Modern Real Analysis, Fall 2005, 1
Ilkka Holopainen2
19. helmikuuta 2015
1These lecture notes are based mainly on the notes of Saksman: Moderni reaalianalyysi (1998) and Astala: Moderni reaalianalyysi (2002) 2Please inform about errors [email protected] 2 Modern Real Analysis, Fall 2005
Sis¨alt¨o
1 Measure and integration theory: review 3 1.1 Measures ...... 3 1.11 Metricoutermeasure...... 5 1.20 Regularity of measures, Radon-measures ...... 7 1.31 Uniquenessofmeasures ...... 10 1.36 Extensionofmeasures ...... 12 1.45 Productmeasures ...... 15 1.52 Fubini’stheorem ...... 17
2 Hausdorff measures 23 2.1 Some basic properties of the Hausdorff measure ...... 23 2.12 Hausdorffdimension ...... 25 2.17 Hausdorff measures in Rn ...... 27 2.33 Rectifiable and nonrectifiable 1-sets ...... 34
3 Compactness and convergence theorem for Radon measures of Rn 43 3.1 Riesz’ representation theorem ...... 43 3.13 Weak convergence of measures ...... 49 3.17 Compactnessofmeasures ...... 50
4 On the Hausdorff dimension of fractals 52 4.1 The principle of mass distribution and Frostman’s lemma ...... 52 4.16 Selfsimilarfractals ...... 57 4.54 Drawing a selfsimilar fractal by way of random walk ...... 67 4.68Riesz’capacity ...... 71 printed: 2015-2-19, 0.01 3
1 Measure and integration theory: review
1.1 Measures Let X be a set and let (X)= A: A X P { ⊂ } be the power set of X.
Definition 1.2. The collection (X) is a σ-algebra “sigma algebra”) in X, if M ⊂ P (1) ; ∅ ∈ M (2) A Ac = X A ; ∈M ⇒ \ ∈ M (3) A , i N ∞ A . i ∈ M ∈ ⇒ i=1 i ∈ M Example 1.3. 1. (XS) is the largest σ -algebra in X; P 2. , X is the smallest σ -algebra in X; {∅ } 3. Leb(Rn) is the class of Lebesgue measurable sets of = Rn.
4. If is a σ-algebra in X and A X, then M ⊂ A = B A: B M| { ∩ ∈ M} is a σ-algebra in A.
5. If is a σ-algebra in X and A , then M ∈ M = B X : B A MA { ⊂ ∩ ∈ M} is a σ-algebra in X.
Definition 1.4. If (X) is a family of subsets of X, then F ⊂ P σ( )= : is a σ-algebra in X, F {M M F ⊂ M} \ is the σ-algebra spanned by . It is the smallest σ-algebra, which contains . F F Example 1.5. Recall that the set I Rn is an open n-interval, if it is of the form ⊂ I = (x ,...,x ): a Observe that all open subsets of Rn, closed sets, -sets (countable intersections of open sets), Gδ -sets (countable unions of closed sets), -sets, -sets (etc.) are Borel-sets. Thus for example Fσ Fσδ Gδσ the set of rational numbers Q is Borel. 4 Modern Real Analysis, Fall 2005 Remark 1.6. In every topological space X one can define Borel-sets Bor(X)= σ( A: A X open ). { ⊂ } Definition 1.7. Let be a σ-algebra in X. A mapping µ: [0, + ] is a measure, if there M M → ∞ holds: (i) µ( ) = 0, ∅ (ii) µ ∞ A = ∞ µ(A ), if the sets A are disjoint. i=1 i i=1 i i ∈ M The triple S (X, ,µ)P is called a measure space and the elements of measurable sets. M M The condition (ii) is called complete additivity. It follows from the definition that a measure is monotone: If A, B and A B, then µ(A) µ(B). ∈ M ⊂ ≤ Remark 1.8. 1. If µ(X) < , the measure µ is finite. ∞ 2. If µ(X) = 1, then µ is a probability measure. 3. If X = ∞ A , where µ(A ) < i, the measure µ is σ-finite. Then we shall say that X is i=1 i i ∞ ∀ σ-finite with respect to µ. S 4. If A and µ(A) = 0, then A is of measure zero. ∈ M 5. If X is a topological space and Bor(X) (i.e. every Borel-set is measurable), then µ is a ⊂ M Borel-measure. Example 1.9. 1. X = Rn, = Leb Rn = the family of Lebesgue-measurable sets and µ = M mn = the Lebesgue measure. 2. X = Rn, = Bor Rn = the family of Borel-sets and µ = m Bor Rn = the restriction of M n| the Lebesgue measure to the family of Borel-sets. 3. Let X = be any set. Fix x X and set for all A X 6 ∅ ∈ ⊂ 1, if x A; µ(A)= ∈ (0, if x A. 6∈ Then µ: (X) [0, + ] is a measure (so called Dirac measure of x X). We often write P → ∞ ∈ µ = δx. 4. If f : Rn [0, + ] is Lebesgue-measurable, then µ: Leb(Rn) [0, + ], → ∞ → ∞ µ(E)= f(x)dmn(x), ZE is a measure. (Cf. Mitta ja integraali, Lause 3.22) 5. If (X, ,µ) is a measure space and A , then the mapping µxA: [0, + ], M ∈ M MA → ∞ (µxA)(B)= µ(B A), ∩ is a measure. It is called the restriction of µ to A. printed: 2015-2-19, 0.01 5 Theorem 1.10. Let (X, ,µ) be a measure space and A , A ,... . M 1 2 ∈ M (a) If A1 A2 A3 , then ⊂ ⊂ · · · ∞ µ Ai = lim µ(Ai). i→∞ i=1 [ (b) If A A A and µ(A ) < for some k, then 1 ⊃ 2 ⊃ 3 ⊃··· k ∞ ∞ µ Ai = lim µ(Ai). i→∞ i=1 \ Proof. Mitta ja integraali. 1.11 Metric outer measure Definition 1.12. A mappingµ ˜ : (X) [0, + ] is an outer measure in X, if the following holds: P → ∞ (i)µ ˜( ) = 0; ∅ (ii)µ ˜(A) ∞ µ˜(A ), if A ∞ A X. ≤ i=1 i ⊂ i=1 i ⊂ Remark 1.13.P 1. An outer measureS is defined for all subsets of X. 2. Condition (ii) implies that an outer measure is monotone, i.e.µ ˜(A) µ˜(B), if A B X. ≤ ⊂ ⊂ 3. In several books (eg. Evans-Gariepy, Mattila, ...) an outer measure is called a measure. 4. Letµ ˜ be an outer measure in X and A X. Then the restriction of µ˜ to A, defined by ⊂ (˜µxA)(B) =µ ˜(B A) ∩ is an outer measure in X. Every outer measure defines the σ- algebra of “measurable” sets in terms of the Carath´eodory condition. Definition 1.14. Letµ ˜ be an outer measure in X. A set E X isµ ˜-measurable, or briefly ⊂ measurable, if µ˜(A) =µ ˜(A E) +µ ˜(A E) ∩ \ for all A X. ⊂ Theorem 1.15. Let µ˜ be an outer measure in X and = = E X : E is µ˜-measurable M Mµ˜ { ⊂ } Then (a) is a σ-algebra and M (b) µ =µ ˜ is a measure (i.e. µ is completely additive). |M Proof. Mitta ja integraali. 6 Modern Real Analysis, Fall 2005 Definition 1.16. We say that an outer measureµ ˜ of a topological space X is a Borel outer measure, if every Borel-set of X isµ ˜-measurable (i.e. if the measure defined byµ ˜ is a Borel-measure). We shall next study the question when an outer measureµ ˜ of a topological space X is Borel. Definition 1.17 (Carath´eodory). An outer measureµ ˜ of a metric space (X, d) is a metric outer measure , if µ˜(A B) =µ ˜(A) +µ ˜(B) ∪ for all A, B X, for which dist(A, B) = inf d(a, b): a A, b B > 0. ⊂ { ∈ ∈ } Theorem 1.18. An outer measure µ˜ of a metric space (X, d) is a Borel outer measure, if and only if µ˜ is a metric outer measure. We first formulate and prove the following lemma. Lemma 1.19. Let µ˜ be a metric outer measure, A X and G an open set s.t. A G. If ⊂ ⊂ A = x A: dist(x, Gc) 1/k , k N, k { ∈ ≥ } ∈ then µ˜(A) = limk→∞ µ˜(Ak). Proof. Because G is open, A ∞ A . Thus A = ∞ A . Let ⊂ k=1 k k=1 k S B = A A S. k k+1 \ k Then ∞ ∞ A = A2n B2k B2k+1 , ∪ ! ∪ ! k[=n k[=n and thus ∞ ∞ µ˜(A) µ˜(A )+ µ˜(B ) + µ˜(B ) . ≤ 2n 2k 2k+1 kX=n kX=n =(I) =(II) Let now n . | {z } | {z } →∞ (1) If the sums (I), (II) 0, when n , then → →∞ µ˜(A) lim µ˜(A2n) µ˜(A) ≤ n→∞ ≤ and the claim is true. (2) If (I) 0, when n , then 6→ →∞ µ˜(B )= . 2n ∞ n X On the other hand n−1 A A B , ⊃ 2n ⊃ 2k k[=1 where 1 1 dist(B ,B ) > 0. 2k 2k+2 ≥ 2k + 1 − 2k + 2 printed: 2015-2-19, 0.01 7 Becauseµ ˜ is a metric outer measure, then n−1 n−1 µ˜(B ) =µ ˜ B µ˜(A ) µ˜(A). 2k 2k ≤ 2n ≤ k=1 k=1 X [ When n , we obtain →∞ µ˜(A) = lim µ˜(Ak)= . k→∞ ∞ The argument goes in the same way, if the sum (II) 0, when n . 6→ →∞ Proof of Theorem 1.18.. Suppose first thatµ ˜ is a metric outer measure and prove thatµ ˜ is a Borel outer measure. Because Bor(X) = σ( F : F X closed ) and is a σ-algebra, then it { ⊂ } Mµ˜ is enough to show that every closed set F X isµ ˜-measurable. ⊂ Let E X be an arbitrary test set in the Carath´eodory condition. We apply Lemma 1.19 for ⊂ the sets A = E F and G = X F . Let A = x E F : dist(x, Gc) 1/k , k N, then \ \ k { ∈ \ ≥ } ∈ dist(A , F ) 1/k k ≥ and lim µ˜(Ak) =µ ˜(E F ). k→∞ \ Becauseµ ˜ is metric, µ˜(E) µ˜ (E F ) A =µ ˜(E F ) +µ ˜(A ). ≥ ∩ ∪ k ∩ k Letting k we get →∞ µ˜(E) µ˜(E F ) +µ ˜(E F ). ≥ ∩ \ On the other hand from the monotonicity of the outer measure it follows that µ˜(E) µ˜(E F ) +µ ˜(E F ). ≤ ∩ \ Thus F isµ ˜-measurable andµ ˜ is a Borel outer measure. The proof of the converse implication is left as an exercise. 1.20 Regularity of measures, Radon-measures Among outer measures particularly useful are those with a large class of measurable sets. These outer measures are called regular. Definition 1.21. We say that an outer measureµ ˜ of X is regular, if for every A X there exists ⊂ aµ ˜-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. (a) We say that an outer measureµ ˜ of X is Borel-regular, if µ is a Borel-measure and for every A X there exists a Borel-set B Bor(X) such that A B and µ(B) =µ ˜(A). ⊂ ∈ ⊂ (b) If (X, ,µ) is a measure space such that Bor(X) , then the measure µ is called Borel- M ⊂ M regular, if for every A there exists B Bor(X) such that A B and µ(A)= µ(B). ∈ M ∈ ⊂ Lemma 1.23. If µ˜ is a Borel-regular outer measure in X and A X is µ˜-measurable s.t. µ(A) < ⊂ , then µ˜xA is Borel-regular. If A Bor(X), then the assumption µ(A) < is not needed. ∞ ∈ ∞ Proof. (HT 1/5) 8 Modern Real Analysis, Fall 2005 Theorem 1.24. Let µ˜ be a Borel-regular outer measure in a metric space X, A X µ˜-measurable ⊂ and ε> 0. (a) If µ(A) < , then there exists a closed set C A s.t. µ(A C) < ε. ∞ ⊂ \ (b) If there exist open sets V ,V ,... X s.t. A ∞ V and µ(V ) < i, then there exists 1 2 ⊂ ⊂ i=1 i i ∞ ∀ an open set V X s.t. A V and µ(V A) < ε. ⊂ ⊂ \ S Proof. (a): By replacingµ ˜ with a Borel-regular outer measureµ ˜xA (see Lemma 1.23) we may assume thatµ ˜(X) < . We first prove the claim for a Borel set A. Let ∞ = A X : ε> 0 closed C A and open V A s.t. µ(V C) < ε . D { ⊂ ∀ ∃ ⊂ ⊃ \ } We easily see that satisfies condition (1) and (2) in the definition of the σ-algebra. Suppose that D A , A ,... and let ε > 0. Then there exists closed sets C and open sets V s.t. C A V 1 2 ∈ D i i i ⊂ i ⊂ i and µ(V C ) < ε/2i. Now V = V is open and i \ i i i µSV C µ(V C ) < ε. \ i ≤ i \ i i i [ X ⊂Si(Vi\Ci) On the other hand by Theorem 1.10| (b){z } n ∞ lim µ V Ci = µ V Ci , n→∞ \ \ i=1 i=1 [ [ and hence there exists n s.t. n µ V C < ε. \ i i=1 [ Because n C is closed, satisfies also the condition (3) of σ-algebra. We next show that i=1 i D D contains closed sets. Let C be closed and S V = x X : dist(x,C) < 1/i . i { ∈ } Then V is open V V and C = V . Therefore lim µ(V )= µ(C) and lim µ(V i 1 ⊃ 2 ⊃··· i i i→∞ i i→∞ i \ C) = 0. This implies that C . Thus is σ-algebra containing all closed sets. In particular, ∈ D TD Bor(X) . Therefore part (a) holds for all Borel-sets. ⊂ D Suppose next that A isµ ˜-measurable and µ(A) < . Becauseµ ˜ is Borel-regular, there exists a ∞ Borel-set B A s.t. µ(A)= µ(B). Then µ(B A) = 0. Further there exists a Borel-set D B A ⊃ \ ⊃ \ s.t. µ(D) = 0. Now E = B D is Borel, E A and \ ⊂ µ(A E) = 0. \ ⊂D On the basis of what was proved earlier, for| each{z } ε > 0 there exists a closed set C E = B D ⊂ \ ( A) s.t. µ(E C) < ε. But then ⊂ \ µ(A C) µ(A E)+ µ(E C) < ε, \ ≤ \ \ and hence (a) holds for the set A. (b) By applying part (a) to the set V A we obtain closed sets C V A s.t. i \ i ⊂ i \ µ(V A C ) < ε2−i. i \ \ i Then V = (V C ) is open, A V and µ(V A) < ε. i i \ i ⊂ \ S printed: 2015-2-19, 0.01 9 Remark 1.25. The Borel-regularity of the outer measure was not needed to prove claims (a) and (b) for Borel-sets A. Therefore Theorem 1.24 holds for all Borel outer measures, if we, furthermore, assume that A is Borel. In what follows important are Radon-measures, which will be defined next. Recall that a topo- logical space X is locally compact, if every point x X has a neighbourhood with compact closure. ∈ A topological space is Hausdorff , if its distinct points have disjoint neighbourhoods. Definition 1.26. Let X be a locally compact Hausdorff space. We say that a measure µ is a Radon-measure, if µ is a Borel-measure and (a) µ(K) < for all compact K X; ∞ ⊂ (b) µ(V ) = sup µ(K): K V compact for all open V X; { ⊂ } ⊂ (c) µ(B) = inf µ(V ): B V and V X open for all Borel-sets B Bor(X). { ⊂ ⊂ } ∈ Remark 1.27. 1. In general, a Borel-regular measure (in a locally compact Hausdorff space) need not be a Radon-measure (ks. HT 2/2). 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 µ˜-measurable . |{ ⊂ } Then µ is a Radon-measure , but not Borel-regular. (HT) In some cases Radon measures can be easily characterised. Theorem 1.28. Let µ be a Borel-measure in Rn. Then µ is a Radon-measure, if and only if µ is locally finite, i.e. x Rn, µ B(x, r) < , when 0 < r < r . ∀ ∈ ∞ x Proof. Because of the definition 1.26 part (a) all Radon-measures are locally finite. Suppose next that µ is locally finite Borel-measure in Rn. If K Rn is compact, then choose ⊂ for every x K an open ball with centre at x with a finite measure. Because K is compact, it can ∈ be covered with finitely many such balls. Therefore the measure of K is finite and (a) holds. We next prove conditions (b) and (c) for every Borel-set A Rn. By applying part (a) of ⊂ Theorem 1.24 (see also Remark 1.25) for Borel sets of finite measure A = A B(0, i), B(0, i)= x Rn : x i , i ∩ { ∈ | |≤ } we find closed sets C A s.t. i ⊂ i µ(A C ) < 1/i. i \ i n Then Ci is as a closed and bounded set a compact set (in R ). Now 1.10(a) µ(A) µ(A ) µ(C ) >µ(A ) 1/i µ(A) . ≥ i ≥ i i − −−−−→ This implies (b). Because A B(0, i) and µ(B(0, i)) < , it follows from Theorem 1.24 part ⊂ i ∞ (b) that there exists open sets V Rn s.t. A V and µ(V A) < 1/j. Then Sj ⊂ ⊂ j j \ µ(A) µ(V )= µ(A)+ µ(V A) <µ(A) + 1/j, ≤ j j \ This implies (c). 