Real Analysis II, Winter 2018

Transcription

1 Real Analysis II, Winter 2018 From the Finnish original Moderni reaalianalyysi 1 by Ilkka Holopainen adapted by Tuomas Hytönen January 18, Version dated September 14, 2011

2 Contents 1 General theory of measure and integration Measures Metric outer measures Regularity of measures, Radon measures Uniqueness of measures Extension of measures Product measure Fubini s theorem Hausdorff measures Basic properties of Hausdorff measures Hausdorff dimension Hausdorff measures on R n

3 Chapter 1 General theory of measure and integration 1.1 Measures Let be a set and P() = {A : A } its power set. Definition 1.2. A collection M P () is a σ-algebra of if 1. M; 2. A M A c = \ A M; 3. A i M, i N A i M. Example P() is the largest σ-algebra of ; 2. {, } is the smallest σ-algebra of ; 3. Leb(R n ) = the Lebesgue measurable subsets of R n ; 4. If M is a σ-algebra of and A, then is a σ-algebra of A. M A = {B A : B M} 5. If M is a σ-algebra of and A M, then is a σ-algebra of. M A = {B : B A M} Definition 1.4. If F P() is any family of subsets of, then σ(f) = {M : M is a σ-algebra of, F M} is the σ-algebra generated by F. It is the smallest σ-algebra that contains F. 2

4 Example 1.5. A subset I R n is an open n-interval if it has the form where a j < b j +. Then I = {(x 1,..., x n ) : a j < x j < b j }, σ({i : I is an n-interval}) = σ({a : A R n is open}) =: Bor(R n ) is the σ-algebra of Borel sets of R n. (Think why there is equality on the left.) We observe that, in R n, all open sets, closed sets, G δ sets (countable intersections of open sets), F σ sets (countable unions of closed sets), F σδ sets, G δσ 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, one can define the Borel sets Bor() = σ({a : A open}. Definition 1.7. Let M be a σ-algebra of. A mapping µ : M [0, + ] is a measure, if 1. µ( ) = 0, 2. µ( A i) = µ(a i), if A i M are disjoint. The triplet (, M, µ) is called a measure space and the elements of M measurable 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 If µ() <, the measure µ is finite. 2. If µ() = 1, then µ is a probability measure. 3. If = A i, where µ(a i ) < for all i, the measure µ is σ-finite. We also say that 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 is a topological space and Bor() M (i.e., every Borel set is measurable), then µ is a Borel measure. Example = R n, M = Leb R n, µ = m n = Lebesgue measure. 2. = R n, M = Bor R n, µ = m n Bor R n = the restriction of the Lebesgue measure on Borel sets. 3. Let be any set. Fix x and define for all A { 1, if x A; µ(a) = 0, if x / A. Then µ : P() [0, + ] is a measure (so called Dirac measure at x). It is often denoted by µ = δ x. 3

5 4. If f : R n [0, ] is Lebesgue measurable, then µ : Leb(R n ) [0, ] defined by µ(e) = f(x)dm n (x) is a measure. (See course Measure and integral.) E 5. If (, M, µ) is a measure space and A M, then µ A : M A [0, ] defined by (µ A)(B) = µ(b A) is a measure. It is called the restriction of µ on A. Theorem Let (, M, µ) be a measure space and A 1, A 2,... M. 1. If A 1 A 2..., then µ( A i ) = lim i µ(a i ). 2. If A 1 A 2... and µ(a k ) < for some k, then µ( Proof. Measure and integral Metric outer measures A i ) = lim i µ(a i ). Definition A mapping µ : P() [0, + ] is an outer measure (or exterior measure) on if 1. µ( ) = 0; 2. µ(a) µ(a i), if A A i. Remark The outer measure is thus defined on all subsets of. 2. Condition (2) implies the monotonicity of the outer measure, i.e., µ(a) µ(b) if A B. 3. Many books (e.g. Evans Gariepy, Mattila,... ) refer to outer measures simply as measures. 4. Let µ be an outer measure on and A. Then its restriction to A, is an outer measure on. ( µ A)(B) = µ(b A), Every outer measure defines a σ-algebra of measurable sets through the so called Carathéodory condition: 4

6 A set E is µ- Definition Let µ be an outer measure on. measurable, or just measurable, if µ(a) = µ(a E) + µ(a \ E) for all A. Theorem Let µ be an outer measure on and Then 1. M is a σ-algebram and M = M µ = {E : E is µ-measurable}. 2. µ = µ M is a measure (i.e., µ is countably additive). Proof. Measure and integral. Definition An outer measure µ on a topological space is called a Borel outer measure if every Borel set of 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 has this property. Definition 1.17 (Carathéodory). An outer measure µ on a metric space (, d) is a metric outer measure if µ(a B) = µ(a) + µ(b) for all A, B such that dist(a, B) = inf{d(a, b) : a A, b B} > 0. Theorem An outer measure µ on a metric space (, d) is a Borel outer measure, if and only if µ is a metric outer measure. We begin with a lemma: Lemma Let µ be a metric outer measure and A G, where G is open. If A k = {x A : dist(x, G c ) 1/k}, k N, then µ(a) = lim k µ(a k ). Proof. Since G is open, we have A = k=1 A k. Let B k = A k+1 \ A k. Then ( A = A 2n k=n k=n ) ( B 2k k=n B 2k+1 ), so that µ(a) µ(a 2n ) + µ(b 2k ) + µ(b 2k+1 ) = µ(a 2n ) + (I) n + (II) n. k=n 5

