX = lim µ × ν((Xi × Yi) ∩ E) = µ × ν(E). i→∞ This completes the proof in the general case. Theorem 1.56 (Tonelli). Let f : X × Y → [0, ∞] be µ × ν-measurable. Then:

1. For all (x0, y0) ∈ X × Y ,

x 7→ f(x, y0) is µ-measurable, y 7→ f(x0, y) is ν-measurable.

18 2.

x 7→ f(x, y)dν(y) is µ-measurable, ˆY y 7→ f(x, y)dµ(x) is ν-measurable. ˆX

3.   f(x, y)dν(y) dµ(x) = f(x, y)d(µ × ν)(x, y) ˆX ˆY ˆX×Y   = f(x, y)dµ(x) dν(y). ˆY ˆX

Proof. Part (1) was already proved in Lemma 1.54. If f = χE and E ∈ σ(S), the other parts are also true by the previous lemmas. By linearity, it remains true for all σ(S)-simple functions. In the general case, the we concentrate on the first halves of (2) and (3), as the other halves are completely analogous. For a given f as in the assumptions, we find a sequence an increasing sequence of simple functions such that 0 ≤ si % f pointwise. Then monotone convergence shows that

f(x, y)dν(y) = lim si(x, y)dν(y) = lim si(x, y)dν(y), ˆY ˆY i→∞ i→∞ ˆY which also proves that the left side is a µ-measurable function of x, as a (mono- tone) limit of such functions. Another application of monotone convergence shows that     f(x, y)dν(y) dµ(x) = lim si(x, y)dν(y) dµ(x) ˆX ˆY ˆX i→∞ ˆY   = lim si(x, y)dν(y) dµ(x), i→∞ ˆX ˆY and yet another one (on the product measure space X × Y ) that

f(x, y)d(µ × ν)(x, y) = lim si(x, y)d(µ × ν)(x, y) ˆX×Y ˆX×Y i→∞

= lim si(x, y)d(µ × ν)(x, y). i→∞ ˆX×Y But the right sides of the last two equations are equal by the already established case of simple functions, and thus also the left sides are equal. Remark 1.57. The previous results are formulated for the product measure µ × ν and (µ × ν)-measurable functions. They cannot be directly applied to the n k n+k complete measure space (R × R , Leb(R ), mn+k). For instance, Lemmas n k n+k 1.53 and 1.54 do not hold in (R × R , Leb(R ), mn+k). Namely, even if

19 k A ⊂ R is non-measurable (with respect to mk), the set E = {x} × A is still n ∗ mn+k-measurable for all x ∈ R , since mn+k(E) = 0. However, Ex = A, which shows the failure of Lemma 1.53. A version of Tonelli’s theorem on n k n+k (R × R , Leb(R ), mn+k) has (1) only for almost every x and y. Theorem 1.58 (Fubini). Let f : X × Y → R˙ be µ × ν-measurable and assume that at least one of the three integrals     |f|d(µ × ν), |f(x, y)|dν(y) dµ(x), |f(x, y)|dµ(x) dν(y), ˆX×Y ˆX ˆY ˆY ˆX is finite. Then

1. y 7→ f(x, y) is integrable over Y for almost all x ∈ X; 2. x 7→ f(x, y) is integrable over X for almost all y ∈ Y ;

3. x 7→ f(x, y)dν(y) is integrable over X; ˆY

4. y 7→ f(x, y)dµ(x) is integrable over Y ; ˆX 5. f is integrable over X × Y and     fd(µ × ν) = f(x, y)dν(y) dµ(x) = f(x, y)dµ(x) dν(y). ˆX×Y ˆX ˆY ˆY ˆX Proof. This is a similar deduction from Tonelli’s Theorem 1.56 as the analogous result for the Lebesgue measure in the course “Measure and Integral”.

20 Chapter 2

Hausdorff measures

2.1 Basic properties of Hausdorff measures

The n-dimensional Lebesgue measure mn is well suited for measuring the size n of “large” subsets of R , but it is too rough for “small” sets. For instance, m2 cannot distinguish a point from a line in R2, since they both have measure zero. In this chapter we introduce a whole scale of “s-dimensional” measures Hs, 0 ≤ s < ∞, which are able to see the fine structure of sets better than the Lebesgue measure. The idea is that a set A ⊂ Rn is “s-dimensional if 0 < Hs(A) < ∞, even if the geometry of A is very complicated. These measure can be defined on any metric space (X, d). We assume, however, that X is separable, i.e., that there is a countable dense subset S = ∞ ¯ {xi}i=1, so that X = S. This assumption guarantees that X has a δ-cover for every δ > 0. Definition 2.2. 1. The diameter of a nonempty set E ⊂ X is d(E) = sup d(x, y). x,y∈E

∞ 2. For δ > 0, a δ-cover of a set A ⊂ X is a countable collection {Ei}i=1 of subsets of X such that ∞ [ A ⊂ Ei, d(Ei) ≤ δ ∀i ∈ N. i=1 Let us fix a “dimension” s ∈ [0, ∞) and δ > 0. For A ⊂ X, we define

∞ s n X s ∞ o (2.3) Hδ(A) = inf d(Ei) : {Ei}i=1 is a δ-cover of A , i=1 where we agree that d({x})0 = 1 for all x ∈ X and d(∅)s = 0 for all s ≥ 0. It follows at once from the definition that Hs (A) ≥ Hs (A) if 0 < δ ≤ δ . δ1 δ2 1 2 Thus the limit (2.5) below exists and we can make the following definition:

21 Definition 2.4. The s-dimensional Hausdorff (outer) measure of A ⊂ X is

s s  s  (2.5) H (A) := lim Hδ(A) = sup Hδ(A) . δ→0 δ>0

s Theorem 2.6. 1. Hδ : P(X) → [0, ∞] is an outer measure for every δ > 0. 2. Hs : P(X) → [0, ∞] is a metric outer measure.

s S∞ Proof. 1. Clearly Hδ (∅) = 0. Let then A ⊂ i=1 Ai ⊂ X, where we assume s without loss of generality that Hδ(Ai) < ∞. Given ε > 0, we can find for ∞ every Ai a δ-cover {Eij}j=1 such that

∞ X s s −i d(Eij) ≤ Hδ(Ai) + ε2 . j=1

∞ S∞ Then {Eij}i,j=1 is δ-cover of i=1 and hence of A, and thus

∞ ∞ ∞ s X s X s −i
X s Hδ(A) ≤ d(Eij) ≤ Hδ(Ai) + ε2 ≤ Hδ(Ai) + ε. i,j=1 i=1 i=1

s As ε → 0, this implies the countable semiadditivity of the Hδ. s s 2. Clearly H (∅) = limδ→0 Hδ(∅) = limδ→0 0 = 0. S∞ s If A ⊂ i=1, since we already proved that Hδ is an outer measure, it follows that ∞ ∞ s X s X s Hδ(A) ≤ Hδ(A) ≤ H (A). i=1 i=1 As δ → 0, we see that Hs is an outer measure as well.

Let then A1,A2 ⊂ X be sets with dist(A1,A2) > 0. We want to show that

s s s H (A1 ∪ A2) = H (A1) + H (A2).

Since ≤ is clear (we already checked that Hs is an outer measure), it suffices to prove ≥, which in turn will follow in the limit δ → 0 if we can show that

s s s (2.7) Hδ(A1 ∪ A2) ≥ Hδ(A1) + Hδ(A2)

s for all δ ≤ dist(A1,A2)/3. We may also assume that Hδ(A1 ∪ A2) < ∞. ∞ Let us then fix a δ-cover {Ei}i=1 of A1 ∪ A2 such that

∞ X s s d(Ei) ≤ Hδ(A1 ∪ A2) + ε. i=1

22 Since δ ≤ dist(A1,A2)/3, each Ei intersects at most one of A1 or A2. Those that intersect Ak form a δ-cover of Ak. Thus we can split

2 ∞ ∞ G k ∞ [ k {Ei}i=1 = {Ei }i=1,Ak ⊂ Ei , k=1 i=1 and hence

2 2 ∞ ∞ X s X X k s X s s Hδ(Ak) ≤ d(Ei ) = d(Ei) ≤ Hδ(A1 ∪ A2) + ε. k=1 k=1 i=1 i=1 With ε → 0, we deduce (2.7), which completes the proof.

By Theorem 1.18 and the previous Theorem 2.6, every Borel set of X is Hs-measurable. We denote the restriction of Hs to Hs-measurable sets by the same symbol Hs. Thus: Theorem 2.8. Hs is a Borel measure. From Corollary 1.29 we now get:

n s s s Corollary 2.9. If A ⊂ R is H -measurable and H (A) < ∞, then H xA is a Radon measure. Theorem 2.10. On a separable metric space X, the outer measure Hs is Borel regular. Proof. The previous theorem shows that Hs is a Borel measure, so it remains to prove that for everyăA ⊂ X there is B ∈ Bor(X) such that A ⊂ B and Hs(A) = Hs(B). If Hs(A) = ∞, we can choose B = X. Suppose that Hs(A) < ∞. Then for i ∞ every i we can find a 1/i-cover {Ej}j=1 of A such that

∞ X i s s d(Ej) ≤ H1/i(A) + 1/i. j=1

¯ i Since d(E) = d(E) for every E ⊂ X, we may assume that the sets Ej are closed. Then ∞ ∞ \ [ i B = Ej i=1 j=1 i ∞ is a Borel set that contains A. For each i, the collection {Ej}j=1 is a 1/i-cover of B, and hence

∞ s s X i s s H1/i(A) ≤ H1/i(B) ≤ d(Ej) ≤ H1/i(A) + 1/i. j=1 With i → ∞ we deduce Hs(A) = Hs(B), as claimed.

23 Remark 2.11. 1. Hs is the counting measure.

s 2. Hδ is usually not a metric outer measure. 3. Roughly speaking, H1 ∼ length, H2 ∼ area, etc.