10 Modern Real Analysis, Fall 2005 Corollary 1.29. Let µ˜ be a locally finite metric outer measure in Rn. Then the measure µ = µ˜ , = A Rn : A µ˜-measurable , determined by µ˜ is a Radon-measure. |M M { ⊂ } Remark 1.30. Theorem 1.28 holds also more generally. For example, if X is a locally compact metric space, whose topology has a countable base. 1.31 Uniqueness of measures Next we study when two measures ν,µ: σ( ) [0, + ] agree, i.e. ν(A) = µ(A) A (cf. HT F → ∞ ∀ ∈ F 2/1). Definition 1.32. 1. A collection (X), = , is π-system, if A B for all A, B . F ⊂ P F 6 ∅ ∩ ∈ F ∈ F 2. A collection (X) is d-system (or Dynkin’s class), if there holds: D ⊂ P (i) X ; ∈ D (ii) A, B and A B B A ; ∈ D ⊂ ⇒ \ ∈ D (iii) A A and A k ∞ A . 1 ⊂ 2 ⊂··· k ∈D∀ ⇒ k=1 k ∈ D 3. The d-system spanned by the collection S is the smallest d-system, which contains , i.e. A A : d-system, . {D D A⊂D} \ Theorem 1.33 (Dynkin’s lemma). Let (X) be a π-system. If is a d-system and , A ⊂ P D A ⊂ D then σ( ) . A ⊂ D Proof. It is enough to prove the claim in the case, when is the d-system spanned by . Then D A it is enough to show that is a σ-algebra. D 1. A Ac : This follows directly from the conditions (i) and (ii) of a d-system. ∈D ⇒ ∈ D 2. X (condition (i)). ∈ D 3. A ,B A B : This follows if we show that ∈ D ∈A ⇒ ∩ ∈ D = A : A B B D1 { ∈ D ∩ ∈D ∀ ∈ A} is a d-system, which contains . Then it must hold that = . The inclusion holds A D1 D A ⊂ D1 because is a π-system and . Condition (i) is also clear. If A , A , A A , A A ⊂ D 1 2 ∈ D1 1 ⊂ 2 then for all B ∈A (A A ) B = (A B) (A B) , 2 \ 1 ∩ 2 ∩ \ 1 ∩ ∈ D ∈D ∈D and hence A A and the condition (ii)| {z holds.} Also| {z the} condition (iii) is valid, because 2 \ 1 ∈ D if A A and A k, then 1 ⊂ 2 ⊂··· k ∈ D1 ∀ ∞ ∞ A B = (A B) k ∩ k ∩ ∈ D k=1 k=1 [ [ ∈D for all B and therefore A . | {z } ∈A ∪k k ∈ D1 printed: 2015-2-19, 0.01 11 4. A, B A B : In the same way as before we see that ∈D ⇒ ∩ ∈ D = A : A B B D2 { ∈ D ∩ ∈D ∀ ∈ D} is a d-system. Furthermore by part (3) and thus = . A ⊂ D2 D2 D On the basis of the above arguments is closed with respect to intersection. Furthermore, D because is closed with respect to taking complement of a set, we see that is closed with respect D D to union (i.e. is an algebra). Finally we notice that satisfies the condition (iii) of a σ-algebra, D D because if A1, A2,..., , then ∈ D ∞ ∞ A = A′ , k k ∈ D k[=1 k[=1 where A′ = A A A and A′ A′ . Therefore is a σ-algebra. k 1 ∪ 2 ∪···∪ k ∈ D k ⊂ k+1 D Theorem 1.34. Let be a π-system, = σ( ) and let ν,µ: [0, 1] be probability measures F M F M→ such that ν(A)= µ(A) for all A . Then ν µ. ∈ F ≡ Proof. We write = A : ν(A) = µ(A) and show that = . On the basis of D { ∈ M } D M Theorem1.33 it is enough to show that is a d-system, because then = σ( ) . D M F ⊂D⊂M (i) Because ν(X)=1= µ(X), then X . ∈ D (ii) If A, B and A B, then ∈ D ⊂ ν(B A)= ν(B) ν(A)= µ(B) µ(A)= µ(B A), \ − − \ and therefore B A . \ ∈ D (iii) Let A , A ,... and A A . Then by Theorem 1.10 (a) 1 2 ∈ D 1 ⊂ 2 ⊂··· ∞ ∞ ν Ai = lim ν(Ai) = lim µ(Ai)= µ Ai , i→∞ i→∞ i=1 i=1 [ [ and hence A . ∪i i ∈ D Therefore is a d-system and theorem is proved. D Corollary 1.35. Let ν and µ be finite Borel-measures in Rn. If ν(I) = µ(I) for every finite n- interval I, then ν(B)= µ(B) for all Borel sets B. Proof. Because ∞ Rn = ] j, j[n, − j[=1 then the assumption (ν(I) = µ(I)) and Theorem 1.10 (a) imply that ν(Rn) = µ(Rn) = c < . 0 ∞ If c = 0, then ν 0 µ and the claim is clear. Otherwise by dividing with c , we may assume 0 ≡ ≡ 0 that ν and µ are probability measures. Because = I : I finite n-interval is a π-system and F { } σ( ) = Bor(Rn), the claim follows from Theorem 1.34. F 12 Modern Real Analysis, Fall 2005 1.36 Extension of measures Suppose that we are given a collection (X) and a mapping µ: [0, + ], which is F ⊂ P F → ∞ σ-additive in , i.e., F A disjoint ∞ ∞ k ∈ F (1.37) µ Ak = µ(Ak). ∞ ⇒ A k=1 k=1 k=1 k ∈ F [ X We next study conditions underS which the measure µ can be extended to a measure defined in σ( ). In view of the previous section we suppose that is a π-system. In the general case this is F F not enough for extension, but if we require a bit more, extension will be possible. Definition 1.38. 1. A family (X) is an algebra, if A ⊂ P (i) ; ∅∈A (ii) A, B A B ; ∈A ⇒ ∩ ∈A (iii) A X A . ∈A ⇒ \ ∈A 2. A family (X) is a semialgebra, if P ⊂ P (a) ; ∅ ∈ P (b) A, B A B ; ∈P ⇒ ∩ ∈ P (c) A X A = n A , where the sets A are disjoint. ∈P ⇒ \ k=1 k k ∈ P We introduce the notation S Ai iG=1 to denote the union of disjoint sets Ai. Lemma 1.39. Let be semialgebra. Then P n ¯ = A : A , i N P { i i ∈ P ∈ } iG=1 is an algebra (so called algebra spanned by ). P Proof. (i) ¯. ∅∈ P (ii) Let A, B ¯. Then ∈ P n m A = A , B = B , A ,B i j i j ∈ P iG=1 jG=1 and A B = A B ∩ i ∩ j Gi,j and A B , and therefore A B ¯. i ∩ j ∈ P ∩ ∈ P printed: 2015-2-19, 0.01 13 (iii) Let A = n A ¯, A . Because is a semialgebra, then X A = m B , B , i=1 i ∈ P i ∈ P P \ i j=1 j j ∈ P and thus X A ¯. Therefore by part (ii) F \ i ∈ P F n n X A = (X A ) ¯. \ i \ i ∈ P i=1 i=1 G \ ∈P¯ | {z } Definition 1.40. If is an algebra and µ: [0, + ] is σ-additive s.t. µ( ) = 0, then µ is a A A → ∞ ∅ measure on the algebra . A Lemma 1.41. Let be a semialgebra and let µ: [0, + ] be σ-additive s.t. µ( ) = 0. Then P P → ∞ ∅ there exists a unique measure µ¯ on the algebra ¯, P µ¯ : ¯ [0, + ], P → ∞ for which µ¯(A)= µ(A) A . ∀ ∈ P Proof. Let A ¯. Then ∈ P n A = A , A . i i ∈ P iG=1 Set n (1.42)µ ¯(A)= µ(Ai). Xi=1 n m 1.µ ¯ is well defined: Let A = i=1 Ai = j=1 Bj. Then F F n B = B A = (B A ) and j j ∩ j ∩ i i=1 Gm A = A A = (A B ), i i ∩ i ∩ j jG=1 and hence µ(B ) σ-add.= µ(B A ) σ-add.= µ(A ). j j ∩ i i Xj Xj,i Xi 2.µ ¯ is a measure on the algebra ¯: Clearlyµ ¯( ) = 0. Let A ¯, i N, s.t. A = ∞ A ¯. P ∅ i ∈ P ∈ i=1 i ∈ P We must prove that ∞ F µ¯(A)= µ¯(Ai). Xi=1 Write A = ni A , where A . Thus i j=1 ij ij ∈ P F A = A , A . ij ij ∈ P Gi,j 14 Modern Real Analysis, Fall 2005 On the other hand A ¯, and hence ∈ P n A = B , B , k k ∈ P kG=1 and thus B = A B = A B . k ∩ k ij ∩ k Gi,j Here A B , because is a semialgebra. Now it follows from the σ-additivity of µ that ij ∩ k ∈ P P ∞ ni n µ(B )= µ(A B ) and µ(A )= µ(A B ). k ij ∩ k ij ij ∩ k Xi=1 Xj=1 Xk=1 and thus (1.42) µ¯(A) = µ(Bk) Xk=1 n ∞ ni = µ(A B ) ij ∩ k Xk=1 Xi=1 Xj=1 ∞ ni n (∗) = µ(A B ) ij ∩ k Xi=1 Xj=1 Xk=1 =µ(Aij ) ∞ n i | {z } = µ(Aij) Xi=1 Xj=1 ∞ ni (1.42) = µ¯(A ), because A = A , A . i i ij ij ∈ P Xi=1 jG=1 (The equality ( ) holds, because the series in question have positive terms.) ∗ 3. Uniqueness: Letν ¯ andµ ¯ be measures on the algebra ¯ s.t. P (1.43)ν ¯(A) =µ ¯(A)= µ(A) A . ∀ ∈ P We claim thatν ¯(A) =µ ¯(A) for all A ¯. Let A ¯ and write ∈ P ∈ P n A = A , A . i i ∈ P iG=1 Then from the σ-additivity ofν ¯ and ofµ ¯ and from (1.43) it follows that n n n ν¯(A)= ν¯(Ai)= µ(Ai)= µ¯(Ai) =µ ¯(A). Xi=1 Xi=1 Xi=1 Theorem 1.44 (Carath´eodory-Hahn extension theorem). Let (X) be an algebra and let A ⊂ P µ: [0, + ] be a measure on the algebra . A→ ∞ A printed: 2015-2-19, 0.01 15 (a) Then there exists a measure ν : σ( ) [0, + ] s.t. ν = µ. A → ∞ |A (b) If µ is σ-finite with respect to (i.e. X = ∞ A , A , µ(A ) < ), then ν is unique. A i=1 i i ∈A i ∞ Proof. We defineν ˜: (X) [0, + ] by settingS P → ∞ ∞ ∞ ν˜(E) = inf µ(A ): E A , A i . { i ⊂ i i ∈A∀ } Xi=1 i[=1 1. Thenν ˜ is an outer measure andν ˜(A)= µ(A) for all A . (HT 2/4) ∈A 2. Every A isν ˜-measurable. (HT 2/5) ∈A 3. If is the set ofν ˜-measurable sets, thenν ˜ is a measure (Theorem 1.15). On the basis of M |M item 2. we see that σ( ) and thus ν =ν ˜ σ( ) is the desired measure. A ⊂ M | A 4. For the uniqueness suppose that ν ,ν : σ( ) [0, + ] are measures s.t. 1 2 A → ∞ ν1(A)= ν2(A)= µ(A) for all A . If A and µ(A) < , then by Theorem 1.34 ∈A ∈A ∞ ν (A B)= ν (A B) 1 ∩ 2 ∩ for all B σ( ). Choose the sets A s.t. A A ,µ(A ) < and X = ∞ A . ∈ A i ∈ A 1 ⊂ 2 ⊂··· i ∞ i=1 i Because ν and ν are measures in σ( ), then for all B σ( ) there holds 1 2 A ∈ A S ν1(B) = lim ν1(B Ai) = lim ν2(B Ai)= ν2(B) i→∞ ∩ i→∞ ∩ in other words ν1 = ν2. 1.45 Product measures We now apply the Carath´eodory-Hahn extension theorem for the construction of a product measure. Let (X, ,µ) and (Y, ,ν) be measure spaces and M N = A B : A , B (X Y ) S { × ∈ M ∈N}⊂P × be the family of so called measurable rectangles. Lemma 1.46. is a semialgebra. S Proof. (a) Clearly . ∅ ∈ S (b) If A B and A′ B′ , then (A B) (A′ B′) = (A A′) (B B′) . × ∈ S × ∈ S × ∩ × ∩ × ∩ ∈ S (c) If A B , then its complement × ∈ S (A B)c = (Ac B) (A Bc) (Ac Bc) × × ⊔ × ⊔ × ∈S ∈S ∈S is of the required type. | {z } | {z } | {z } 16 Modern Real Analysis, Fall 2005 Define λ: [0, + ] by setting S → ∞ (1.47) λ(A B)= µ(A)ν(B). × Clearly λ( ) = 0. We next prove that λ is σ-additive in . ∅ S Lemma 1.48. If A B , i N, are disjoint s.t. i × i ∈ S ∈ ∞ (A B )= A B , i × i × ∈ S iG=1 then ∞ λ(A B )= λ(A B). i × i × Xi=1 Proof. We must show that ∞ µ(A)ν(B)= µ(Ai)ν(Bi). Xi=1 Fix x A and denote I(x)= i N: x A . Then ∈ { ∈ ∈ i} B = Bi i∈GI(x) and thus ν(B)= ν(Bi). i∈XI(x) Therefore for all x X there holds ∈ χA(x)ν(B)= χA(x) ν(Bi) i∈XI(x) = χA(x) χAi (x)ν(Bi) i∈XI(x) ∞ = χA(x) χAi (x)ν(Bi) i=1 ∞ X = χAi (x)ν(Bi). Xi=1 Because a series with positive terms can be termwise integrated (Mitta ja integraali, Lause 3.20), then µ(A)ν(B)= χA(x)ν(B) dµ(x) X Z ∞ = χAi (x)ν(Bi) dµ(x) ZX i=1 ∞ X = χAi (x)ν(Bi) dµ(x) i=1 ZX X∞ = µ(Ai)ν(Bi). Xi=1 printed: 2015-2-19, 0.01 17 Theorem 1.49. Let (X, ,µ) and (Y, ,ν) be σ-finite measure spaces and let (X Y ) be M N S ⊂ P × the family of measurable rectangles. Then the measure space X Y,σ( ) has a unique measure × S µ ν, so called product measure of µ and ν, such that × (µ ν)(A B)= λ(A B)= µ(A)ν(B) × × × for all A and B . ∈ M ∈N Proof. Because λ is σ-additive in the semialgebra and λ( ) = 0, then by Lemman 1.41 there S ∅ exists a unique measure λ¯ in the algebra ¯ s.t. λ¯(A B)= λ(A B) for all A B . In addition, S × × × ∈ S it follows from the σ-finiteness of µ and ν that λ¯ is σ-finite with respect to ¯. The claim follows S now from the Carath´eodory-Hahn extension theorem. Recall from Real Analysis I that a measure space (X, ,µ) is complete, if the subsets of sets M of measure zero are measurable. In other words, if A B and µ(B) = 0, then A . A ⊂ ∈ M ∈ M measure can always be completed (see Real Analysis I, Theorem 1.12): Theorem 1.50. Let (X, ,µ) be a measure space. Define ¯ (X) by setting M M ⊂ P ¯ = A F : A and F E for some E , µ(E) = 0 . M { ∪ ∈ M ⊂ ∈ M } and define µ¯ : ¯ [0, + ], M→ ∞ µ¯(A F )= µ(A) , ∪ where A and F as above. Then (1) ¯ is a σ-algebra in X; M (2) µ¯ is a complete measure ; (3) µ =µ ¯ . |M The measure µ¯ is called the completion of µ. In the same way (X, ¯ , µ¯) is the completion of the M measure space (X, ,µ). M Remark 1.51. 1. Ifµ ˜ is an outer measure in X, then the measure space (X, ,µ) determined Mµ˜ by it is complete. 2. Ifµ ˜ is a Borel-regular metric outer measure, then the measure space (X, ,µ) determined Mµ˜ by it is the completion of (X, Bor(X),µ). 3. If (X, ,µ) and (Y, ,ν) are complete σ-finite measure spaces, then the product measure M N space (X Y,σ( ),µ ν) is not necessarily complete. For example, if A , A = , µ(A) = 0 × S × ∈ M 6 ∅ and B Y, B , then A B A Y, (µ ν)(A Y ) = 0, but A B σ( ) (see ⊂ 6∈ N × ⊂ × × × × 6∈ S Lemma 1.53). In particular, the product measure of the Lebesgue measures mn and mk (Rn Rk,σ( ),m m ) is not complete, and hence m m = m . Instead m is the × S n × k n × k 6 n+k n+k completion of m m . n × k 1.52 Fubini’s theorem Let (X, ,µ) and (Y, ,ν) be σ-finite measure spaces and µ ν the product measure given by M N × Theorem 1.49. We now study the question for what kind of functions f : X Y R˙ there holds: × → f(x,y) d(µ ν)(x,y)= f(x,y) dν(y) dµ(x)= f(x,y) dµ(x) dν(y)? × ZX×Y ZX ZY ZY ZX 18 Modern Real Analysis, Fall 2005 Let E X Y . Denote ⊂ × E = y Y : (x,y) E , x X, x { ∈ ∈ } ∈ Ey = x X : (x,y) E , y Y. { ∈ ∈ } ∈ Lemma 1.53. If E σ( ), then E and Ey for all x X for and y Y . ∈ S x ∈N ∈ M ∈ ∈ Proof. Let = E σ( ): E x X . It is enough to show that is a σ-algebra s.t. C { ∈ S x ∈N ∀ ∈ } C . Then, in fact, there holds = σ( ) (and the claim about the set E holds), because σ( ) S⊂C C S x S is the smallest σ-algebra containing . The claim about the sets Ey is proved in the same way. S Let A B . Because (A B) = B , if x A, and (A B) = , if x A, then × ∈ S × x ∈ N ∈ × x ∅∈N 6∈ A B . Therefore . We next show that is a σ-algebra. × ∈C S⊂C C (i) Clearly . ∅∈C c c (ii) Observe that (E )x = (Ex) . Thus E E x X ∈C ⇒ x ∈N ∀ ∈ (E )c x X ⇒ x ∈N ∀ ∈ (Ec) x X ⇒ x ∈N ∀ ∈ Ec . ⇒ ∈C (iii) We now see that ( E ) = (E ) . Therefore ∪i i x ∪i i x E , i N E . i ∈C ∈ ⇒ i ∈C [i Let f : X Y R˙ . Define for all x X and all y Y the functions f : Y R˙ and f y : X R˙ × → ∈ ∈ x → → by setting fx(y)= f(x,y), f y(x)= f(x,y). Lemma 1.54. If f : X Y R˙ is µ ν-measurable, then × → × (a) f is ν-measurable for all x X, x ∈ (b) f y is µ-measurable for all y Y . ∈ Proof. If U R is open, then f −1U σ( ), because f is µ ν-measurable. In the same way ⊂ ∈ S × we have f −1(+ ) σ( ) and f −1( ) σ( ). If E is any one of the above sets f −1U, f −1(+ ), ∞ ∈ S −∞ ∈ S ∞ or f −1( ), then by Lemma 1.53 f −1U, f −1(+ ), or f −1( ) = E is ν-measurable. In the −∞ x x ∞ x −∞ x same way (f y)−1U, (f y)−1(+ ), or (f y)−1( ) = Ey is µ-measurable. ∞ −∞ Lemma 1.55. If E σ( ), then ∈ S g : X [0, + ], g(x)= ν(E ), → ∞ x is µ-measurable and f : Y [0, + ], f(y)= µ(Ey), → ∞ is ν-measurable. Furthermore g(x) dµ(x)= f(y) dν(y) = (µ ν)(E). × ZX ZY printed: 2015-2-19, 0.01 19 Proof. We show that the family = E σ( ): x ν(E ) µ-measurable and ν(E ) dµ(x) = (µ ν)(E) F ∈ S 7→ x x × Z is a σ-algebra, containing . Then = σ( ) and the claims about the function g, g(x) = ν(E ), S F S x hold true. In the same way we prove the claims about f. If E = A B , then × ∈ S B, if x A; Ex = ∈ ( , if x A, ∅ 6∈ and hence ν(Ex)= ν(B)χA(x) and therefore x ν(E ) is µ-measurable and 7→ x ν(E ) dµ(x)= ν(B) χ dµ = (µ ν)(A B). x A × × ZX ZX (Remark. x χ (x) is µ-measurable, because A .) Thus . By Dynkin’s lemma (Theorem 7→ A ∈ M S ⊂ F 1.33) it is enough to show that is a d-system, because then σ( )= . In other words one should F S F prove that (i) X Y , × ∈ F (ii) E, F , E F F E , ∈ F ⊂ ⇒ \ ∈ F (iii) E E , E ∞ E . i ⊂ 2 ⊂··· i ∈F ⇒ i=1 i ∈ F We prove these in the order (i), (iii),S (ii). (i) We saw above that , and hence X Y . S ⊂ F × ∈ F (iii) Let E E , E , and E = ∞ E . The functions x ν(E ) are measurable. i ⊂ 2 ⊂ ··· i ∈ F i=1 i 7→ ix In addition E E and E = ∞ E , and hence ν(E ) = lim ν(E ) and 1x ⊂ 2x ⊂ ··· x S i=1 ix x i→∞ ix x ν(E ) is measurable as the limit function of measurable functions. Further ν(E ) 7→ x S 1x ≤ ν(E ) , and by the monotone convergence theorem (MCT) 2x ≤··· 1.10 (µ ν)(E) = lim (µ ν)(Ei) × i→∞ × Ei∈F = lim ν(Eix) dµ(x) i→∞ ZX MCT = lim ν(Eix) dµ(x) i→∞ ZX = ν(Ex) dµ(x). ZX Therefore E . ∈ F (ii) Because X and Y are σ-finite, there exist sets ∞ X X , X , µ(X ) < , X = X , and 1 ⊂ 2 ⊂··· i ∈ M i ∞ i i=1 ∞ [ Y Y , Y , ν(Y ) < , Y = Y . 