7 Consider the limit n. (1) If (I) n, (II) n 0 as n, then µ(a) lim n µ(a 2n) lim n µ(a 2n+1) µ(a), and the claim of the lemma is true. (2) If (I) n 0 as n, then n µ(b 2n ) =. On the other hand, A A 2n n 1 k=1 B 2k, where dist(b 2k, B 2k+2 ) 1 2k k + 2 > 0. Since µ is a metric outer measure, it follows that n 1 k=1 µ(b 2k ) = µ ( n 1 ) B 2k µ(a 2n ) µ(a 2n+1 ) µ(a). k=1 As n, we deduce that µ(a) = lim k µ(a k ) =, and the claim of the lemma is true again. If (II) n 0 as n, the argument is entirely similar. Proof of Theorem Let us first assume that µ is a metric outer measure and prove that µ is a Borel outer measure. Since Bor() = σ({f : F closed}) and M µ is a σ-algebra, it suffices to prove that every closed set F is µ-measurable. Let E be an arbitrary set, for which we check the Carathéodory condition µ(e) = µ(e F ) + µ(e \ F ). We apply Lemma 1.19 with A = E \ F \ F = G. When A k A is as in the lemma, we have dist(a k, F E) dist(a k, F ) 1/k > 0, lim µ(a k) = µ(a) = µ(e \ F ). k By the assumption that µ is a metric outer measure and observing that A k A E, we have µ(a k ) + µ(f E) = µ(a k (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

8 1.20 Regularity of measures, Radon measures Among all outer measures, particularly useful are those that have many measurable sets. These have a name: (We denote by µ the measure determined by the outer measure µ.) Definition An outer measure µ on is called regular, if for every A, there is a µ-measurable set E such that A E and µ(e) = µ(a). (E is a measurable cover of A.) Definition Let be a topological space. 1. An outer measure µ on is called Borel regular, if µ is a Borel measure and for every A, there is a Borel set B Bor() such that A B and µ(b) = µ(a). 2. If (, M, µ) is a measure space such that Bor() M, the measure µ is called Borel regular if for every A M there is B Bor() such that A B and µ(a) = µ(b). Lemma Let µ be a Borel regular outer measure on and A be µ-measurable. If µ(a) < or A Bor(), then µ A is Borel regular. Proof. Exercise. Theorem Let µ be a Borel regular outer measure on a metric space, let A be µ-measurable and ε > If µ(a) <, there is a closed set C A such that µ(a \ C) < ε. 2. If there are open sets V 1, V 2,... such that A V i and µ(v i ) < 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 A in place of µ (see Lemma 1.23), we may assume without loss of generality that µ() <. Let D = {A : 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 A c D. Let then A 1, A 2,... D, let A = A i and ε > 0. There are closed sets C i and open sets V i such that C i A i V i and µ(v i \ C i ) < ε/2 i. Then V = V i A is open and, by Theorem 1.10, lim µ( n V \ n n ( C i ) = µ(v \ C i ) µ ) (V i \ C i ) µ(v i \ C i ) < ε. and thus µ(v \ n C i) < ε for some n, where C := n C i A is closed. Thus C A V and µ(v \ C) < ε, and thus A D. Hence D is indeed a σ-algebra. 7

9 We check that D contains all closed sets. So let C be closed and V i := {x : dist(x, C) < 1/i}. Then each V i is open, V 1 V 2... and i V i = C. Hence lim i µ(v i ) = µ(c) and lim i µ(v i \ C) = 0. (Here we use the assumption that µ() <.) 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 V i \A to find closed sets C i V i \A such that µ((v i \ A) \ C i ) < ε2 i. Then V = (V i \ C i ) is open, A V, and µ(v \ A) < ε. Remark 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 is locally compact if every x has a neighbourhood whose closure is compact. A topological space is a Hausdorff space if its distinct points have disjoint neighbourhoods. Definition Let 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 ; 2. µ(v ) = sup{µ(k) : K V compact} for all open V ; 3. µ(b) = inf{µ(v ) : B V, V open} for all Borel sets B Bor(). Remark 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 A and µ = µ {E R : E is µ-measurable}. Then µ is a Radon measure but it is not Borel regular. (Exercise.) 8