4. By results below, it is easy to see that (e.g.) H1 is not σ-finite on R2.

2.12 Hausdorff dimension

Let (X, d) be a separable metric space. We now define for all subsets A ⊂ X a dimension that represents the (metric) size of A. In contrast to, say, the topological dimension, this dimension need not be an integer. Lemma 2.13. Let A ⊂ X and s ≥ 0. 1. If Hs(A) < ∞, then Ht(A) = 0 for all t > s. 2. If Hs(A) > 0, then Ht(A) = ∞ for all 0 ≤ t < s. Proof. It is enough to prove the first claim, since the second one follows from ∞ it. Let t > s and {Ei}i=1 be a δ-cover of A for some δ > 0. Then

∞ ∞ X t t−s X s d(Ej) ≤ δ d(Ej) , j=1 j=1 and taking the infimum of both sides over all δ-covers shows that

t t−s s Hδ(A) ≤ δ Hδ(A). Letting δ → 0 proves that first claim. The previous lemma shows that the graph of s 7→ Hs(A) is very simple: at a unique point s0 ∈ [0, ∞], it drops from ∞ to zero. Definition 2.14. The Hausdorff dimension of a set A ⊂ X is the number

s dimH(A) := inf{s > 0 : H (A) = 0}.

We observe:

t 1. If t < dimH(A), then H (A) = ∞.

t 2. If t > dimH(A), then H (A) = 0.

s In general, we cannot say anything about the value of H (A), when s = dimH(A): it can be any value in the interval [0, ∞]. However:

s (2.15) 0 < H (A) < ∞ ⇒ dimH(A) = s.

A set A ⊂ X that satisfies the left side of (2.15) is called an s-set.

24 Lemma 2.16. 1. If A ⊂ B, then dimH(A) ≤ dimH(B).

2. If Ak ⊂ X for all k ∈ N, then

∞  [  dimH Ak = sup dimH(Ak). k k=1 Proof. Exercise.

Thus e.g. dimH Q = 0.

2.17 Hausdorff measures on Rn

Recall that a mapping T : Rn → Rn is an isometry if |T x − T y| = |x − y| for n n all x, y ∈ R . Every isometry of R is known to be affine, i.e. T x = a0 + Ux, n n n where a0 ∈ R and U : R → R is a linear isometry. A mapping R : Rn → Rn is a similarity if |Rx − Ry| = c|x − y| for all x, y ∈ Rn, where c > 0 is a constant (dilation or scaling factor). Such a similarity takes the form Rx = a0 + cUx, where U is again a linear isometry.

Theorem 2.18. Let A ⊂ Rn and s ≥ 0. Then: s s n 1. H (A + x0) = H (A) for all x0 ∈ R ; 2. Hs(U(A)) = Hs(A) for every linear isometry U : Rn → Rn; 3. Hs(R(A)) = csHs(A), if R : Rn → Rn is a similarity with scaling c > 0. Proof. These are immediate from the observation that d(R(E)) = cd(E) for every E ⊂ Rn, when R is a similarity with scaling factor c > 0.

Let (X, d1), (Y, d2) be metric spaces. Recall that f : X → Y is a Lipschtitz mapping with constant L > 0 (for short L-Lipschitz) if

d2(f(x), f(y)) ≤ Ld1(x, y) for all x, y ∈ X. Similarly, g : X → Y is a bi-LIpschitz mapping with constant L > 0 (for short L-bi-Lipschitz) if 1 d (x, y) ≤ d (g(x), g(y)) ≤ Ld (x, y) L 1 2 1 for all x, y ∈ X. In particular, a bi-Lipschitz mapping is always an injection (one-to-one) by the left-hand inequality above.

Lemma 2.19. Let (X, d1) and (Y, d2) be metric spaces. 1. If f : X → Y is a Lipschitz mapping of constant L, then

Hs(fA) ≤ LsHs(A) ∀A ⊂ X.

25 2. If g : X → Y is a bi-Lipschitz mapping, then

dimH(gA) = dimH(A) ∀A ⊂ X.

∞ ∞ Proof. (1): If {Ej}j=1 is a δ-cover of A, then {fEj}j=1 is an Lδ-cover of fA. We can choose the cover so that ∞ X s s d(Ej) ≤ Hδ(A) + ε, j=1 and hence ∞ ∞ s X s X s s s HLδ(fA) ≤ d(fEj) ≤ [Ld(Ej)] ≤ L (Hδ(A) + ε). j=1 j=1 The claim follows by letting first ε → 0 and then δ → 0. (2): We apply part (1) to both g : X → Y and g−1 : g(A) → X to see that

Hs(g(A)) ≤ LsHs(A), Hs(A) = Hs(g−1g(A)) ≤ LsHs(g(A)), and hence dimH(gA) = dimH(A).

n We next investigate the connection of H and the Lebesgue measure mn. Theorem 2.20. For all A ∈ Bor(Rn), we have n H (A) = cnmn(A) for some constant cn ∈ (0, ∞). n n Proof. If Q ⊂ R is a cube of sidelength√ `(Q), we can divide it into N cubes of sidelength N −1`(Q) and diameter nN −1`(Q). Hence √ n √n n −1
n/2 n n/2 H nN −1`(Q)(Q) ≤ N nN `(Q) = n `(Q) = n mn(Q).

n n/2 As N → ∞, this shows that H (Q) ≤ n mn(Q) < ∞ for all cubes Q. If E ⊂ Rn is nonempty, then E ⊂ B(x, d(E)) for any x ∈ E, and hence ∗ n mn(E) ≤ mn(B(x, d(E))) = cnd(E) ; this bound is obvious if E = ∅. Thus, if ∞ {Ei}i=1 be any δ-cover of A, we have ∞ ∞ X n X ∗ ∗ d(Ei) ≥ cnmn(Ei) ≥ cnmn(A) i=1 i=1 ∗ by the subadditivity of mn. Taking the infimum over all δ-covers, we get n ∗ n ∗ n Hδ (A) ≥ cnmn(A), and then also H (A) ≥ cnmn(A). In particular, H (Q) ≥ cnmn(Q) > 0 for all cubes Q. n Since both H and mn are invariant under translations (and by elementary n comparisons between cubes and balls), we see that both H and mn are uni- formly distributed Borel measures in the sense of Definition 2.24. The result then follows from the general Theorem 2.25 below.

26 Corollary 2.23. n ∗ n H (A) = cnmn(A) ∀A ⊂ R . n ∗ Proof. Exercise. (Use the fact that both H and mn are Borel regular outer measures.) Definition 2.24. A Borel measure µ on a metric space X is uniformly dis- tributed if 0 < µ(B(x, r)) = µ(B(y, r)) < ∞ for all x, y ∈ X and 0 < r < ∞. Theorem 2.25. If µ, ν are uniformly distributed Borel measures on a separable metric space X, then there is a constant c0 ∈ (0, ∞) such that

µ(A) = c0ν(A) for all A ∈ Bor(X). Proof. Denote g(r) = µ(B(x, r)) and h(r) = ν(B(x, r)) for 0 < r < ∞ and x ∈ X. Let U ⊂ X be open and bounded. We apply Fubini’s theorem (see Remark 2.26 for justification) to the function f(x, y) = 1B(x,r)(y) = 1B(y,r)(x) to see that

ν(B(x, r) ∩ U) dµ(x) = 1B(x,r)(y) dν(y) dµ(x) ˆU ˆU ˆU

= 1B(y,r)(x) dµ(x) dν(y) = µ(B(y, r) ∩ U) dν(y). ˆU ˆU ˆU If x ∈ U, then B(x, r) ∩ U = B(x, r) when r is small enough, and hence µ(B(x, r) ∩ U) ν(B(x, r) ∩ U) lim = 1 = lim , ∀x ∈ U. r→0+ g(r) r→0+ h(r) Moreover, both ratios are clearly bounded by 1 for every r > 0. Thus we can use dominated convergence to see that ν(B(x, r) ∩ U) 1 µ(U) = lim dµ(x) = lim ν(B(x, r) ∩ U)dµ(x) ˆU r→0+ h(r) r→0+ h(r) ˆU and similarly 1 ν(U) = lim µ(B(x, r) ∩ U)dν(x). r→0+ g(r) ˆU Observing that the two right-hand integrals above are equal, it follows that µ(U) g(r) = lim =: c0 ν(U) r→0+ h(r) where the existence of the limit is part of the conclusion. Since 0 < µ(U), ν(U) < ∞, we see that c0 ∈ (0, ∞) as well. Thus claim holds for all open bounded U ⊂ X. This is easily extended to all A ∈ Bor(X) via Dynkin’s lemma.

27 Remark 2.26. Let us justify the use of Fubini’s theorem above. First, the S∞ measures are σ-finite since X = i=1 B(x, i) and µ(B(x, i)) < ∞. The function f is the indicator of {(x, y) ∈ X × X : |x − y| < r}, which is an open set, and therefore Borel set in X × Y . Finally, we can use a general result: If Y and Z are topological spaces whose topologies have countable bases (i.e., they are so-called N2 spaces), then

(2.27) Bor(Y × Z) = σ(Bor(Y ) × Bor(Z)).

In the theorem, Y = Z = X is a separable metric space and hence an N2 space ∞ (a basis for the topology is given e.g. by B(xi, qj), where {xi}i=1 is dense in X ∞ and {qj}j=1 = Q+). Thus a Borel function on X × X is also measurable with respect to the product σ-algebra, and Fubini’s (or in fact Tonelli’s) theorem can be used. Let us justify the inclusion ⊂ in (2.27); this is what we need for product- measurability above. The other direction is left as an exercise. If A ⊂ Y × Z is open, using the countable topological bases of Y and Z, it can be expressed as a countable union [ A = Ai × Bj, i,j where Ai ⊂ Y,Bj ⊂ Z are open, hence Ai × Bj ∈ Bor(Y ) × Bor(Z), and hence the countable union belongs to σ(Bor(Y ) × Bor(Z)). This proves the claimed inclusion. We next construct sets whose Hausdorff dimension is not an integer. We recall the construction of Cantor sets from Real Analysis I: Let 0 < λ < 1/2. Let I0,1 := [0, 1], I1,1 := [0, λ] and I1,2 := [1 − λ, 1]. In n general, if intervals In,i, i = 1,..., 2 , are already defined, let In+1,2i−1 be the interval that has the same left end-point as In,i, and λ times its length, while In+1,2i is the interval that has the same right end-point as In,i, and λ times its n length. Thus dist(In+1,2i−1,In+1,2i) = (1 − 2λ)λ and

n n d(In,i) = λ ∀n, ∀i = 1,..., 2 .

