Measure and Integration Summary

Transcription

1 Measure and Integration Summary Jacob Shapiro September 4, 2013 Abstract This is a (very rough) translation of Prof. Michael Struwe s German lecture Notes into English, as preparation for the final test (held in August 2013). Struwe followed rather closely the construction presented in the boo of Evans and Gariepy and also had some inspiration from Amann and Escher. Contents 1 σ -Algebras and Measures The Abstract Space D Measure E R σ-subadditivity R Monotonicity R D Measurability of a Set (Caratheodory) R E R D Algebra R E

2 CONTENTS CONTENTS T D Measure Spaces T Properties of µ on Σ σ-additivity Continuity from inside Continuity from outside E D µ-measure Zero Sets (Null Sets) T Sets of Measure Zero are Measurable Construction of Measures D Covering Classes E Different Covering Classes Of R n Trivial Covering Class T Forming Measures from Covering Classes E Generating the Measure defined in D Pre-measure σ-finite pre-measure R T Carathéodory-Hahn Extension Theorem T Uniqueness of Caratheodory Measure R

3 CONTENTS CONTENTS Overlapping Σ s Cannot Generalize Beyond A E to The Lebesgue Measure D Elementary Volume Premeasure n-dimensional Box (-cell) The Volume of an n-dimensional Box An elementary Set R R D The Lebesgue Measure L n T Open Sets are Countable Unions of -cells C Borel σ-algebra D Borel Measures T Approximation from Outside by Open Sets D Regular Borel Measure C The Lebesgue Measure is a regular Borel measure T Another Characterization of Measurability C Another Characterization of Measurability C Another Characterization of Measurability

4 CONTENTS CONTENTS Comparison with the Jordan Measure D The Jordan Measure T Comparison with Lebesgue-measure T Invariance for Rigid Transformations The Limits of Measurability E Vitali Sets Banach-Tarsi The Lebesgue-Stieltjes Measure D Distance between two sets D Metric Measure T Caratheodory Criterion for Borel Measures T The Lebesgue-Stieltjes Measure is a regular Borel measure T Lebesgue-Stieltjes measure on half-open intervals E Familiar Examples The Hausdorff Measure D The Hausdorff Measure T The Hausdorff Measure is a regular Borel measure on R n Step Step Step

5 CONTENTS CONTENTS T Preparation for Hausdroff Dimension E Compact Sets R Connection between Lebesgue and Hausdorff measures E Different Dimension Hausdorff Measure D Hausdorff Dimension E The Cantor Dust Radon Measures D The Radon Measure E T Approximation of Radon Measures by Open or Compact Sets Topological Definitions D Locally Compact Space D Semicontinuous Functions Lower Semincontinuous Upper Semicontinuous D Support D Continuous Functions with Compact Support Measurable Functions Definition and Elementary Characteristics

6 CONTENTS CONTENTS T Approximation with Step Functions R Monotone Convergence and Real Functions Lusin s Theorem and Egoroff s Theorem Egorov s (Egoroff) Theorem Lusin s Theorem Convergence in Measure D Convergence In Measure T Convergence Almost Everywhere Implies In Measure T Convergence in Measure gives a subsequence that Converges Almost Everywhere Integration Definition and Elementary Characteristics D σ-step Function D Integral for σ-step Functions Convergence Theorems Absolute continuity of the Integral Vitali s Theorem D Uniformly Integrable T Vitali s Theorem L p (Ω, µ) spaces Complex Measures (Rudin RCA Chapter 6) 47 6

7 CONTENTS CONTENTS 4.1 Total Variation D A Partition D A Complex Measure D Total Variation Measure of µ T µ is a positive measure on M T D Positive and Negative Variations Jordan Decomposition of µ Absolute Continuity σ-finite Measures The Lebesgue-Radon-Niodym Theorem Theorem Theorem Theorem The Hahn Decomposition Theorem Intergration on Product Measures (Rudin RCA Chapter 8) Measurability on Cartesian Products A Rectangle in X Y A Measurable Rectangle The Class of all Elementary Sets S T A monotone Class x-section and y-section

8 CONTENTS CONTENTS Theorem Theorem Theorem Product Measures Theorem The Product of Measures The Fubini Theorem The Fubini Theorem Counterexamples Completion of Product Measures Theorem Convolutions Theorem Convolutions Product Measures, Multiple Integrals Fubini s Theorem Convolution Differentiation (Rudin RCA) Derivatives of Measures Lemma Theorem Wea L The Maximal Function (Hardy-Littlewood) Lebesgue Points Example Theorem Lebesgue Point a.e Theorem The Radon-Niodym Derivative Nicely Shrining Sets

9 CONTENTS CONTENTS Theorem Theorem Metric Density Theorem Theorem Theorem Theorem The Fundamental Theorem of Calculus Counter Examples a b Denjoy and Perron Absolutely Continuous Functions Theorem Total Variation Function of a Function Theorem Theorem Theorem Differentiation of Measures Differentiability of the Lebesgue Integral Differentiation of Radon Measures Differentiation of Absolutely Continuous Measures on R Lebesgue Decomposition The Radon-Niodym Theorem

10 1 σ -ALGEBRAS AND MEASURES Conventional Notation The numbers in parenthesis after every heading are a reference to Struwe s own number codes, which I wanted to retain for convenience. C c (X) = { f C (X, C) : closure X ( f 1 (C\ {0}) ) is compact } D : definition R : remar P : proof E : example T : theorem A : application C : Corollary 1 σ -Algebras and Measures 1.1 The Abstract Space Let X be a set, and let 2 X be the power set of X D Measure (strw def ) A map µ : 2 X [0, ] is a measure on X, in case the following conditions hold: 1. µ ( ) = 0 2. A =1 A = µ (A) =1 µ (A ) E (strw ex ) 1. Trivial Measure Let X be a non-empty set. Define µ : 2 X [0, ] by { 0 A = µ (A) := 1 otherwise µ is a measure on X. 2. The Counting Measure µ (A) := A where A is the number of elements in A. 10

11 1.1 The Abstract Space 1 σ -ALGEBRAS AND MEASURES R σ-subadditivity (strw rem ) 2 implies in particular that µ ( =1 A ) =1 µ (A ), a characteristic of set functions that is called σ-subadditivity R Monotonicity (strw rem ) Claim:A B µ (A) µ (B). P : Choose A 1 := B and A = 2 in R (strw rem ) The external Jordan measure on R (denoted here by µ) is not a measure according to Consider the following set, A, defined by: A = Q [0, 1] = {q ; N} with the cover A = {q } N. Observe that A =1 A and µ (A) = 1 =1 µ (A ) = D Measurability of a Set (Caratheodory) (strw def ) A set A 2 X is called µ-measurable if B 2 X, µ (B) = µ (B A) + µ (B\A) R (strw rem ) Due to subadditivity of the measure µ and the fact that B B = ((B A) (B\A)), it suffices to show only: in order to satisfy µ (B) µ (B A) + µ (B\A) E (strw ex ) 1. Let X be a nonempty set a let µ be defined as in 1. Claim:A X is measurable iff A {, X}. 11