10 In some cases, Radon measures can be easily characterised. Theorem Let µ be a Borel measure on R n. Then µ is a Radon measure if and only if µ is locally finite, i.e. x R n r x > 0 : µ((b(x, r)) < for 0 < r < r x. Proof. By definition, Radon measures are locally finite, so it remains to prove the converse. Let µ be a locally finite Borel measure on R n. If K R n 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 We will prove both (2) and (3) of Definition 1.26 for all Borel sets A R n. Consider the Borel sets A i = A B(0, i) of finite measure. By Theorem 1.24 and Remark 1.25, there are closed sets C i A i such that µ(a i \ C i ) < 1/i. Since C i A i B(0, i), the set C i is also bounded and thus compact. (Here we use the fact that we are in R n.) Now µ(a) µ(a i ) µ(c i ) > µ(a i ) 1/i µ(a) by Theorem 1.10, and thus (2) of Definition 1.26 holds. Since A B(0, i), where V i = B(0, i) are open sets with µ(v i ) <, 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 Corollary Let µ be a locally finite metric outer measure on R n. Then its restriction µ to µ-measurable subsets is a Radon measure. Remark Theorem 1.28 is valid under more general assumptions, e.g. with R n replaced by a locally compact metric space whose topology has a countable basis Uniqueness of measures We next investigate conditions for a collection F P() so that two measures µ, ν : σ(f)) [0, + ] coincide, as soon as they coincide on F. Definition A nonempty collection F P() is a π-system if A B F whenever A, B F. 2. A collection D P() is a d-system (or Dynkin class) if (a) D; (b) if A, B D and A B, then B \ A D; (c) if A 1, A 2,... D and A 1 A 2..., then k=1 A k D. 9

11 3. The d-system generated by a collection A P() 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() 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 D, and hence A c = \ A D. Thus = c 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 = (A c B c ) c ). Finally, if A 1, A 2,... D, then A k := k j=1 A j D are increasing, and k=1 A k = k=1 A k 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 := {A D : A F D F F}. It follows directly from the definition that for any F, G D: ( ) G F G F D F F, G G F G. In particular, we see that A A (since A is a π-system) and we need to prove that D D. For this purpose, we prove that for any F D, the collection F is a d-system. Assume this for a moment. Since D is the smallest d-system that contain A, it implies that ( ) A F D F and allows us to conclude that A A ( ) D A ( ) A D ( ) D D, 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 is a d-system. For this, let F F be arbitrary. Then F = F D, and thus F. If A, B F 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. If A 1, A 2,... F is ( an increasing sequence, then A 1 F, A 2 F D is an increasing sequence, and ) k=1 A k F = k=1 (A k F ) D so that k=1 A k F. Thus F satisfies all three properties of a d-system, and we are done. 10

12 Theorem 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() = 1 = ν(), we have 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). 3. If A 1, A 2,... D and A 1 A 2..., then also k=1 A k D, since ( µ k=1 ) ( A k = lim µ(a k) = lim ν(a k) = ν k k Thus D is a d-system, and the theorem is proved. k=1 A k ). Corollary If µ, ν : Bor(R n ) [0, + ) are finite measures and µ(i) = ν(i) for every n-interval I, then µ ν. Proof. Since R n = j=1 I j for the n-intervals I j = ( j, j) n, the assumption implies that µ(r n ) = ν(r n ) = c, say. If c = 0, then µ 0 ν, and the result is clear. Otherwise, c 1 µ, c 1 ν : Bor(R n ) [0, 1] are probability measures that coincide on all n-intervals, which clearly form a π-system F. Since σ(f) = Bor(R n ), the result follows from Theorem Extension of measures Consider a collection F P() and a mapping µ : F [0, + ] that is σ- additive on F, i.e., ( (1.37) µ k=1 A k ) = µ(a k ) if A k F are disjoint and k=1 A k F. 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 A family A P() is an algebra if (a) A; (b) A, B A implies A B A; (c) A A implies \ A A. 11

13 2. A family A P() is a semialgebra if it satisfies (1a) and (1b) above, but (1c) is replaced by n \ A = k=1 for some n N and disjoint A k A, whenever A A. A k Let us introduce the notation i A i for the union of disjoint sets A i. Lemma Let P be a semialgebra. Then { P n = } A i : A : i P, i N is an algebra (called the algebra generated by P). Proof. Clearly P. If A = n m A i, B = B j P, A i, B j P, j=1 then A B = i,j A i B j, where A i B j P and thus A B P. Moreover, since A i P, we have \ A i P by the semialgebra property, and thus \ A = n ( \ A i ) P by the closedness of P under intersections, which we just proved. Definition A measure on an algebra A is a σ-additive mapping µ : A [0, ] such that µ( ) = 0. Lemma 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 we set A = n A i P, A i P, (1.42) µ(a) = n µ(a i ). 12