1 ⊂ 2 ⊂··· i ∈N i ∞ i i[=1 20 Modern Real Analysis, Fall 2005 Let = E σ( ): E (X Y ) i N E { ∈ S ∩ i × i ∈F ∀ ∈ } and we prove that is a σ-algebra, which contains . Then = σ( ) or E (X Y ) E S E S ∩ i × i ∈ F for all i N and all E σ( ). If E = A B , then E (X Y ) for all i. By the ∈ ∈ S × ∈ S ∩ i × i ∈ S first part of the proof E (X Y ) for all i, and hence . In particular X Y . ∩ i × i ∈ F S⊂E × ∈E We prove that is a d-system. If E , E E , then by the part (iii) of the proof E j ∈E j ⊂ j+1 E (X Y )= E (X Y ) i, ∪j j ∩ i × i ∪j j ∩ i × i ∈F ∀ and hence E . We next suppose that E, F , E F. Denote ∪j j ∈E ∈E ⊂ E = E (X Y ) and i ∩ i × i F = F (X Y ), i ∩ i × i and then E , F i and i i ∈F ∀ F E = (F E) (X Y ). i \ i \ ∩ i × i We next show that F E all i N. For all x X there holds i \ i ∈ F ∈ ∈ ν((F E ) )= ν(F E )= ν(F ) ν(E ), i \ i x ix \ ix ix − ix because E F and ν(E ) < . From these facts we conclude that ix ⊂ ix ix ∞ x ν((F E ) ) 7→ i \ i x is µ-measurable and ν((F E ) ) dµ(x)= ν(F ) dµ(x) ν(E ) dµ(x) i \ i x ix − ix ZX ZX ZX = (µ ν)(F ) (µ ν)(E ) × i − × i = (µ ν)(F E ). × i \ i Thus F E i. We have shown that = σ( ) and thus . i \ i ∈F ∀ E S F⊂E Let next E, F , E F . We want to show that F E . As soon as that is done we ∈ F ⊂ \ ∈ F have finally shown that = σ( ). Because F E F E and F S i \ i ⊂ i+1 \ i+1 ∞ F E = F E , \ i \ i i=1 [ ∈F it follows from part (iii) that F E . | {z } \ ∈ F Theorem 1.56 (Tonelli). Let f : X Y [0, + ] be µ ν-measurable. Then there holds: × → ∞ × (1) For all (x ,y ) X Y 0 0 ∈ × x f(x,y ) is µ-measurable and 7→ 0 y f(x ,y) is ν-measurable. 7→ 0 printed: 2015-2-19, 0.01 21 (2) x f(x,y) dν(y) is µ-measurable and 7→ ZY y f(x,y) dµ(x) is ν-measurable. 7→ ZX (3) f(x,y) dν(y) dµ(x)= f(x,y) d(µ ν)(x,y) × ZX ZY ZX×Y = f(x,y) dµ(x) dν(y). ZY ZX Proof. (a) If E σ( ), then the claim holds for (µ ν) -measurable function f = χ by Lemmas 1.54 ∈ S × E and 1.55. (b) By linearity the claim holds for simple functions (with respect to σ( )). S k f = a χ , E σ( ). i Ei i ∈ S Xi=1 (c) If f : X Y [0, + ] is µ ν-measurable, then there is an increasing sequence of simple × → ∞ × functions (with respect to σ( )) s f s.t. S i ր s d(µ ν) f d(µ ν). i × → × ZX×Y ZX×Y Because s f, then s f and sy f y for all (x,y) X Y . Therefore by the monotone i ր ix ր x i ր ∈ × convergence theorem (MCT) s dν(y) f dν(y) and ix → x ZY ZY sy dµ(x) f y dµ(x). i → ZX ZX “From the measurability of the limit function” it now follows that the functions x f y(x)= f(x,y), 7→ y f (y)= f(x,y), 7→ x x f (y) dν(y)= f(x,y) dν(y) and 7→ x ZY ZY y f y(x) dµ(x)= f(x,y) dµ(x) 7→ ZX ZX are measurable and the claims (1) and (2) hold. If gi(x)= six dν(y), ZY 22 Modern Real Analysis, Fall 2005 then g g , so by the MCT i ≤ i+1 s dν(y) dµ(x) f dν(y) dµ(x)= f(x,y) dν(y) dµ(x). ix → x ZX ZY ZX ZY ZX ZY On the other hand (b) s dν(y) dµ(x) = s d(µ ν) f d(µ ν), ix i × → × ZX ZY ZX×Y ZX×Y and hence the left equality of (3) holds. In the same way one proves the right hand side of (3). Remark 1.57. The above results have been formulated for the product measure µ ν and (µ ν)- × × measurable functions. These results cannot be directly applied to the complete measure space (Rn Rk, Leb(Rn+k),m ). For example Lemmas 1.53 and 1.54 do not hold for the measure -space × n+k (Rn Rk, Leb(Rn+k),m ). Indeed, if B Rk is non-measurable (w.r.t. m ), then E = x B × n+k ⊂ k { }× is m -measurable for all x Rn, because m∗ (E) = 0. On the other hand if B = E , we see n+k ∈ n+k x that the claim of Lemma 1.53 is not valid. In Tonelli’s theorem we can require in the case of the measure space (Rn Rk, Leb(Rn+k),m ) condition (1) only for a.e. x and a.e. y . × n+k 0 0 Theorem 1.58 (Fubini). Let f : X Y R˙ be µ ν-measurable and suppose that at least one of × → × the integrals f d(µ ν), f(x,y) dν(y) dµ(x), or | | × | | ZX×Y ZX ZY f(x,y) dµ(x) dν(y) | | ZY ZX is finite. Then (1) y f(x,y) is integrable in Y for almost all x X; 7→ ∈ (2) x f(x,y) is integrable in X for almost all y Y ; 7→ ∈ (3) x f(x,y) dν(y) is integrable in X, i.e. 7→ Y R f(x,y) dν(y) dµ(x) < ; X Y ∞ Z Z (4) y f(x,y) dµ(x) is integrable in Y ; 7→ X (5) f is integrableR in X Y and × f d(µ ν)= f(x,y) dν(y) dµ(x)= f(x,y) dµ(x) dν(y) < . × ∞ ZX×Y ZX ZY ZY ZX Proof. In the same way as in measure theory, Theorem 4.3 (Fubini 2.) printed: 2015-2-19, 0.01 23 2 Hausdorff measures 2.1 Some basic properties of the Hausdorff measure The Lebesgue n-dimensional measure mn is well suited for the measurement of the size of “large” n n subsets of R , but it is too crude for measuring “small” subsets of R . For example, m2 cannot distinguish a singleton of R2 from a line, because both have measure zero. In this chapter we introduce a whole spectrum of “s-dimensional” measures s, 0 s < , H ≤ ∞ which are able to see the fine structure of sets, better than the Lebesgue measure. The key idea is that a set A Rn is “s-dimensional”, if 0 < s(A) < , even if the geometric structure of A were ⊂ H ∞ very complicated. These measures can be defined in any metric space (X, d). We suppose, however, that X is separable, i.e. X has a countable dense subset S = x ∞ , and hence X = S¯. This assumption is { i}i=1 only needed to guarantee that X has so called δ-covering for all 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. A countable collection E ∞ of subsets of X is a δ-covering, δ > 0, of A X if { i}i=1 ⊂ ∞ A E and d(E ) δ i N. ⊂ i i ≤ ∀ ∈ i[=1 We fix “dimension” s [0, ) and δ > 0. For A X, we define ∈ ∞ ⊂ ∞ (2.3) s(A) = inf d(E )s : E is a δ-covering of A , Hδ i { i} i=0 X where we make the convention that d( x )0 = 1 x X and d( )s = 0 s 0. { } ∀ ∈ ∅ ∀ ≥ We readily see from the definition that s (A) s (A), Hδ1 ≥Hδ2 if 0 < δ δ . Therefore the following limit (2.5) exists and we can set the definition. 1 ≤ 2 Definition 2.4. The s-dimensional Hausdorff (outer)measure of a set A X is ⊂ s s s (2.5) (A) = lim δ(A) = sup δ(A) . H δ→0 H H δ>0 Theorem 2.6. (i) s : (X) [0, + ] is an outer measure for all δ > 0. Hδ P → ∞ (ii) s : (X) [0, + ] is a metric outer measure. H P → ∞ Proof. (i) (a) Clearly s( ) = 0. Hδ ∅ 24 Modern Real Analysis, Fall 2005 (b) Let then A ∞ A X. We may suppose that s(A ) < i. Let ε> 0 and choose ⊂ i=1 i ⊂ Hδ i ∞∀ for every i a δ-covering Ei ∞ of the set A s.t. S { j}j=1 i ∞ d(Ei)s s(A )+ ε2−i. j ≤Hδ i Xj=1 i ∞ Then i,j Ej is a δ-covering of the union i=1 Ai and thus also of A and therefore S s(A) Sd(Ei)s Hδ ≤ j Xi,j ∞ s(A )+ ε2−i ≤ Hδ i i=1 X ∞ ε + s(A ). ≤ Hδ i Xi=1 Letting ε 0 the desired conclusion follows. → (ii) Clearly s( )=0. If A ∞ A X, then by part (i) and the definition of s we obtain H ∅ ⊂ i=1 i ⊂ H ∞ ∞ S s(A) s(A ) s(A ). Hδ ≤ Hδ i ≤ H i Xi=1 Xi=1 Letting δ 0 we see that s is an outer measure. Let then A , A X be sets, for which → H 1 2 ⊂ dist(A1, A2) > 0. We wish to show that s(A A )= s(A )+ s(A ). H 1 ∪ 2 H 1 H 2 It is enough to show that (2.7) s(A A ) s(A )+ s(A ), Hδ 1 ∪ 2 ≥Hδ 1 Hδ 2 if δ dist(A , A )/3. We may assume that s(A A ) < . Let ε > 0 and choose a ≤ 1 2 Hδ 1 ∪ 2 ∞ δ-covering E ∞ of the set A A such that { i}i=1 1 ∪ 2 ∞ d(E )s s(A A )+ ε. i ≤Hδ 1 ∪ 2 Xi=1 Because δ dist(A , A )/3, every E intersects at most one of the sets A or A . Therefore ≤ 1 2 i 1 2 we may write the sum as E ∞ = E′ ∞ E′′ ∞ , { i}i=1 { i}i=1 ⊔ { i }i=1 where ∞ ∞ A E′ and A E′′. 1 ⊂ i 2 ⊂ i i[=1 i[=1 Therefore ∞ ∞ s(A )+ s(A ) d(E′)s + d(E′′)s Hδ 1 Hδ 2 ≤ i i i=1 i=1 X∞ X s = d(Ei) Xi=1 s(A A )+ ε. ≤Hδ 1 ∪ 2 Because ε> 0 was arbitrary, we obtain (2.7). printed: 2015-2-19, 0.01 25 On the basis of Theorems 1.18 and 2.6 every Borel-set of X is s-measurable. We denote the H restriction of s to s-measurable sets with the same symbol s. Now there holds: H H H Theorem 2.8. s is a Borel-measure. H Corollary 1.29 yields now: Corollary 2.9. If A Rn is s-measurable and s(A) < , then sxA is a Radon-measure. ⊂ H H ∞ H Theorem 2.10. The outer measure s of a separable metric space X is Borel-regular. H Proof. Because by the previous theorem the outer measure defined by s is Borel, it is enough H to show that for all A X there exists B Bor(X) s.t. A B and s(A)= s(B). ⊂ ∈ ⊂ H H Let A X. If s(A) = , we may choose B = X and the claim holds. If there holds ⊂ H ∞ s(A) < , then we choose an 1/i-covering Ei ∞ of A for each i N s.t. H ∞ { j}j=1 ∈ ∞ d(Ei)s s (A) + 1/i. j ≤H1/i Xj=1 Because d(E)= d(E¯) for all E X, we may suppose that the sets Ei are closed. Then ⊂ j ∞ ∞ i B = Ej i\=1 j[=1 is a Borel set and A B. Furthermore Ei is a 1/i-covering of B for all i N, and hence ⊂ { j} ∈ ∞ s (A) s (B) d(Ei)s s (A) + 1/i. H1/i ≤H1/i ≤ j ≤H1/i Xj=1 Letting i the claim s(A)= s(B) follows. →∞ H H Remark 2.11. 1. 0 is the counting measure. H 2. s is not, in general, a metric outer measure. Hδ 3. Roughly speaking 1 is a length measure, 2 is area, etc. H ∼ H ∼ 4. It is easily seen (applying results that will be proven soon), that (e.g.) the plane R2 is not σ-finite with respect to 1. H 2.12 Hausdorff dimension Let (X, d) be a separable metric space. In this chapter we shall define a dimension for sets A X, ⊂ which reflects the metric size of the set A. Unlike the topological dimension, this dimension need not be an integer. Lemma 2.13. Let A X and s 0. ⊂ ≥ (i) If s(A) < , then t(A) = 0 for all t>s. H ∞ H (ii) If s(A) > 0, then t(A)= for all 0 t Proof. It is enough to prove (i), because the claim (ii) follows from (i). Let δ > 0 and E ∞ { j}j=1 be a δ-covering of A s.t. ∞ d(E )s s(A) + 1 s(A) + 1 < . j ≤Hδ ≤H ∞ Xj=1 Then for all t>s ∞ t (A) d(E )t Hδ ≤ j Xj=1 ∞ δt−s d(E )s ≤ j Xj=1 δt−s ( s(A) + 1) . ≤ H The claim follows by letting δ 0. → By virtue of the preceding lemma the function s s(A) has a simple graph: There exists a 7→ H unique s [0, ], a point, such that on the right side of this point the graph is 0 and on the left 0 ∈ ∞ side of this point it is infinity. Definition 2.14. The Hausdorff dimension of a subset A X is a number ⊂ dim (A) = inf s> 0: s(A) = 0 . H { H } Summarizing what was said above: 1. If t< dim (A), then t(A)= . H H ∞ 2. If t> dim (A), then t(A) = 0. H H In general, about the number s(A), for s = dim (A), we cannot say anything: it can be any H H number in [0, ]. Nevertheless: ∞ (2.15) 0 < s(A) < dim (A)= s . H ∞ ⇒ H A set A X, for which 0 < s(A) < holds, is called an s-set. ⊂ H ∞ Lemma 2.16. (i) If A B, then dim (A) dim (B). ⊂ H ≤ H (ii) If Ak X, k N, then ⊂ ∈ ∞ dimH Ak = sup dimH(Ak). k=1 k [ Proof. (HT) Thus, for example, dimH(Q) = 0. printed: 2015-2-19, 0.01 27 2.17 Hausdorff measures in Rn Next we evaluate (or estimate) Hausdorff measures and dimensions of Cantor type fractal sets in Rn. To this end we study the invariance properties of Hausdorff measures. Other, more efficient, methods for the determination of the Hausdorff dimension will be given later. Recall first that a mapping T : Rn Rn is an isometry, if → Tx Ty = x y x,y Rn. | − | | − | ∀ ∈ It is well-known that every isometry of Rn is an affine mapping, i.e. of the form Tx = a0 + Ux, where a Rn and U : Rn Rn is a linear isometry. 