12 1.1 The Abstract Space 1 σ -ALGEBRAS AND MEASURES P Tae B := X. Then µ (B) = 1 because B = X is assumed to be nonempty. But for any A 2 X, X A = A, so in order for a set A to be measurable we need that 1 µ (A)+µ (X\A), which is the case only when A = or A = X so that at least one of the terms will be zero. QED 2. With respect to the the counting measure (2) every set is measurable. This is clear as for any two sets A, B, B = B A + B\A (where there is only one ind of infinity for the counting measure, by definition) R (strw rem ) Some authors denote a σ-subadditive set-function as an outer measure, whose restriction to a family of measurable sets is called simply measure D Algebra (strw def ) A 2 X is called an algebra in case the following holds: 1. X A 2. A A (X\A) A 3. If A i A i {1, 2,..., m} where m N, then ( m i=1 A i) A. A is called a σ-algebra if the last condition holds for a countable union rather than only for a finite union R (strw rem ) If A is an algebra (σ-algebra), using de Morgan s laws, we can show that also a finite (countable) intersection would lie in A: ( ) A = X\ (X\A ) E (strw ex ) =1 =1 1. Let X be a nonempty set. Then {, X} is an algebra. 2. For any set X, 2 X is a σ-algebra. 12

13 1.1 The Abstract Space 1 σ -ALGEBRAS AND MEASURES T (strw thrm ) If µ is a measure, then Σ := { A 2 X : A is µ measurable } is a σ-algebra. (A is considered to be µ-measurable in the sense of 1.1.6). P Step 1: X Σ: Tae any B 2 X. Since B X, B\X = and B X = B. But µ ( ) := 0. Step 2: (X\A) Σ A Σ: Tae any A Σ. Thus B 2 X, µ(b) = µ (B A) + µ (B\A). Observe that B A = B\ (X\A) and B\A = (X\A) B. Thus µ (B) = µ (B\ (X\A)) + µ (B (X\A)). Thus (X\A) Σ. Step : If A i Σ i {1, 2,..., m} where m N, then ( m i=1 A i) Σ: Apply induction on m. The case where m = 1: 1 i=1 A i = A 1 Σ. Assume the case for m 1 holds: Define S := m 1 i=1 A i. Thus we assume that S Σ. Verify the case for m: Using the notation for the previous case, we now also have A m Σ. We would lie to show next that (S A m ) Σ. Tae some B 2 X. We now that µ (B) = µ (B S) + µ (B\S) because S Σ. We also now that µ (B\S) = µ ((B\S) A m ) + µ ((B\S) \A m ), because A m Σ. Plugging the second equation into the first we get: µ (B) = µ (B S) + µ ((B\S) A m ) + µ ((B\S) \A m ). Observe that B (S A m ) = (B S) ((B\S) A m ). By subadditivity: µ (B (S A m )) µ (B S)+µ ((B\S) A m ). In addition, we could write (B\S) \A m = B\ (S A m ). Plugging this in we get: µ (B) µ (B (S A m ))+µ (B\ (S A m )). Thus (S A m ) Σ. Step 3: If A i Σ i N, then ( i N A i) Σ: Disjoint Decomposition: First we construct a mutually disjoint sequence of sets, {C i }, which will obey similar properties as the given {A i }. Note: This is a procedure we shall often carry out. 13

14 1.1 The Abstract Space 1 σ -ALGEBRAS AND MEASURES Thus, define: { A1 i = 1 C i := ( i 1 ) A i \ j=1 A j i N\ {1} Claim: C i C j = whenever i j. Proof: Tae i N, j N\ {i}, where WLOG i > j. Now assume the contrary: x (C i C j ). That is, x A i (x / A i 1) and x A j (x / A j 1). But j i 1 by assumption, so we get x / A j from the first assertion and x A j from the second assertion. contradiction. Claim: i N A i = i N C i Proof of : Tae some x 0 i N C i. i 0 N : x 0 C i0. x A i0. x i N A i. Proof of : Tae some x 0 i N A i. i 0 N : x 0 A i0. If x 0 / A j j {1, 2,..., i 0 1} then x 0 C i0. Otherwise x 0 C min({j {1,2,...,i0 1} x 0 A j}). In either case x 0 i N C i. Claim: C i Σ i N (given that A i Σ i N) Proof: Tae some i N. If i = 1 we are done. Otherwise, using the steps 2 and we now that Σ is closed under finite unions ( and complements. i 1 ) ( But C i = A i \ j=1 A j = A i Thus C i Σ. X\ i 1 j=1 A j So we can wor with the disjoint {C i } instead of {A i }. We shall now (finally) proceed to show that ( i N C i) Σ: Tae any B 2 X. Pic any n N. ) ( ( i 1 )) = X\ (X\A i ) j=1 A j. Claim (Addivitiy with any set):µ (B ( n i=1 C i)) = n i=1 µ (B C i). Proof: Since C n Σ, apply on it with the set B ( n i=1 C i): µ (B ( n i=1 C i)) = µ ((B ( n i=1 C i)) C n )+µ ((B ( n i=1 C i)) \C n ) Using the fact that {C i } are mutually disjoint we get: µ (B ( ( ( n i=1 C n 1 )) i)) = µ (B C n ) + µ B i=1 C i Carry out this procedure now n 1 times on µ until we reach the desired result. Since n i=1 C i Σ, using 1.1.6, ( B ( n 1 i=1 C i )), 14

15 1.1 The Abstract Space 1 σ -ALGEBRAS AND MEASURES QED µ (B) = µ (B ( n i=1 C i)) + µ (B\ ( n i=1 C i)). Using the above claim and the fact that B\ ( n i=1 C i) B\ ( i=1 C i), which allows us to use the monotonicity of µ, we arrive at: µ (B) n i=1 µ (B C i) + µ (B\ ( i=1 C i)) In the limit where n : µ (B) i=1 µ (B C i) + µ (B\ ( i=1 C i)) Using σ-subadditivity of µ we get: µ (B) µ ( i=1 B C i) + µ (B\ ( i=1 C i)) = µ (B ( i=1 C i)) + µ (B\ ( i=1 C i)) D Measure Spaces (strw def ) If µ is a measure on X and Σ := { A 2 X : A is µ measurable }, the triplet (X, Σ, µ) is called a measure space T Properties of µ on Σ (strw thrm ) Let A Σ N σ-additivity Claim: If A A m = m then µ ( =1 A ) = =1 µ (A ). P Using σ-subadditivity, µ ( =1 A ) =1 µ (A ). QED Using with B = X, we get µ ( n i=1 A i) = n i=1 µ (A i) for all n N. In the limit where n : ( n ) lim µ A i = µ (A i ) n i=1 But using monotonicity, µ ( i=1 A i) µ ( n i=1 A i) for all n N, thus µ ( i=1 A i) lim n µ ( n i=1 A i). i=1 15