We denote 2n ∞ [ \ Cn(λ) := In,i,C(λ) := Cn(λ). i=1 n=1 Then C(λ) is a compact uncountable set without interior points. It is also “self- similar” and m1(C(λ)) = 0. The case λ = 1/3 is a prominent special case that is frequently considered. Theorem 2.28. For all 0 < λ < 1/2, we have log 2 dim C(λ) = . H log(1/λ)

In particular, dimH C(λ) can take any value in (0, 1).

28 2n n Proof. Since {In,i}i=1 is a λ -cover of C(λ), we have

n s X s n ns n −n Hλn (C(λ)) ≤ d(In,i) = 2 λ = 2 2 = 1. i=1

Since λn → 0 as n → ∞, this shows that Hs(C(λ)) ≤ 1 < ∞. A proof of Hs(C(λ)) > 0 is left as an exercise.

29 Chapter 3

Compactness and convergence of Radon measures

3.1 Riesz representation theorem

In this section we investigate Radon measures of Rn from a new angle by study- ing the following problem: Which operations (“functionals”) on continuous func- tions correspond to integration with respect to a Radon measure? This will be used to provide useful notions of compactness and convergence for Radon mea- sures in the following sections. n We denote (Other texts use the notation Cc(R ).) n n n C0(R ) = {f ∈ C(R ) : supp(f) ⊂ R is compact}.

Here C(Rn) is the space of continuous functions f : Rn → R and

supp(f) = {x ∈ Rn : f(x) 6= 0} n is the support of f. Since the elements of C0(R ) are bounded functions, we n may equip C0(R ) with the norm kfk = sup |f(x)|. n x∈R The following result is crucial for numerous applications. The theorem can be proved in a larger generality of locally compact Hausdorff spaces, but the case of Rn below is easier, since we can use the machinery of Chapter 1, in particular the notion of metric outer measures.

n Definition 3.2. A mapping Λ: C0(R ) → R is a linear and positive functional if

n 1. Λ(αf1 + βf2) = αΛ(f1) + βΛ(f2) for all f1, f2 ∈ C0(R ) and all α, β ∈ R. n n 2. Λ(f) ≥ 0 whenever f ∈ C0(R ) and f(x) ≥ 0 for all x ∈ R .

30 If µ is a Radon measure on Rn, then

n Λµ(f) = fdµ, f ∈ C0(R ), ˆ n R is a positive linear functional. Indeed, f is integrable, since it is bounded and compactly supported, and µ(supp(f)) < ∞, since µ is a Radon measure. The converse to this observation is a deep result:

n Theorem 3.3 (Riesz representation theorem). Let Λ: C0(R ) → R be a posi- tive linear functional. Then there is a unique Radon measure µ such that

n Λ(f) = f(x)dµ(x), ∀f ∈ C0(R ). ˆ n R For the proof, we need a lemma:

Lemma 3.4. 1. Let K ⊂ V ⊂ Rn, where K is compact and V is open. Then n there exists f ∈ C0(R ) such that

n supp(f) ⊂ V, χK (x) ≤ f(x) ≤ 1 ∀x ∈ R .

Sm n 2. Let K ⊂ j=1 Vj ⊂ R , where K is compact and each Vj is open. Then n there exist functions hj ∈ C0(R ) such that

m X 0 ≤ hj ≤ 1, supp(hj) ⊂ Vj, χK ≤ hj ≤ 1. j=1

Proof. 1. Let δ = dist(K, ∂V ). Since K is compact, δ > 0. Then 2 f(x) = max(0, 1 − dist(x, K)) δ satisfies the required properties.

2. For every x ∈ K, we can find a radius rx > 0 such that B(x, 2rx) ⊂ Vj for some j. Since Kăis compact, it can be covered by finitely many of the

balls B(x, rx), say B(xi, rxi ), where i = 1, . . . , k, and then also by their ¯ closures B(xi, rxi ). Let [ ¯ Kj = B(xi, rxi ) ⊂ Vj.

i:B(xi,2rxi )⊂Vj ¯ Then each Kj is compact (as a finite union of the compact sets B(xi, rxi )) Sm and K ⊂ j=1 Kj. n By the first part, we can pick functions gj ∈ C0(R ) so that

χKj ≤ gj ≤ 1, supp(gj) ⊂ Vj.

31 We define

h1 = g1, hj = (1 − g1) ··· (1 − gj−1)gj, j = 2, . . . , m. Then clearly

0 ≤ hj ≤ 1, supp hj ⊂ supp gj ⊂ Vj.

Writing fi = 1 − gi and using telescopic summation, we find that

m m j−1 m j−1 X X Y X Y hj = gj (1 − gi) = (1 − fj) fi j=1 j=1 i=1 j=1 i=1 m j−1 j m m X  Y Y  Y Y = fi − fi = 1 − fi = 1 − (1 − gi) ≤ 1. j=1 i=1 i=1 i=1 i=1 Sm Moreover, if x ∈ K ⊂ j=1 Kj, then gj(x) = 1 for at least one j; hence Qm Pm i=1(1 − gi(x)) = 0 and thus j=1 hj(x) = 1.

Proof of the Riesz representation theorem. We define

n µ˜(V ) = sup{Λ(f): f ∈ C0(R ), 0 ≤ f ≤ 1, supp(f) ⊂ V } for every open V ⊂ Rn. (In particular, µ˜(∅) = 0, since f ≡ 0 is the only admissible function in the supremum.) It follows that

(3.5) 0 ≤ µ˜(V1) ≤ µ˜(V2),

n for all open sets V1 ⊂ V2 ⊂ R . We can thus define n (3.6) µ˜(A) = inf{µ˜(V ): A ⊂ V ⊂ R ,V open} for all A ⊂ Rn, and this agrees with the original definition for open sets. We shall prove that µ˜ is a metric outer measure. This implies that Borel sets of Rn are µ˜-measurable by Theorem 1.18. 1. The monotonicity µ˜(A1) ≤ µ˜(A2) if A1 ⊂ A2

is immediate from the definition (3.6). (µ(A2) is an infimum over a smaller set than µ(A1).) 2. We prove subadditivity for open sets first, i.e.,

∞ ∞  [  X (3.7) µ˜ Vj ≤ µ˜(Vj) j=1 j=1

n n S∞ for open sets Vj ⊂ R . Let f ∈ C0(R ) with K = supp(f) ⊂ j=1 Vj. Sm Since K is compact, we have K ⊂ j=1 Vj for some finite m. We can then

32 Pm apply Lemma 3.4 to find functions hj as in the lemma. Since j=1 hj = 1 Pm n on K, we have f = j=1 hjf, where hjf ∈ C0(R ) has supp(hjf) ⊂ Vj and 0 ≤ hjf ≤ 1, so that Λ(hjf) ≤ µ˜(Vj) by definition. Thus

m m m ∞  X  X X X Λ(f) = Λ hjf = Λ(hjf) ≤ µ˜(Vj) ≤ µ˜(Vj). j=1 j=1 j=1 j=1

Taking the supremum over all such f proves (3.7).

For general sets Aj, we choose open Vj ⊃ Aj such that µ˜(Vj) ≤ µ˜(Aj) + 2−jε. Then

∞ ∞ ∞ ∞  [  (3.6)  [  (3.7) X X µ˜ Aj ≤ µ˜ Vj ≤ µ˜(Vj) ≤ µ˜(Aj) + ε, j=1 j=1 j=1 j=1

and subadditivity follows as ε → 0. Thus µ˜ is an outer measure.

n n 3. Let V1,V2 ⊂ R be open and dist(V1,V2) > 0. If fj ∈ C0(R ) has supp(fj) ⊂ Vj and 0 ≤ fj ≤ 1, then 0 ≤ f1 + f2 ≤ 1 and supp(f1 + f2) ⊂ V1 ∪ V2, so that

Λ(f1) + Λ(f2) = Λ(f1 + f2) ≤ µ˜(V1 ∪ V2)

and taking the supremum over all such f1, f2 we get

(3.8) µ˜(V1) +µ ˜(V2) ≤ µ˜(V1 ∪ V2).

n If A1,A2 ⊂ R are general sets with dist(A1,A2) > 0, we can find open n sets Vj ⊃ Aj with dist(V1,V2) > 0, say Vj = {x ∈ R : dist(x, Aj) < 1 3 dist(A1,A2)}. Let also V ⊃ A1 ∪A2 be open with µ˜(V ) ≤ µ˜(A1 ∪A2)+ε. Then dist(V1 ∩ V,V2 ∩ V ) ≥ dist(V1,V2) > 0 and hence

µ˜(A1) +µ ˜(A2) ≤ µ˜(V1 ∩ V ) +µ ˜(V2 ∩ V )

≤ µ˜((V1 ∪ V2) ∩ V ) ≤ µ˜(V ) ≤ µ˜(A1 ∪ A2) + ε.

As ε → 0 we see that µ˜ is a metric outer measure.

4. We check that µ˜ is locally finite. Given any ball B(x, r) ⊂ Rn, we choose n some f0 ∈ C0(R ) such that χB(x,r) ≤ f0 ≤ 1. Thus f0 ≥ f for all admissible f in the definition of µ˜(B(x, r)), and hence Λ(f0) − Λ(f) = Λ(f0 − f) ≥ 0 for all such f. Thus

µ˜(B(x, r)) = sup Λ(f) ≤ Λ(f0) < ∞. f

By Corollary 1.29, n µ =µ ˜| Bor(R ) is a Radon measure.