0 ∈ → In the same way, a mapping R: Rn Rn is said to be a similarity, if → Rx Ry = c x y x,y Rn, | − | | − | ∀ ∈ where c> 0 is a constant (stretching factor, scaling factor, etc.). Then R is of the form Rx = a0 + cUx, where U is again a linear isometry. Theorem 2.18. Let A Rn. For the outer measure s, s 0, there holds: ⊂ H ≥ (a) s(A + x )= s(A) x Rn, H 0 H ∀ 0 ∈ (b) s U(A) = s(A) for all linear isometries U : Rn Rn, H H → (c) s R(A) = cs s(A), if R: Rn Rn is a similarity map, with scaling factor c> 0. H H → Proof. The claims follow from the observation that d R(E) = c d(E) E Rn, where R is as ∀ ⊂ in part (c). Let (X, d ) and (Y, d ) be metric spaces. Recall next that the mapping f : X Y is L-Lipschitz 1 2 → (with a constant L> 0), if d f(x),f(y) L d (x,y) 2 ≤ 1 for all x,y X. In the same way, a mapping g : X Y is L-bilipschitz, if ∈ → 1 d (x,y) d f(x),f(y) L d (x,y) L 1 ≤ 2 ≤ 1 for all x,y X. We observe that an L-bilipschitz mapping is always an injection because of the ∈ inequality on the left hand side. Lemma 2.19. Let (X, d1) and (Y, d2) be separable metric spaces. (i) If f : X Y is L-Lipschitz, then → s(fA) Ls s(A) A X. H ≤ H ∀ ⊂ (ii) If g : X Y is L-bilipschitz, then → dim (gA) = dim (A) A X. H H ∀ ⊂ 28 Modern Real Analysis, Fall 2005 Proof. (i) We may supppose that s(A) < . Fix ε> 0, δ> 0 and choose a δ-covering E ∞ of A H ∞ { j}j=1 s.t. ∞ d(E )s s(A)+ ε. j ≤Hδ Xj=1 Then f(E ) ∞ is a Lδ-covering of fA and hence { j }j=1 ∞ s (fA) d f(E ) s HLδ ≤ j Xj=1 ∞ Ls d(E )s ≤ j Xj=1 Ls( s(A)+ ε). ≤ Hδ The claim follows by letting ε 0 and δ 0. → → (ii) Applying part (i) to the mapping g−1 : g(A) X we obtain → L−s s(A) s(gA). H ≤H Thus L−s s(A) s(gA) Ls s(A), H ≤H ≤ H which yields the claim. We next study the connection of the Lebesgue measure m and the Hausdorff measure n in n H Rn. Theorem 2.20. For all A Bor(Rn) there holds ∈ n(A)= c m (A), H n n where c (0, ) is a constant. n ∈ ∞ Proof. Let Q = [0, 1)n. We show first that (2.21) 0 < n(Q) < . H ∞ n Clearly Q can be covered by jn equal subcubes Q j of Q, with edge length equal to 1/j and { i}i=1 diameter d(Qi)= √n/j. If δj = √n/j, then jn √n n n (Q) = nn/2. Hδj ≤ j Xi=1 Letting δ = √n/j 0, we get j → n(Q) nn/2. H ≤ printed: 2015-2-19, 0.01 29 Let next δ > 0 and suppose that E ∞ is a δ-covering of Q. Clearly E I , where I is a suitably { i}i=1 i ⊂ i i chosen cube with edge length 2d(Ei). Then ∞ ∞ ∞ n −n n −n d(Ei) = 2 2d(Ei) = 2 mn(Ii) i=1 i=1 i=1 X X ∞ X 2−nm I 2−nm (Q) ≥ n i ≥ n i=1 [ = 2−n, and hence n(Q) 2−n. Hδ ≥ Thus n(Q) 2−n H ≥ and the claim (2.21) holds. We write c = n(Q) and shall show that n H (2.22) n(A)= c m (A) H n n for all Borel-sets A Rn. The claim (2.22) holds for A = [0, 1)n. By part (c) of Theorem 2.18 ⊂ the relation (2.22) holds if A is an arbitrary “halfopen to the right” cube of Rn, i.e. of the form A = [a , b ) [a , b ), 0 < b a = = b a < . We next observe that every halfopen 1 1 ×···× n n 1 − 1 · · · n − n ∞ to the right n-interval of Rn can be represented as a union of disjoint cubes, halfopen to the right. Therefore (2.22) holds for all halfopen to the right n-intervals of Rn. If Q = [ m,m)n, m N, m − ∈ then the intersections of halfopen to the right n-intervals and Qm form a π-system and span the Borel sets of Q . By Theorem 1.34 n(A) = c m (A) for all A Bor([ m,m)n). Because this m H n n ∈ − holds for all m N, then (2.22) holds for all A Bor(Rn). ∈ ∈ Both n and m∗ are Borel-regular outer measures of Rn, and hence there holds: H n Corollary 2.23. n(A)= c m∗ (A) A Rn. H n n ∀ ⊂ Proof. (HT) We also give another proof for Theorem 2.20, which is based on uniformly distributed measures. Definition 2.24. We say that a Borel-measure µ of a metric space X is uniformly distributed, if 0 <µ B(x, r) = µ B(y, r) < ∞ for all x,y X and all 0 Proof. Denote g(r)= µ B(x, r) and h(r)= ν B(x, r) , for 0 ν B(x, r) U µ(U)= lim ∩ dµ(x) r→0+ h(r) ZU 1 lim inf ν B(x, r) U dµ(x). ≤ r→0+ h(r) ∩ ZU By Fubini’s theorem (see Remark 2.26) it now follows that ν B(x, r) U dµ(x)= χ (y) dν(y) dµ(x) ∩ B(x,r) ZU ZU ZU = χB(y,r)(x) dµ(x) dν(y) ZU ZU µ B(y, r) dν(y) ≤ ZU = g(r)ν (U). Thus g(r) 0 <µ(U) lim inf ν(U). ≤ r→0+ h(r) In the same way we get h(r) 0 <ν(U) lim inf µ(U). ≤ r→0+ g(r) We see that the following limit exists g(r) lim = c0 r→0+ h(r) and that µ(U)= c ν(U) for all open U X and therefore for all U Bor(X). 0 ⊂ ∈ Remark 2.26. We now motivate the application of Fubini’s theorem in the preceding proof. First, both µ and ν are σ-finite. Secondly, the following general result holds: If Y and Z are topological spaces, with a countable base (i.e. are N2), then (2.27) Bor(Y Z)= σ Bor(Y ) Bor(Z) . × × In the preceding theorem Y = Z = X is a separable metric space, and hence it is N2 and thus (2.27) holds. Therefore the open set ∆ = (x,y) X X : d(x,y) < r is as a Borel-set of X X r { ∈ × } × µ ν-measurable. This implies further that the mapping f : X X R, × × → f(x,y)= χB(x,r)(y)= χB(y,r)(x)= χ∆r (x,y) is µ ν-measurable and therefore Fubini’s theorem applies. × printed: 2015-2-19, 0.01 31 We next show the inclusion relation of (2.27) Bor(Y Z) σ Bor(Y ) Bor(Z) , which we × ⊂ × applied above. The other direction is left as an exercise. Let A Y Z be open. Because Y and ⊂ × Z are N2 spaces, then A can be represented as a countable union A = A B , i × j N i,j[∈ where A Y,B Z are open. Then A B Bor(Y ) Bor(Z), and hence A σ Bor(Y ) i ⊂ j ⊂ i × j ∈ × ∈ × Bor(Z) . Therefore Bor(Y Z) σ Bor(Y ) Bor(Z) . × ⊂ × We next construct sets with noninteger Hausdorff dimension. Recall the construction of the Cantor set from Real Analysis I. (We use slightly different notation and consider only a special case.) Let 0 <λ< 1/2. Denote I = [0, 1], I = [0, λ] and I = [1 λ, 1]. In other words, I 0,1 1,1 1,2 − 1,1 and I is obtained from I by removing its middle interval with length 1 2λ. Next we remove 1,2 0,1 − open interval of length (1 2λ)λ from the middle of closed intervals I1,i and continue the process − n inductively. Suppose that the intervals In,i, i = 1,..., 2 of step n have been defined. Then the n+1 intervals In+1,j, j = 1,..., 2 of the step (n + 1) are obtained by removing an open interval of length (1 2λ)λn from the middle of the intervals of step n. Thus − d(I )= λn, n and i = 1,..., 2n. n,i ∀ ∀ Denote 2n Cn(λ)= In,i i[=1 (“approximation of the nth step”) and ∞ C(λ)= Cn(λ). n=1 \ Then C(λ) is compact, uncountable set, without interior points. Furthermore C(λ) is “selfsimilar” and m1 C(λ) = 0. Cantor’s 1/3-set C(1/3) is a special case of this construction, recurrent in literature. I0,1 C0 I1,1 I1,2 C1 I2,1 I2,4 C2 . . . . Theorem 2.28. For all 0 <λ< 1/2 log 2 dim C(λ)= . H log(1/λ) In particular, dimH C(λ) can attain all values in the interval (0, 1). 32 Modern Real Analysis, Fall 2005 Proof. It is enough to show that (2.29) 1/2 s C(λ) 1, ≤H ≤ if log 2 s = . log(1/λ) (i) We first give a heuristic argument for finding the exponent s: Clearly C(λ)= C C , 1 ∪ 2 where C1 and C2 are disjoint and similar to C(λ) with the scaling factor λ. If C(λ) were an s-set, then by part (c) of Theorem 2.18 s C(λ) = s(C )+ s(C ) H H 1 H 2 = 2λs s C(λ) . H Thus 1 = 2λs, Solving this for s yields log 2 s = . log(1/λ) (ii) A rigorous proof of (2.29): If δ > 0 is given, then choose n N so large that λn < δ. Then n ∈ I 2 is a δ-covering of C(λ) and further { n,i}i=1 2n 2n s C(λ) (λn)s = (1/2)n = 1. Hδ ≤ i=1 i=1 X X Thus s C(λ) 1. H ≤ We give a proof for the lower bound (2.29) only in the special case λ = 1/3, because later on we shall give more efficient methods to determine the Hausdorff dimension. The general case λ (0, 1/2) would not bring any essential changes to the proof. Suppose that E ∞ is such ∈ { j}j=1 a δ-covering of C(1/3) that ∞ log 2 d(E )s s C(1/3) + δ, s = . j ≤H log 3 j=1 X For each j choose a closed interval I (= [a, b]) s.t. E int I (=]a, b[) and d(I ) < (1 + j j ⊂ j j δ)d(E ). Then int I ∞ is an open covering of C(1/3), and hence by the compactness of j { j}j=1 C(1/3) we can choose a finite subcover. By relabelling the intervals Ij , we may suppose that m C(1/3) I ⊂ j j[=1 printed: 2015-2-19, 0.01 33 and ∞ s C(1/3) + δ d(E )s H ≥ j Xj=1 m (1 + δ)−s d(I )s. ≥ j Xj=1 To prove the lower bound (2.29) it is enough to prove that m 1 (2.30) d(I )s , j ≥ 2 Xj=1 if I m is a covering of C(1/3) with finitely many closed intervals I . For each j choose { j}j=1 j k = k(j) N, with ∈ (2.31) 3−(k+1) d(I ) < 3−k. ≤ j Let k0 be the largest one of the numbers k(j), j = 1,...,m. On the basis of the construction and the choice of the number k = k(j) each Ij can intersect only one of the intervals Ik,i of k0−k(j) step k. Therefore Ij intersects at most 2 of the intervals Ik0,i. Thus the number of such intervals of step k0 is at most m 2k0−k(j). Xj=1 k0 On the other hand, there are 2 intervals of step k0. Every one of these contains points of C(1/3) and C(1/3) m I , and hence ⊂ j=1 j S m 2k0 2k0−k(j). ≤ Xj=1 Now we can compute m m 2k0 2k0−k(j) = 2k0 2−k(j) ≤ Xj=1 Xj=1 m = 2k0 (3−k(j))s Xj=1 m s 2k0 3d(I ) . ≤ j j=1 X Simplification yields m d(I )s 3−s = 1/2. j ≥ Xj=1 Remark 2.32. Refining the above argument one can show that log 2 s C(λ) = 1, s = , H log(1/λ) (cf. [Fa1, s. 14-15]). 34 Modern Real Analysis, Fall 2005 2.33 Rectifiable and nonrectifiable 1-sets Consider first rectifiable arcs in Rn. Let ψ : [a, b] Rn, a < b, be a continuous injection. Then → ψ[a, b]=Γis an arc. We define that the length of Γ is m (Γ) = sup ψ(t ) ψ(t ) : a = t ψ(a) ψ(ti−1) ψ(ti) ψ(b) We say that Γ is rectifiable, if (Γ) < . L ∞ Theorem 2.34. Let Γ Rn be a rectifiable arc. Then ⊂ (Γ) = 1(Γ). L H Proof. Every rectifiable arc Γ can be represented in terms of the arc length (HT). In other words, there exists a homeomorphism ψ : [0, (Γ)] Γ, with (ψ [s,t]) = t s for all 0 s t (Γ). 0 L → L 0 − ≤ ≤ ≤L By the definition of the arc length ψ (t) ψ (s) (ψ [a, b]) = t s , 0 s t (Γ), | 0 − 0 |≤L 0 | − | ≤ ≤ ≤L and hence ψ0 is 1-Lipschitz. Thus 1(Γ) = 1(ψ [0, (Γ)]) 1([0, (Γ)]) = (Γ), H H 0 L ≤H L L where the last equality follows from the fact that 1 = m in R. H 1 To prove the reverse inequality it is enough to show (by the definition of (Γ)), that L 1(ψ [s,t]) ψ (s) ψ (t) H 0 ≥ | 0 − 0 | for all 0 s t (Γ). We may suppose that ψ (s) = ψ (t). Let P be an orthogonal projection ≤ ≤ ≤ L 0 6 0 onto the line S through the points ψ0(s) and ψ0(t). Then P is 1-Lipschitz and P (ψ [s,t]) J, 0 ⊃ where J S is the interval on S between the points ψ (s) and ψ (t). Thus ⊂ 0 0 1 ψ [s,t] 1 P (ψ [s,t]) 1(J)= ψ (s) ψ (t) . H 0 ≥H 0 ≥H | 0 − 0 | In the last equality we used Theorem 2.18 and the fact that 1 = m in R. H 1 printed: 2015-2-19, 0.01 35 Corollary 2.35. Let Γ Rn be an arc and ψ : [0, (Γ)] Rn its parametrization with respect to ⊂ L → arc length. If E [0, (Γ)] is Lebesgue-measurable, then ψE Rn is 1-measurable and 1(ψE)= ⊂ L ⊂ H H m1(E). Remark 2.36. If Γ Rn is a rectifiable arc and ψ : [0, (Γ)] Rn its parametrization with ⊂ L → respect to arc length then ψ = (ψ1,...,ψn), where every ψi, i = 1,...,n, is 1-Lipschitz and thus absolutely continuous. Therefore ψ′(t) exists a.e. t [0, (Γ)] and further the derivative i ∈ L ψ′(t) = (ψ′ (t),...,ψ′ (t)) Rn 1 n ∈ exists for almost all t [0, (Γ)]. ∈ L By Theorem 2.34 every rectifiable arc is a 1-set, i.e. 0 < 1(Γ) < . Sets of this type also H ∞ include finite unions of rectifiable arcs and also countable unions Γ , if 1(Γ ) < . ∪i i i H i ∞ Are there 1-sets of essentially different character? P We approach this question in view of examples. 0 2 Example 2.37. We start with the closed unit square of the plane J0 = Q1 = [0, 1] . Let J1 1 be the union of four closed squares Qi , i = 1,..., 4 , in its corners, each with edge length 1/4. 1 Then each of the four squares Qi , i = 1,..., 4 of J1, will be replaced with four corner squares 2 n Qj , j = 1,..., 16, each with edge length 1/16. Continuing in this way in the step n we have 4 n n −n squares Qj , j = 1,..., 4 , each with egde length 4 . Let 4n n Jn = Qj j[=1 and ∞ J = Jn. n=0 \ Then J is a set of Cantor type. As a matter of fact J = C(1/4) C(1/4). × J0 J1 J2 1 1 0 Q1 Q2 Q1 1 1 2 Q3 Q4 Qj J3 3 Qi ······ What is the Hausdorff dimension dimH J of J? We find a suitable candidate for it using similarities. Observe that 4 J = J˜i, iG=1 36 Modern Real Analysis, Fall 2005 where J˜ is similar to J with the scaling factor 1/4, and hence s(J˜ ) = (1/4)s s(J) and further i H i H s(J) = 4(1/4)s s(J). H H If J is an s-set, then we get from the above consideration the condition 4(1/4)s = 1 n n which gives s = 1. Let us give some further details. Fix δ > 0. Observe first that d(Qj )= √2/4 . n If n N is so large that √2/4n δ, then Qn 4 is a δ-covering of J and thus ∈ ≤ { j }j=1 4n 1(J) d(Qn) 4n√2/4n = √2. Hδ ≤ j ≤ Xj=1 Therefore 1(J) √2 < . An argument similar to the proof of Theorem 2.28 one can show that H ≤ ∞ 1(J) > 0. Thus H dim (J)=1 and 0 < 1(J) < , H H ∞ in other words J is a 1-set. However, its geometric structure is very different from that of a rectifiable curve. Remark.: A positive lower bound can also be found by using an orthogonal projection P : R2 S → onto a line S with slope 2. Then the image set P (J) is a segment with length 3/√5. Because the − projection P is 1-Lipschitz, then 1(J) 1(P (J)) = 3/√5. As a matter of fact, it can be shown H ≥H that 1(J)= √2. H P (J) It was pointed out above that J = C(1/4) C(1/4). Observe that × dimH(J) = 1 = 2 log 2/ log 4 = dimH C(1/4) + dimH C(1/4). Example 2.38. Let q ,q ,... be those points of the closed unit disk D¯ = x R2 : x 1 whose 1 2 { ∈ | |≤ } both coordinates are rational numbers. These points form a countable dense subset of D¯ . Let ∞ E = Sj, j[=1 where S = x R2 : x q = 2−j . j { ∈ | − j| } Now ∞ 0 < 1(E) 2π 2−j = 2π < . H ≤ ∞ Xj=1 Thus E is a 1-set and dim E = 1. However, E is dense in D¯, E¯ D¯ = D¯ , and therefore E is “very H ∩ big”. In which sense does E resemble a rectifiable arc? printed: 2015-2-19, 0.01 37 These and other similar examples raise several questions: In which sense the Cantor-set of Example 2.37 is different from a rectifiable arc? • What kind of set is a general 1-set? Is there a way to distinguish between “Cantor-type” and • “rectifiable” parts and how these parts could be defined? Rectifiable arcs have tangent lines a.e. Does this property have a counterpart for sets such as • in Example 2.38? Definition 2.39. We say that an 1-measurable set E Rn is 1-rectifiable (or briefly rectifiable), H ⊂ if there exist countably many rectifiable arcs Γ , i N, s.t. i ∈ ∞ 1 E Γ = 0. H \ i i=1 [ A set F Rn is purely 1-unrectifiable (or briefly unrectifiable), if 1(F Γ) = 0 for every rectifiable ⊂ H ∩ arc Γ Rn. ⊂ The set E in Example 2.38 set is 1-rectifiable, whereas the set J in Example 2.37 is purely 1-unrectifiable. (HT). Remark 2.40. 