Measure Theory Summary - ISG D-PHYS Personal web space · 2013. 9. 4. · Measure and Integration...

Measure and Integration Summary

Jacob Shapiro

September 4, 2013

Abstract

This is a (very rough) translation of Prof. Michael Struwe’s Germanlecture Notes into English, as preparation for the final test (held in August2013). Struwe followed rather closely the construction presented in thebook of Evans and Gariepy and also had some inspiration from Amannand Escher.

Contents

1 σ -Algebras and Measures 101.1 The Abstract Space . . . . . . . . . 10

1.1.1 D Measure . . . . . . . . . . 101.1.2 E . . . . . . . . . . . . . . . 101.1.3 R σ-Subadditivity . . . . . . 111.1.4 R Monotonicity . . . . . . . 111.1.5 R . . . . . . . . . . . . . . . 111.1.6 D Measurability of a Set (Caratheodory) 111.1.7 R . . . . . . . . . . . . . . . 111.1.8 E . . . . . . . . . . . . . . . 111.1.9 R . . . . . . . . . . . . . . . 121.1.10 D Algebra . . . . . . . . . . 121.1.11 R . . . . . . . . . . . . . . . 121.1.12 E . . . . . . . . . . . . . . . 12

1

mailto:mailto:[email protected]

CONTENTS CONTENTS

1.1.13 T . . . . . . . . . . . . . . . 131.1.14 D Measure Spaces . . . . . 151.1.15 T Properties of µ on Σ . . . 15

1.1.15.1 σ-additivity . . . . 151.1.15.2 Continuity from in-

side . . . . . . . . . 161.1.15.3 Continuity from out-

side . . . . . . . . . 161.1.16 E . . . . . . . . . . . . . . . 171.1.17 D µ-Measure Zero Sets (Null

Sets) . . . . . . . . . . . . . . 171.1.18 T Sets of Measure Zero are

Measurable . . . . . . . . . . 171.2 Construction of Measures . . . . . . 17

1.2.1 D Covering Classes . . . . . 171.2.2 E Different Covering Classes 17

1.2.2.1 Of Rn . . . . . . . . 171.2.2.2 Trivial Covering Class 18

1.2.3 T Forming Measures fromCovering Classes . . . . . . . 18

1.2.4 E Generating the Measuredefined in 1 . . . . . . . . . . 19

1.2.5 D Pre-measure . . . . . . . 191.2.5.1 σ-finite pre-measure 20

1.2.6 R . . . . . . . . . . . . . . . 201.2.7 T Carathéodory-Hahn Ex-

tension Theorem . . . . . . . 201.2.8 T Uniqueness of Caratheodory

Measure . . . . . . . . . . . . 211.2.9 R . . . . . . . . . . . . . . . 23

2

CONTENTS CONTENTS

1.2.9.1 Overlapping Σ’s . . 231.2.9.2 Cannot Generalize

Beyond A . . . . . 231.2.10 E to 1.2.9.2 . . . . . . . . . 23

1.3 The Lebesgue Measure . . . . . . . . 231.3.1 D Elementary Volume Pre-

measure . . . . . . . . . . . . 231.3.1.1 n-dimensional Box

(k-cell) . . . . . . . 231.3.1.2 The Volume of an

n-dimensional Box 241.3.1.3 An elementary Set 24

1.3.2 R . . . . . . . . . . . . . . . 241.3.3 R . . . . . . . . . . . . . . . 241.3.4 D The Lebesgue Measure Ln 241.3.5 T Open Sets are Countable

Unions of k-cells . . . . . . . 241.3.6 C Borel σ-Algebra . . . . . 251.3.7 D Borel Measures . . . . . . 251.3.8 T Approximation from Out-

side by Open Sets . . . . . . 251.3.9 D Regular Borel Measure . 251.3.10 C The Lebesgue Measure is

a regular Borel measure . . . 261.3.11 T Another Characterization

of Measurability . . . . . . . 261.3.12 C Another Characterization

of Measurability . . . . . . . 261.3.13 C Another Characterization

of Measurability . . . . . . . 26

3

CONTENTS CONTENTS

1.3.14 Comparison with the JordanMeasure . . . . . . . . . . . . 261.3.14.1 D The Jordan Mea-

sure . . . . . . . . . 261.3.14.2 T Comparison with

Lebesgue-measure . 271.3.15 T Invariance for Rigid Trans-

formations . . . . . . . . . . . 271.4 The Limits of Measurability . . . . 27

1.4.1 E Vitali Sets . . . . . . . . . 271.4.2 Banach-Tarski . . . . . . . . 27

1.5 The Lebesgue-Stieltjes Measure . . 271.5.1 D Distance between two sets 271.5.2 D Metric Measure . . . . . 281.5.3 T Caratheodory Criterion for