33 5. We prove that Λ(f) = fdµ ˆ n R n for all f ∈ C0(R ). By splitting f = f+ − f−, where f+ = max(f, 0) and f− = max(−f, 0) and using the linearity of both Λ and the integral, it suffices to consider f ≥ 0. Fix an ε > 0 and consider  0, if f(x) ≤ (k − 1)ε,  fk(x) = f(x) − (k − 1)ε, if (k − 1) < f(x) < kε, ε, if kε ≤ f(x).

n Then 0 ≤ fk ≤ ε, fk ∈ C0(R ), and fk ≡ 0 if (k − 1)ε ≥ kfk. Thus Pm f(x) = k=1 fk(x) if m ≥ kfk/ε + 1. Let K(k) = {f ≥ kε} if k ≥ 1 and K(0) = supp(f). Thus

(3.9) εχK(k) ≤ fk ≤ εχK(k−1), and hence m m m X X X (3.10) ε µ(K(k)) ≤ fkdµ = fdµ ≤ ε µ(K(k − 1)). ˆ n ˆ n k=1 k=1 R R k=1

−1 On the other hand, if δ > 0, then (3.9) implies that ε (1 + δ)fk ≥ 1 in −1 some neighbourhood W ⊃ K(k), and in particular ε (1 + δ)fk ≤ g for every g appearing in the definition of µ(W ). Thus

εµ(K(k)) ≤ εµ(W ) = ε sup Λ(g) ≤ (1 + δ)Λ(fk). g

Since δ > 0 is arbitrary, it follows that εK(k) ≤ Λ(fk). −1 Similarly, ε fk is an admissible function for µ(V ) for any open V ⊃ −1 K(k − 1), and thus ε Λ(fk) ≥ infV ⊃K(k−1) µ(V ) = µ(K(k − 1)). Hence

εµ(K(k)) ≤ Λ(fk) ≤ εµ(K(k − 1)), and thus K m m X X X (3.11) ε µ(K(k)) ≤ Λ(fk) = Λ(f) ≤ ε µ(K(k − 1)). k=1 k=1 k=1

A comparison of (3.10) and (3.11) shows that m X fdµ − Λ(f) ≤ ε µ(K(k − 1)) − µ(K(k))
ˆ n R k=1 = εµ(K(0)) − µ(K(k))
≤ εµ(supp(f)), where µ(supp(f)) < ∞ since supp(f) is compact and µ is a Radon mea-

sure. Letting ε → 0 we see that Λ(f) = n fdµ. R ´ 34 6. Let us finally prove uniqueness. Let µ1 be any Radon measure such that n n Λ(f) = n fdµ for all f ∈ C0(R ). Let V ⊂ R be open. By Lemma 3.4, R n we can find´ a sequence fj ∈ C0(R ) such that 0 ≤ f1 ≤ f2 ≤ · · · ≤ fj → χV . Then monotone convergence implies that

µ1(V ) = χV dµ1 = lim fjdµ1 = lim Λ(fj), ˆ n j→∞ j→∞ R

and hence µ1(V ) is uniquely determined for all open sets V . But by the outer regularity of Radon measures this uniquely determines µ1(B) = inf{µ1(V ): V ⊃ B open} for all Borel sets B.

The support supp(µ) of a Radon measure µ on Rn is the smallest closed set, whose complement has zero measure. More precisely,

n [ n supp(µ) = R \ {V : V ⊂ R open , µ(V ) = 0}.

Corollary 3.12. Let K ⊂ Rn be compact and Λ: C(K) → R be a positive and linear functional. Then there is a unique finite Borel measure µ on K such that

Λ(f) = f(x)dµ(x) ∀f ∈ C(K). ˆK Proof. Exercise.

3.13 Weak convergence of measures

n Definition 3.14. Let µk, k ∈ N, be Radon measures on R . We say that the sequence (µk) converges weakly to a Radon measure µ if

lim fkdµ = fdµ k→∞ ˆ n ˆ n R R n w for all f ∈ C0(R ). This is denoted by µk * µ or µk → µ.

Example 3.15. 1. δk * 0. −1 Pk 2. Let µk = k j=1 δj/k. Then for all f ∈ C0(R),

k 1 X j 1 fdµ = f( ) → f(x)dx, ˆ k k k ˆ R j=1 0 since the sums are Riemann sums of the function f over finer and finer partitions of [0, 1]. Thus µk * m1x[0, 1]

Easy example (like 3.15) show that µk * µ does not always imply that µ(Ak) → µ(A). However:

35 n Theorem 3.16. Let µ, µk, k ∈ N, be Radon measures on R such that µk * µ. Then

n 1. lim sup µk(K) ≤ µ(K), if K ⊂ R is compact, k→∞ n 2. lim inf µk(V ) ≥ µ(V ), if V ⊂ R is open. k→∞ Proof. 1. Given a compact K ⊂ Rn, let V ⊃ K be an arbitrary open set n containing it. By Lemma 3.4, there is f ∈ C0(R ) such that χK ≤ f ≤ χV . Hence

µ(V ) ≥ fdµ = lim fdµk ≥ lim sup µk(K). ˆ n k→∞ ˆ n R R k→∞ Since µ(K) = inf{µ(V ): V ⊃ K open}, the claim follows.

2. Given an open V ⊂ Rn, let K ⊂ V be an arbitrary compact set contained n in it. By Lemma 3.4, there is f ∈ C0(R ) such that χK ≤ f ≤ χV . Hence

µ(K) ≤ fdµ = lim fdµk ≤ lim inf µk(V ). ˆ n k→∞ ˆ n k→∞ R R Since µ(V ) = sup{µ(K): K ⊂ V compact}, the claim follows.

3.17 Compactness of measures

Weak convergence of measures is not only a natural but also a very useful concept, since bounded families of Radon measure are sequentially compact. This provides a way — in many situations the only known way — of constructing suitable measures as limits of weakly convergent sequences.

n Theorem 3.18. Let (µk) be a sequence of Radon measures on R such that sup µ(K) < ∞ k

n for all compact K ⊂ R . Then there is a subsequence (µkj ) and a Radon measure µ such that

µkj * µ. For the proof we need:

n Lemma 3.19. The normed space (C0(R ), k · k), where kfk = sup{|f(x)| : x ∈ n ∞ n R }, is separable, i.e., there is a countable dense set F = {fj}j=1 ⊂ C0(R ). n In other words, if f ∈ C0(R ) and ε > 0, then kf − fjk < ε for some fj ∈ F. Proof. Exercise.

36 Proof of Theorem 3.18. Assume first that

n (3.20) sup µ(R ) = A < ∞. k

∞ n Let {fj}j=1 be dense in C0(R ). By assumption (3.20), the sequence n o f1dµk : k ∈ N ⊂ R ˆ n R 1 is bounded, and hence there is a subsequence (µk) of (µk) such that

1 k→∞ f1dµk −→ a1 ˆ n R j for some a1 ∈ R. Recursively, for every j ≥ 2 we choose a subsequence (µk) of j−1 (µk ) such that j k→∞ fjdµk −→ aj ˆ n R k ∞ for some aj ∈ R. Then the diagonal sequence {µk}k=1 satisfies

k (3.21) lim fjdµk = aj k→∞ ˆ n R ∞ for all j ≥ 1. Let L be the vector space spanned by {fj}j=1, i.e.,

m n X o L = g = λjfj : λj ∈ R, m ∈ N . j=1

We define X X Λ(g) = λjaj ∀g = λjfj ∈ L. j j By (3.21), k Λ(g) = lim gdµk ∀g ∈ L. k→∞ ˆ n R In particular, Λ is a well-defined, positive and linear functional on L. In addition, (3.20) implies that

(3.22) |Λ(g)| ≤ Akgk ∀g ∈ L.

n If f ∈ C0(R ) is arbitrary, we choose a sequence (hj) ⊂ L such that khj − fk → 0, and set Λ(f) = lim Λ(hj). j→∞ n Then (3.22) implies that Λ is well defined on C0(R ) and (3.22) continues to n n hold for all g ∈ C0(R ). In addition, Λ is a positive linear functional on C0(R ).

37 + Namely, if f ≥ 0 and kf − hjk → 0, then the nonnegative functions hj = + + + max(hj, 0) satisfy kf − hj k = kf − hj k ≤ kf − hjk → 0, and hence

k Λ(f) = lim Λ(hj) = lim lim hjdµk ≥ 0. j→∞ j→∞ k→∞ ˆ n R By Riesz Representation Theorem, there is a Radon measure µ such that

Λ(f) = fdµ ˆ n R n k n for all f ∈ C0(R ). We then prove that µk * µ. Let ε > 0. If f ∈ C0(R ), ε we pick g ∈ L such that kf − gk ≤ 2A . When k is large enough, we have k |Λ(g) − gdµk| ≤ ε, and hence ´ k k k Λ(f) − fdµk ≤ |Λ(f − g)| + Λ(g) − gdµk + (g − f)dµk ˆ n ˆ n ˆ n R R R ≤ Akf − gk + ε + Akg − fk ≤ 2ε.

k Hence µk * µ. Finally, we remove the extra assumption (3.20). From the assumption n supk µk(K) < ∞ for all compact K ⊂ R and the previous argument it fol- m m−1 lows that for every m ∈ N, there is a subsequence (µk ) ⊂ (µk ) ⊂ ... ⊂ (µk) and a Radon measure µm such that

m k→∞ m µk xB(0, m) * ν .

k Then the diagonal sequence (µk) satisfies

k k→∞ m µkxB(0, m) * ν

n for all m ∈ N. If f ∈ C0(R ), then supp(f) ⊂ B(0, m) for some (all) large enough m, and hence

k k k fdµk = fdµk = fd(µkxB(0, m)) ˆ n ˆ ˆ n R B(0,m) R has a limit (in fact, fdµm), as k → ∞. So we can define ´ k Λ(f) := lim fdµk, k→∞ ˆ n R and this is clearly a positive linear functional, hence given by Λ(f) = n fdµ R for some Radon measure µ by the Riesz representation theorem. ´

38 Chapter 4

On the Hausdorff dimension of fractals

4.1 Mass distribution and Frostman’s lemma

In general it is difficult to estimate the Hausdorff measure and dimension from below. The following simple principle of mass distribution will often solve the problem:

Theorem 4.2. Let K ⊂ Rn be compact. Suppose that there is a Radon measure µ on Rn such that µ(K) > 0 and s µ(C) ≤ c0d(C) n for all compact C ⊂ R with d(C) ≤ δ0, where s > 0 and δ0 > 0 are constants. Then dimH(K) ≥ s.