16 1.1 The Abstract Space 1 σ -ALGEBRAS AND MEASURES Continuity from inside Claim: If A A +1 N then µ ( N A ) = lim µ (A ). P Define a disjoint decomposition of {A }: C 1 := A 1, C := A \A 1 N\ {1}. Note that =1 C = =1 A. Since {C } are disjoint, using the first part of this theorem, we now that µ ( =1 C ) = =1 µ (C ). Since A = ( j=1 C ) j by construction, µ (A ) = µ j=1 C j but again, ( ) since {C } are disjoint and using the first part we get µ j=1 C j = j=1 µ (C j). Thus, µ (A ) = j=1 µ (C j). In the limit where, lim µ (A ) = j=1 µ (C j) = µ ( =1 C ) = µ ( =1 A ). QED Continuity from outside Claim: If A A +1 N and µ (A 1 ) < then µ ( N A ) = lim µ (A ). P Define C := A 1 \A N, where = C 1 C 2..., and observe that µ (A 1 ) = µ (C ) + µ (A ) N. Thus in the limit of n, µ (A 1 ) lim µ (A ) = lim µ (C ). But {C } satisfies the conditions for , so: µ (A 1 ) lim µ (A ) = µ ( N C ) = µ ( N (A 1\A ) ) Using de Morgan s laws we get: µ ( N (A 1\A ) ) = µ ( A 1 \ ( N A )) But observe that (due to measurability of N A ): µ (A 1 ) = µ (( N A )) + µ ( A1 \ (( N A ))). Thus: µ ( A 1 \ ( N A )) = µ (A1 ) µ (( N A )). QED 16

17 1.2 Construction of Measures 1 σ -ALGEBRAS AND MEASURES E (strw ex ) Let X = N and let µ be the counting measure. Define A := {n N n } N. Observe that µ (A ) = N. Thus, lim µ (A ) =. However, N A =, so, µ ( N A ) = µ ( ) = 0. So clearly the assumption that µ (A 1 ) < is necessary D µ-measure Zero Sets (Null Sets) (strw def ) N 2 X is called a µ-measure zero set iff µ (N) = T Sets of Measure Zero are Measurable (strw thrm ) Claim: A µ-null set, N 2 X, is µ-measurable. P Tae any B 2 X. By monotonicity, µ (B N) + µ (B\N) µ (N) + µ (B) = 0 + µ (B). Thus µ (B N) + µ (B\N) µ (B). 1.2 Construction of Measures Let X be any set D Covering Classes (strw def ) K 2 X is called covering class of X in case: 1. K 2. {K j } K : X = j=1 K j E Different Covering Classes (strw ex ) Of R n The open n-dimensional-intervals, I := n =1 (a, b ) such that a b and a, b R for all N, form a covering class for R n, as well as the closed intervals, and the half-open intervals. 17

18 1.2 Construction of Measures 1 σ -ALGEBRAS AND MEASURES Trivial Covering Class Every algebra A 2 2X on a set X is a covering class of X because, X A T Forming Measures from Covering Classes (strw thrm ) Claim: If K is a covering class of X, and λ : K [0, ] such that λ ( ) = 0 then the following is a measure on X: P µ (A) := inf λ (K j ) K j K, A j=1 j=1 K j A 2 X First of all, since K, and, and λ ( ) = 0, µ ( ) = 0. Now, tae some A, A 2 X N and assume that A N A. { } By the definition of the infimum, N, ε > 0 K such that A j N K(),ε j j=1 ( λ Pic ε := 2 δ where δ > 0. { } Thus K (),2 δ j and: K (),ε j ) < µ (A ) + ε K such that A j N K(),2 δ j K (),ε j and j=1 K(),2 δ j < µ (A ) + 2 δ. Sum up m equations lie that where {1,..., m} and tae the limit as m : =1 j=1 ( λ K (),2 δ j ) µ (A ) + δ { } But K (),2 δ j is a sequence of sets in K whose countable union covers A, as assumed: A N A N =1 j N K (),2 δ j And so, µ (A), being an infimum over the set of all such countable covers, must be smaller than ) N j N (K λ (),2 δ j. Thus: 18

19 1.2 Construction of Measures 1 σ -ALGEBRAS AND MEASURES QED µ (A) µ (A ) + δ =1 Since we have not assumed anything about δ other than that it is positive, we might as well tae the limit as δ 0 and arrive at the desired result that µ fulfills E Generating the Measure defined in 1 (strw ex ) Let X be a nonempty set, and let K = {, X}. Let there be a map λ : K [0, ] given by { 0 A = λ (A) = 1 A = X A K Then the measure induced by λ according to is the same as the measure defined in 1. P Part 1:µ ( ) = 0 Since λ ( ) = 0 and K, by definition of the infimum µ ( ) = 0. Part 2: µ (A) = 1 A 2 X \ { } Tae some A 2 X \ { }. The only member of K which covers A has to be X. But λ (X) = 1. So we have an infimum on {1}, which must equal D Pre-measure (strw def ) If A 2 2X is an algebra, λ : A [0, ] is called a pre-measure in case: 1. λ ( ) = 0 2. λ ( =1 A ) = =1 λ (A ) where ( =1 A ), A A N and A A j = j. Note: Thus, =1 A is required to lie in A in addition to all of the A. This requirement has to be posed because A is merely an algebra and not a sigmaalgebra. 19

20 1.2 Construction of Measures 1 σ -ALGEBRAS AND MEASURES σ-finite pre-measure λ is called σ-finite in case a cover of X exists (X = =1 S ) such that S A and λ (S ) < for all N. WLOG we can tae the sets S to be disjoint, following the same construction in the proof of R (strw rem ) We may always assume that when taing the infimum we are dealing with a mutually disjoint cover, because if we define the disjoint decomposition to be as in , then we will always have that λ (C ) λ (A ) due to λ s sigma-additivity on A, of which C are also members of due to A being an algebra T Carathéodory-Hahn Extension Theorem (strw thrm ) Let λ : A [0, ] be a pre-measure on X, where A is an algebra, and let µ be defined by Claim: P where A is the covering class. 2. µ (A) = λ (A) A A. 3. A is µ-measurable A A. 1. By , since every algebra is itself a covering class, we may invoe on λ and A. 2. Tae any A A. By definition of the infimum, since A A, µ (A) λ (A). By the definition of the infimum, ε > 0 a cover of A in A let s call it {A ε } such that A N Aε and N λ (Aε ) < µ (A) + ε. By we may assume that {A ε } is a mutually disjoint cover. Define a new sequence of sets, C ε := Aε A N. Observe that {C ε} is mutually disjoint and that N Cε = A. In addition, since A ε = Cε (Aε \A), which is a disjoint union, using the additivity of λ we get that λ (A ε ) = λ (Cε ) + λ (Aε \A) so that λ (A ε ) λ (Cε ). Summing up those equations and going to the limit we get: N λ (Cε ) N λ (Aε ). 20