Borel Measures . . . . . . . . 281.5.4 T The Lebesgue-Stieltjes Mea-

sure is a regular Borel measure 291.5.5 T Lebesgue-Stieltjes measure

on half-open intervals . . . . 301.5.6 E Familiar Examples . . . . 31

1.6 The Hausdorff Measure . . . . . . . 321.6.0.1 D The Hausdorff

Measure . . . . . . 321.6.1 T The Hausdorff Measure

is a regular Borel measureon Rn . . . . . . . . . . . . . . 321.6.1.1 Step 1 . . . . . . . . 331.6.1.2 Step 2 . . . . . . . . 331.6.1.3 Step 3 . . . . . . . . 33

4

CONTENTS CONTENTS

1.6.2 T Preparation for HausdroffDimension . . . . . . . . . . . 34

1.6.3 E Compact Sets . . . . . . . 351.6.4 R Connection between Lebesgue

and Hausdorff measures . . . 351.6.5 E Different Dimension Haus-

dorff Measure . . . . . . . . . 361.6.6 D Hausdorff Dimension . . 361.6.7 E The Cantor “Dust” . . . . 36

1.7 Radon Measures . . . . . . . . . . . 371.7.1 D The Radon Measure . . . 371.7.2 E . . . . . . . . . . . . . . . 381.7.3 T Approximation of Radon

Measures by Open or Com-pact Sets . . . . . . . . . . . 39

1.8 Topological Definitions . . . . . . . . 401.8.1 D Locally Compact Space . 401.8.2 D Semicontinuous Functions 40

1.8.2.1 Lower Semincontin-uous . . . . . . . . . 40

1.8.2.2 Upper Semicontin-uous . . . . . . . . . 40

1.8.3 D Support . . . . . . . . . . 401.8.4 D Continuous Functions with

Compact Support . . . . . . 40

2 Measurable Functions 402.1 Definition and Elementary Charac-

teristics . . . . . . . . . . . . . . . . . 40

5

CONTENTS CONTENTS

2.1.1 T Approximation with StepFunctions . . . . . . . . . . . 40

2.1.2 R Monotone Convergence andReal Functions . . . . . . . . 41

2.2 Lusin’s Theorem and Egoroff’s The-orem . . . . . . . . . . . . . . . . . . 412.2.1 Egorov’s (Egoroff) Theorem 422.2.2 Lusin’s Theorem . . . . . . . 43

2.3 Convergence in Measure . . . . . . . 442.3.1 D Convergence In Measure 442.3.2 T Convergence Almost Ev-

erywhere Implies In Measure 442.3.3 T Convergence in Measure

gives a subsequence that Con-verges Almost Everywhere . 45

3 Integration 453.1 Definition and Elementary Charac-

teristics . . . . . . . . . . . . . . . . . 453.1.1 D σ-Step Function . . . . . 453.1.2 D Integral for σ-Step Func-

tions . . . . . . . . . . . . . . 453.2 Convergence Theorems . . . . . . . 463.3 Absolute continuity of the Integral . 463.4 Vitali’s Theorem . . . . . . . . . . . 46

3.4.1 D Uniformly Integrable . . 463.4.2 T Vitali’s Theorem . . . . . 46

3.5 Lp (Ω, µ) spaces . . . . . . . . . . . . 47

4 Complex Measures (Rudin RCA Chapter6) 47

6

CONTENTS CONTENTS

4.1 Total Variation . . . . . . . . . . . . 474.1.1 D A Partition . . . . . . . . 474.1.2 D A Complex Measure . . . 484.1.3 D Total Variation Measure

of µ . . . . . . . . . . . . . . . 484.1.4 T |µ| is a positive measure

on M . . . . . . . . . . . . . . 484.1.5 T . . . . . . . . . . . . . . . 484.1.6 D . . . . . . . . . . . . . . . 484.1.7 Positive and Negative Variations–