∞ Proof. If {Ej}j=1 is a δ-cover of K with 0 < δ ≤ δ0, then

∞ ∞ ∞ ∞  [  X X s X s 0 < µ(K) ≤ µ E¯j ≤ µ(E¯j) ≤ c0d(E¯j) = c0d(Ej) . j=1 j=1 j=1 j=1

Thus s s H (K) ≥ Hδ(K) ≥ µ(K)/c0 > 0 and hence dimH(K) ≥ s. The same proof also gives:

39 Theorem 4.3. Let X be a separable metric space and K ⊂ X. If there is an outer measure µ on X such that µ(K) > 0 and

s µ(U) ≤ c0d(U) for all U ⊂ X with d(U) ≤ δ0, where s > 0 and δ0 > 0 are constants, then dimH(K) ≥ s. 1 1 Example 4.4. Let C( 3 ) be the usual Cantor 3 -set, and

2n 1 [ Cn( 3 ) = In,i i=1 its nth level approximation, so that

∞ 1 \ 1 C( 3 ) = Cn( 3 ). n=1

1 Then there is a Radon probability measure µ supported on C( 3 ) such that −n µ(In,i) = 2 . While it can be constructed via the methods of Chapter 1, we now indicate another method based on the results of Chapter 3. We define a sequence of Radon measures (µn) by setting 3n µ = m C ( 1 ), n ∈ . n 2n 1x n 3 N Then −n −n 2 n µn(In,i) = 2 , µnxIn,i = m1xIn,i ∀i = 1,..., 2 ; m1(In,i) 1 1 supp(µn) = Cn( 3 ), µn(Cn( 3 )) = 1. It follows that

n−m −n −m m (4.5) µn(Im,i) = 2 · 2 = 2 ∀m ≤ n, ∀i = 1,..., 2 .

n−m (Here 2 is the number of intervals In,j contained in Im,i.) Thus µn divides its mass “evenly” on the approximating intervals of the Cantor set. Since µn(R) = 1,

Theorem 3.18 guarantees the existence of a subsequence (µnk ) that converges weakly to some Radon measure µ:

µnk * µ, as k → ∞.

Consider an approximating interval Im,i and an open U ⊃ R such that U ∩ 1 Cm( 3 ) = Im,i. Then Theorem 3.16 guarantees that −m −m 2 = lim sup µnk (Im,i) ≤ µ(Im,i) ≤ µ(U) ≤ lim inf µnk (U) = 2 , k→∞ k→∞

−m 1 and thus µ(Im,i) = 2 . It is also easy to check that supp(µ) = C( 3 ).

40 If E ⊂ R is closed and 3−n ≤ d(E) < 3−n+1 for some n ∈ N, then E can intersect at most two intervals In,i. Thus

−n −n s s µ(E) ≤ 2µ(In,i) = 2 · 2 = 2 · (3 ) ≤ 2 · d(E) , where s = log 2/ log 3. Thus Theorem 4.2 guarantees that

log 2 dim (C( 1 )) ≥ s = . H 3 log 3

s 1 1 With the (easy) estimate H (C( 3 )) ≤ 1, this shows that dimH(C( 3 )) = s. Similar considerations can be made in other situations. But we first prove the following fundamental result: Theorem 4.6 (Frostman’s lemma). Let K be compact and s > 0. Then

Hs(K) > 0 if and only if there exists a Radon measure µ on Rn such that

s n supp(µ) ⊂ K, µ(K) > 0, µ(B(x, r)) ≤ c0r ∀x ∈ R , ∀r > 0. Proof. The direction “⇐” follows from (the proof of) Theorem 4.2. Namely, if C ⊂ Rn is compact, then C ⊂ B(x, d(C)) for all x ∈ C and hence

s µ(C) ≤ c0d(C) by assumption. The proof of “⇒” is more difficult and uses a so-called stopping time argument. Such arguments are very useful in a variety of contexts, e.g. in harmonic analysis. By scaling and translation (cf. Theorem 2.18), we may assume that K ⊂ [0, 1)n and Hs(K) > 0. We say that [0, 1)n is a dyadic cube of generation 0. In general, dyadic cubes of generation m are all sets of the form hj − 1 j  hj − 1 j  hj − 1 j  Q = 1 , 1 × 2 , 2 × n , n , 2m 2m 2m 2m 2m 2m

m −m for integers 1 ≤ j1, j2, . . . , jn ≤ 2 . Thus the side-length of these cubes is 2 , and they are obtained by splitting dyadic cubes of generation m − 1 into 2n equal cubes by bisecting each of their sides. Let

n Dm = {Q : Q is a dyadic cube of generation m in [0, 1) }.

For every Q ∈ Dm, m ≥ 1, there is a unique Q˜ ∈ Dm−1 that contains Q (i.e., Q ⊂ Q˜). We may assume (by further scaling and translation if necessary) that K 6⊂ Q for any Q ∈ D1.

41 s s We now use the hypothesis that H (K) > 0. This means that also Hδ(K) > 0 for some δ > 0, and hence

∞ X s (4.7) d(Qj) ≥ a0 > 0 j=1 S∞ whenever K is covered by a sequence of dyadic cubes Qj ∈ m=0 Dm. Indeed, we may take s s a0 = min{δ , Hδ(K)}. m m n n For every m ∈ N, we define a measure µm by setting µmx(R \ [0, 1) ) = 0 and for every Q ∈ Dm:

( m 2−ms µmxQ = mnxQ, if Q ∩ K 6= ∅; (4.8) mn(Q) m µmxQ = 0, if Q ∩ K = ∅.

n n Since ∪{Q : Q ∈ Dm} = [0, 1) , this defines a Radon measure on R with m n supp(µm) ⊂ [0, 1] . m m We next modify µm to define a sequence of measures µk , k = m − 1,..., 0, m by downward recursion on k. If µk+1 is already defined, we define for every Q ∈ Dk:

( m m m −ks µk xQ = µk+1xQ, if µk+1(Q) ≤ 2 ; (4.10) m 2−ks m m −ks µk Q = m µk+1 Q, if µk+1(Q) > 2 . x µk+1(Q) x

m m Finally we let µ = µ0 . Note that, at each step of the recursion, we either keep the previous measure, or multiply it by a factor smaller than one; thus the measure of a cube can never increase in the construction, and thus

m −ks s (4.11) µ (Q) ≤ 2 = `(Q) ∀Q ∈ Dk, ∀k = 0, 1, . . . , m.

We next prove that the measures µm(Rn) have a positive lower bound. To this end, we observe that by construction, for every x ∈ K, there is a a cube Sm Q ∈ k=0 Dm such that x ∈ Q and (cf. (4.8) and (4.10)) (4.12) µm(Q) = `(Q)s = c(n, s)d(Q)s.

Let Qi, i = 1, . . . , `, be the maximal (with respect to set inclusion) dyadic cubes with this property. Then they are pairwise disjoint, and

` G K ⊂ Qi. i=1 Thus (4.7) implies that

` ` m n X m X s (4.13) µ (R ) = µ (Qi) = c(n, s) d(Qi) ≥ c(n, s)a0 > 0. i=1 i=1

42 We can then define 1 νm = µm. µm(Rn) Since νm(Rn) = 1, by Theorem 3.18, there is a subsequence (νmj ) and a Radon measure µ such that νmj * µ as j → ∞. We prove that µ is the required measure. First of all, since supp(νm) ⊂ [0, 1] and νm(Rn) = 1, the same is true for the limit measure (see Theorem 3.16). Second, (4.8) implies that supp(νm) is contained in a c(n)2−m-neighbourhood of K, and thus supp(µ) ⊂ K since 2−mj → 0. In particular, µ(K) = µ(Rn) = 1. n 1 Let then x ∈ R and 0 < r < 2 . Consider dyadic cubes Qi ∈ D`, where 2−`−1 ≤ 2r < 2−`. At most 2n such cubes can intersect B(x, r). Thus

2n 3.16   mj mj [ µ(B(x, r)) ≤ lim inf ν (B(x, r)) ≤ lim inf ν Qi j→∞ j→∞ i=1 (4.11) (2−`)s (4.13) n mj s ≤ 2 lim inf ν ≤ c0r . j→∞ µmj (Rn)

1 If r ≥ 2 , then n s s µ(B(x, r)) ≤ µ(R ) = 1 ≤ 2 r . Hence all properties required in Frostman’s lemma have been verified.

n Remark 4.14. 1. Frostman’s lemma extends easily to all Fσ-sets K ⊂ R . 2. Frostman’s lemma actually holds for all Borel sets K ⊂ Rn, but in this generality the proof is much harder.

Corollary 4.15. Let K ⊂ Rn be compact. Then

dimH(K) = sup{s : ∃ a probability measure µ such that supp(µ) ⊂ Kand s n µ(B(x, r)) ≤ c0r ∀r > 0, ∀x ∈ R }. 4.16 Self-similar fractals

Let X be a metric space, ∅ 6= S ⊂ X and ε > 0. The set [ S(ε) = {x ∈ X : dist(x, S) < ε} = B(x, ε). x∈S is called the ε-neighbourhood of S. Let ∅(ε) = ∅.

43 Definition 4.17. The Hausdorff distance of nonempty sets A, B ⊂ X is

dH (A, B) = inf{r > 0 : A ⊂ B(r),B ⊂ A(r)}.

Moreover, let dH (∅, ∅) = 0 and dH (A, ∅) = dH (∅,A) = ∞ if A 6= ∅. n n For example, if A = {0} ⊂ R and B = B(0, r) ⊂ R , then dH (A, B) = r. There is also a “triangle inequality” for all A, B, C ⊂ X:

Lemma 4.18. If A, B, C ⊂ X, then dH (A, B) ≤ dH (A, C) + dH (C,B). Proof. Exercise.

However, dH is not a metric in general, since it can happen that dH (A, B) = 2 ∞ (e.g. if X = R and A and B are the x- and y-axes), or dH (A, B) = 0 even if A 6= B (e.g. if A = X = R and B = Q). Let us define

C(X) = {K : K ⊂ X is compact and nonempty}.

Then:

Lemma 4.19. 1. If A, B ⊂ X are closed and A 6= B, then dH (A, B) > 0.

2. dH defines a metric on C(X). Proof. Exercise.

We henceforth consider the case X = Rn: n Lemma 4.20. The metric space (C(R ), dH ) is complete. n Proof. Let (Aj) be a Cauchy sequence in (C(R ), dH ). We will prove that it converges to

∞ ∞ \ [ (4.21) A = Bk, where Bk = Ai, k=1 i=k

∞ in the metric dH . Since (Aj) is a Cauchy sequence, the sequence dH (Ai,A1)i=1 is bounded. Thus Ai ⊂ A1(r) for some r > 0 and all i. Thus Bk ⊂ A1(r) is compact as a closed subset of the (closed and bounded, hence) compact set A1(r). Then A is the intersection of a decreasing sequence of nonempty compact sets, hence it is also nonempty and compact, i.e., A ∈ C(Rn). To prove that dH (Ai,A) → 0, fix ε > 0 and let i0 be so large that dH (Aj,Ak) < ε for all j, k ≥ i0. Then for i ≥ k, we have Ai ⊂ Aj(ε), hence

∞ [ Ai ⊂ Aj(ε) i=k and thus Bk ⊂ Aj(ε) ⊂ Aj(2ε). In particular, A ⊂ Aj(2ε) for j ≥ i0.