21 1.2 Construction of Measures 1 σ -ALGEBRAS AND MEASURES Using the fact again that λ is σ-additive and that N Cε = A is a disjoint union and that A A, we can write that N λ (Cε ) = λ ( N Cε ) = λ (A). So finally we arrive at λ (A) µ (A) + ε, ε > Following 1.1.7, we tae any B 2 X, and try to show that µ (B) µ (B A) + µ (B\A). By the definition of the infimum, ε > 0 a cover of B in A let s call it {B ε} such that B N Bε and N λ (Bε ) < µ (B) + ε. By we may assume that {B ε } is a mutually disjoint cover. But N, B ε = (Bε A) (Bε \A), (a disjoint union) and since B ε, A A, we conclude that λ (Bε ) = λ (Bε A) + λ (Bε \A). In addition, observe that B A N Bε A and B\A N Bε \A, so that by the definition of µ for these two sets, we get that µ (B A) N λ (Bε A) and µ (B\A) N λ (Bε \A). Putting everything together, we conclude that ε > 0: µ (B)+ε > N λ (Bε ) = N (λ (Bε A) + λ (Bε \A)) = N λ (Bε A)+ N λ (Bε \A) µ (B A) + µ (B\A) QED T Uniqueness of Caratheodory Measure (strw thrm ) Let X be a set and A be an algebra on it. Claim: If λ : A [0, ] is a σ-finite pre-measure, then the induced measure is unique for measurable sets (w.r.t. the induced measure). P Let µ be the induced measure by λ (accordng to 1), and let Σ be the σ-algebra of µ-measurable sets in X. Pic any measure ν : 2 X [0, ] such that ν (A) = λ (A) A A. 1. Claim: ν (A) µ (A) A Σ Proof: Tae some A 2 X. By the definition of the infimum, ε > 0 a cover of A in A let s call it {A ε } such that A N Aε and N λ (Aε ) < µ (A) + ε. But A ε A, so λ (Aε ) = ν (Aε ), so N λ (Aε ) = N ν (Aε ). But ν is a measure and A N Aε, so according to 2, N λ (Aε ) = N ν (Aε ) ν (A). 2. Claim: ν (A) µ (A) A Σ Proof: Tae some A Σ. 21

22 1.2 Construction of Measures 1 σ -ALGEBRAS AND MEASURES Case 1: S A such that A S and λ (S) <. By step 1, ν (S\A) µ (S\A). But by monotonicity, µ (S\A) µ (S). But on A, µ = λ, so µ (S) = λ (S), which is finite. Thus ν (S\A) <. By step 1 again, ν (A) + ν (S\A) µ (A) + µ (S\A), which is equal to µ (S) because A Σ and A S. But S A, so µ (S) = λ (S) = ν (S), which is smaller than or equal to (by sub-additivity) to: ν (A) + ν (S\A). Thus because the quantities are finite, equality must apply on all expressions involved. ν (A) + ν (S\A) = µ (A) + µ (S\A) Thus ν (A) µ (A) ν (S\A) µ (S\A) For which the only consistent possibility is ν (A) = µ (A). Case 2: Since λ is σ-finite, we now that {S }, a mutually disjoint sequence in A, such that X N S, and λ (S ) <. Define a disjoint decomposition of A = N A where A := A S N Tae some n N. Observe that ( n =1 A ) Σ because A Σ and A Σ. In addition, ( n =1 A ) ( n =1 S ) A (since finite intersections are in A) and λ ( n =1 S ) = n =1 λ (S ) < because {S } is disjoint. So we may apply Case 1 on n =1 A to get: ( n ) ( n ) ν A = µ A =1 =1 Due to monotonicity of ν: ( ) ( n ) ν (A) = ν A ν A Thus: =1 ( n ) ν (A) µ A =1 In the limit of n, we get: ν (A) lim n µ ( n =1 A ). Observe that n =1 A n+1 =1 A for all n N, so we may invoe to conclude lim n µ ( n =1 A ) = µ ( n=1 ( n =1 A )) = µ ( =1 A ) = µ (A). Thus finally ν (A) µ (A). =1 22

23 1.3 The Lebesgue Measure 1 σ -ALGEBRAS AND MEASURES QED R (strw rem 1.2.2) Overlapping Σ s It is not clear in wether the Σ of µ-measurable sets contains also all the ν-measurable sets Cannot Generalize Beyond A It is not possible to improve the generality of the uniqeness statement in E to (strw ex ) Let X = [0, 1] R, let A := {, X} and let a pre-measure be given as in 1.2.4, which generates 1. Claim: Then the set of measurable sets for µ, the induced measure, is Σ = {, X} = A. Proof: That both these sets are measurable is clear. That no other set is measurable is clear because for any non-empty set, 1 2. The Lebesgue measure will then not agree with µ. 1.3 The Lebesgue Measure D Elementary Volume Premeasure (strw def ) a 1 b n-dimensional Box (-cell) Tae a = a 2..., b = b 2... Rn. a n b n Define { n i=1 I := (a, b) := (a i, b i ) a i < b i i {1,..., n} otherwise Analogously, we define the n-dimensional closed boxes and half-open boxes: [a, b], (a, b], [a, b). For open-end points we allow a i = or b i =. 23

24 1.3 The Lebesgue Measure 1 σ -ALGEBRAS AND MEASURES The Volume of an n-dimensional Box Define a function vol : 2 Rn [0, ] by vol (I) := vol ((a, b)) := vol ([a, b]) := vol ([a, b)) := vol ([a, b)) := An elementary Set An elementary set is a finite union of disjoint -cells. Extend the definition of vol to such sets: ( m ) m vol I := vol (I ) R =1 1. The set of elementary sets in R n forms an algebra (but not a σ-algebra) denote it by E. 2. vol is additive. 3. vol is σ-finite. 4. vol is non-negative. 5. vol is monotone R vol is a pre-measure on E. TODO D The Lebesgue Measure L n (def ) The Lebesgue-measure L n is the Caratheodory-Hahn extension of vol. = T Open Sets are Countable Unions of -cells (strw thrm ) Let G Open (R n ). Claim: {I } E such that I I j j and G = N I Proof: TODO But according to 3 I are L n -measurable, and so according to G = N I is L n -measurable. { n i=1 b i a i a i < b i i {1,..., n} 0 otherwise 24

25 1.3 The Lebesgue Measure 1 σ -ALGEBRAS AND MEASURES C Borel σ-algebra The σ-algebra of L n -measurable sets contains the Borel σ-algebra (smallest σ-algebra containing all open sets) D Borel Measures A measure µ on R n is called Borel if every Borel set is µ-measurable. Thus can be restated as L n is a Borel measure T Approximation from Outside by Open Sets (strw thrm ) Let A 2 Rn. Claim: L n (A) = inf ({L n (G) G Open (R n ) G A}) Proof: Step 1: The result holds for any member of the set: indeed, due to monotonicity, for any G Open (R n ) such that G A, L n (A) L n (G). Thus the result must hold for the infimum of the set as well. Step 2: Approximate L n (A): δ > 0 { } I δ E such that N Iδ A and Ln (A)+δ > N vol ( I) δ. Construct another sequence of elementary sets, { J} δ such that J δ Open (R n ), J δ Iδ and vol ( ( ) J) δ < vol I δ + δ 2 for all N. Thus N vol ( ) J δ δ + N vol ( I) δ. By subadditivity, L ( n N J ) δ N vol ( ) J δ Thus, L ( n N J ) δ L n (A) + 2δ. Claim: inf ({L n (G) G Open (R n ) G A}) L ( n N J ) δ Proof: N J δ Open (Rn ) and N J δ N Iδ A. Thus in the limit of δ 0, inf ({L n (G) G Open (R n ) G A}) L n (A) D Regular Borel Measure (strw def ) If µ is a Borel measure on R n that is both inner regular and outer regular, it is called a regular Borel measure. According to Struwe: A borel measure is regular if A 2 Rn a Borel set B A such that µ (A) = µ (B). 25