14 1. µ is well defined: If A = n A i = m j=1 B j, then A i = A i A = m A i B j, B j = A B j = j=1 n A i B j, where A i B j P, and hence, by the σ-additivity of µ on P, µ(a i ) = i i µ(a i B j ) = j j µ(a i B j ) = i j µ(b j ). 2. µ is a measure on P: Clearly µ( ) = 0. For σ-additivity, consider A i P with A = A i P. Thus and hence A = n B k (B k P), A i = k=1 n i j=1 A ij (A ij P), B k = A B k = i,j A ij B k, A ij = A ij A = k A ij B k, A ij B k P. By the definition of µ, and the σ-additivity of µ on P, µ(a) = µ(b k ) = µ(a ij B k ) = µ(a ij ) = k k i,j i,j i µ(a 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 A = n A i P. Theorem 1.44 (Carathéodory Hahn extension theorem). Let A P() 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. 2. If µ is σ-finite with respect to A (i.e., = A i, where A i A and µ(a i ) < ), then ν is unique. Proof. We define ν : P() [0, ] by { ν(e) = inf µ(a i ) : E } A i, A i A for all i. 13

15 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 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 ν i : σ(a) [0, ] (i = 1, 2) such that ν 1 (A) = ν 2 (A) = µ(a) for all A A. If A A satisfies 0 < µ(a) <, then B µ(a) 1 ν i (A B) are two probability measures on σ(a) that agree on the π-system A, and hence ν 1 (A B) = ν 2 (A B) for all B σ(a) by Theorem If µ(a) = 0, the same conclusion is obvious, since 0 ν i (A B) ν i (A) = µ(a) = 0 for both i = 1, 2 in this case. By σ-finiteness, we can then choose sets A 1 A 2... with A 1, A 2,... A and =. From the continuity of measure it then follows that ν 1 (B) = lim k ν 1(B A k ) = lim k ν 2(B A k ) = ν 2 (B) for all B σ(a), and thus ν 1 ν Product measure We apply the Carathéodory Hahn extension theorem for the construction of the product measure. Let (, M, µ) and (, N, ν) be measure spaces and S = {A B : A M, B N } P( ) the collection of so called measurable rectangles. Lemma S is a semialgebra. Proof. Clearly = S. If A B, A B S, then (A B) (A B ) = (A A ) (B B ) S. Moreover, (A B) c = (A c B) (A B c ) (A c B c ), 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

16 Lemma If A i B i S are disjoint sets such that A i B i = A B S, then λ(a i B i ) = λ(a B). Proof. From the assumed equality of sets, we get B = B i x A, i:x A i and hence ν(b) = This can be further written as χ A (x)ν(b) = i:x A i ν(b i ) x A. χ Ai (x)ν(b i ). Integrating the positive series with respect to µ, we deduce µ(a)ν(b) = χ A (x)ν(b)dµ(x) = χ Ai (x)ν(b i )dµ(x) = µ(a i )ν(b i ), which was the claim, recalling the definition of λ. Theorem 1.49 (Existence of product measure). Let (, M, µ) and (, N, ν) be σ-finite measure spaces, and S P( ) the family of measurable rectangles. Then the measurable space (, 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 (, 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 Let (, 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

17 1. M is a σ-algebra of ; 2. µ is a complete measure; 3. µ = µ M. Here µ is called the completion of µ, and (, M, µ) the completion of (, M, µ). Remark If µ is an outer measure on, its associated measure space (, M µ, µ) is always complete. 2. If µ is a Borel regular metric outer measure on, its associated measure space (, M µ, µ) is the completion of (, Bor(), µ). 3. Even if (, M, µ) and (, N, ν) are complete σ-finite measure spaces, the product measure space (, σ(s), µ ν) need not be complete. For example, if A M and µ(a) = 0 and B P( ) \ N, then A B A, where µ(a ) = 0, but A B / σ(s) (see Lemma 1.53). In particular, the product measure space (R n R k, σ(s), m n m k ) of Lebesgue measures m n and m k is not complete, and hence m n m k m n+k. Instead, m n+k is the completion of m n m k Fubini s theorem Let (, M, µ) and (, N, ν) be σ-finite measure spaces, and µ ν the product measure constructed above. We study conditions under which f : Ṙ satisfies ( ) f(x, y)d(µ ν)(x, y) = f(x, y)dν(y) dµ(x) If E, we denote = ( ) f(x, y)dµ(x) dν(y). E x = {y : (x, y) E}, E y = {x : (x, y) E}. Lemma If E σ(s), then E x N and E y M for all x, y. Proof. Let C = {E σ(s) : E x N x }. If E = A B S, then E x = B N if x A, and E x = N if x / A, and thus S C. It suffices to prove that C is a σ-algebra, since then σ(s) C, and E x N for allăx holds for all E σ(s) as claimed. (The result concerning the sets E y is completely analogous.) The main point is that the operation E E x commutes with the standard set operations, namely ( ) (E c ) x = (E x ) c, E i = (E i ) x. x i i 16