44 Conversely, if i ≥ i0 and x ∈ Ai, then x ∈ Aj(ε) for every j ≥ i0, and hence there is xj ∈ Aj ∩ B(x, ε). In particular, the sequence (xj)j≥i0 is bounded, and hence has an accumulation point y. This point must be in Bk for all k, and hence in A. On the other hand, |x − y| ≤ ε, hence x ∈ A(2ε), and thus Ai ⊂ A(2ε) when i ≥ i0. We have proved that dH (Ai,A) ≤ 2ε for i ≥ i0. Remark 4.22. For a metric space X, there holds:

1. If X is compact, then (C(X), dH ) is compact. 2. If X is complete and B(X) = {A ⊂ X : A is closed, bounded and nonempty},

then (B(X), dH ) is complete. We aim to apply the Banach fixed point theorem in the complete metric space n (C(R ), dH ). We recall this result from Topology. Let us use the notation f ◦m = f ◦ f ◦ · · · ◦ f (m times) for mappings f : Y → Y . Theorem 4.23 (Banach fixed point theorem). Let (X, d) be a complete metric space and f : X → X a strict contraction, i.e., a Lipschitz function of constant L < 1. Then f has a unique fixed point x0 ∈ X, i.e., f(x0) = x0. In addition, ◦m x0 = lim f (y) m→∞ for any y ∈ X. Proof. We note that d(f ◦(m+1)(y), f ◦m(y)) = d(f ◦m(f(y)), f ◦m(y)) ≤ Lmd(f(y), y) and hence if m > k ≥ 1,

m−1 X d(f ◦m(y), f ◦k(y)) ≤ d(f ◦(j+1)(y), f ◦j(y)) j=k m−1 X d(f(y), y) ≤ Ljd(f(y), (y)) ≤ Lk → 0 1 − L j=k ◦m ∞ as k → ∞, since L < 1. Thus (f (y))m=1 is a Cauchy sequence, and hence convergent to some x0 ∈ X. But then

◦(m+1) ◦m  ◦m  x0 = lim f (y) = lim f(f (y)) = f lim f(f (y) = f(x0), m→∞ m→∞ m→∞ since f is continuous. If f(xi) = xi for i = 1, 2, then

d(x1, x2) = d(f(x1), f(x2)) ≤ Ld(x1, x2), hence d(x1, x2) = 0, and thus x1 = x2.

45 n n Let then ψj : R → R , j = 1, . . . , k, be strict contractions of constant c0 < 1, i.e., n |ψj(x) − ψj(y)| ≤ c0|x − y| ∀x, y ∈ R , ∀j = 1, . . . , k. We use them to define a mapping Ψ: C(Rn) → C(Rn) by setting k [ (4.24) Ψ(A) = ψjA j=1 for all compact A. Since the mappings ψj are continuous, each ψjA and hence their finite union Ψ(A) is compact, thus Ψ(A) ∈ C(Rn) as required. Since

dH (Ψ(A), Ψ(B)) ≤ max dH (ψjA, ψjB) ≤ c0dH (A, B) j for all A, B ∈ C(Rn), the mapping Ψ is also a strict contraction. By the Banach Fixed Point Theorem 4.23, there is a unique compact set F such that

k [ F = Ψ(F ) = ψjF. j=1

◦m Moreover, we have dH (F, Ψ (A)) → 0 as m → ∞, no matter how we chose the k compact A 6= ∅! The set F is called the invariant set for the family {ψj}j=1. 1 1 2 Example 4.25. Let ψ1(x) = 3 x and ψ2(x) = 3 x + 3 for x ∈ R. Then the 1 Cantor set C( 3 ) is the invariant set for the family {ψ1, ψ2}. Namely, it is easy to see that 1 1 1 1 Ψ(C( 3 )) = ψ1(C( 3 )) ∪ ψ2(C( 3 )) = C( 3 ), so the claim follows from the uniqueness of the fixed point. n n Let us next consider a family of contractive similarities ψj : R → R , i.e., n |ψj(x) − ψj(y)| = rj|x − y| ∀x, y ∈ R , where 0 < rj < 1 for j = 1, . . . , k. In this case the invariant set F for the family k {ψj}j=1 is called a self-similar fractal. Let s > 0 be the unique number such that k X s (4.26) rj = 1. j=1

Pk 0 Such a unique s exists, since the sum is a decreasing function of s and j=1 rj = Pk s k > 1 andă j=1 rj → 0 as s → ∞. Theorem 4.27. There is a Radon measure µ such that µ(F ) = µ(Rn) = 1 and k X s −1 (4.28) µ(A) = ri µ(ψi A) i=1 for all Borel sets A ⊂ Rn.

46 Proof. We begin by observing that raising (4.26) to any power m ∈ N we obtain

X s (4.29) (rj1 rj2 ··· rjm ) = 1.

1≤j1,...,jm≤k Fix an x ∈ F and denote

xj1,...,jm = ψj1 ◦ · · · ◦ ψjm (x), where m ∈ N and 1 ≤ ji ≤ k. For every m ∈ N, we define a Radon measure X µ = (r r ··· r )sδ . m j1 j2 jm xj1,...,jm 1≤j1,...,jm≤k

n Then (4.29) implies that µm(R ) = 1 for all m ∈ N. Since ◦m ◦m xj1,...,jm ∈ Ψ ({x}) ⊂ Ψ (F ) = F, we have supp(µm) ⊂ F . By Theorem 3.18, (µm) has a weakly convergent subsequence

(4.30) µmj * µ, j → ∞, where µ is a Radon measure on Rn. Moreover, by Theorem 3.16,

1 = lim sup µmj (F ) ≤ µ(F ) ≤ µ(U) ≤ lim inf µmj (U) = 1, j→∞ j→∞ when U is an open neighbourhood of F . Hence µ(F ) = 1 and supp(µ) ⊂ F . In fact, the full sequence satisfies µm * µ as m → ∞. To prove this, it suffices by (4.30) to check that  ∞ fdµm ˆ n m=1 R n n is a Cauchy sequence for every f ∈ C0(R ). Fix f ∈ C0(R ) and ε > 0. By uniform continuity, we can find an η > 0 such that

(4.31) |f(x) − f(y)| < ε whenever |x − y| < η.

Writing r = maxj rj < 1, we observe that

m (4.32) d(ψj1 ◦ · · · ψjm (F )) ≤ r d(F ) < η, when m ≥ m0 and m0 is large enough. If now ` > m ≥ m0, then

fdµ` − fdµm ˆ n ˆ n R R X X = (r ··· r )sf(x ) − (r ··· r )sf(x ) j1 j` j1,...,j` j1 jm j1,...,jm j1,...,j` j1,...,j` X = (r ··· r )sf(x ) − f(x )
, j1 j` j1,...,j` j1,...,jm j1,...,j`

47 where the last equality follows from

X s (rj1 ··· rj` ) f(xj1,...,jm )

j1,...,j` X s X s = (rj1 ··· rjm ) f(xj1,...,jm ) (rjm+1 ··· rj` )

j1,...,jm jm+1,...,j` X s = (rj1 ··· rjm ) f(xj1,...,jm ) · 1.

j1,...,jm Since

xj1,...,j` = ψj1 ◦ · · · ◦ ψjm ◦ ψjm+1 ◦ · · · ◦ ψj` (x) ∈ ψj1 ◦ · · · ◦ ψjm (F ) and

xj1,...,jm ∈ jm+1 ◦ · · · ◦ ψj` (x), we get from (??) that

|xj1,...,j` − xj1,...,jm | < η. Hence (4.31) implies that

X s fdµ` − fdµm ≤ (rj1 ··· rj` ) ε = ε. ˆ n ˆ n R R j1,...,j`

We have proved the weak convergence µm * µ of the full sequence. n We finally observe that for all f ∈ C0(R ),

X s fdµm = f(xj1,...,jm )(rj1 ··· rjm ) ˆ n R j1,...,jm k X  X s s = f ◦ ψi(xj2,...,jm )ri (rj2 ··· rjm ) j2,...,jm i=1 k  X s  = ri (f ◦ ψi) dµm−1. ˆ n R i=1 n Since µm * µ and both f and all f ◦ ψi belong to C0(R ), taking the limit m → ∞ of both sides gives

k  X s  (4.33) fdµ = ri (f ◦ ψi) dµ. ˆ n ˆ n R R i=1

By approximation we see that (4.33) holds for the characteristic function χA of any Borel set A. Thus

k  X s  µ(A) = χAdµ = ri (χA ◦ ψi) dµ ˆ n ˆ n R R i=1 k k X s X s −1 = ri χψ−1Adµ = ri µ(ψi A) ˆ n i i=1 R i=1

48 and (4.28) is hence proved.

1 1 1 2 Example 4.34. Let F = C( 3 ), ψ1(x) = 3 x and ψ2(x) = 3 x + 3 as in Example −1 1 1 1 −1 1 1 4.25. Since ψ1 (C( 3 ) ∩ [0, 3 ]) = C( 3 ) and ψ2 (C( 3 ) ∩ [0, 3 ]) ⊂ [−2, −1] ⊂ 1 R \ C( 3 ), the measure µ provided by Theorem 4.27 satisfies 1 1 1 s 1 1 1 µ(C( 3 ) ∩ [0, 3 ]) = ( 3 ) µ(C( 3 )) = 2 · 1 = 2 , 1 s 1 1 s 1 s where ( 3 ) = 2 if s is satisfies (4.26), which in this case reads as ( 3 ) +( 3 ) = 1. 1 Thus s = dimH C( 3 ).Similarly 1 2 1 µ(C( 3 ) ∩ [ 3 , 1]) = 2 . By iteration, we see that µ is the “evenly distributed” measure constructed in s 1 Example 4.4. In fact in can be shown that µ = H xC( 3 ). The previous example gives rise to the question: If F is the self-similar k fractal determined by a family {ψj}j=1, is dimH F given by the solution of the equation k X s rj = 1, j=1 where rj is the scaling factor of ψj? It is easy to see that the answer is no, if the sets ψjF overlap too much:

Example 4.35. If ψ1(x) = (1 − ε)x and ψ2(x) = (1 − ε)x + ε for x ∈ R, where 1 ε ∈ (0, 2 ], then

[0, 1] = [0, 1 − ε] ∪ [ε, 1] = ψ1[0, 1] ∪ ψ2[0, 1],

s s so that [0, 1] is the self-similar fractal determined by {ψ1, ψ2}. Now r1 + r2 = 2(1 − ε)s, and this is equal to 1 if and only if

log 1 s = 2 . log(1 − ε) We see that s → ∞ as ε → 0. In order to avoid situations like Example 4.35, we define the following con- cept:

k Definition 4.36. A family of similarities {ψj}j=1 satisfies the open set condi- tion, if there exists a bounded open V ⊂ Rn such that

k G (4.37) Ψ(V ) = ψjV ⊂ V, j=1 where the notation implies the requirement that the sets ψjV are disjojnt, i.e.,

(4.38) ψiV ∩ ψjV = ∅ if i 6= j.