26 1.3 The Lebesgue Measure 1 σ -ALGEBRAS AND MEASURES C The Lebesgue Measure is a regular Borel measure (strw cor ) P By 1.3.8, we can build a Borel measure composed of unions of open intervals that has a measure which converges to that of a given set T Another Characterization of Measurability (strw thrm ) A set A 2 Rn is L n -measurable iff ε > 0 G Open (R n ) : G A : L n (G\A) < ε. P TODO TODO C Another Characterization of Measurability (strw cor ) A set A 2 Rn is L n -measurable iff ε > 0 G Open (R n ), F Closed (R n ) : G A F : L n (G\A) < ε L n (A\F ) < ε. P TODO C Another Characterization of Measurability (strw cor ) A set A 2 Rn is L n -measurable iff ε > 0 G Open (R n ), F Closed (R n ) : G A F : L n (G\F ) < ε. P TODO Comparison with the Jordan Measure D The Jordan Measure A bounded set A R n is Jordanmeasurable in case µ J (A) = µ J (A) =: µ J (A), where µ J (A) := sup ({vol (E) E A, A E}) µ J (A) := inf ({vol (E) E A, A E}) where E is the algebra of elementary sets as defined in 1. 26

27 1.4 The Limits of Measurability 1 σ -ALGEBRAS AND MEASURES T Comparison with Lebesgue-measure (strw thrm 1.3.4) Let A R n be bounded. Then µ J (A) L n (A) µ J (A) and if A is Jordan-measurable then µ J (A) = L n (A). P TODO T Invariance for Rigid Transformations Claim: The Lebesgue measure is invariant under translations and orthogonal transformations. Proof: Let Φ : R n R n be defined by Φ (x) := x 0 + Rx for all x R n where x 0 R n and R is an orthogonal n n matrix. TODO Due to Jordan measure invariance under such transformations, and we show that Lebesgue measure is the Jordan measure for intervals, and we approximate open sets with intervals and then an arbitrary set with open sets. 1.4 The Limits of Measurability E Vitali Sets (strw ex 1.4.1) Let ξ R\Q and define G := { + lξ, l Z}, which is an additive group generated by 1 and ξ. For x, y R set x y x y G. Observe that defines an equivalence relation on R. Let X := {[x] = x + G x R}, which is the set of equivalence classes of. TODO Banach-Tarsi 1.5 The Lebesgue-Stieltjes Measure Let F be a monotonically { increasing, continuous from the left function. F (b) F (a) b a Define λ F ([a, b)) := a, b R. 0 otherwise Observe that {[ a, b )} is a covering class of R (but not an algebra on R). Define Λ F to be the measure induced by λ F and {[ a, b )} according to 1.2.3, that is: Λ F (A) := inf ({ =1 λ F ([a, b )) : A =1 [a, b )}) D Distance between two sets dist (A, B) := inf ({ a b a A, b B}) 27

28 1.5 The Lebesgue-Stieltjes Measure 1 σ -ALGEBRAS AND MEASURES D Metric Measure (strw def ) A measure µ on R n is metric in case A, B 2 Rn such that dist (A, B) > 0, µ (A B) = µ (A) + µ (B) T Caratheodory Criterion for Borel Measures (strw thrm ) A metric measure is a Borel measure. P Let µ be a metric measure on R n. 1. Step 1: If F Closed (R n ) then F is µ-measurable. Pic any B 2 Rn. According to 1.1.7, we would lie to show that µ (B) µ (B F ) + µ (B\F ). Assume µ (B) <, otherwise we are done. Define F := { } x R n dist ({x}, F ) 1 N. Observe that dist (B F, B\F ) 1 > 0 N, so by 1.5.2: µ ((B F ) (B\F )) = µ (B F ) + µ (B\F ). But µ is a measure, so by 1.1.4, Because B (B F ) (B\F ), µ ((B F ) (B\F )) µ (B). Finally we get that µ (B) µ (B F ) + µ (B\F ) N. Claim: lim µ (B\F ) = µ (B\F ) Proof: Define R := (F \F +1 ) B N. Observe that {R } are disjoint, and that B\F = (B\F ) ( l= R l) N. By 1.1.4, µ (B\F ) µ (B\F ). By 1.1.3, µ (B\F ) µ (B\F ) + l= µ (R l). Claim: l=1 µ (R l) < Proof: Observe that dist (R i, R j ) > 0 j i + 2. Since µ is metric, µ (R i R j ) = µ (R i ) + µ (R j ) j i + 2. Thus µ (R 1 R 3 ) = µ (R 1 ) + µ (R 3 ) Thus by induction: µ ( m =1 R 2) = m =1 µ (R 2) and µ ( m =1 R 2+1) = m =1 µ (R 2+1). But by 1.1.4, µ (B) µ ( m =1 R 2) and µ (B) µ ( m =1 R 2+1). So adding these two inequalities we get that 2µ (B) m =1 µ (R ). So that in the limit m, > 2µ (B) =1 µ (R ). Thus lim l= µ (R l) = 0. Thus lim µ (B\F ) µ (B\F ) lim µ (B\F ) and so finally µ (B\F ) = lim µ (B\F ). 28

29 1.5 The Lebesgue-Stieltjes Measure 1 σ -ALGEBRAS AND MEASURES 2. Step 2: Generalize from closed set to Borel sets. The set of measurable sets is a sigma-algebra, and the sigma-algebra of Borel sets is generated by closed sets T The Lebesgue-Stieltjes Measure is a regular Borel measure (strw thrm ) P Claim: The Lebesgue-Stieltjes measure is metric. Proof: Tae any A, B 2 R such that dist (A, B) > 0. To satisfy 1.5.2, we would lie to show that Λ F (A B) = Λ F (A)+Λ F (B). Follows immediately from Define δ := dist (A, B). Λ F (A B) := inf ({ =1 λ F ([a, b )) : A =1 [a, b )}) Approximate Λ F (A B): ε > 0 a cover {[ a ε, bε )} N of A B by half-open intervals such that N λ F ([ a ε, bε )) < Λ F (A B) + ε. Claim: WLOG b a < δ N Proof: If 0 N such that b ε 0 a ε 0 δ, define a sequence of points on R, {x i } m i=1, such that x 0 := a ε 0, x m := b ε 0 and 0 < x i x( i 1 < δ i {1,..., m}. Then λ F [ a ε 0, b ε 0 ) ) = F ( ) ( ) b ε 0 F a ε 0 = F (xm ) F (x 0 ) = F (x m ) F (x m 1 ) + F (x m 1 ) F (x 1 ) + F (x 1 ) F (x 0 ) = λ F ([ x m 1, x m )) + + λ F ([ x 0, x 1 )) = m i=1 λ F ([ x i 1, x i )) So we might as well redefine our cover of A B to be: {[ a ( ε, bε )} N {[ a ε, b ε )} N \ { [ a ε 0, b ε 0 ) }) {[ x i 1, x i )} i {1,...,m}. Repeat this procedure for every such 0. But recall what we defined δ to be: δ := dist (A, B). So that necessarily means that N, either [ a ε, bε ) A = or [ a ε, bε ) B =. Thus {[ aε, bε )} N breas down into two separate covers, one of A and the other of B. By the definition of the Lebesgue-Stieltjes measure as the infimum, it must be smaller than the sum on λ F on any cover. So we get: { ΛF (A) N such that [ a ε, bε ) A λ F ([ a ε, bε )) Λ F (B) N such that [ a ε, bε ) B λ F ([ a ε, bε )) Summing up these two inequalities, we get: Λ F (A) + Λ F (B) N λ F ([ a ε, bε )) < Λ F (A B) + ε 29