Jordan Decomposition of µ . 494.2 Absolute Continuity . . . . . . . . . 49

4.2.1 σ-finite Measures . . . . . . . 494.2.2 The Lebesgue-Radon-Nikodym

Theorem . . . . . . . . . . . . 494.2.3 Theorem . . . . . . . . . . . . 504.2.4 Theorem . . . . . . . . . . . . 504.2.5 Theorem . . . . . . . . . . . . 504.2.6 The Hahn Decomposition The-

orem . . . . . . . . . . . . . . 50

5 Intergration on Product Measures (RudinRCA Chapter 8) 515.1 Measurability on Cartesian Products 51

5.1.1 A Rectangle in X × Y . . . . 515.1.2 A Measurable Rectangle . . 515.1.3 The Class of all Elementary

Sets . . . . . . . . . . . . . . 515.1.4 S × T . . . . . . . . . . . . . 515.1.5 A monotone Class . . . . . . 515.1.6 x-section and y-section . . . 51

7

CONTENTS CONTENTS

5.1.7 Theorem . . . . . . . . . . . . 525.1.8 Theorem . . . . . . . . . . . . 525.1.9 Theorem . . . . . . . . . . . . 52

5.2 Product Measures . . . . . . . . . . 525.2.1 Theorem . . . . . . . . . . . . 525.2.2 The Product of Measures . . 52

5.3 The Fubini Theorem . . . . . . . . . 535.3.1 The Fubini Theorem . . . . . 535.3.2 Counterexamples . . . . . . . 53

5.4 Completion of Product Measures . 535.4.1 Theorem . . . . . . . . . . . . 53

5.5 Convolutions . . . . . . . . . . . . . . 535.5.1 Theorem–Convolutions . . . 53

6 Product Measures, Multiple Integrals 546.1 Fubini’s Theorem . . . . . . . . . . . 546.2 Convolution . . . . . . . . . . . . . . 54

7 Differentiation (Rudin RCA) 547.1 Derivatives of Measures . . . . . . . 54

7.1.1 Lemma . . . . . . . . . . . . . 547.1.2 Theorem . . . . . . . . . . . . 547.1.3 Weak L1 . . . . . . . . . . . . 55

7.1.3.1 The Maximal Func-tion (Hardy-Littlewood) 55

7.1.4 Lebesgue Points . . . . . . . 557.1.4.1 Example . . . . . . 55

7.1.5 Theorem–Lebesgue Point a.e. 557.1.6 Theorem–The Radon-Nikodym

Derivative . . . . . . . . . . . 557.1.7 Nicely Shrinking Sets . . . . 56

8

CONTENTS CONTENTS

7.1.8 Theorem . . . . . . . . . . . . 567.1.9 Theorem . . . . . . . . . . . . 567.1.10 Metric Density . . . . . . . . 567.1.11 Theorem . . . . . . . . . . . . 577.1.12 Theorem . . . . . . . . . . . . 577.1.13 Theorem . . . . . . . . . . . . 577.1.14 Theorem–The Fundamental

Theorem of Calculus . . . . . 577.1.14.1 Counter Examples 57

7.1.14.1.1 a . . . . . 577.1.14.1.2 b–Denjoy and

Perron . . 577.1.15 Absolutely Continuous Func-

tions . . . . . . . . . . . . . . 587.1.16 Theorem . . . . . . . . . . . . 587.1.17 Total Variation Function of

a Function . . . . . . . . . . . 587.1.18 Theorem . . . . . . . . . . . . 587.1.19 Theorem . . . . . . . . . . . . 597.1.20 Theorem . . . . . . . . . . . . 59

8 Differentiation of Measures 598.1 Differentiability of the Lebesgue In-

tegral . . . . . . . . . . . . . . . . . . 598.2 Differentiation of Radon Measures . 598.3 Differentiation of Absolutely Con-

tinuous Measures on R . . . . . . . . 598.4 Lebesgue Decomposition . . . . . . . 598.5 The Radon-Nikodym Theorem . . . 59

9

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.

• Cc (X) ={f ∈ C (X, C) : closureX

(f−1 (C\ {0})

)is compact

}• D : definition

• R : remark

• P : proof

• E : example

• T : theorem

• A : application

• C : Corollary

1 σ -Algebras and Measures

1.1 The Abstract SpaceLet X be a set, and let 2X be the power set of X.

1.1.1 D Measure

(strw def. 1.1.1)A map µ : 2X → [0, ∞] is a measure on X, in case the following conditionshold:

1. µ (∅) = 0

2. A ⊂⋃∞

k=1 Ak=⇒µ (A) ≤∑∞

k=1 µ (Ak).

1.1.2 E

(strw ex. 1.1.1)

1. Trivial MeasureLet X be a non-empty set. Define µ : 2X → [0, ∞] by

µ (A) :=

{0 A = ∅1 otherwise

µ is a measure on X.