49 1 Example 4.39. Let F = C( 3 ) and {ψ1, ψ2} be as in Example 4.34. By choosing (for example) V = (0, 1)? we observe that

1 2 Ψ(V ) = ψ1V t ψ2V = (0, 3 ) t ( 3 , 1) ⊂ V, so that {ψ1, ψ2} satisfies the open set condition. k Theorem 4.40. Let {ψj}j=1 be a family of contractive similarities that satisfy k the open set condition. Let F be the self-similar fractal determined by {ψj}j=1. If rj is the scaling factor of ψj and

k X s (4.41) rj = 1, j=1 then 0 < Hs(F ) < ∞, i.e., F is an s-set, where s is determined by (4.41). In particular, dimH F = s. For the proof of this result we need several lemmas.

k Lemma 4.42. Let {ψj}j=1 be a family of contractive similarities that satisfy the open set condition with V ⊂ Rn. k ¯ 1. If F is the self-similar fractal determined by {ψj}j=1, then F ⊂ V . 0 2. Let V = {ψj1 ◦ · · · ◦ ψj` (V ) : 1 ≤ ji ≤ k; ` ∈ N}. If U, U ∈ V and U ∩ U 0 6= ∅, then U ⊂ U 0 or U 0 ⊂ U. Proof. 1. From (4.37) it follows that Ψ(V¯ ) ⊂ Ψ(V ) ⊂ V¯ , and hence

V¯ ⊃ Ψ(V¯ ) ⊃ Ψ◦2(V¯ ) ⊃ · · · ⊃ Ψ◦i(V¯ ) ⊃ · · · .

◦i Since dH (Ψ (V¯ ),F ) → 0, it follows that F ⊂ V¯ . 2. We denote

Vj1,j2,...,j` = ψj1 ◦ ψj2 ◦ · · · ◦ ψj` (V ).

The open set condition (4.37) implies in particular that ψjV ⊂ V for all

j = 1, . . . , k, and hence Vj1,j2,...,j` ⊂ V . If m ≥ `, this implies that

Vj ,j ,...,j = ψj ◦ · · · ◦ ψj (ψj ◦ · · · ◦ ψj (V )) (4.43) 1 2 m 1 ` `+1 m ⊂ ψj1 ◦ · · · ◦ ψj` (V ) = Vj1,j2,...,j` It is enough to prove: If

(4.44) Vj1,j2,...,j` ∩ Vk1,k2,...,k` 6= ∅,

then ji = ki for all i = 1, . . . , `. Namely, if Vj1,j2,...,jm ∩ Vk1,k2,...,k` 6= ∅ for some m ≥ `, by (4.43) it follows that also (4.44) holds. Suppose to the contrary that ji 6= ki for some i ∈ {1, 2, . . . , `}, and let N be the smallest such i. Then

ψj1 ◦ · · · ◦ ψjN −1 = ψk1 ◦ · · · ◦ ψkN −1 = f

50 is injective, since each ψi is, and

Vj1,j2,...,j` ⊂ Vj1,j2,...,jN = f(ψjN V ),

Vk1,k2,...,k` ⊂ Vk1,k2,...,kN = f(ψkN V ).

If the sets on the left intersect, then f(ψjN V ) ∩ f(ψkN V ) 6= ∅, hence

ψjN V ∩ ψkN V 6= ∅, since f is injective, and thus jN = kN by (4.38). But this is a contradiction to jN 6= kN , which proves that ji = ki for all i in (4.44).

n Lemma 4.45. Let {Vj : j ∈ J} be a collection of disjoint open subsets of R such that, for some positive constants α, β and R, for every j ∈ J there are 0 points xj, xj such that

0 B(xj, αR) ⊂ V ⊂ B(xj, βR).

n −n Then any ball B(x, R) can intersect at most q sets Vj, where q ≤ (1 + 2β )α . ¯ Proof. If B(x, R) ∩ Vj 6= ∅, then

B(xj, αR) ⊂ Vj ⊂ B(x, (1 + 2β)R), and the sets on the left are disjoint. If there are N sets Vj like this, then

n X n cn (1 + 2β)R = mn(B(x, (1 + 2β)R)) ≥ mn(B(xj, αR)) = cn(αR) N, j which gives the claimed upper bound for N.

51 Chapter 5

Differentiation of measures

5.1 Besicovitch and Vitali covering theorems

The following Besicovitch covering theorem differs from the so called basic cov- ering theorem in that the balls need not be disjoint; however, the amount of overlap of the balls is under control. Theorem 5.2 (Besicovitch covering theorem). For every dimension n? there are natural numbers P (n) and Q(n) such that the following holds: Let A ⊂ Rn be bounded and B be a family of closed balls of Rn such that every x ∈ A is the centre of some B¯(x, r) ∈ B. Then:

1. There is a countable (possibly finite) collection of balls Bi ∈ B that cover n A and every point of R belongs to at most P (n) balls Bi, i.e., X χA ≤ χBi ≤ P (n). i

2. There are countable families B1,..., BQ(n) ⊂ B such that

Q(n) [  [  A ⊂ B

i=1 B∈Bi

and the balls in each Bi, i = 1,...,Q(n), are disjoint, i.e.,

0 0 0 B ∩ B = ∅ if B,B ∈ Bi,B 6= B .

The proof is based on geometric properties of the Euclidean space Rn.

Lemma 5.3. Let x, y ∈ Rn \{0} satisfy max(|x|, |y|) ≤ |x − y|. Then x y − ≥ 1. |x| |y|

52 Proof. By symmetry, we may assume that |y| ≥ |x|. Then

x y |y|x − |x|y |y|(x − y) + (|y| − |x|)y − = = |x| |y| |x||y| |x||y|

|y||x − y| − |y| − |x| |y| |x − y| − |y| − |x| ≥ = |x||y| |x| |y| − (|y| − |x|) |x| ≥ = = 1. |x| |x|

Lemma 5.4. For every dimension n, there is a number N(n) such that: When- n ever r1, . . . , rk > 0 and x1, . . . , xk ∈ R satisfy

1. xj ∈/ B¯(xi, ri) when j 6= i, and Tk ¯ 2. i=1 B(xi, ri) 6= ∅, then k ≤ N(n). Tk ¯ Proof. By translation, we may assume that 0 ∈ i=1 B(xi, ri); whence |xi| ≤ ri and (by (1)) xi 6= 0 for all i = 1, . . . , k. Thus

0 < |xi| ≤ ri < |xi − xj| if i 6= j, and Lemma 5.3 implies that the vectors ui := xi/|xi| ∈ B¯(0, 1) satisfy ¯ n |ui − uj| ≥ 1 when i 6= j. Since the unit ball B(0, 1) ⊂ R is compact, there can be at most N(n) points ui like this. Proof of the Besicovitch covering theorem.

1. For each x ∈ A, we choose some B¯(x, r(x)) ∈ B, and denote

M1 = sup r(x). x∈A

Since A is bounded, we may assume that M1 < ∞, for otherwise we simply choose a single large enough ball B ∈ B that contains A, and we are done. 1 Choose x1 ∈ A such that r1 > 2 M1 and then inductively

j [ ¯ 1 xj+1 ∈ A \ B(xi, r(xi)) such that r(xj+1) > 2 M1 i=1

1 as long as this is possible. Since A is bounded and |xi − xj| > 2 M1 when i 6= j, the process terminates after finitely many, say k1, steps. Let Bi = B¯(xi, r(xi)) for these chosen points with i = 1, . . . , k1.

53 We then denote

k n [1 o M2 = sup r(x): x ∈ A \ Bi . i=1

1 Clearly M2 ≤ 2 M1. We then continue the induction for j ≥ k1 to choose

j [ ¯ 1 xj+1 ∈ A \ B(xi, r(xi)) such that r(xj+1) > 2 M2 i=1 until the process terminates after finitely many steps for the same reason as before. Continuing in this way, we obtain

• indices 0 = k0 < k1 < k2 < . . ., 1 • numbers M1 > M2 > . . . such that Mi+1 ≤ 2 Mi, • balls Bi = B¯(xi, r(xi)) ∈ B, and

• index sets Ij = {kk−1 + 1, . . . , kj}, j = 1, 2,....

We prove that the balls Bi satisfy the required conditions. For this we observe that:

1 1 (5.6) 2 Mj < r(xi) ≤ Mj ≤ 2 Mj−1 ∀i ∈ Ij,

j [ (5.7) xj+1 ∈ A \ Bi ∀j = 1, 2,..., i=1

[ [ (5.8) xi ∈ A \ Bj ∀i ∈ Ik, ∀k = 1, 2,...,

m6=k j∈Im

Properties (5.6) and (5.7) are immediate from the selection process. To prove (5.8), let i ∈ Ik, m 6= k, and j ∈ Im. If m < k, then (5.7) implies that xi ∈/ Bj. If m > k, then (5.6) implies that

j∈Mm m>k i∈I 1 1 k r(xj) ≤ 2 Mm−1 ≤ 2 Mk < r(xi).

By (5.7) we know that xj ∈/ Bi, thus |xj − xi| > r(xi) > r(xj), and hence xi ∈/ Bj. This completes the verification of (5.8). Clearly (5.6) implies that

(5.9) Mi → 0, r(xi) → 0 as i → ∞.