30 1.5 The Lebesgue-Stieltjes Measure 1 σ -ALGEBRAS AND MEASURES Thus: Λ F (A) + Λ F (B) Λ F (A B) + ε for all ε > 0. Thus Λ F (A) + Λ F (B) = Λ F (A B) and so the Lebesgue-Stieltjes measure is metric. Thus by 1.5.3, the Lebesgue-Stieltjes measure is Borel. Claim: The measure is Borel regular. Proof: Following 1.3.9, pic any A 2 R. If Λ F (A) =, then R would be a Borel set that has a Lebesgue-Stieltjes measure equal to that of A. Approximate the Lebesgue-Stieltjes measure of A: ε > 0 a cover {[ a ε, bε )} N of A by half-open intervals such that N λ F ([ a ε, bε )) < Λ F (A) + ε. In particular, we can pic ε := 1 j. Define B := j N ( N [ a 1 j, b 1 j ) Observe that B is a Borel set such that A B ) ( N [ a 1 j, b 1 j by 1.1.4, Λ F (A) Λ F (B) Λ F (( N [ a 1 j, b 1 j ) )) N λ F Λ F (A) + 1 j. ) ). Thus In the limit of j, Λ F (A) = Λ F (B) and thus the Lebesgue-Stieltjes measure is Borel regular T Lebesgue-Stieltjes measure on half-open intervals (strw thrm ) Despite the set of half-open intervals not being an algebra: a < b, Λ F ([ a, b )) = λ F ([ a, b )). P This direction stems immediately from the definitions combined with the fact that [ a, b ) [ a, b ). Approximate the Lebesgue-Stieltjes measure of [ a, b ): ε > 0 a cover {[ a ε, bε )} N of A by half-open intervals such that N λ F ([ a ε, bε )) < Λ F ([ a, b )) + ε. That is: ε > 0 a cover {[ a ε, bε )} N of A by half-open intervals such that N F (bε ) F (aε ) < Λ F ([ a, b )) + ε. 30 ( [ a 1 j, b 1 j ) ) <

31 1.5 The Lebesgue-Stieltjes Measure 1 σ -ALGEBRAS AND MEASURES QED F is continuous from the left (in particular, at b and at any a ). Thus: α > 0 δ (α, b) > 0 such that F (b) F (b δ (α)) < α. β > 0 γ (β, ) > 0 such that F (a ε ) F (aε γ (β, )) < β 2 for any N. WLOG we may assume that a ε γ (β, ) < bε 1 for all N, for, if that is not the case, we can always choose a finer cover in our approximation, and the result for λ F should be the same according to the calculation of the preceding proof: Observe that [a, b δ (α, b)] is compact. Observe that {(a ε γ (β, ), bε )} N is an open cover for [a, b δ (α, b)]. Thus a finite cover: {(a ε γ (β, ), bε )} {1,..., m} for some m N. Recall { that F is monotonically increasing, so: F (b δ (α)) F (b ε m) F (a ε 1 γ (β, 1)) F (a) Adding up these two inequalities we get: F (b δ (α)) F (a) F (b ε m) F (a ε 1 γ (β, 1)) = Using the fact that F is monotonically increasing and that we assumed a ε γ (β, ) < bε 1 for all N, we can eep adding and removing terms from the expression lie so: = F (b ε m) F (a ε m γ (β, m))+f (a ε m γ (β, m)) F (a ε 1 γ (β, 1)) F (b ε m) F (a ε m γ (β, m)) + F ( bm 1) ε F (a ε 1 γ (β, 1)) m =1 F (bε ) F (aε γ (β, )). Plugging in the results from F being left continuous we get: F (b) F (a) < F (b δ (α)) F (a)+α m =1 F (bε ) F (aε γ (β, ))+ α m ( =1 F (b ε ) F (a ε ) + β 2 ) +α ( =1 F (b ε ) F (a ε ) + β 2 ) + α = =1 F (bε ) F (aε ) + β + α = N λ F ([ a ε, bε )) + β + α < Λ F ([ a, b )) + ε + β + α Thus finally: λ F ([ a, b )) = F (b) F (a) < Λ F ([ a, b )) + ε + β + α for all ε > 0, α > 0, β > E Familiar Examples (ex ) 1. If F (x) := x x R the resulting Lebesgue-Stieltjes measure is the Lebesgue measure. { { 1 x (0, ) 2. If F (x) := 0 x (, 0 ] x R then λ 1 x [ a, b ) F ([a, b)) := 0 otherwise a, b R and the resulting Lebesgue-Stieltjes measure is the Dirac measure, which 31

32 1.6 The Hausdorff Measure 1 σ -ALGEBRAS AND MEASURES would be just the indicator function of a set at 0: Λ F (A) = χ A (0) A 2 Rn. 1.6 The Hausdorff Measure Let s > 0, δ > 0, A 2 Rn. Define ({ }) Hδ s (A) := inf r s A B r (x ) r < δ =1 Note that in the definition, we are intetionally unspecific about where the open balls are located {x }. They can be somewhere, just as long as the above conditions are met. The infimum is taen over all such sets of open balls. Claim: Hδ s is a measure on Rn. Proof: The open balls in R n form a covering class: since we can cover R n by a countable number of open balls, and we can tae our class to include. B r (x ) r s is a non-negative set function, and so by Hs δ is a measure on R n. Claim: Hδ s 1 (A) Hδ s 2 (A) whenever δ 1 δ 2 and for all A 2 Rn and s > 0. Proof: Hδ s 1 (A) =1 rs for all {B r (x )} such that A =1 B r (x ) r < δ 1. On the other hand, ε > 0, { B ρ ε (y ) } such that A =1 B ρ ε (y ) ρ ε < δ 2 and =1 (ρε )s < Hδ s 2 (A) + ε. Observe that δ 2 δ 1, so that actually ρ ε < δ 2 δ 1, and { B ρ ε (y ) } is also a cover in the set of covers over which the infimum would be invoed to compute Hδ s 1 (A). Thus we must have Hδ s 1 (A) =1 (ρε )s < Hδ s 2 (A) + ε for all ε > 0. As a result we conclude that Hδ s is monotonically decreasing in δ, and so the limit lim δ 0 + Hδ s (A) = sup ({Hs δ (A) δ > 0}) must exist (though it may be ) A 2 Rn D The Hausdorff Measure (def ) For any A 2 Rn and s 0 define the s-dimensional Hausdorff Measure to be { H s lim (A) := δ 0 + Hδ s (A) s > 0 A s = 0 = T The Hausdorff Measure is a regular Borel measure on R n (thrm 1.6.1) P 32