1. If A is 1-rectifiable and E A is 1-measurable, then E is 1-rectifiable. ⊂ H 2. If the sets A , A ,..., are 1-rectifiable, then A is 1-rectifiable. 1 2 ∪j j 3. One can also define m-rectifiable sets for all m N, cf. Mattila’s book [Mat]. ∈ Theorem 2.41. Every 1-measurable set A Rn with 1(A) < , is of the form A = E F , H ⊂ H ∞ ∪ where E is 1-rectifiable and F is purely 1-unrectifiable. Moreover, E and F are unique up to a set of zero 1-measure. H Proof. HT 6/5. We next discuss geometric properties of rectifiable/unrectifiable sets. Recall Lebesgue’s density theorem from Real Analysis I: If E Leb(Rn), then ∈ m (E B(x, r)) lim n ∩ = 1 r→0+ mn(B(x, r)) for a.e. x E. It is a natural question whether Hausdorff measures have some similar properties. ∈ One can show that n(B¯(x, r)) = (2r)n, H for B(x, r) Rn, cf. e.g. [EG], [Mat]. Keeping this in mind we define: ⊂ Definition 2.42. Let 0 s < , A Rn and a Rn. The upper and lower s-densities at the ≤ ∞ ⊂ ∈ point a are s ¯ ∗s (A B(a, r)) Θ (A, a) = limsup H ∩ s , r→0+ (2r) s ¯ s (A B(a, r)) Θ∗(A, a) = lim inf H ∩ . r→0+ (2r)s s ∗s If Θ∗(A, a)=Θ (A, a), then this value is called the (s-dimensional) density of a set A at a and denoted by Θs(A, a). 38 Modern Real Analysis, Fall 2005 We study densities using covering theorems. Recall from Real Analysis I the following basic covering theorem and the notion of a Vitali covering of a set. If B is an open (closed) ball centered at x with radius r, then 5B is an open (closed) ball centered at x with radius 5r. Theorem 2.43 (Basic covering theorem). Let be an arbitrary family of balls of Rn s.t. F D = sup d(B): B < . { ∈ F} ∞ Then there exists a countable (possibly finite) family s.t. G ⊂ F B B = B ,B , B = B , i.e. the balls of are disjoint; and i ∩ j ∅ ∀ i j ∈ G i 6 j G B 5B . ⊂ B[∈F B[∈G Definition 2.44. Let be a family of balls in Rn. We say that is a Vitali covering of a set V V E Rn if for every x E and every ε> 0 there exists B s.t. x B and d(B) < ε. The family ⊂ ∈ ∈V ∈ is a closed (open) Vitali covering, if every B is closed (open) ball. V ∈V Theorem 2.45 (Vitali’s covering theorem for Hausdorff measures). Let 0 (2.46) d(B )s = i ∞ Xi=1 or (2.47) s E B = 0. H \ i i=1 [ Remark 2.48. Applying so called Besicovitch covering theorem we obtain a counterpart of Theo- rem 2.45 for a general Radon-measure of Rn(see Theorems ??, ??). Proof. We may supppose that 0 < d(B) < 1 for all B . We choose the balls inductively: ∈ V Let B be arbitrary. Suppose that disjoint balls B ,...,B have been chosen. Let 1 ∈V 1 m ∈V d = sup d(B): B , B B = 1 i m . m { ∈V ∩ i ∅ ∀ ≤ ≤ } If dm = 0, then it must be m E B , ⊂ i i[=1 and the claim is proven ((2.47) holds). Indeed, if there existed x E m B , then ∈ \∪i=1 i dist(x, m B ) > 0, ∪i=1 i because m B is compact. Because is a Vitali covering of E, there would exist B s.t. x B ∪i=1 i V ∈V ∈ and B m B = and therefore d > 0. ∩∪i=1 i ∅ m If d > 0, then choose B such that d(B ) > d /2. If this selection process will not m m+1 ∈V m+1 m end for any m, we obtain disjoint balls B ∞ . Therefore we must show: If { i}i=1 ⊂V ∞ d(B )s < , i ∞ Xi=1 printed: 2015-2-19, 0.01 39 then the condition (2.47) holds. For this purpose we show first that k ∞ (2.49) E B 5B k N. \ i ⊂ j ∀ ∈ i[=1 j=[k+1 Indeed, if k x E B , ∈ \ i i[=1 then x B˜ , where B˜ B = for all 1 j k. Because d(B ) 0, when m , ∈ ∈ V ∩ i ∅ ≤ ≤ m → → ∞ then d(B˜) > 2d(Bm+1) for some m. Then B˜ must intersect one of the sets Bk+1,...,Bm, because otherwise d d(B˜) > 2d(B ) > d . m ≥ m+1 m Let Bj be the first one of the balls Bk+1,...,Bm, which B˜ intersects. Then B˜ B = , d(B ) > d /2 d(B˜)/2 and k + 1 j m. ∩ j 6 ∅ j j−1 ≥ ≤ ≤ Thus B˜ 5B and (2.49) holds. ⊂ j B˜ Bj Finally let δ > 0. When k is large enough, then d(5B ) δ for all j k. Thus j ≤ ≥ ∞ k s E B s E B Hδ \ i ≤Hδ \ i i=1 i=1 [ ∞ [ s 5B ≤Hδ j j=[k+1 ∞ d(5B )s ≤ j j=Xk+1 ∞ = 5s d(B )s 0, j → j=Xk+1 when k . Thus →∞ ∞ s E B = 0 Hδ \ i i=1 [ for all δ > 0 and (2.47) holds. The next theorem is useful when studying the local properties of s-sets. Theorem 2.50. Let 0 (a) 2−s Θ∗s(A, a) 1 s-a.e. a A. ≤ ≤ H ∈ (b) If A is s-measurable, then Θ∗s(A, a) = 0 s-a.e. a Rn A . H H ∈ \ Proof. We first prove the left inequality of part (a). Observe first that ∞ x A:Θ∗s(A, x) < 2−s = x A: s(A B¯(x, r)) < (1 1/k)rs 0 Then s(E)+ ε s(A U) H ≥H ∩ s(A B ) ≥ H ∩ j Xj (2.51) t d(B )s ≥ j Xj t s(E B ) ≥ Hδ ∩∪j j t s(E), ≥ Hδ where the last inequality follows from (2.52) and the subadditivity of s. Letting ε 0 and δ 0, Hδ → → we obtain t s(E) s(E) < . Therefore s(E) = 0, because t> 1. H ≤H ∞ H Finally we prove (b): Let t> 0 and B = x Rn A:Θ∗s(A, x) >t { ∈ \ } and we prove that s(B) = 0. Fix δ > 0 and ε > 0. We apply part (b) of Theorem 1.24 to the H Borel-regular outer measure sxA (see Lemma 1.23). Because ( sxA)(B) = 0, then by 1.24 there H H exists an open U Rn s.t. B U and s(A U) < ε. For every x B there exists a radius ⊂ ⊂ H ∩ ∈ 0 < r(x) < δ s.t. B¯(x, r(x)) U and ⊂ s A B¯(x, r(x)) >t 2r(x) s. H ∩ From the basic covering theorem 2.43 it follows that there exist disjoint balls Bi = B¯(xi, r(xi)) s.t. B 5B . Thus ⊂∪i i t s (B) t d(5B )s H10δ ≤ i Xi s s = 5 t d(Bi) Xi 5s s(A B ) ≤ H ∩ i Xi 5s s(A U) ≤ H ∩ 5sε. ≤ Letting ε 0 we see that s (B) = 0, which implies further that s(B) = 0, when δ 0. → H10δ H → Corollary 2.53. Let A, B Rn be s-measurable s.t. B A and s(A) < . Then for s-a.e. ⊂ H ⊂ H ∞ H x B there holds: ∈ ∗s ∗s s s Θ (A, x)=Θ (B,x) and Θ∗(A, x)=Θ∗(B,x). Proof. s A B¯(x, r) s (A B) B¯(x, r) s B B¯(x, r) H ∩ H \ ∩ H ∩ s = s + s . (2r) (2r) (2r) r→0+ 0 Hs-a.e. x∈B −−−−→ | {z } 42 Modern Real Analysis, Fall 2005 Definition 2.54. We say that a point x E is a s-regular point of a set E Rn, if ∈ ⊂ ∗s s Θ (E,x)=Θ∗(E,x) = 1. If x E is not a s-regular point, then we say that it is an s-nonregular point of E. ∈ For what kind of sets s-a.e. points are s-regular? H Theorem 2.55. (a) If E Rn is 1-rectifiable and 1(E) < , then ⊂ H ∞ Θ1(E,x)=Θ∗1(E,x) = 1for 1-a.e. x E. ∗ H ∈ (b) If F R2 is purely 1-unrectifiable, then ⊂ Θ1(F,x) 3/4for 1-a.e. x F. ∗ ≤ H ∈ Proof. We skip the proof of part (b) (see Falconer’s book [Fa1, s. 40-44]) and we only prove part (a). 1. Let E = Γ be a rectifiable arc and ψ its parametric representation. Suppose that x = ψ(t ) Γ 0 ∈ is not an endpoint of Γ. Then for all small enough r > 0 there exists t1 1(ψ[t ,t ] B¯(x, r)) 1 P (ψ[t ,t ] B¯(x, r)) r. H 1 0 ∩ ≥H 1 0 ∩ ≥ ψ(t2) x ψ(t1) By considering, in the same way, the projection onto the line through x and ψ(t2), we observe that 1 Γ B¯(x, r) 2r H ∩ ≥ for all small enough r> 0. Thus Θ1(Γ,x ) 1, if xis not an endpoint of Γ. From part (a) of ∗ ≥ Theorem 2.50 it now follows that 1 Θ1(E,x) Θ∗1(E,x) 1 ≤ ∗ ≤ ≤ for 1-a.e. x E = Γ. H ∈ 2. If E Γ, where Γ is a rectifiable arc, then the claim of part (a) follows from what was proved ⊂ above and Corollary 2.53. printed: 2015-2-19, 0.01 43 3. In the general case the upper bound Θ∗1(E,x) 1 for 1-a.e. x E follows from part (a) of ≤ H ∈ Theorem 2.50. For this purpose we assume that 1(E) < . Because E is 1-rectifiable, then H ∞ E = E E , 0 ∪ j [j where 1(E )=0 and E = E Γ Γ , where Γ is a rectifiable arc. Thus 1-almost very H 0 j ∩ j ⊂ j j H point of E is a point of some Ej and therefore Θ1(E,x) Θ1(E ,x) 1 ∗ ≥ ∗ j ≥ for 1-a.e. x E . Hence Θ1(E,x) 1 for 1-a.e. x E. H ∈ j ∗ ≥ H ∈ 3 Compactness and convergence theorem for Radon measures of Rn 3.1 Riesz’ representation theorem In this section we only study Radon measures of Rn but from a new point of view: We shall study what kind of operations of continuous functions (”functionals”) correspond to integration with respect to Radon measures. This will give in the next sections natural compactness and convergence notions for the Radon measures of Rn. Write C (Rn)= f C(Rn): supp(f) Rn compact . 0 { ∈ ⊂ } Here C(Rn) is the space of continuous functions f : Rn R of Rn and → supp(f)= x Rn : f(x) = 0 { ∈ 6 } n n is the support of f. Because all elements of C0(R ) are bounded functions, we may equip C0(R ) with the norm f = sup f(x) . k k x∈Rn| | The next result is crucial for numerous applications. This theorem can be proved for every locally compact Hausdorff space (cf. Rudin [Ru]), but we consider here only the case of Rn, which is easier. Indeed, in this case we may apply the machinery of Chapter 1, in particular, metric outer measures. Definition 3.2. A mapping Λ: C (Rn) R is linear and positive functional, if 0 → (i) Λ(αf + βf )= αΛ(f )+ βΛ(f ) for all f ,f C (Rn) and all α, β R. 1 2 1 2 1 2 ∈ 0 ∈ (ii) Λ(f) 0 always, when f C (Rn) and f(x) 0 x Rn. ≥ ∈ 0 ≥ ∀ ∈ If µ is a Radon-measure in Rn, then n Λµ(f)= fdµ, f C0(R ) n ∈ ZR is a positive linear functional. Indeed f is integrable because f is bounded and of compact support and µ(supp(f)) < , because µ is a Radon-measure. ∞ It is a deep fact that also converse holds: 44 Modern Real Analysis, Fall 2005 Theorem 3.3 (Riesz’ representation theorem ). Let Λ: C (Rn) R be a positive and linear func- 0 → tional. Then there exists a unique Radon measure µ, more precisely, a measure space (Rn, Bor(Rn),µ) for which the following holds: Λ(f)= f(x) dµ(x) n ZR for all f C (Rn). ∈ 0 For the proof of Riesz’ representation theorem we need an auxiliary result. Lemma 3.4. (a) Let V Rn be open and K V compact. Then there exists f C (Rn) s.t. ⊂ ⊂ ∈ 0 supp(f) V and χ (x) f(x) 1 x Rn. ⊂ K ≤ ≤ ∀ ∈ (b) Let V Rn, j = 1,...,m, be open and K m V compact. Then there exists functions j ⊂ ⊂ j=1 j h C (Rn), with j ∈ 0 S m 0 h 1, supp(h ) V and χ h 1. ≤ j ≤ j ⊂ j K ≤ j ≤ Xj=1 Proof. (a) Write δ = dist(K, ∂V ). Because K is compact, it follows that δ > 0. Then the function 2 f(x) = max 0, 1 dist(x, K) − δ satisfies the conditions of part (a). (b) For every x K there exists a ball B(x, r ), with B(x, 2r ) V for some j. Because K is ∈ x x ⊂ j compact, it can be covered by finitely many such balls, i.e. k K B(x , r ). ⊂ i xi i[=1 Let A be the union of those closed balls B¯(x , r ) for which B(x , 2r ) V . Then j i xi i xi ⊂ j m A V and K A . j ⊂ j ⊂ j j[=1 By part (a) we choose functions g C (Rn) s.t. j ∈ 0 χ g 1 and supp(g ) V . Aj ≤ j ≤ j ⊂ j Then define h1 = g1, h = (1 g )g , 2 − 1 2 . . h = (1 g ) (1 g )g . m − 1 · · · − m−1 m printed: 2015-2-19, 0.01 45 Then clearly 0 h 1 and supp(h ) V . ≤ j ≤ j ⊂ j Induction with respect to m shows that m h = 1 (1 g ) (1 g ). Furthermore, j=1 j − − 1 · · · − m P m χ h 1, K ≤ j ≤ Xj=1 because if x K, then x A for some j and further 1 g (x) = 0 and then m h (x) = 1. ∈ ∈ j − j j=1 j P Proof of Riesz’ representation theorem. Define µ˜( ) = 0 ∅ and set µ˜(V ) = sup Λ(f): f C (Rn), 0 f 1 and supp(f) V { ∈ 0 ≤ ≤ ⊂ } for every open set V Rn. Then it follows from the definition that ⊂ (3.5) 0 µ˜(V ) µ˜(V ), ≤ 1 ≤ 2 if V ,V Rn are open and V V . 1 2 ⊂ 1 ⊂ 2 Next define (3.6)µ ˜(A) = inf µ˜(V ): A V Rn,V open { ⊂ ⊂ } for all A Rn and show thatµ ˜ is a metric outer measure. Then all Borel-sets of Rn areµ ˜-measurable ⊂ by Theorem 1.18. 1. Monotonicity µ˜(A ) µ˜(A ), if A A , 1 ≤ 2 1 ⊂ 2 follows directly from (3.5) and the definition (3.6). 2. We prove first subadditivity for open sets. In other words, if V Rn, j N, are open, then j ⊂ ∈ ∞ ∞ (3.7)µ ˜ V µ˜(V ). j ≤ j j=1 j=1 [ X To prove this let f C (Rn) s.t. 0 f 1 and supp(f) ∞ V . Because of the compact- ∈ 0 ≤ ≤ ⊂ j=1 j ness of supp(f) m S supp(f) V . ⊂ j j[=1 Write K = supp(f). Lemma 3.4, part (b), implies that there exist functions h C (Rn) with j ∈ 0 m 0 h 1, supp(h ) V and χ h 1. ≤ j ≤ j ⊂ j K ≤ j ≤ Xj=1 46 Modern Real Analysis, Fall 2005 Then m f = hjf, Xj=1 supp(h f) V and 0 h f 1 j = 1,...,m, j ⊂ j ≤ j ≤ ∀ and hence m m ∞ Λ(f)= Λ(h f) µ˜(V ) µ˜(V ). j ≤ j ≤ j Xj=1 Xj=1 Xj=1 Taking sup over all “admissible” functions f in the definition ofµ ˜ V we obtain (3.7). Let ∪j j then A Rn, j N, be arbitrary sets. Fix ε> 0 and choose open sets V Rn s.t. A V j ⊂ ∈ j ⊂ j ⊂ j and µ˜(V ) µ˜(A )+ ε/2j . j ≤ j Then ∞ ∞ A V , j ⊂ j j[=1 j[=1 and hence by monotonicity and (3.7) we obtain ∞ ∞ ∞ ∞ µ˜ A µ˜ V µ˜(V ) µ˜(A )+ ε, j ≤ j ≤ j ≤ j j=1 j=1 j=1 j=1 [ [ X X which implies subadditivity for all sets by letting ε 0. We have proved thatµ ˜ is an outer → measure. 3. Let V ,V Rn be open sets and dist(V ,V ) > 0. Let further f C (Rn) s.t. 0 f 1 1 2 ⊂ 1 2 j ∈ 0 ≤ j ≤ and supp(f ) V , j = 1, 2. Then 0 f + f 1 and supp(f + f ) V V , and hence j ⊂ j ≤ 1 2 ≤ 1 2 ⊂ 1 ∪ 2 Λ(f ) + Λ(f ) = Λ(f + f ) µ˜(V V ). 1 2 1 2 ≤ 1 ∪ 2 Taking sup over all admissible functions f1 and f2 we obtain (3.7) (3.8)µ ˜(V ) +µ ˜(V ) µ˜(V V ) µ˜(V ) +µ ˜(V ). 1 2 ≤ 1 ∪ 2 ≤ 1 2 Let then A , A Rn be arbitrary sets with dist(A , A ) > 0. Fix ε> 0 and choose an open 1 2 ⊂ 1 2 set V Rn s.t. A A V and ⊂ 1 ∪ 2 ⊂ µ˜(V ) µ˜(A A )+ ε. ≤ 1 ∪ 2 Choose then open sets V Rn, j = 1, 2, s.t. A V and dist(V ,V ) > 0. (We may choose j ⊂ j ⊂ j 1 2 for instance V = x Rn : dist(x, A ) < 1 dist(A , A ) .) Now A V V and j { ∈ j 3 1 2 } j ⊂ j ∩ dist(V V,V V ) > 0, 1 ∩ 2 ∩ and hence by (3.8) µ˜(A ) +µ ˜(A ) µ˜(V V ) +µ ˜(V V ) 1 2 ≤ 1 ∩ 2 ∩ =µ ˜(V (V V )) ∩ 1 ∪ 2 µ˜(V ) ≤ µ˜(A A )+ ε. ≤ 1 ∪ 2 Letting now ε 0 we see thatµ ˜ is a metric outer measure. → printed: 2015-2-19, 0.01 47 4. We prove next thatµ ˜ is locally finite: If B(x, r) Rn, then choose f C (Rn) s.t. 0 f 1 ⊂ 0 ∈ 0 ≤ 0 ≤ and χ f . B(x,r) ≤ 0 Then f f for all functions f admissible in the definition ofµ ˜ B(x, r) . Because Λ(f f) ≤ 0 0 − ≥ 0, then Λ(f) Λ(f ). ≤ 0 Taking sup over all such functions f we obtain µ˜ B(x, r) = sup Λ(f) Λ(f0) < . f ≤ ∞ Corollary 1.29 implies that µ =µ ˜ Bor(Rn) | is a Radon-measure. 