18 Hence, if E C, then E x N by definition, hence (E x ) c N, i.e., (E c ) x N? and thus E c C. The proof concerning unions is similar, and C is obvious; thus C is a σ-algebra as required. If f : Ṙ, we denote f x (y) = f y (x) = f(x, y). Lemma If f : Ṙ is µ ν-measurable, then f x : Ṙ is ν-measurable for every x, f y : Ṙ is µ-measurable for every y. Proof. The main point is that {f x U} = {f U} x for any set U. If U = (t, ], t Ṙ, then {f U} σ(s) by the assumed measurability of f, thus {f U} x N by the previous lemma, hence {f x U} by the observation just made, and thus f x is measurable. The case of f y is analogous. Lemma If E σ(s), then x ν(e x ) is µ-measurable, y µ(e y ) is ν-measurable, and ν(e x )dµ(x) = µ(e y )dµ(y) = (µ ν)(e). Proof. We consider the family F = { E σ(s) : x ν(e x ) is µ-measurable, } ν(e x )dµ(x) = (µ ν)(e). We need to prove that F = σ(s). (The claim concerning E y is analogous.) If E = A B S, then { B, x A, E x = ν(e x ) = χ A (x)ν(b), ν(e x )dµ(x) = µ(a)ν(b),, x / A, 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 S F. Let then E 1, E 2,... F satsify E 1 E 2... and E = E i. Then E ix N, E 1x E 2x... and E x = E ix so that x ν(e x ) = lim k ν(e kx ) is µ-measurable as an increasing limit of µ-measurable functions, and monotone convergence theorem and continuity of measure imply that ν(e x )dµ(x) = lim k Thus E = E i F. ν(e kx )dµ(x) = lim k µ ν(e k) = µ ν(e). 17

19 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(), ν( ) <. Then x ν((f \ E) x ) = ν(f x \ E x ) = ν(f x ) ν(e x ) is µ-measurable as a difference of two such functions. (The assumption ν( ) <, makes sure that we don t encounter any undefined here.) These functions are also pointiweise bounded by ν( ) <, and hence integrable over the finite measure space (, M, µ). Then ν((f \ E) x )dµ(x) = ν(f x )dµ(x) ν(e x )dµ(x) = 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 µ A and ν B in place of µ and ν. This gives ν B(E x )d(µ A)(x) = (µ A) (ν B)(E) E σ(s). Both (µ A) (ν B) and (µ ν) (A B) are measures on σ(s) that coincide on the π-system S (which contains ), and hence they are equal. Moreover, fd(µ A) = fdµ is clear when f is a characteristic function and follows for A simple functions by linearity and for positive measurable functions by monotone convergence. Hence the previous identity can be rewritten as χ A (x)ν(b E x )dµ(x) = (µ ν)((a B) E) for all E σ(s) and A, B respectively of finite µ and ν measure. We finally use σ-finiteness to write = i, = i, with increasing sequences i and i such that µ( i ), ν( i ) <. We use the previous identity with A = i, B = i. Then χ i (x) 1 and ν( i E x ) ν(e x ) as increasing limits, and hence monotone convergence implies that ν(e x )dµ(x) = lim χ i (x)ν( i E x )dµ(x) i This completes the proof in the general case. = lim χ i (x)ν( i E x )dµ(x) i = lim µ ν(( i i ) E) = µ ν(e). i Theorem 1.56 (Tonelli). Let f : [0, ] be µ ν-measurable. Then: 1. For all (x 0, y 0 ), x f(x, y 0 ) is µ-measurable, y f(x 0, y) is ν-measurable. 18

20 2. 3. ( x y f(x, y)dν(y) is µ-measurable, f(x, y)dµ(x) is ν-measurable. ) f(x, y)dν(y) dµ(x) = = f(x, y)d(µ ν)(x, y) ( ) f(x, y)dµ(x) dν(y). Proof. Part (1) was already proved in Lemma If f = χ E 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 s i f pointwise. Then monotone convergence shows that f(x, y)dν(y) = lim s i(x, y)dν(y) = lim s i (x, y)dν(y), i i which also proves that the left side is a µ-measurable function of x, as a (monotone) limit of such functions. Another application of monotone convergence shows that ( ) ( f(x, y)dν(y) dµ(x) = lim i ( = lim i ) s i (x, y)dν(y) dµ(x) ) s i (x, y)dν(y) dµ(x), and yet another one (on the product measure space ) that f(x, y)d(µ ν)(x, y) = lim s i(x, y)d(µ ν)(x, y) i = lim s i (x, y)d(µ ν)(x, y). i 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 The previous results are formulated for the product measure µ ν and (µ ν)-measurable functions. They cannot be directly applied to the complete measure space (R n R k, Leb(R n+k ), m n+k ). For instance, Lemmas 1.53 and 1.54 do not hold in (R n R k, Leb(R n+k ), m n+k ). Namely, even if 19