2. The Counting Measureµ (A) := |A| ≤ ∞ where |A| is the number of elements in A.

10

1.1 The Abstract Space 1 σ -ALGEBRAS AND MEASURES

1.1.3 R σ-Subadditivity

(strw rem. 1.1.1)2 implies in particular that µ (

⋃∞k=1 Ak) ≤

∑∞k=1 µ (Ak), a characteristic of set

functions that is called σ-subadditivity.

1.1.4 R Monotonicity

(strw rem. 1.1.2)

• Claim:A ⊂ B ⇒ µ (A) ≤ µ (B).

• P : Choose A1 := B and Ak = ∅ ∀k ≥ 2 in 2.

1.1.5 R

(strw rem. 1.1.3)The external Jordan “measure” on R (denoted here by µ̄) is not a measureaccording to 1.1.1.Consider the following set, A, defined by:

A = Q ∩ [0, 1] = {qk; k ∈ N}

with the cover Ak = {qk}∀ k ∈ N.Observe that A ⊂

⋃∞k=1 Ak and µ̄ (A) = 1 �

∑∞k=1 µ̄ (Ak) = 0.

1.1.6 D Measurability of a Set (Caratheodory)

(strw def. 1.1.2)A set A ∈ 2X is called µ-measurable if ∀B ∈ 2X , µ (B) = µ (B ∩A) + µ (B\A).

1.1.7 R

(strw rem. 1.1.4)Due to subadditivity of the measure µ and the fact that B ⊂ B = ((B ∩A) ∪ (B\A)),it suffices to show only:

µ (B) ≥ µ (B ∩A) + µ (B\A)

in order to satisfy 1.1.6.

1.1.8 E

(strw ex. 1.1.2)

1. Let X be a nonempty set a let µ be defined as in 1.

• Claim:A ⊂ X is measurable iff A ∈ {∅, X}.

11


• P Take B := X.Then µ (B) = 1 because B = X is assumed to be nonempty.But for any A ∈ 2X , X ∩ A = A, so in order for a set A to bemeasurable we need that 1 ≥ µ (A)+µ (X\A), which is the case onlywhen 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 thereis only one kind of infinity for the counting measure, by definition).

1.1.9 R

(strw rem. 1.1.5)Some authors denote a σ-subadditive set-function as an outer measure, whoserestriction to a family of measurable sets is called simply measure.

1.1.10 D Algebra

(strw def. 1.1.3)A ⊂ 2X is called an algebra in case the following holds:

1. X ∈ A

2. A ∈ A ⇒ (X\A) ∈ A

3. If Ai ∈ A∀ i ∈ {1, 2 , . . . , m} where m ∈ N, then (⋃m

i=1 Ai) ∈ A.

A is called a σ-algebra if the last condition holds for a countable union ratherthan only for a finite union.

1.1.11 R

(strw rem. 1.1.6)If A is an algebra (σ-algebra), using de Morgan’s laws, we can show that also afinite (countable) intersection would lie in A:

∞⋂k=1

Ak = X\

( ∞⋃k=1

(X\Ak)

)

1.1.12 E

(strw ex. 1.1.3)

1. Let X be a nonempty set. Then {∅, X} is an algebra.

2. For any set X, 2X is a σ-algebra.

12


1.1.13 T

(strw thrm. 1.1.1)If µ is a measure, then Σ :=

{A ∈ 2X : 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:Take any B ∈ 2X . Since B ⊂ X, B\X = ∅ and B ∩ X = B. Butµ (∅) := 0.

– Step 2: (X\A) ∈ Σ∀A ∈ Σ:Take any A ∈ Σ.Thus ∀B ∈ 2X , 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 2 12 : If Ai ∈ Σ∀ i ∈ {1, 2 , . . . , m} where m ∈ N, then (⋃m

i=1 Ai) ∈Σ:Apply induction on m.∗ The case where m = 1:⋃1

i=1 Ai = A1 ∈ Σ.∗ Assume the case for m− 1 holds:

Define S :=⋂m−1

i=1 Ai. Thus we assume that S ∈ Σ.∗ Verify the case for m:

Using the notation for the previous case, we now also have Am ∈Σ.We would like to show next that (S ∪Am) ∈ Σ.Take some B ∈ 2X .We know that µ (B) = µ (B ∩ S) + µ (B\S) because S ∈ Σ.We also know that µ (B\S) = µ ((B\S) ∩Am)+µ ((B\S) \Am),because Am ∈ Σ.Plugging the second equation into the first we get:µ (B) = µ (B ∩ S) + µ ((B\S) ∩Am) + µ ((B\S) \Am).Observe that B ∩ (S ∪Am) = (B ∩ S) ∪ ((B\S) ∩Am).By subadditivity: µ (B ∩ (S ∪Am)) ≤ µ (B ∩ S)+µ ((B\S) ∩Am).In addition, we could write (B\S) \Am = B\ (S ∪Am) .Plugging this in we get: µ (B) ≥ µ (B ∩ (S ∪Am))+µ (B\ (S ∪Am)).Thus (S ∪Am) ∈ Σ.