33 1.6 The Hausdorff Measure 1 σ -ALGEBRAS AND MEASURES Step 1 Claim: H s is a measure. Proof: If s = 0 then H s is the counting measure, which we have already observed in 2 to be a measure. If s > 0: Following 1.1.1, If A =, then clearly (as we included in our covering class of open balls) Hδ s ( ) = 0 and so we get Hs ( ) = 0. This satisfies 1. Next, assume that {A } R n is a countable sequence of sets such that A N A. Because Hδ s is a measure, we now that Hs δ (A) N Hs δ (A ). On the other hand, because H s is the supremum over all Hδ s with δ > 0, Hδ s (A ) H s (A ) for all N, N Hs δ (A ) N Hs (A ). So that Hδ s (A) N Hs (A ), δ > 0. Thus in the limit of δ 0 + we get Step 2 Claim: H s is a metric measure. Proof: Tae any A, B 2 Rn such that dist (A, B) > 0. If s = 0 then H s is the counting measure, which is evidently metric. If s > 0: Since we just showed that H s is a measure, we now by that H s (A B) H s (A) + H s (B). Firstly, we now that δ > 0, Hδ s (A B) Hs (A B). Next, given any δ > 0, ε > 0 { B r ε (x ε )} such that A B N B r ε (xε ), r ε < δ and N (rε )s < Hδ s (A B) + ε. Thus given any δ > 0, ε > 0 { B r ε (x ε )} such that A B N B r ε (xε ), r ε < δ and N (rε )s < H s (A B) + ε. What if we pic some δ such that 0 < δ < dist (A, B)? Then our cover { B r ε (x ε )} decomposes into two disjoint covers, one which covers only A and another which covers only B: A N A B r ε (x ε ) and B N B B r ε (x ε ) where N A := { N B r ε (x ε ) A } and similarly for N B. So that Hδ s (A) N A (r ε)s and Hδ s (B) N B (r ε)s. Thus: Hδ s (A) + Hs δ (B) Hs (A B) + ε for all 0 < δ < dist (A, B) and ε > 0. In the limit ε 0 and δ 0 we get the desired result. Thus by H s is Borel Step 3 Claim: H s is Borel regular. Proof: Tae any A 2 Rn. We need to find a Borel set, B, such that B A and H s (A) = H s (B). 33

34 1.6 The Hausdorff Measure 1 σ -ALGEBRAS AND MEASURES If s = 0, and A is infinite we can just tae the whole space, which is a Borel set. If A is finite then it is itself a Borel set. If s > 0: Assume that H s (A) <, otherwise we can just tae the Borel set to be the whole space. Given any δ > 0, ε > 0 { B r ε (x ε )} such that A N B r ε (xε ), rε < δ and N (rε )s < Hδ s (A) + ε < Hs (A) + ε. Pic ε := 1 j. Define our Borel set to be B := ( j N N B r 1/j ( Observe that B is a Borel set such that A B N B r 1/j ( x 1/j ( x 1/j ))) )). ( x 1/j )). Thus ( ) s < (( by 1.1.4, Hδ s (A) Hs δ (B) Hs δ N B r 1/j N H s (A) + 1 j. In the limit of j and δ 0, H s (A) = H s (B) and thus the Hausdorff measure is Borel regular T Preparation for Hausdroff Dimension (lemm ) Let A 2 Rn and 0 s < t <. Then: 1. H s (A) < H t (A) = 0 P Assume that H s (A) <. Thus Hδ s (A) < for any δ > 0. Pic some δ > 0. ε > 0 { B r ε (x ε )} such that A N B r ε (xε ), rε < δ and N (rε )s < H s δ (A) + ε < Hs (A) + ε. In addition as { B r ε (x ε )} is a bona fida cover of A with r ε < δ, we now as well that: H t δ (A) N (rε )t But observe that (r ε )t = (r ε )t s (r ε )s < δ t s (r ε )s. Thus N (rε )t δ t s N (rε )s. Thus H t δ (A) δt s (H s (A) + ε) for all δ > 0, and for all ε > 0. r 1/j In the limit δ 0, as H s (A) < and ε can be chosen as we wish, we get H t (A) = H t (A) > 0 H s (A) = P This is the logical negation of the previous claim. 34

35 1.6 The Hausdorff Measure 1 σ -ALGEBRAS AND MEASURES Note: This theorem says that the range of values H s (A) can tae for a given A is basically only three: for s = 0, some finite value at some s 0 [0, ), and then 0 for all s > s E Compact Sets (ex ) Let Q := [ 1, 1] n R n. 2 n L n (Q) H n (Q) 2 n n n/2 L n (Q) 1 H n (Q) n n/2 P By its definition, Hδ n (Q) N (r ) n for all {r } such that r < δ N and Q N B r (x ) for some {x }. Define the following cover: Define m δ := log 2 ( n Thus m δ > log 2 ( n 2δ 2δ ) + 1. ). Thus 2 m δ 1 n < δ. Next define a cover of 2 (m δ+1)n balls, each of radius r := 2 m δ 1 n. If we divided Q into n-dimensional cubes, each of side-length 2 m δ, it is clear that the cover we just defined will have each cube contained in each ball. ThusH n δ (Q) N (r ) n = 2 (m δ+1)n r n = n n/2. The claim follows from the fact that L n (Q) = 2 +n. Define ω n := L n (B 1 (0)). For example, ω 3 = 4π 3. Thus Ln (B r (x)) = ω n r n. Thus by the σ-subadditivity of L n, for each cover of B 1 (0) with {B r (x )}: L n (B 1 (0)) N Ln (B r (x )) = ω n N rn Thus we get the relation: 1 N rn. Going on to the Hausdorff measure, ε > 0 { B r ε (x ε )} such that B 1 (0) N B r ε (xε ), rε < δ and N (rε )n < H n δ (B 1 (0)) + ε. Thus we get the relation H s δ (B 1 (0)) 1. At any rate, since H n δ is monotone and Q B 1 (0), H n δ (Q) Hn δ (B 1 (0)). Thus we get our result in the limit of δ R Connection between Lebesgue and Hausdorff measures (rem ) In fact A 2 Rn which are Lebesgue-measurable, L n (A) = ω n H n (A). 35

36 1.6 The Hausdorff Measure 1 σ -ALGEBRAS AND MEASURES E Different Dimension Hausdorff Measure (ex ) A 2 Rn and s > n, H s (A) = 0. P Cover R n with n-dimensional cubes Q l. By 1.6.3, each cube has H n (Q l ) n n/2 <. Thus for each cube by 1, H s (Q l ) = 0. Thus using , H s (R n ) = 0. Thus by monotonicity, H s (A) = D Hausdorff Dimension (def ) Define dim H (A) := inf ({s R s > 0 H s (A) = 0}) to be the Hausdorff dimension of A 2 Rn E The Cantor Dust (ex ) (Amann-Escher-AnalysisIII-2009-English-Ed-pg. 39) Define A 0 := [0, 1] 2 R 2. Define A 1 by dividing A 0 into 16 identical squares (labelling the top left one 1 all the way to the bottom right one 16), and removing from it 12 open squares so that in the end only squares 2, 5, 12, 15 are left (as closed squares). Define A 2 is a similar fashion within each of the four remaining squares of A 1 (so that it would have in the end 16 smaller squares), and so on. Define A := N A. Claim: dim H (A) = 1. Proof: Step 1: H 1 (A) 2 2 Each A consists of 4 squares, each of edge-length 4. Thus A can be covered by 4 balls, each of radius, (this is in fact the smallest radius we could imagine for a ball to cover a square of edge-length 4 ). Thus for a given δ > 0, Hδ 1 (A ) 4 j= = 2 2, for all N such that < δ. But A A, so by monotonicity of Hδ 1 (which is a measure), we get that: Hδ 1 (A) 2 2 so that in the limit of δ, we get the required result. Step 2: H 1 (A) 1 2 Estimate H 1 (A): 36