5. We must still show that Λ(f)= f dµ n ZR for all f C (Rn). ∈ 0 Let f C (Rn). We may suppose that f 0, because f = f f , where f = max(0,f) ∈ 0 ≥ + − − + ∈ C (Rn) and f = max(0, f) C (Rn). Fix ε> 0 and set for all k N 0 − − ∈ 0 ∈ f (x) = max (k 1)ε, min(f(x), kε) (k 1)ε. k − − − f kε (k − 1)ε fk Clearly 0 f ε and f C (Rn) for all k. Because f 0, if (k 1)ε f , then ≤ k ≤ k ∈ 0 k ≡ − ≥ k k∞ m f = fk Xk=1 for some m N. Let K(k)= x Rn : f(x) kε , k N and K(0) = supp(f). Then ∈ { ∈ ≥ } ∈ (3.9) εχ f εχ , K(k) ≤ k ≤ K(k−1) and hence m m m (3.10) ε µ K(k) fk dµ = f dµ ε µ K(k 1) . ≤ Rn Rn ≤ − k=1 k=1 Z Z k=1 X X X 48 Modern Real Analysis, Fall 2005 On the other hand, if δ > 0, then by (3.9) 1 (1 + δ)f 1 ε k ≥ in some neighbourhood W of K(k). In particular, 1 (1 + δ)f g ε k ≥ for every function g C (Rn) admissible in the definition of µ(W ). Thus ∈ 0 Λ (1 + δ)f εµ(W ) εµ K(k) , k ≥ ≥ and further Λ(f ) εµ K(k) k ≥ letting δ 0. In the same way, f /ε is admissible in the definition of µ(V ) for every neigh- → k bourhood V of K(k 1), and hence Λ(f ) εµ(V ). Then by the definition − k ≤ Λ(f ) εµ K(k 1) . k ≤ − Combining these inequalities we obtain m m (3.11) εµ K(k) Λ(f) ε µ K(k 1) . ≤ ≤ − k=1 k=1 X X The inequalities (3.10) and (3.11) imply that m Λ(f) f dµ ε µ K(k 1) µ K(k) − Rn ≤ − − Z k=1 X = ε µ K(0) µ K(m) − εµ supp(f) . ≤ <∞ Letting ε 0 we see that | {z } → Λ(f)= f dµ n ZR and thus µ is the desired Radon-measure. 6. Finally, we prove the uniqueness of µ. Let µ1 also be a Radon-measure, for which Λ(f)= f dµ1 n ZR for all f C (Rn). Let V Rn be open and bounded. By Lemma 3.4 there exists a sequence ∈ 0 ⊂ f C (Rn) s.t. j ∈ 0 0 f (x) f (x) f (x) χ (x) ≤ 1 ≤ 2 ≤···≤ j → V printed: 2015-2-19, 0.01 49 for all x Rn. According to Monotone Convergence Theorem ∈ µ1(V ) = lim fj dµ1 j→∞ Z = lim Λ(fj) j→∞ = lim fj dµ j→∞ Z = µ(V ). n n Because Bor(R ) is a σ-algebra spanned by open and bounded sets, then µ1 = µ (Bor(R )). For a Radon-measure µ of Rn we define support, supp(µ), as the smallest closed set such that µ vanishes outside it. More precisely, supp(µ)= Rn V : V Rn open and µ(V ) = 0 . \ { ⊂ } [ Corollary 3.12. Let K Rn be compact and let Λ: C(K) R be a positive and linear functional. ⊂ → Then there exists a unique finite Borel-measure µ in the set K s.t. Λ(f)= f(x) dµ(x) ZK for all f C(K). ∈ Proof. (HT) 3.13 Weak convergence of measures Definition 3.14. Let µ , k N, be Radon-measures in Rn. We say that a sequence (µ ) converges k ∈ k weakly to a Radon-measure a µ, if lim f dµk = f dµ k→∞ n n ZR ZR all f C (Rn). This is denoted by µ ⇀µ or µ wµ. ∈ 0 k k → Example 3.15. (i) In R there holds δk ⇀ 0. (ii) Let µ = 1 δ + δ + + δ . Then for all f C (R) k k 1/k 2/k · · · k/k ∈ 0 k 1 1 k→∞ f dµk = f(j/k) f(x) dx, R k −−−→ 0 Z Xj=1 Z because the sums are Riemann’s sums of the function f on [0, 1]. Thus µk ⇀m1x[0, 1]. Easy examples (such as 3.15) show that it is not always true that µ (A) µ(A), if µ ⇀ µ. k → k However, there holds: Theorem 3.16. Let µ, µ , k N, be Radon-measures in Rn s.t. µ ⇀µ. Then k ∈ k (a) lim sup µ (K) µ(K), if K Rn is compact, k→∞ k ≤ ⊂ 50 Modern Real Analysis, Fall 2005 (b) lim inf µ (V ) µ(V ), if V Rn is open. k→∞ k ≥ ⊂ Proof. (a) Let K Rn be compact and V Rn be open s.t. K V . Choose by part (a) of Lemma 3.4 ⊂ ⊂ ⊂ a function f C (Rn) with χ f 1 and supp(f) V . Then ∈ 0 K ≤ ≤ ⊂ µ(V ) f dµ ≥ n ZR = lim f dµk k→∞ n ZR lim sup µk(K). ≥ k→∞ Because this holds for all open V K, the claim in part (a) is proven. ⊃ (b) If V Rn is open, then let K V compact. In the same way as above we obtain ⊂ ⊂ µ(K) lim inf µk(V ). ≤ k→∞ Because µ is a Radon-measure, µ(V ) = sup µ(K): K V compact , { ⊂ } and part (b) is proven. 3.17 Compactness of measures The weak convergence of measures is not merely natural but also a very useful notion. The families of bounded Radon-measures are sequentially compact. In many cases this is the only way to construct measures (as limiting measures of weakly convergent sequences). n Theorem 3.18. Let (µk) be a sequence of Radon-measures in R with sup µk(K) < k ∞ for all compact K Rn. Then there exists a subsequence (µ ) and a Radon-measure µ with ⊂ kj µkj ⇀ µ. For the proof of this theorem we need the following auxiliary result. Lemma 3.19. The norm space C (Rn), , where f = sup f(x) : x Rn , is separable, i.e. 0 k·k k k {| | ∈ } there is a countable dense set = f ∞ C (Rn). In other words, if f C (Rn) and ε > 0, F { j}j=1 ⊂ 0 ∈ 0 then f f < ε for some f . k − jk j ∈ F Proof. (HT) Proof of Theorem 3.18. Suppose first that n (3.20) sup µk(R )= A< . k ∞ printed: 2015-2-19, 0.01 51 Let f ∞ be a dense set in C (Rn). It follows from the assumption (3.20) that { j}j=1 0 f1 dµk : k N n ∈ ZR 1 is a bounded subset of R, and hence there is a subsequence (µk) of (µk) s.t. 1 k→∞ f1 dµk a1 n −−−→ ZR for some a R. Choose inductively for all j 2 a subsequence µj of the sequence µj−1 s.t. 1 ∈ ≥ { k} { k } j k→∞ fj dµk aj n −−−→ ZR for some a R. Then the diagonal sequence µk ∞ satisfies j ∈ { k}k=1 k (3.21) lim fj dµk = aj k→∞ n ZR for all j 1. Let L be the vector space spanned by the functions f , ≥ j m L = g = λ f : λ R, m N . j j j ∈ ∈ j=1 X Set Λ(g)= λjaj, Xj when g = λjfj. Xj Then by (3.21) we see that k Λ(g) = lim g dµk k→∞ n ZR for all g L. In particular, Λ is well defined, positive and linear functional in L. Moreover, it follows ∈ from (3.20) that (3.22) Λ(g) A g | |≤ k k for all g L. If f C (Rn) is arbitrary, then choose a sequence (h ), h L, s.t. ∈ ∈ 0 j j ∈ j→∞ f h 0, k − jk −−−→ and set Λ(f) = lim Λ(hj). j→∞ Then it follows from (3.22) that Λ is well defined in C (Rn) and (3.22) holds for all g C (Rn). 0 ∈ 0 Furthermore, Λ is a positive linear functional in C (Rn). In fact, if f 0 and f h 0, then 0 ≥ k − jk→ lim inf (min h ) 0, and hence j→∞ j ≥ k Λ(f) = lim lim hj dµk 0, j→∞ k→∞ n ≥ ZR 52 Modern Real Analysis, Fall 2005 because h dµk A min(0, min h ). By Riesz’ representation theorem there exists a Radon- j k ≥ j measure µ s.t. R Λ(f)= f dµ n ZR for all f C (Rn). We prove next that µk ⇀µ. Let ε> 0. For f C (Rn), we choose g L such ∈ 0 k ∈ 0 ∈ that f g ε . Then for large values of k k − k≤ 2A k k k Λ(f) f dµk Λ(f g) + Λ(g) g dµk + (g f) dµk − n ≤ | − | − n n − ZR ZR ZR ≤ε A f g + ε + A f g ≤ k − k | k −{z k } 2ε. ≤ k Therefore µk ⇀µ. Finally, we give up the hypothesis (3.20). From the assumption that sup µk(K) < k ∞ for all compact K Rn and the above argument there follows that for every m N there exists a ⊂ ∈ subsequence (µm) of (µ ) s.t. µm : k N µm−1 : k N and k k { k ∈ } ⊂ { k ∈ } mx m µk B(0,m) ⇀ν , where νm is a Radon-measure with supp(νm) B¯(0,m). Then for the diagonal sequence µk there ⊂ k holds kx m µk B(0,m) ⇀ν for all m N. Thus νmxB(0,ℓ)= νℓ, when ℓ m. Therefore we may define a Radon-measure µ in ∈ ≤ Rn setting µ(E)= ν1 E B(0, 1) + νm E B(0,m) B(0,m 1) , E Bor(Rn). ∩ ∩ \ − ∈ m≥2 X Because supp(f) B(0,m ), it follows that ⊂ 0 m0 k f dµ = f dν = lim f dµk, n n k→∞ n ZR ZR ZR k and hence µk ⇀µ. 4 On the Hausdorff dimension of fractals 4.1 The principle of mass distribution and Frostman’s lemma Finding a lower bound for the Hausdorff measure (and dimension) is, generally speaking, difficult. We give next a simple method, the so called principle of mass distribution, which often resolves this problem. Theorem 4.2. Let K Rn be compact. Suppose that there exists a Radon-measure µ in Rn s.t. ⊂ µ(K) > 0 printed: 2015-2-19, 0.01 53 and µ(C) c d(C)s ≤ 0 for all compact C Rn with d(C) δ , where s> 0 and δ > 0 are constants. Then ⊂ ≤ 0 0 dim (K) s. H ≥ Proof. If E ∞ is a δ-covering of K and 0 < δ δ , then { j}j=1 ≤ 0 ∞ ∞ ∞ ∞ 0 <µ(K) µ E¯ µ(E¯ ) c d(E¯ )s = c d(E )s. ≤ j ≤ j ≤ 0 j 0 j j=1 j=1 j=1 j=1 [ X X X Then s(K) µ(K)/c > 0 H ≥ 0 and therefore dim (K) s. H ≥ The above argument also proves the following: Theorem 4.3. Let X be a separable metric space and K X. Suppose that there exists an outer ⊂ measure µ in X s.t. µ(K) > 0 and µ(U) c d(U)s ≤ 0 for all U X with d(U) δ , where s> 0 and δ > 0 are constants. Then dim (K) s. ⊂ ≤ 0 0 H ≥ Example 4.4. Let C(1/3) be the Cantor 1/3-set, 2n Cn(1/3) = In,i i[=1 the approximation at the step n and hence ∞ C(1/3) = Cn(1/3). n=1 \ We next construct for C(1/3) a “natural” measure µ. Define a sequence of Radon-measures (µn) by setting 3n µ = m xC (1/3), n N. n 2n 1 n ∈ Then µ (I ) = 2−n i = 1,..., 2n; n n,i ∀ −n 2 n µnxIn,i = m1xIn,i i = 1,..., 2 ; m1(In,i) ∀ supp(µn)= Cn(1/3); µn Cn(1/3) = 1. The first equality implies that (4.5) µ (I ) = 2n−m 2−n = 2−m m n, i = 1,..., 2m. n m,i · ∀ ≤ ∀ 54 Modern Real Analysis, Fall 2005 n−m (Here 2 is the cardinality of the interval In,j contained in Im,i.) Therefore µn divides the measure “uniformly” to the intervals approximating the Cantor-set. Because µ (R) 1, we can by Theorem n ≡ 3.18 choose a subsequence (µnk ), which converges weakly to a Radon-measure µ, µ ⇀ µ, kun k . nk →∞ If I is an interval of step m, then let U R be such an open set that U C (1/3) = I . By m,i ⊂ ∩ m m,i Theorem 3.16 −m (4.5) 2 = limsup µnk (Im,i) µ(Im,i) k→∞ ≤ µ(U) lim inf µn (U) ≤ ≤ k→∞ k (4.5) = 2−m. Thus −m µ(Im,i) = 2 , and hence (4.5) holds also for µ and hence also µ is “uniformly distributed” at all steps n. We easily see that supp(µ)= C(1/3). If E R is closed s.t. ⊂ 3−n d(E) < 3−n+1, ≤ for some n N, then E can intersect at most two intervals I of step n. Therefore ∈ n,i µ(E) 2 µ(I ) = 2 2−n = 2 (3−n)s 2 d(E)s, ≤ n,i · · ≤ · where s = log 2/ log 3. Thus by Theorem 4.2 log 2 dim C(1/3) s = . H ≥ log 3 Combining these with (the easily proven) estimate s C(1/3) 1 one obtains dim C(1/3) = s. H ≤ H Similar arguments can be carried out also in other contexts. Before these we prove the following fundamental fact. Theorem 4.6 (Frostman’s lemma). Let K Rn be compact and s> 0. Then ⊂ s(K) > 0, H if and only if there exists a Radon-measure µ in Rn s.t. supp(µ) K, µ(K) > 0 and µ B(x, r) c rs x Rn, r> 0. ⊂ ≤ 0 ∀ ∈ ∀ Proof. The implication “ ” follows from Theorem 4.2. Indeed if C Rn is compact, then ⇐ ⊂ C B x, 2d(C) for all x C and thus ⊂ ∈ µ(C) c d(C)s. ≤ The proof of the implication “ ” is much more difficult. For this purpose one makes use of so ⇒ called “stopping time” arguments. These arguments are very useful in many different contexts, e.g. in harmonic analysis. printed: 2015-2-19, 0.01 55 We may suppose that (cf. Theorem 2.18 (c)) that K [0, 1)n and s(K) > 0. ⊂ H We call the cube [0, 1)n a dyadic cube of order 0 . Generally, dyadic cubes of order m are sets of the form j 1 j j 1 j j 1 j Q = 1 − , 1 2 − , 2 n − , n , 2m 2m × 2m 2m ×···× 2m 2m h h h where 1 j , j ,...,j 2m, j N. The edge length of cubes of order m is 2−m and these are ≤ 1 2 n ≤ i ∈ obtained by subdiving cubes of order m 1 into 2n cubes of equal size by halving the edges. Let − = Q: Q is a dyadic cube of order m in [0, 1)n . Dm { } Q1 ∈D1 Q2 ∈D2 Q ∈D0 For every cube Q , m N, there exists a unique cube Q˜ of order m 1 containing ∈ Dm ∈ ∈ Dm−1 − Q, (Q Q˜). ⊂ Q Q˜ We may suppose (by scaling if necessary) that K Q, if Q . 6⊂ ∈ D1 Next we make use of the assumption s(K) > 0. This implies that s(K) > 0 for some δ > 0 and H Hδ hence ∞ (4.7) d(Q )s a > 0, j ≥ 0 Xj=1 always when K is covered with dyadic cubes Q . For example, one may choose a to be j ∈∪mDm 0 a = min δs, s(K) . 0 { Hδ } For all m N we define a measure µm by setting µmxRn [0, 1)n = 0 and in every cube Q : ∈ m m \ ∈ Dm − m 2 ms µ xQ = mnxQ, if Q K = ; m mn(Q) ∩ 6 ∅ (4.8) µmxQ = 0, if Q K = . m ∩ ∅ Because Q: Q = [0, 1)n, we see that (4.8) defines a Radon-measure µm in Rn with ∪{ ∈ Dm} m supp(µm) [0, 1]n. Next we modify the measure µm by defining a measure µm setting m ⊂ m m−1 µm xRn [0, 1)n = 0 m−1 \ 56 Modern Real Analysis, Fall 2005 and in every cube Q : ∈ Dm−1 µm xQ = µmxQ, if µm(Q) 2−(m−1)s; m−1 m m ≤ (4.9) − − m 2 (m 1)s m m −(m−1)s µ xQ = m µ xQ, if µ (Q) > 2 . m−1 µm(Q) m m m m Continue in the same way; the measure µm−j−1 is obtained from the measure µm−j by setting in each cube Q : ∈ Dm−j−1 µm xQ = µm xQ, if µm (Q) 2−(m−j−1)s; m−j−1 m−j m−j ≤ (4.10) − − − m x 2 (m j 1)s m x m −(m−j−1)s µ Q = m µ Q, if µ (Q) > 2 . m−j−1 µm−j (Q) m−j m−j m m Finally we define µ = µ0 . The idea of the construction is that in none of the steps the measure of none of the dyadic cubes increases. Therefore (4.11) µm(Q) 2−js Q and j = 0, 1,...,m. ≤ ∀ ∈ Dj ∀ We next show that the measure µm(Rn) has a positive lower bound. We observe that by the construction for each x K there exists a dyadic cube Q , j = j(x), with x Q and ∈ ∈ Dj ∈ µm(Q) = 2−js (cf. (4.8) and (4.10)). For each x K we choose as large such a cube as possible ∈ Q = Q(x). Then (4.12) µm Q(x) = 2−js = c(n,s)d Q(x) s. Further, if x,x′ K, then either Q (x)=Q(x′) or Q(x) Q (x′)= . Thus ∈ ∩ ∅ ℓ Q(x)= Q K i ⊃ x[∈K iG=1 and (4.7) implies that ℓ d(Q )s a . i ≥ 0 Xi=1 m n Because the cubes Qi are disjoint, we obtain a lower bound for µ (R ) ℓ ℓ (4.13) µm(Rn)= µm(Q )= c(n,s) d(Q )s c(n,s)a > 0. i i ≥ 0 Xi=1 Xi=1 Now we can set 1 νm = µm. µm(Rn) Because νm(Rn) 1, it follows from Theorem 3.18 that there exists a subsequence (νmj ) and a ≡ Radon-measure µ s.t. νmj ⇀ µ, when j . →∞ We next show that µ is the desired measure. Firstly, because supp(νm) [0, 1]n and νm(Rn) 1, then the same holds for the limit measure ⊂ ≡ µ, see Theorem 3.16. printed: 2015-2-19, 0.01 57 Secondly, (4.8) implies that supp(νm) is contained in the c(n)2−m-neighborhood of K. Thus supp(µ) K, ⊂ because 2−mj 0. In particular, µ(K)= µ(Rn) = 1. → Let then x Rn and 0 3.16 µ B(x, r) lim inf νmj B(x, r) ≤ j→∞ 2n mj lim inf ν Qi ≤ j→∞ i=1 [ (4.11) (2−ℓ)s 2n lim inf ≤ j→∞ µmj (Rn) (4.13) c rs. ≤ 0 If r> 1/4, then µ B(x, r) µ(Rn) = 1 4srs. ≤ ≤ Therefore all the requirements of Frostman’s lemma for the measure µ are valid. Remark 4.14. 