21 A R k is non-measurable (with respect to m k ), the set E = {x} A is still m n+k -measurable for all x R n, since m n+k (E) = 0. However, E x = A, which shows the failure of Lemma A version of Tonelli s theorem on (R n R k, Leb(R n+k ), m n+k ) has (1) only for almost every x and y. Theorem 1.58 (Fubini). Let f : Ṙ 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), is finite. Then 1. y f(x, y) is integrable over for almost all x ; 2. x f(x, y) is integrable over for almost all y ; 3. x f(x, y)dν(y) is integrable over ; 4. y f(x, y)dµ(x) is integrable over ; 5. f is integrable over and ( ) fd(µ ν) = f(x, y)dν(y) dµ(x) = ( ) f(x, y)dµ(x) dν(y). 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

22 Chapter 2 Hausdorff measures 2.1 Basic properties of Hausdorff measures The n-dimensional Lebesgue measure m n is well suited for measuring the size of large subsets of R n, but it is too rough for small sets. For instance, m 2 cannot distinguish a point from a line in R 2, since they both have measure zero. In this chapter we introduce a whole scale of s-dimensional measures H s, 0 s <, which are able to see the fine structure of sets better than the Lebesgue measure. The idea is that a set A R n is s-dimensional if 0 < H s (A) <, even if the geometry of A is very complicated. These measure can be defined on any metric space (, d). We assume, however, that is separable, i.e., that there is a countable dense subset S = {x i }, so that = S. This assumption guarantees that has a δ-cover for every δ > 0. Definition The diameter of a nonempty set E is d(e) = sup d(x, y). x,y E 2. For δ > 0, a δ-cover of a set A is a countable collection {E i } of subsets of such that A E i, d(e i ) δ i N. Let us fix a dimension s [0, ) and δ > 0. For A, we define { } (2.3) Hδ(A) s = inf d(e i ) s : {E i } is a δ-cover of A, where we agree that d({x}) 0 = 1 for all x and d( ) s = 0 for all s 0. It follows at once from the definition that H s δ 1 (A) H s δ 2 (A) if 0 < δ 1 δ 2. Thus the limit (2.5) below exists and we can make the following definition: 21

23 Definition 2.4. The s-dimensional Hausdorff (outer) measure of A is ( ) (2.5) H s (A) := lim H s δ 0 δ(a) = sup Hδ(A) s. δ>0 Theorem Hδ s : P() [0, ] is an outer measure for every δ > H s : P() [0, ] is a metric outer measure. Proof. 1. Clearly Hδ s( ) = 0. Let then A A i, where we assume without loss of generality that Hδ s(a i) <. Given ε > 0, we can find for every A i a δ-cover {E ij } j=1 such that d(e ij ) s Hδ(A s i ) + ε2 i. j=1 Then {E ij } i,j=1 is δ-cover of and hence of A, and thus Hδ(A) s d(e ij ) s ( H s δ (A i ) + ε2 i) Hδ(A s i ) + ε. i,j=1 As ε 0, this implies the countable semiadditivity of the H s δ. 2. Clearly H s ( ) = lim δ 0 H s δ ( ) = lim δ 0 0 = 0. If A, since we already proved that Hs δ is an outer measure, it follows that Hδ(A) s Hδ(A) s H s (A). As δ 0, we see that H s is an outer measure as well. Let then A 1, A 2 be sets with dist(a 1, A 2 ) > 0. We want to show that H s (A 1 A 2 ) = H s (A 1 ) + H s (A 2 ). Since is clear (we already checked that H s is an outer measure), it suffices to prove, which in turn will follow in the limit δ 0 if we can show that (2.7) H s δ(a 1 A 2 ) H s δ(a 1 ) + H s δ(a 2 ) for all δ dist(a 1, A 2 )/3. We may also assume that H s δ (A 1 A 2 ) <. Let us then fix a δ-cover {E i } of A 1 A 2 such that d(e i ) s Hδ(A s 1 A 2 ) + ε. 22

24 Since δ dist(a 1, A 2 )/3, each E i intersects at most one of A 1 or A 2. Those that intersect A k form a δ-cover of A k. Thus we can split and hence 2 Hδ(A s k ) k=1 {E i } = k=1 2 {Ei k }, A k Ei k, k=1 2 d(ei k ) s = d(e i ) s Hδ(A s 1 A 2 ) + ε. 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 is H s -measurable. We denote the restriction of H s to H s -measurable sets by the same symbol H s. Thus: Theorem 2.8. H s is a Borel measure. From Corollary 1.29 we now get: Corollary 2.9. If A R n is H s -measurable and H s (A) <, then H s A is a Radon measure. Theorem On a separable metric space, the outer measure H s is Borel regular. Proof. The previous theorem shows that H s is a Borel measure, so it remains to prove that for everyăa there is B Bor() such that A B and H s (A) = H s (B). If H s (A) =, we can choose B =. Suppose that H s (A) <. Then for every i we can find a 1/i-cover {Ej i} j=1 of A such that j=1 d(ej) i s H1/i s (A) + 1/i. Since d(e) = d(ē) for every E, we may assume that the sets Ei j are closed. Then B = j=1 is a Borel set that contains A. For each i, the collection {Ej i} j=1 is a 1/i-cover of B, and hence H s 1/i (A) Hs 1/i (B) j=1 E i j d(ej) i s H1/i s (A) + 1/i. With i we deduce H s (A) = H s (B), as claimed. 23