37 1.7 Radon Measures 1 σ -ALGEBRAS AND MEASURES { ([ x ε Given any δ > 0, ε > 0 B r ε y ε ])} such that A N B r ε ([ x ε r ε < δ and N rε < H1 δ (A) + ε < H1 (A) + ε. ([ ]) x Let π := x be the projection map of the first coordinate. y Observe that π (A) = [0, 1] (why?). Observe that since A N B r ε ([ x ε y ε y ε ]), ]), π (A) ( ([ ])) x N π ε B r ε y ε. Thus [0, 1] N (xε rε, xε + rε ). So by monotonicity of the Lebesgue-measure, we get: L 1 ([0, 1]) L 1 ( N (xε rε, xε + rε )) N L1 ((x ε rε, xε + rε )) But since these are just intervals, the Lebesgue measure of them is simply their length: 1 N 2rε So that finally we get the fact that: H 1 (A) ε for any ε > 0. Thus 1 2 H1 (A) 2 2. Step 4: dim H (A) 1. Tae any s > 1. By 1, H s (A) = 0. Thus H s (A) = 0 s > 1. So by 1.6.6, dim H (A) 1. Step 5: dim H (A) 1. Tae any s < 1. By 2, H s (A) =. Thus H s (A) = s < 1. So by 1.6.6, dim H (A) Radon Measures Let µ be a measure on R n D The Radon Measure (def ) µ is a Radon measure if both the following conditions hold: 1. µ is Borel regular. 2. µ (K) < K 2 Rn such that K is compact. This second condition is equivalent to µ being locally finite. 37

38 1.7 Radon Measures 1 σ -ALGEBRAS AND MEASURES E (ex ) 1. The Lebesgue and Lebesgue-Stieltjes measures on R n are Radon measures. Proof: TODO 2. The Hausdorff measure of dimension s < n is not a Radon measure. Proof: TODO 3. In case µ is a Radon measure, A 2 Rn is µ-measurable, then the measure ν, defined by ν (B) := µ (A B) B 2 Rn is also a Radon measure. Proof: Step 1: ν is a measure. ν ( ) = µ (A ) = µ ( ) = 0 Let B =1 B. ν (B) = µ (B A) But note that B A =1 A B. So that µ (B A) N µ (A B ) = N ν (B ). Step 2: ν (K) < K 2 Rn such that K is compact. ν (K) = µ (K A) µ (K) <, by virtue of µ s monotonicity. Step 3: ν is Borel. Let C 2 Rn be a Borel set, and tae any B 2 Rn. Then: ν (B) = µ (B A). Since µ is Borel, we now that C is measurable. So we may invoe on B A: µ (B A) = µ ((B A) C) + µ ((B A) \C) = µ ((B C) A) + µ (A (B\C)) = ν (B C) + ν (B\C) Thus C is ν-measurable, and so ν is Borel. Step 4: ν is Borel regular. Tae any B 2 Rn. We need to find some C B such that C is Borel and ν (C) = ν (B). Case 1: µ (B) < µ is Borel regular, so we can choose two Borel sets, C and D, such that: C A B µ (C) = µ (A B) D B\A µ (D) = µ (B\A) Due to A being µ-measurable, invoe on C to get: µ (C) = µ (C A) + µ (C\A) Observe that µ (B) = µ (B A)+µ (B\A) = µ (C)+µ (B\A) µ (C) µ (B) <. 38

39 1.7 Radon Measures 1 σ -ALGEBRAS AND MEASURES Thus we move µ (C) to the other side of the equation to get: µ (C\A) = µ (C) µ (C A) = µ (A B) µ (C A) But C A B and so C A B A, so via monotonicity we get: µ (A B) µ (C A) To summarise, we achieved at the fact that µ (C\A) 0 which means that µ (C\A) = 0 (because µ is a non-negative set function). Thus µ (A B) µ (C A) ν (B) = ν (C). Next, we follow the same procedure for D: µ (D) = µ (D A) + µ (D\A) Observe that µ (B) = µ (B A)+µ (B\A) = µ (B A)+µ (D) µ (D) µ (B) < Since µ (D) <, we may move it to the other side of the equation: µ (D A) = µ (D) µ (D\A) = µ (B\A) µ (D\A). But D\A B\A, so we conclude finally that µ (B\A) µ (D\A) which means that µ (D A) 0 µ (D A) = 0 ν (D) = 0. Finally, observe that B C D and then invoing and 1.1.3, ν (B) ν (C D) ν (C)+ν (D) = ν (B)+0 = ν (B) = ν (C D) = ν (B). Which is great since C D is a Borel set, as C and D are. Case 2: µ (B) = Divide R n into a countable union of compact n-dimensionalcubes: R n = l N Q l, which only overlap on their faces. Thus µ (Q l ) < becuase µ is a Radon measure. By monotonicity then µ (Q l B) <. So we may apply Case 1 on each Q l B to get a Borel set C l such that C l Q l B and ν (C l ) = ν (Q l B). Thus l N C l is a Borel set, l N C l B. Define {R l } to be the same n-dimensional-cubes as {Q l }, only instead of being closed they are half-closed, so that they are disjoint, and their disjoint union now forms R n. Observe that ν (Q l B) = ν (R l B). Thus: ν ( l N C l ) = l N ν (C l) = l N ν (Q l B) and also that B = l N B R l thus ν (B) = ν ( l N B R ) l = l N ν (B R l) = l N ν (B Q l) T Approximation of Radon Measures by Open or Compact Sets (thrm ) Let µ be a Radon measure on R n. Similar to 1.3.8, we have the same theorem for Radon measures. TODO 39

40 1.8 Topological Definitions 2 MEASURABLE FUNCTIONS 1.8 Topological Definitions D Locally Compact Space A topological space X is locally compact if x X V Open (X) : x V closure X (V ) is compact D Semicontinuous Functions Let f be a real (or extended-real in which case the following intervals are halfclosed to include the infinities) function on a topological space Lower Semincontinuous If f 1 ((α, )) Open (X) α R then f is lower semicontinuous. Equivalently, x 0 X, ɛ > 0 a neighborhood U of x 0 such that f (x) f (x 0 ) ɛ for all x U Upper Semicontinuous If f 1 ((, α)) Open (X) α R then f is lower semicontinuous. Equivalently, x 0 X, ɛ > 0 a neighborhood U of x 0 such that f (x 0 ) f (x) ɛ for all x U D Support The support of a complex function f on a topological space X is closure X ( f 1 (C\ {0}) ) D Continuous Functions with Compact Support C c (X) := {f : X C f is continuous and support (f) is compact} Observe that C c (X) is a vector space. 2 Measurable Functions 2.1 Definition and Elementary Characteristics T Approximation with Step Functions (strw thrm ) Let f [0, 1] Ω be µ-measurable. Claim: µ-measurable sets A Ω N such that f = N 1 χ A P Define A 1 := f 1 ([1, ]). A 1 is µ-measurable. { Define inductively A := x Ω : f (x) 1 + } 1 j=1 1 j χ A j (x) N\ {1}. 40