1. Clearly Frostman’s lemma holds for all -sets K Rn. Fσ ⊂ 2. Frostman’s lemma holds also for all Borel-sets K in Rn but in that case the proof is much more difficult. Corollary 4.15. Let K Rn compact. Then ⊂ dim (K) = sup s: probab.-measure µ, for which supp(µ) K and µ B(x, r) c rs r> 0, x Rn H { ∃ ⊂ ≤ 0 ∀ ∀ ∈ } 4.16 Selfsimilar fractals Let X be a metric space, = S X and ε> 0. The set ∅ 6 ⊂ S(ε)= x X : dist(x,S) < ε { ∈ } is an ε-neighborhood of S. In other words S(ε)= B(x, ε). x[∈S Set (ε)= . ∅ ∅ Definition 4.17. The Hausdorff-distance between the sets A X and B X, A = = B, is ⊂ ⊂ 6 ∅ 6 d (A, B) = inf r> 0: A B(r),B A(r) . H { ⊂ ⊂ } Set d ( , )=0 and d (A, )= d ( , A)= , if A = . H ∅ ∅ H ∅ H ∅ ∞ 6 ∅ For example, if A = 0 Rn and B = B(0, r) Rn, then d (A, B)= r. It is easily seen that { }⊂ ⊂ H “the triangle inequality” holds for all A,B,C X: ⊂ 58 Modern Real Analysis, Fall 2005 Lemma 4.18. If A,B,C X, then d (A, B) d (A, C)+ d (C,B). ⊂ H ≤ H H Proof. (HT) However, d is not necessarily a metric, because it is possible that d (A, B) = (e.g. X = H H ∞ R2, A is the line y = x and B is the line y = x), or d (A, B) = 0, although A = B (e.g. − H 6 X = R = A, B = Q). We denote the family of compact nonempty subsets of X by (X), i.e. C (X)= K : K X compact and nonempty . C { ⊂ } Then there holds: Lemma 4.19. If A, B X are closed and A = B, then d (A, B) > 0. In particular, d defines a ⊂ 6 H H metric in (X). C Proof. (HT) In what follows we only study the case of Rn. Lemma 4.20. The metric space ( (Rn), d ) is complete. C H Proof. Let (A ) be a Cauchy sequence in ( (Rn), d ). Denote j C H ∞ Bk = Ai i[=k and prove that ∞ (4.21) A = Bk k\=1 is the limit of the sequence (Aj) in the metric dH . Because (Aj) is a Cauchy-sequence, then dH (Ai, A1), i = 1, 2,..., is a bounded sequence. Then A A (r) i ⊂ 1 for some r> 0 and all i. Because A1(r) is compact as a bounded and closed set and B A (r) k ⊂ 1 is closed, then Bk is compact. In particular, A is compact and nonempty as the intersection of a decreasing sequence of compact sets, i.e. A (Rn). We must prove that ∈C d (A , A) 0, H i → when i . Let ε> 0 and i so large that →∞ 0 dH (Aj, Ak) < ε/2 for all j, k i . Then for all i k ≥ 0 ≥ A A (ε), i ⊂ j and hence ∞ A A (ε) i ⊂ j i[=k printed: 2015-2-19, 0.01 59 and B A (2ε). k ⊂ j In particular, A A (2ε), ⊂ j for j i . Conversely, if i i and x A , then by the definition of the Hausdorff-metric for ≥ 0 ≥ 0 ∈ i each j i there exists x A s.t. x x < ε. Then the sequence (x ) is bounded. Let y be its ≥ 0 j ∈ j | − j| j accumulation point. Because y B for all k, then y A. On the other hand x y ε, and hence ∈ k ∈ | − |≤ A A(2ε), i ⊂ for i i . we have proved that d (A , A) 2ε, for i i . ≥ 0 H i ≤ ≥ 0 Remark 4.22. In the case of a metric space X there holds (cf. e.g. [Ho3, Theorem 3.7, Lause 3.10]): 1. If X is a compact metric space, then ( (X), d ) is compact. C H 2. If (X)= A X : A closed, bounded and nonempty B { ⊂ } and X is complete, then ( (X), d ) is complete. B H Our goal is next to apply Banach’s fixed point theorem in the complete metric space ( (Rn), d ). C H Recall first this theorem from Topology (or see [Ho4, Theorem 0.8]). Before that we introduce the notation f ◦ m = f f f ◦ ◦···◦ m times for mappings f : Y Y . → | {z } Theorem 4.23. Let (X, d) be a complete metric space and f : X X a proper contraction, in → other words f is L-Lipschitz with the constant L< 1. Then f has a unique fixed point x X, i.e. 0 ∈ f(x0)= x0. Moreover ◦m x0 = lim f (y) m→∞ for all y X. ∈ Let now the mappings ψ : Rn Rn, j = 1,...,k, be proper contractions with the constant j → c0 < 1, i.e. ψ (x) ψ (y) c x y x,y Rn, j = 1,...,k. | j − j |≤ 0 | − | ∀ ∈ ∀ With the help of this family of contractions we define a mapping Ψ: (Rn) (Rn) by setting C →C k (4.24) Ψ(A)= ψjA, j[=1 when A Rn is compact. Because the mappings ψ are continuous, then each ψ A is compact and ⊂ j j thus Ψ(A) is compact (in other words Ψ(A) (Rn)). Because ∈C dH (Ψ(A), Ψ(B)) max dH (ψjA, ψjB) c0 dH (A, B) ≤ j ≤ 60 Modern Real Analysis, Fall 2005 for all A, B (Rn), it follows that Ψ is a proper contraction. Then there exists a unique compact ∈C set F , with k F = Ψ(F )= ψjF. j[=1 Furthermore d (F, Ψ◦ m(A)) 0, when m , no matter how the compact set A = is chosen! H → →∞ 6 ∅ The set F is called the invariant set corresponding to the family ψ k . { j }j=1 Example 4.25. If ψ (x) = 1 x and ψ (x) = 1 x + 2 , x R, then the invariant set corresponding 1 3 2 3 3 ∈ to the family ψ , ψ is Cantor’s 1/3-set C(1/3). Indeed, it is easily seen that { 1 2} Ψ(C(1/3)) = ψ (C(1/3)) ψ (C(1/3)) = C(1/3), 1 ∪ 2 and hence the claim follows from the uniqueness of fixed points. jne ... ◦ A = {0, 1} Ψ(A) Ψ 2(A) Consider then a family of contracting similarity mappings ψ : Rn Rn, i.e. j → ψ (x) ψ (y) = r x y x,y Rn, | j − j | j| − | ∀ ∈ and 0 < r < 1 j = 1,...,k. Then the invariant set F corresponding to this family ψ k is j ∀ { j}j=1 called a selfsimilar fractal. Let s> 0 be the unique number with k s (4.26) rj = 1. Xj=1 k 0 Such a unique number s exists because the sum is strictly decreasing as a function of s j=1 rj = k> 1 and k rs 0, when s 0. j=1 j → → P TheoremP 4.27. There exists a Radon-measure µ, with µ(F )= µ(Rn) = 1 and k s −1 (4.28) µ(A)= ri µ(ψi (A)) Xi=1 for all Borel-sets A Rn. ⊂ Proof. Observe first that raising the both sides of the equation (4.26) to any power m N we ∈ obtain (4.29) (r r r )s = 1. j1 j2 · · · jm 1≤j1X,...,jm≤k Fix x F and denote ∈ x = ψ ψ (x), j1···jm j1 ◦··· jm where m N and 1 j k. For every m N define a Radon-measure ∈ ≤ i ≤ ∈ s µm = (rj rj rj ) δx ··· . 1 2 · · · m j1 jm 1≤j1X,...,jm≤k printed: 2015-2-19, 0.01 61 Then equality (4.29) implies that µ (Rn) = 1 m N. m ∀ ∈ Because x Ψ◦ m( x ) Ψ◦ m(F )= F, j1···jm ∈ { } ⊂ it must hold that supp(µ ) F. By Theorem 3.18 the sequence (µ ) has a weakly convergent m ⊂ m subsequence, (4.30) µ ⇀µ, j , mj →∞ where µ is a Radon-measure in Rn. Moreover by Theorem 3.16 1 = limsup µmj (F ) µ(F ) µ(U) lim inf µmj (U) = 1, j→∞ ≤ ≤ ≤ j→∞ when U is an open neighborhood of F . Thus µ(F ) = 1 and supp(µ) F . ⊂ As a matter of fact, the whole sequence µ ⇀ µ, when m . To prove this it is by (4.30) m → ∞ enough to show that ∞ f dµm Rn m=1 Z is a Cauchy-sequence for each f C (Rn). Fix f C (Rn) and ε> 0. By uniform continuity there ∈ 0 ∈ 0 exists η > 0 s.t. (4.31) f(x) f(y) < ε, when x y < η. | − | | − | We note that (4.32) d ψ ψ (F ) (max r )md(F ) < η, j1 ◦···◦ jm ≤ j <1 when m m and m is large enough. If now ℓ>m| {zm}, then ≥ 0 0 ≥ 0 s s f dµℓ f dµm = (rj1 rjℓ ) f(xj1···jℓ ) (rj1 rjm ) f(xj1···jm ) Rn − Rn · · · − · · · Z Z j1,...,jℓ j1,...,jm X X = (r r )s f(x ) f(x ) , j1 · · · jℓ j1···jℓ − j1···jm j1,...,jℓ X where the latter equality holds because (r r )sf(x )= (r r )sf(x ) (r r r )s j1 · · · jℓ j1···jm j1 · · · jm j1···jm jm+1 jm+1 · · · jℓ j1X,...,jℓ j1X,...,jm jm+1X,...,jℓ =1 = (r r )sf(x ). j1 · · · jm j1···jm | {z } j1X,...,jm Because x = ψ ψ ψ ψ (x) ψ ψ (F ) j1···jℓ j1 ◦··· jm ◦ jm+1 ◦···◦ jℓ ∈ j1 ◦··· jm ∈F | {z } 62 Modern Real Analysis, Fall 2005 and x ψ ψ (F ), j1···jm ∈ j1 ◦··· jm then by (4.32) we obtain x x < η. | j1···jℓ − j1···jm | Therefore by (4.31) s f dµℓ f dµm ε (rj1 rjℓ ) = ε. Rn − Rn ≤ · · · Z Z j1,...,jℓ X We have proved that the whole sequence µm ⇀µ . Finally we observe that for all f C (Rn), ∈ 0 s f dµm = f(xj1···jm )(rj1 rjm ) Rn · · · Z j1X,...,jm k = f ψ (x )rs (r r )s ◦ i j2···jm i j2 · · · jm j2X,...,jmXi=1 k s = ri (f ψi) dµm−1. Rn ◦ Z Xi=1 Because the whole sequence µ ⇀µ, then this equality implies, letting m that m →∞ k s (4.33) f dµ = ri (f ψi) dµ. Rn Rn ◦ Z Z Xi=1 By approximation we see that (4.33) holds for the characteristic function χA of every Borel-set A Rn. Therefore ⊂ µ(A)= χA dµ n ZR k s = ri (χA ψi) dµ Rn ◦ Z Xi=1 k s − = ri χψ 1A dµ Rn i Xi=1 Z k s −1 = ri µ(ψi A) Xi=1 and the equality (4.28) is proven. 1 1 2 Example 4.34. Let F = C(1/3), ψ1(x) = 3 x and ψ2(x) = 3 x + 3 as in Example 4.25. Because ψ−1 C(1/3) [0, 1/3] = C(1/3) and ψ−1 C(1/3) [0, 1/3] [ 2, 1] R C(1/3), then the 1 ∩ 2 ∩ ⊂ − − ⊂ \ measure µ given by Theorem 4.27 satisfies µ C(1/3) [0, 1/3] = (1/3)sµ C(1/3) = 1/2. ∩ s s s Observe that (1/3) = 1/2, if s satisfies the condition (4.26) i.e. r1 + r2 = 1, rj = 1/3. Then s = dimH C(1/3). In the same way µ C(1/3) [2/3, 1] = 1/2. ∩ printed: 2015-2-19, 0.01 63 By iterating we see that µ is the “uniformly distributed” measure of Example 4.4. As a matter of fact one can show that µ = sxC(1/3). H The preceding example raises the question: If F is the selfsimilar fractal given by the family ψ k , is dim F determined as the solution of the equation { j}j=1 H k s rj =1? Xj=1 (Here rj is the scaling factor of ψj.) It is easily seen that the answer is no, if the sets ψjF overlap too much: Example 4.35. If ψ (x) = (1 ε)x and ψ (x) = (1 ε)x + ε, when x R and ε (0, 1/2], then 1 − 2 − ∈ ∈ [0, 1] = ψ [0, 1] ψ [0, 1] . 1 ∪ 2 [0,1−ε] [ε,1] Therefore [0, 1] is the selfsimilar fractal determined| {z } by| {zψ }, ψ . Now rs + rs = 2(1 ε)s = 1, if { 1 2} 1 2 − (and only if) log 1/2 s = . log(1 ε) − We see that s , when ε 0. →∞ → In order to exclude the situation in Example 4.35 we define the following notion. Definition 4.36. Similarities ψ k satisfy the open set condition, if there exists a bounded and { j}j=1 open set V Rn s.t. ⊂ k (4.37) Ψ(V )= ψ V V j ⊂ j[=1 and sets ψjV are disjoint, i.e. (4.38) ψ V ψ V = , for i = j. i ∩ j ∅ 6 Example 4.39. Let F = C(1/3) and ψ , ψ as in Example 4.34. By choosing (e.g.) V = (0, 1) { 1 2} we see that Ψ(V ) = (0, 1/3) (2/3, 1) V, ∪ ⊂ =ψ1V =ψ2V and hence ψ1, ψ2 satisfies the open set condition.| {z } | {z } { } Theorem 4.40. Let ψ k be a family of contracting similarities, which satisfy the open set { j}j=1 condition. Let F be the selfsimilar fractal determined by the family ψ k . If r is the scaling { j}j=1 j factor of ψj and k s (4.41) rj = 1, Xj=1 then 0 < s(F ) < . In other words F is an s-set, where s is determined by the condition (4.41). H ∞ In particular, dimH F = s. 64 Modern Real Analysis, Fall 2005 For the proof we need several auxiliary results. Lemma 4.42. Suppose that ψ k satisfies the open set condition wit the set V Rn. { j}j=1 ⊂ (a) If F is the self similar fractal determined by ψ k , then F V¯ . { j}j=1 ⊂ (b) Let = ψ ψ (V ): 1 j k,ℓ N . If U, U ′ and U U ′ = , then either V { j1 ◦···◦ jℓ ≤ i ≤ ∈ } ∈ V ∩ 6 ∅ U U ′ or U ′ U. ⊂ ⊂ Proof. (a) Condition (4.37) implies that Ψ(V¯ ) V¯ . ⊂ Therefore V¯ Ψ(V¯ ) Ψ◦ 2(V¯ ) Ψ◦ i(V¯ ) . ⊃ ⊃ ⊃···⊃ ⊃··· Because d (Ψ◦ i(V¯ ), F ) 0, we obtain F V¯ . H → ⊂ (b) Write V = ψ ψ ψ (V ). j1j2···jℓ j1 ◦ j2 ◦···◦ jℓ The condition (4.37) implies, in particular, that ψ V V for all j = 1,. . . ,k and thus j ⊂ V V . If m ℓ, then the previous observation implies j1j2···jℓ ⊂ ≥ (4.43) V = ψ ψ ψ ψ (V ) V . j1j2···jm j1 ◦···◦ jℓ jℓ+1 ◦···◦ jm ⊂ j1j2···jℓ ⊂V Therefore it is enough to show: If | {z } (4.44) V V = , j1j2···jℓ ∩ k1k2···kℓ 6 ∅ then j = k i = 1,...,ℓ. Indeed, if V V = for some m ℓ, then by (4.43) i i ∀ j1j2···jm ∩ k1k2···kℓ 6 ∅ ≥ also (4.44) holds. Suppose, on the contrary, that j = k for some i 1, 2,...,ℓ and let N i 6 i ∈ { } the least one of these indices i 1, 2,...,ℓ . Then ∈ { } ψ ψ = ψ ψ , j1 ◦···◦ jN−1 k1 ◦···◦ kN−1 V V = ψ ψ (ψ V ) and j1···jℓ ⊂ j1···jN j1 ◦···◦ jN−1 jN V V = ψ ψ (ψ V ), k1···kℓ ⊂ k1···kN k1 ◦···◦ kN−1 kN where ψ V ψ V = by condition (4.38). But this yields a contradiction, because this jN ∩ kN ∅ would mean V V = because of the injectivity of the mapping ψ ψ = j1···jℓ ∩ k1···kℓ ∅ j1 ◦···◦ jN−1 ψ ψ . k1 ◦···◦ kN−1 Lemma 4.45. Let V : j J be a collection of disjoint open subsets of Rn s.t. there exist positive { j ∈ } constants α, β and R and for every j J points x ,x′ with ∈ j j B(x , αR) V B(x′ , βR). j ⊂ j ⊂ j Then any ball B(x, R) with radius R intersects at most q sets V¯ , where q (1 + 2β)nα−n. j ≤ Proof. If B(x, R) V¯ = , then ∩ j 6 ∅ B(x , αR) V B x, (1 + 2β)R . j ⊂ j ⊂ printed: 2015-2-19, 0.01 65 αR βR xj ′ xj x Vj R (1 + 2β)R The claim now follows by comparing the volume of balls with radii αR and (1 + 2β)R (i.e. mn- measures) and making use of the disjointness of the sets Vj (and hence also the balls B(xj, αR) are disjoint). In the next lemma the mappings ψj need not be similarities. n Lemma 4.46. Let ψ1,...,ψk be proper contraction in R , ψ (x) ψ (y) r x y , x,y Rn, j = 1,...,k. | j − j |≤ j| − | ∀ ∈ ∀ k s If j=1 rj = 1, then s(F ) < , P H ∞ when F is the invariant set corresponding to the family ψ k . In particular, dim F s. { j}j=1 H ≤ Proof. Recall first that k (4.47) F = Ψ(F )= ψjF. j[=1 Let δ > 0. Denote F = ψ ψ (F ). j1···jℓ j1 ◦···◦ jℓ Let ℓ be so large that (max r , . . . , r )ℓd(F ) < δ. { 1 k} Then d(F ) r r d(F ) < δ, j1···jℓ ≤ j1 · · · jℓ and hence F : 1 j , j ,...,j k { j1···jℓ ≤ 1 2 ℓ ≤ } is a δ-covering of the set F because iterating the formula (4.47) we obtain k ◦ ℓ F =Ψ (F )= Fj1···jℓ . j1,...,j[ℓ=1 Thus k ℓ s(F ) (r r )sd(F )s = d(F )s rs = d(F )s. Hδ ≤ j1 · · · jℓ j 1≤j1,jX2,...,jℓ≤k Xj=1 Letting δ 0 we see that s(F ) d(F )s < . → H ≤ ∞ 66 Modern Real Analysis, Fall 2005 Proof of Theorem 4.40. Suppose that the open set condition holds for the family ψ k of { j}j=1 contracting similarities with an open bounded set V . Let F be the selfsimilar fractal corresponding to this family. By Lemma 4.46 it is enough to show that s(F ) > 0. H By Frostman’s lemma (Theorem 4.6) it is enough to show that the measure µ constructed in Theorem 4.27 satisfies the estimate (4.48) µ B(x, ̺) c ̺s ≤ for all x Rn and all ̺ > 0. Fix a ball B (x, ̺), where ̺ < 1. Write r = min r , . . . , r . Let ∈ min { 1 k} x V and α,β > 0 be such that 0 ∈ (4.49) B(x , α) V B(x , β). 