We use this to prove that ∞ [ A ⊂ Bi, i=1

54 which is the first half of (1) of the theorem. Suppose for contradiction that there exists a point ∞ [ x ∈ A \ Bi. i=1 1 For large enough i, we have r(x) > 2 Mi. But then we should not have terminated the process of choosing the points xj, j ∈ Ii, without also choosing x. This is a contradiction, and hence no such x can exist. We then prove that P (n) in the theorem can be chosen as P (n) = 8nN(n), where N(n) is as in Lemma 5.4. Suppose that \ x ∈ Bi. i∈I

0 0 Let J := {j : I ∩ Ij 6= ∅}, and let I ⊂ I be chosen so that I ∩ Ij T has exactly one element for each j ∈ J . Then i∈I0 Bi 6= ∅ and, by 0 0 0 (5.8), xi ∈/ Bi0 if i, i ∈ I and i 6= i . Thus Lemma 5.4 guarantees that card J = card I0 ≤ N(n). Hence it suffices to show that

n (5.10) card(Ij ∩ I) ≤ 8 ∀j = 1, 2,....

0 By (5.6) and (5.7), if i, i ∈ Ij, then

1 |xi0 − xi| > r(xmin(i,i0)) > 2 Mj,

1 and the balls B(xi, 4 Mj) are disjoint for all i ∈ Ij. On the other hand, for i ∈ Ij ∩ I, we have |xi − x| ≤ r(xi) ≤ Mj, and hence all these balls are contained in B(x, 2Mj). Thus, writing ωn = mn(B(0, 1)), we have

n X 1 ωn(2Mj) = mn(B(x, 2Mj)) ≥ mn(B(xi, 4 Mj) i∈Ij ∩I 1 n = card(Ij ∩ I)ωn( 4 Mj) ,

n n and thus card(Ij ∩ I) ≤ (2 · 4) = 8 .

2. Let Bi be the balls constructed in part (1). Besides the property stated in the theorem, we will only use (5.9) (and no other details of the construc- tion). The convergence ri = r(xi) → 0 allows us to reorganise the balls in such an order that r1 ≥ r2 ≥ ... We then define

B1,1 = B1

and inductively

j [ B1,j+1 = Bk for the smallest k such that Bk ∩ B1,i = ∅. i=1

55 This produces a family

B1 = {B1,i : i = 1, 2,...} of pairwise disjoint balls B1,i. If S B does not cover A, we define B∈B1

B2,1 = Bk for the smallest k such that Bk ∈/ B1, and then inductively

j [ B2,j+1 = Bk for the smallest k such that Bk ∩ B2,j = ∅. i=1

Continuing in this way, we obtain families B1, B2,... ⊂ {Bi : i = 1, 2,...} such that each Bi consists of pairwise disjoint balls. We will finally prove that

m [ [ (5.11) A ⊂ B

k=1 B∈Bk for some m ≤ 4nP (n)+1, which proves the theorem with Q(n) = 4nP (n)+ 1. For this, we assume that there is a point

m [ [ x ∈ A \ B,

k=1 B∈Bk

n S∞ and prove that this implies that m ≤ 4 P (n). Since A ⊂ i=1 Bi, we have x ∈ Bi for some i. Thus Bi ∈/ Bk for any k = 1, . . . , m. By the construction of these families, it follows that for all these k, there exists some Bk,ik ∈ ¯ Bk such that Bi ∩ Bk,ik 6= ∅ and ri ≤ rk,ik , where Bi = B(xi, ri) and ¯ 0 Bk,ik = B(xk,ik , rk,ik ). Let zk ∈ Bi ∩ Bk,ik and zk := zk + ρk(xk,ik − zk), 1 where ρk = 2 ri/rk,ik . Then

0 1 |zk − xk,ik | ≤ (1 − ρk)rk,ik = rk,ik − 2 ri, 0 0 3 |zk − xi| ≤ |zk − zk| + ri ≤ ρrk,ik + ri = 2 ri,

0 ¯ 0 1 ¯ and thus Bk = B(zk, 2 ri) ⊂ Bk,ik ∩ 2Bi, where 2Bi = B(xi, 2ri). Now m m X X χ 0 ≤ χ ≤ P (n) Bk Bk,ik k=1 k=1

56 0 by part (1), but also Bk ⊂ 2Bi. Hence m n  [ 0  ωn(2ri) = mn(2Bi) ≥ mn Bk = dmn ˆSm k=1 k=1 Bk m 1 X ≥ χ 0 dm Bk n P (n) ˆSm 0 k=1 Bk k=1 m 1 X 1 = m (B0 ) = mω ( 1 r )n, P (n) n k P (n) n 2 i k=1 and thus m ≤ P (n)(2 · 2)n as claimed.

Theorem 5.12 (Vitali covering theorem for Radon measures on Rn). Let µ be a Radon measure on Rn, let A ⊂ Rn be a Borel set, and B a family of closed balls such that (5.13) inf{r : B¯(x, r) ∈ B} = 0 ∀x ∈ A.

Then there is a countable family of disjoint balls B1,B2,... ∈ B such that ∞  [  µ A \ Bi = 0. i=1 Proof. Let us first assume that A is bounded, and in particular µ(A) < ∞. By the Besicovitch covering theorem, we have

Q(n) [ [ A ⊂ B,

i=1 B∈Bi where each Bi ⊂ B is a countable subcollection of pairwise disjoint balls. Thus

Q(n) X  [  µ(A) ≤ µ B ,

i=1 B∈Bi and hence  [  1 µ B ≥ µ(A) Q(n) B∈Bi for some i. By convergence properties of measures, we then have  [  1 µ B ≥ µ(A) 2Q(n) B∈D1 for some finite subcollection D ⊂ B . Denoting A = A\S B, this implies 1 i 1 B∈D1  1  µ(A ) ≤ 1 − µ(A) < ∞. 1 2Q(n)

57 We next observe that A1 and

1 [ B = {B ∈ B : B ∩ D = ∅} D∈D1 satisfy the same assumptions as A and B, and hence we find a finite disjoint 1 subcollection D2 ⊂ B such that  1  [ µ(A ) ≤ 1 − µ(A ),A = A \ B. 2 2Q(n) 1 2 1 B∈D2

1 1 From D2 ⊂ B and the definition of B it follows that the union D1 ∪ D2 also consist of pairwise disjoint cubes. Repeating the argument, we find that

 [   1 k µ A \ ≤ 1 − µ(A), 2Q(n) B∈D1∪···∪Dk where D1 ∪· · ·∪Dk ⊂ B is a finite subcollection of pairwise disjoint cubes. Then

∞ [ Dk k=1 is a countable collection of disjoint cubes that satisfies the required property. It remains to consider the case when A is not necessarily bounded. Since µ is a Radon measure, we can have

µ(∂B(0, r)) > 0 for at most countable many radii r > 0 (exercise). Thus we can find a sequence of radii 0 < r1 < r2 < . . . < rk → ∞ such that

µ(∂B(0, rk)) = 0 ∀k ≥ 1.

Let us denote

A1 = A ∩ B(0, r1),Ak = {x ∈ A : rk−1 < |x| < rk}, k ≥ 2,

1 B = {B ∈ B : B ⊂ B(0, r1)}, k B = {B ∈ B : B ⊂ B(0, rk) \ B¯(0, rk−1)}, k ≥ 2. k We observe that each of the pairs (Ak, B ) satisfies the same assumptions as (A, B)? and in addition each Ak is bounded. By the first part of the proof, we can find countable disjoint collections Dk ⊂ Bk such that [ µ(Ak \ B) = 0. B∈Dk

58 S∞ k Then also D = k=1 D ⊂ B is a countable disjoint collection, and ∞ [ X  [  µ(A \ B) ≤ µ(Ak \ B) + µ(∂B(0, rk)) = 0. B∈D k=1 B∈Dk This completes the proof in the general case.

Theorem 5.14. Let µ be a Radon measure on Rn and A ⊂ Rn be µ-measurable. Then µ(A ∩ B¯(x, r)) (5.15) lim = 1 for µ-a.e. x ∈ A, r→0+ µ(B¯(x, r)) and µ(A ∩ B¯(x, r)) (5.16) lim = 0 for µ-a.e. x∈ / A. r→0+ µ(B¯(x, r)) Proof. Since µ(A ∩ B¯(x, r)) ≤ 1, µ(B¯(x, r)) it suffices to prove (5.15) with (lim, =) replaced by (lim inf, ≥). Since

∞ n µ(A ∩ B¯(x, r)) o [ x ∈ A : lim < 1 = Ak, r→0+ µ(B¯(x, r)) k=1 where n µ(A ∩ B¯(x, r)) 1 o Ak = x ∈ A : lim < 1 − , |x| < k , r→0+ µ(B¯(x, r)) k it is enough to prove that each Ak is µ-measurable and µ(Ak) = 0. For measurability, we observe that, for each fixed x, the mappings r 7→ µ(A∩B¯(x, r)) and r 7→ µ(B¯(x, r)) are continuous from the right (since B¯(x, r) = ∩jB¯(x, r + hj) whenever hj → 0+). Thus lim infr→0+ in the definition of Ak can be replaced by lim infj→∞ for some countable sequence rj → 0+. On the other hand, for a fixed r > 0, the mappings

x 7→ µ(A ∩ B¯(x, r)), x 7→ µ(B¯(x, r)) are Borel functions (exercise). Thus the function

µ(A ∩ B¯(x, r)) x 7→ lim inf r→0+ µ(B¯(x, r)) is a Borel function as a countable lim inf of Borel functions, and hence Ak is a Borel set. Let us then fix k and write λ = 1 − 1/k. Let ε > 0 and U ⊂ Rn be an open set such that Ak ⊂ U and µ(U) ≤ µ(Ak) + ε. The family

B = {B¯(x, r) ⊂ U : x ∈ Ak, µ(A ∩ B¯(x, r)) < λµ(B¯(x, r))}

59 satisfies the assumptions of the Vitali covering theorem with Akăin place of A. Hence we can find disjoint balls B1,B2,... ∈ B such that

∞  [  µ A \ Bi = 0. i=1 Thus

∞ ∞ ∞  [  X X µ(Ak) = µ Ak ∩ Bi ≤ µ(Ak ∩ Bi) ≤ λ µ(Bi) i=1 i=1 i=1 ∞  [  = λµ Bi ≤ λµ(U) ≤ λ(µ(Ak) + ε) i=1

Letting ε → 0 we deduce µ(Ak) ≤ λµ(Ak) < ∞ (since Ak is bounded), and hence µ(Ak) = 0. This proves (5.15) The other limit (5.16) follows by applying (5.15) to Rn \ A in place of A.

60