25 Remark H s is the counting measure. 2. H s δ is usually not a metric outer measure. 3. Roughly speaking, H 1 length, H 2 area, etc. 4. By results below, it is easy to see that (e.g.) H 1 is not σ-finite on R Hausdorff dimension Let (, d) be a separable metric space. We now define for all subsets A a dimension that represents the (metric) size of A. In contrast to, say, the topological dimension, this dimension need not be an integer. Lemma Let A and s If H s (A) <, then H t (A) = 0 for all t > s. 2. If H s (A) > 0, then H t (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 {E i } be a δ-cover of A for some δ > 0. Then d(e j ) t δ t s j=1 j=1 d(e j ) s, and taking the infimum of both sides over all δ-covers shows that Letting δ 0 proves that first claim. H t δ(a) δ t s H s δ(a). The previous lemma shows that the graph of s H s (A) is very simple: at a unique point s 0 [0, ], it drops from to zero. Definition The Hausdorff dimension of a set A is the number We observe: dim H (A) := inf{s > 0 : H s (A) = 0}. 1. If t < dim H (A), then H t (A) =. 2. If t > dim H (A), then H t (A) = 0. In general, we cannot say anything about the value of H s (A), when s = dim H (A): it can be any value in the interval [0, ]. However: (2.15) 0 < H s (A) < dim H (A) = s. A set A that satisfies the left side of (2.15) is called an s-set. 24

26 Lemma If A B, then dim H (A) dim H (B). 2. If A k for all k N, then Proof. Exercise. Thus e.g. dim H Q = 0. dim H ( k=1 ) A k = sup dim H (A k ). k 2.17 Hausdorff measures on R n Recall that a mapping T : R n R n is an isometry if T x T y = x y for all x, y R n. Every isometry of R n is known to be affine, i.e. T x = a 0 + Ux, where a 0 R n and U : R n R n is a linear isometry. A mapping R : R n R n is a similarity if Rx Ry = c x y for all x, y R n, where c > 0 is a constant (dilation or scaling factor). Such a similarity takes the form Rx = a 0 + cux, where U is again a linear isometry. Theorem Let A R n and s 0. Then: 1. H s (A + x 0 ) = H s (A) for all x 0 R n ; 2. H s (U(A)) = H s (A) for every linear isometry U : R n R n ; 3. H s (R(A)) = c s H s (A), if R : R n R n is a similarity with scaling c > 0. Proof. These are immediate from the observation that d(r(e)) = cd(e) for every E R n, when R is a similarity with scaling factor c > 0. Let (, d 1 ), (, d 2 ) be metric spaces. Recall that f : is a Lipschtitz mapping with constant L > 0 (for short L-Lipschitz) if d 2 (f(x), f(y)) Ld 1 (x, y) for all x, y. Similarly, g : is a bi-lipschitz mapping with constant L > 0 (for short L-bi-Lipschitz) if 1 L d 1(x, y) d 2 (g(x), g(y)) Ld 1 (x, y) for all x, y. In particular, a bi-lipschitz mapping is always an injection (one-to-one) by the left-hand inequality above. Lemma Let (, d 1 ) and (, d 2 ) be metric spaces. 1. If f : is a Lipschitz mapping of constant L, then H s (fa) L s H s (A) A. 25