0 ⊂ ⊂ 0 Let J = j , j ,... 1,...,k N be an infinite sequence. For each such sequence J there corres- { 1 2 } ∈ { } ponds the least index ℓ = ℓ(J) 1, with ≥ (4.50) r ̺ r r r < ̺. min ≤ j1 j2 · · · jℓ Let = (j ,...,j ): J 1,...,k N S { 1 ℓ(J) ∈ { } } be the set of the truncated sequences obtained in this way. By Lemma 4.42 part (b) (in fact, its proof) the sets V , (j , j ,...,j ) , j1j2···jℓ 1 2 ℓ ∈ S are then disjoint. Because the image of a ball with radius s under the similarity is a ball with radius ts, where t is the scaling factor of the similarity, then from the formula (4.49) it follows that each V contains a ball with radius αr r and is contained in a ball with radius βr r . On j1j2···jℓ j1 · · · jℓ j1 · · · jℓ the other hand, by (4.50) r α̺ αr r βr r β̺, min ≤ j1 · · · jℓ ≤ j1 · · · jℓ ≤ and hence from Lemma 4.45 it follows that the ball B(x, ̺) can intersect at most q of the sets V¯ , (j ,...,j ) , where q (1 + 2β)n(αr )−n. j1···jℓ 1 ℓ ∈ S ≤ min Next we show that for all Borel-sets A Rn ⊂ (4.51) µ(A)= µ (ψ ψ ψ )−1(A) (r r )s. j1 ◦ j2 ◦···◦ jℓ j1 · · · jℓ (j ,...,j )∈S 1 Xℓ By Theorem 4.27 k −1 s (4.52) µ(A)= µ(ψi A)ri . Xi=1 Next consider those indices i 1,...,k for which (i, j2,...,jℓ) for some j2,...,jℓ. Apply −1 ∈ { } ∈ S (4.52) with the set ψi A, to obtain µ(ψ−1A)rs = rsrsµ(ψ−1ψ−1A)= µ (ψ ψ )−1A (r r )s. i i i j j i i ◦ j i j j j X X printed: 2015-2-19, 0.01 67 Repeating this process we end up after a finite number of steps to the formula (4.51). Because supp(µ) F V¯ , then ⊂ ⊂ µ (ψ ψ )−1B(x, ̺) = 0, j1 ◦···◦ jℓ 6 only if (ψ ψ )(F ) B(x, ̺) = . j1 ◦···◦ jℓ ∩ 6 ∅ But then also V¯ B(x, ̺) = j1···jℓ ∩ 6 ∅ and by what was said above this can happen only for q sequences of indices and for these there holds by (4.50) r r ̺. j1 · · · jℓ ≤ Substituting this information into (4.51) we obtain µ B(x, ̺) = µ (ψ ψ ψ )−1B(x, ̺) (r r )s j1 ◦ j2 ◦···◦ jℓ j1 · · · jℓ (j ,...,j )∈S 1 Xℓ qµ(Rn)̺s ≤ s = c0̺ for all x Rn and all 0 < ̺ 1. If ̺> 1, then ∈ ≤ µ B(x, ̺) µ(Rn)̺s. ≤ Example 4.53. The Cantor set C(λ) is a selfsimilar fractal determined by the family ψ , ψ , { 1 2} ψ (x)= λx, ψ (x)= λx + 1 λ, x R . 1 2 − ∈ Now log(1/2) log 2 rs + rs = 2λs = 1 s = = . 1 2 ⇐⇒ log λ log(1/λ) Further ψ , ψ satisfies clearly the open set condition and hence { 1 2} log 2 dim C(λ)= . H log(1/λ) Theorem 4.40 covers the case of several known cases of fractal sets. Some of these are, in addition to Cantor sets, e.g. “the Sierpinski triangle” and “the von Koch curve” (“snowflake curve”). These will be studied in home work problems. 4.54 Drawing a selfsimilar fractal by way of random walk Drawing a fractal can be carried out in principle by fixing a point x0 and counting the itera- tions Ψ◦ m( x ), because d (F, Ψ◦ m( x )) 0, when m . However (in the worst case) { 0} H { 0} → → ∞ card Ψ◦ m( x )= km, where k is the number of similarities, and hence in practice the memory of { 0} a computer is easily exhausted. A better (more correct) way to draw is the following stochastic algorithm, which poses a minimal load on the memory of a computer. 68 Modern Real Analysis, Fall 2005 Algorithm 4.55. The invariant set F can be drawn as follows: 1. step Choose an initial point x F (e.g. a fixed point of some similarity which can be 0 ∈ easily found). 2. step: Suppose that the points x0,...,xℓ−1 have been constructed and drawn. Choose a new index j 1,...,k randomly (i.e. we cast a dice with k sides) and set ∈ { } xℓ = ψj(xℓ−1). Repeat the 2.nd step “without a limit”. Result: With probability 1 there holds: (4.56) lim dH ( x0,x1,...,xℓ , F ) = 0. ℓ→∞ { } We next wish to formulate and prove the preceding claim (4.56) precisely. The following result from stochastic is needed: Lemma 4.57. Let Ω= 1, 2,...,k N = 1, 2,...,k and X :Ω 1,...,k be the projection { } ℓ∈N{ } ℓ → { } X (ω)= ω , when Qω = (ω ,ω ,...,ω ,ω ,ω ,...) Ω. ℓ ℓ 1 2 ℓ−1 ℓ ℓ+1 ∈ Let p > 0, j = 1,...,k, s.t. k p = 1. Let be the σ-algebra in Ω spanned by j j=1 j F X−1(j)=P ω Ω: X (ω)= j , j 1,...,k , ℓ N . ℓ { ∈ ℓ } ∈ { } ∈ Then there exists a unique probability measure P in the measure space (Ω, ) s.t. F (4.58) P ω Ω: X (ω)= j , 1 ℓ m = p p ...p { ∈ ℓ ℓ ≤ ≤ } j1 j2 jm whenever m 1 and j , j ,...,j 1,...,k . ≥ 1 2 m ∈ { } Proof. We shall construct a product measure P in the product space Ω= 1,...,k N = 1,...,k { } { } N ℓY∈ s.t. P = µ µ µ , 1 × 2 ×···× ℓ ×··· where µ µ: ( 1,...,k ) [0, 1], µ( j )= p ℓ ≡ P { } → { } j for all 1 j k. When ≤ ≤ G = ω Ω: ω = j ,ω = j ,...,ω = j = (j , j ,...,j ) 1,...,k , m N, m { ∈ 1 1 2 2 m m} { 1 2 m }× { } ∈ ℓ>mY then set (cf. (4.58)) P(Gm)= pj1 pj2 ...pjm and P( ) = 0. Finite unions of the sets G , m N and form an algebra , to which P can ∅ m ∈ ∅ A be extended. One can show that then P: [0, 1] is σ-additive [cf. e.g. Williams: “Probability A → with Martingales”]. Thereafter the claim follows from the Carath´eodory-Hahn extension theorem 1.44. printed: 2015-2-19, 0.01 69 Remark 4.59. The condition (4.58) implies that for every finite subset J N there holds ⊂ (4.60) P ω Ω: X (ω)= j ℓ J = p . { ∈ ℓ ℓ ∀ ∈ } jℓ ℓ∈J Y In the language of probability theory: (Ω, , P) is “a probability space”, • F The measurable sets ( ) of Ω are “events”, • ∈ F According to Lemma 4.57 “the random variables” X are independent and identically distri- • ℓ buted, X is the outcome of the “ℓ:th casting of dice”, • ℓ The event A Ω (i.e. A ) “happens almost surely (a.s.)”, if P(A) = 1. • ⊂ ∈ F Example 4.61. Let Ω = 0, 1 N and p = p = 1/2. The mapping J :Ω [0, 1], { } 0 1 → ∞ −ℓ J(ω)= ωℓ2 , Xℓ=1 is a surjection and P(A)= m (JA). As a matter of fact, by Corollary 4.63 there exists a set B Ω 1 ⊂ s.t. P(B)=1 and J : B [0, 1] is a bijection. → The next lemma says that a random sequence ω of indices 1, 2,...,k as in Lemman 4.57 contains as a segment almost surely any finite sequence of indices 1,...,k. Lemma 4.62. Let (Ω, , P) be as in Lemma 4.57 and a = (a , . . . , a ) 1, 2,...,k m, m 1. F 1 m ∈ { } ≥ Then P ( ω = (ω )∞ Ω: (ω ,ω ,...,ω )= a for some j N ) = 1. { ℓ ℓ=1 ∈ j j+1 j+m−1 ∈ } Proof. We must prove that P(A ) = 1, when A Ω is the event a a ⊂ A = ω = (ω )∞ Ω: (ω ,ω ,...,ω )= a for some j N . a { ℓ ℓ=1 ∈ j j+1 j+m−1 ∈ } When r 1, then we write ≥ B = ω Ω: (ω ,ω ,...,ω ) = a . r { ∈ rm+1 rm+2 (r+1)m 6 } Then according to condition (4.58) the events Br, r = 1, 2,..., are “independent”, in other words there holds P(B B B )= P(B )P(B ) P(B ) 1 ∩ 2 ∩···∩ r 1 2 · · · r = (1 s)r, − where s = p p p (0, 1). a1 a2 · · · am ∈ Because Ac B B for all r 1, then a ⊂ 1 ∩···∩ r ≥ P(Ac ) P(B B ) = (1 s)r 0, a ≤ 1 ∩···∩ r − → when r . →∞ This easily implies a stronger result: 70 Modern Real Analysis, Fall 2005 Corollary 4.63. Let (Ω, , P) as in Lemma 4.57. Then almost surely the sequence ω Ω contains F ∈ every finite sequence a 1, 2,...,k m, m = 1, 2,..., as its segment. In other words, ∈ { } P( ω Ω: ω = (ω )∞ contains as its segment every { ∈ j j=1 finite sequence a 1,...,k m, m N ) = 1. ∈ { } ∀ ∈ } Proof. There are only countably many finite sequences a 1, 2,...,k m, m N. Therefore, ∈ { } ∈ if Aa is as above, then P A = P(Ω) P Ω A a − \ a a ! a ! \ \ = 1 P Ac − a a ! [ 1 P(Ac ) = 1, ≥ − a a X because by Lemma 4.62 there holds P(Ac ) = 0 for every finite sequence a 1,...,k m, m N. a ∈ { } ∈ Now we can prove that the algorithm described at the beginning of this section produces the fractal F . Theorem 4.64. The random algorithm 4.55 converges a.s. to the selfsimilar set F , i.e. the (??) is valid. ∞ More precisely: For each sequence ω = (ωℓ)ℓ=1 there corresponds a sequence of similarities (ψ )∞ and a set x : ℓ N F , jℓ ℓ=1 { ℓ ∈ }⊂ x = xω = ψ (xω ), ℓ 1, xω x . ℓ ℓ ωℓ ℓ−1 ≥ 0 ≡ 0 Then P( ω Ω: d ( x ,xω,xω,...,xω , F ) ℓ→∞ 0 ) = 1. { ∈ H { 0 1 2 ℓ } −−−→ } Proof. Because x F and ψ (y) F , if y F , then xω : ℓ N F for all ω Ω. Thus it 0 ∈ j ∈ ∈ { ℓ ∈ }⊂ ∈ is enough to show that (4.65) P( ω Ω: xω : ℓ N F dense ) = 1. { ∈ { ℓ ∈ }⊂ } By Corollary 4.63 we can consider only such sequences ω Ω, which contain as its segment all ∈ finite sequences. For the sequel, let ω be such a sequence. Let x F and ε> 0. Then ∈ x ψ ψ ψ ψ (x ) < ε/2 | − jℓ ◦ jℓ−1 ◦···◦ j2 ◦ j1 0 | for a suitable sequence of indices (j1, j2,...,jℓ) (HT). Because ω contains as its segment also this sequence, then there exists m N s.t. ∈ (ωm+1,ωm+2,...,ωm+ℓ) = (j1, j2,...,jℓ), and then xω = ψ ψ (xω ), xω F. m+ℓ jℓ ◦···◦ j1 m m ∈ printed: 2015-2-19, 0.01 71 Now x xω x ψ ψ (x ) + ψ ψ (x ) ψ ψ (xω ) | − m+ℓ| ≤ | − jℓ ◦···◦ j1 0 | | jℓ ◦···◦ j1 0 − jℓ ◦···◦ j1 m | ε/2+ d(ψ ψ (F )) ≤ jℓ ◦···◦ j1 ε/2 + (r )ℓd(F ) ≤ max < ε, for large enough ℓ (and independent of ω or m). Remark 4.66. (a) In the proof of Theorem 4.64 it was not required that the probability to choose the similarity ψ were the same for all j = 1,...,k (i.e. p 1/k that the algorithm j j ≡ produces a fractal F , no matter how these fixed probabilities have been chosen under the k constraints pj > 0 and j=1 pj = 1. On the other hand theP choice of pj influences on the speed of convergence of the algorithm to F . One can prove that the best speed of convergence is obtained when we choose k s s (4.67) pj = rj , where rj = 1 and rj is a scaling factor of ψj. Xj=1 (b) It is not important in the algorithm that x F , because the mappings ψ are proper 0 ∈ j contractions and therefore the influence of the choice of the initial points vanishes quickly. Then d ( x ,x ,x ,...,x , F ) can almost surely be made as small as we wish by H { m m+1 m+2 m+ℓ} choosing m,ℓ N in a suitable way. ∈ (c) Theorem 4.64 holds clearly for all contractions, not merely similarities. In the general case the choice of optimal probabilities (cf. (4.67)) is, however, not obvious. 4.68 Riesz’ capacity In this section we shall study another method of measuring the size of sets, so called Riesz’ capacity. Let s> 0 and µ be a non-trivial Radon-measure Rn. The s-energy of the measure µ is d(µ µ)(x,y) (4.69) Is(µ)= × s . n n x y ZR ×R | − | Theorem 4.70. Let A Rn be compact and s> 0. ⊂ (a) If there exists a finite non-trivial Radon-measure µ s.t. supp(µ) A and I (µ) < , then ⊂ s ∞ s(A)= . H ∞ (b) If t(A) > 0, then there exists a Radon-measure µ with supp(µ) A, µ(Rn) = 1 and H ⊂ I (µ) < for all 0 (a) We may suppose that µ(Rn) = µ(A) = 1. From the assumption and Fubini’s theorem it follows that dµ(y) s < µ-a.e. x A. n x y ∞ ∈ ZR | − | Dominated convergence theorem applied e.g. to the sequence of functions χB(x,1/k)(y) y , k N 7→ x y s ∈ | − | implies that for such a point x A there holds ∈ dµ(y) (4.71) lim = 0. r→0 x y s ZB(x,r) | − | Let ε> 0. When r is small enough, then by (4.71) dµ(y) ε. x y s ≤ ZB(x,r) | − | This yields the estimate 1 dµ(y) µ B(x, r) ε, rs ≤ x y s ≤ ZB(x,r) | − | or µ B(x, r) εrs r < δ . ≤ ∀ x Therefore there exists δ > 0 and a subset B A with µ(B) 1/2 and ⊂ ≥ (4.72) µ B(x, r) εrs, when r < δ and x B. ≤ ∈ The sets B = x A: µ B(x, r) εrs r< 1/m are closed and (extra HT), in particular m { ∈ ≤ ∀ } Borel-sets: Further B B , and hence m ⊂ m+1 lim µ(Bm)= µ Bm = 1. m→∞ m [ Choose then a δ/3-covering E for B s.t. B E = i and { i} ∩ i 6 ∅ ∀ ∞ d(E )s s (B) + 1. i ≤Hδ/3 Xi=1 For every i choose a point x B E and write r = d(E ). Then i ∈ ∩ i i i ∞ ∞ 1 (4.72) µ(B) µ B(x , 2r ) ε (2r )s 2 ≤ ≤ i i ≤ i i=1 i=1 X X ε2s s (B) + 1 ≤ Hδ/3 s s ε2 (A) + 1 . ≤ H Letting ε 0 we see that s(A )= . → H ∞ printed: 2015-2-19, 0.01 73 (b) If t(A) > 0, then by Frostman’s lemma there exists a Radon-measure µ of Rn s.t. µ(Rn)= H 1, supp(µ) A and ⊂ (4.73) µ B(x, r) c rt ≤ 0 for all x Rn and all r> 0. We next apply the formula from Real Analysis I ([Ho2, Lemma ∈ 3.15]): ∞ (4.74) f dµ = µ y Rn : f(y) > λ dλ, n { ∈ } ZR Z0 where f 0 is µ-measurable. Applying formula (4.74) to the function f(y) = x y −s we ≥ | − | obtain ∞ x y −s dµ(y)= µ y Rn : x y −s > λ dλ n | − | { ∈ | − | } ZR Z0 ∞ = µ B(x, λ−1/s) dλ (r = λ−1/s, r−s = λ) Z0 ∞ = s µ B(x, r) r−s−1 dr Z0 1 ∞ t−s−1 n −s−1 c0s r dr + s µ(R )r dr ≤ 0 1 c sZ Z = 0 + 1 < . t s ∞ − Finally integrating with respect to x we obtain c s I (µ) 0 + 1 < . s ≤ t s ∞ − When A Rn, then write ⊂ (A)= µ: µ Radon-measure Rn , supp(µ) A compact, µ(Rn) = 1 . M1 { ⊂ } Definition 4.75. Let A Rn and s> 0. The (Riesz) s-capacity of the set A is ⊂ 1 Cs(A) = sup : µ 1(A) . Is(µ) ∈ M Remark 4.76. 1. The s-capacity of the singleton x is zero for all s> 0 (HT). { 0} 2. The s-capacity of every bounded set is finite. 3. C (A) > 0 µ (A) s.t. I (µ) < . s ⇐⇒ ∃ ∈ M1 s ∞ Example 4.77. Let A = [0, 1] and s (0, 1). Then ∈ 1 dm(y) K < x y s ≤ s ∞ Z0 | − | for all x A. 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