27 2. If g : is a bi-lipschitz mapping, then dim H (ga) = dim H (A) A. Proof. (1): If {E j } j=1 is a δ-cover of A, then {fe j} j=1 We can choose the cover so that d(e j ) s Hδ(A) s + ε, j=1 is an Lδ-cover of fa. and hence HLδ(fA) s d(fe j ) s [Ld(E j )] s L s (Hδ(A) s + 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 : and g 1 : g(a) to see that H s (g(a)) L s H s (A), H s (A) = H s (g 1 g(a)) L s H s (g(a)), and hence dim H (ga) = dim H (A). We next investigate the connection of H n and the Lebesgue measure m n. Theorem For all A Bor(R n ), we have for some constant c n (0, ). H n (A) = c n m n (A) Proof. If Q R n is a cube of sidelength l(q), we can divide it into N n cubes of sidelength N 1 l(q) and diameter nn 1 l(q). Hence H n nn 1 l(q) (Q) N n( nn 1 l(q) ) n = n n/2 l(q) n = n n/2 m n (Q). As N, this shows that H n (Q) n n/2 m n (Q) < for all cubes Q. If E R n is nonempty, then E B(x, d(e)) for any x E, and hence m n(e) m n (B(x, d(e))) = c n d(e) n ; this bound is obvious if E =. Thus, if {E i } be any δ-cover of A, we have d(e i ) n c n m n(e i ) c n m n(a) by the subadditivity of m n. Taking the infimum over all δ-covers, we get Hδ n(a) c nm n(a), and then also H n (A) c n m n(a). In particular, H n (Q) c n m n (Q) > 0 for all cubes Q. Since both H n and m n are invariant under translations (and by elementary comparisons between cubes and balls), we see that both H n and m n are uniformly distributed Borel measures in the sense of Definition The result then follows from the general Theorem 2.25 below. 26

28 Corollary H n (A) = c n m n(a) A R n. Proof. Exercise. (Use the fact that both H n and m n are Borel regular outer measures.) Definition A Borel measure µ on a metric space is uniformly distributed if 0 < µ(b(x, r)) = µ(b(y, r)) < for all x, y and 0 < r <. Theorem If µ, ν are uniformly distributed Borel measures on a separable metric space, then there is a constant c 0 (0, ) such that for all A Bor(). µ(a) = c 0 ν(a) Proof. Denote g(r) = µ(b(x, r)) and h(r) = ν(b(x, r)) for 0 < r < and x. Let U be open and bounded. We apply Fubini s theorem (see Remark 2.26 for justification) to the function f(x, y) = 1 B(x,r) (y) = 1 B(y,r) (x) to see that ν(b(x, r) U) dµ(x) = 1 B(x,r) (y) dν(y) dµ(x) U U U = 1 B(y,r) (x) dµ(x) dν(y) = µ(b(y, r) U) dν(y). U U If x U, then B(x, r) U = B(x, r) when r is small enough, and hence µ(b(x, r) U) lim = 1 = lim r 0+ g(r) r 0+ U U ν(b(x, r) U), x U. 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) r 0+ h(r) r 0+ h(r) U and similarly 1 ν(u) = lim µ(b(x, r) U)dν(x). r 0+ g(r) U Observing that the two right-hand integrals above are equal, it follows that µ(u) ν(u) = lim g(r) r 0+ h(r) =: c 0 where the existence of the limit is part of the conclusion. Since 0 < µ(u), ν(u) <, we see that c 0 (0, ) as well. Thus claim holds for all open bounded U. This is easily extended to all A Bor() via Dynkin s lemma. 27

29 Remark Let us justify the use of Fubini s theorem above. First, the measures are σ-finite since = B(x, i) and µ(b(x, i)) <. The function f is the indicator of {(x, y) : x y < r}, which is an open set, and therefore Borel set in. Finally, we can use a general result: If and Z are topological spaces whose topologies have countable bases (i.e., they are so-called N 2 spaces), then (2.27) Bor( Z) = σ(bor( ) Bor(Z)). In the theorem, = Z = is a separable metric space and hence an N 2 space (a basis for the topology is given e.g. by B(x i, q j ), where {x i } is dense in and {q j } j=1 = Q +). Thus a Borel function on 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 productmeasurability above. The other direction is left as an exercise. If A Z is open, using the countable topological bases of and Z, it can be expressed as a countable union A = A i B j, i,j where A i, B j Z are open, hence A i B j Bor( ) Bor(Z), and hence the countable union belongs to σ(bor( ) 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 I 0,1 := [0, 1], I 1,1 := [0, λ] and I 1,2 := [1 λ, 1]. In general, if intervals I n,i, i = 1,..., 2 n, are already defined, let I n+1,2i 1 be the interval that has the same left end-point as I n,i, and λ times its length, while I n+1,2i is the interval that has the same right end-point as I n,i, and λ times its length. Thus dist(i n+1,2i 1, I n+1,2i ) = (1 2λ)λ n and We denote d(i n,i ) = λ n n, i = 1,..., 2 n. C n (λ) := 2 n I n,i, C(λ) := C n (λ). Then C(λ) is a compact uncountable set without interior points. It is also selfsimilar and m 1 (C(λ)) = 0. The case λ = 1/3 is a prominent special case that is frequently considered. Theorem For all 0 < λ < 1/2, we have n=1 dim H C(λ) = log 2 log(1/λ). In particular, dim H C(λ) can take any value in (0, 1). 28

30 Proof. Since {I n,i } 2n is a λn -cover of C(λ), we have H s λ n(c(λ)) n d(i n,i ) s = 2 n λ ns = 2 n 2 n = 1. Since λ n 0 as n, this shows that H s (C(λ)) 1 <. H s (C(λ)) > 0 is left as an exercise. A proof of 29