– Step 3: If Ai ∈ Σ ∀ i ∈ N, then(⋃

i∈N Ai)∈ Σ:

∗ Disjoint Decomposition:First we construct a mutually disjoint sequence of sets, {Ci},which will obey similar properties as the given {Ai}. Note: Thisis a procedure we shall often carry out.

13


Thus, define:

Ci :=

{A1 i = 1

Ai\(⋃i−1

j=1 Aj

)i ∈ N\ {1}

· Claim: Ci ∩ Cj = ∅ whenever i 6= j.· Proof:

Take i ∈ N, j ∈ N\ {i}, where WLOG i > j.Now assume the contrary: ∃x ∈ (Ci ∩ Cj).That is, x ∈ Ai∧(x /∈ Ak∀k ≤ i− 1) and x ∈ Aj∧(x /∈ Ak∀k ≤ j − 1).But j ≤ i− 1 by assumption, so we get x /∈ Aj from the firstassertion and x ∈ Aj from the second assertion.⇒ contradiction.

· Claim:⋃

i∈N Ai =⋃

i∈N Ci

· Proof of ⊃ :Take some x0 ∈

⋃i∈N Ci. ⇒ ∃i0 ∈ N : x0 ∈ Ci0 .

⇒ x ∈ Ai0 .⇒ x ∈

⋃i∈N Ai.

· Proof of ⊂ :Take some x0 ∈

⋃i∈N Ai.⇒ ∃i0 ∈ N : x0 ∈ Ai0 .

If x0 /∈ Aj∀j ∈ {1, 2, . . . , i0 − 1} then x0 ∈ Ci0 .Otherwise x0 ∈ Cmin({j∈{1,2,...,i0−1} | x0∈Aj}).In either case x0 ∈

⋃i∈N Ci.

· Claim: Ci ∈ Σ ∀ i ∈ N (given that Ai ∈ Σ∀i ∈ N)· Proof: Take some i ∈ N.

If i = 1 we are done.Otherwise, using the steps 2 and 2 12 we know that Σ is closedunder finite unions and complements.But Ci = Ai\

(⋃i−1j=1 Aj

)= Ai∩

(X\⋃i−1

j=1 Aj

)= X\

((X\Ai) ∪

(⋃i−1j=1 Aj

)).

Thus Ci ∈ Σ.So we can work with the disjoint {Ci} instead of {Ai}.We shall now (finally) proceed to show that

(⋃i∈N Ci

)∈ Σ:

Take any B ∈ 2X . Pick any n ∈ N.∗ Claim (Addivitiy with any set):µ (B ∩ (

⋃ni=1 Ci)) =

∑ni=1 µ (B ∩ Ci).

∗ Proof:Since Cn ∈ Σ, apply 1.1.6 on it with the set B ∩ (

⋃ni=1 Ci):

µ (B ∩ (⋃n

i=1 Ci)) = µ ((B ∩ (⋃n

i=1 Ci)) ∩ Cn)+µ ((B ∩ (⋃n

i=1 Ci)) \Cn)Using the fact that {Ci} are mutually disjoint we get:µ (B ∩ (

⋃ni=1 Ci)) = µ (B ∩ Cn) + µ

(B ∩

(⋃n−1i=1 Ci

))Carry out this procedure now n−1 times on µ

(B ∩

(⋃n−1i=1 Ci

)),

until we reach the desired result.

Since⋃n

i=1 Ci ∈ Σ, using 1.1.6,

14


µ (B) = µ (B ∩ (⋃n

i=1 Ci)) + µ (B\ (⋃n

i=1 Ci)).Using the above claim and the fact that B\ (

⋃ni=1 Ci) ⊃ B\ (

⋃∞i=1 Ci),

which allows us to use the monotonicity of µ, we arrive at:µ (B) ≥

∑ni=1 µ (B ∩ Ci) + µ (B\ (

⋃∞i=1 Ci))

In the limit where n → ∞:µ (B) ≥

∑∞i=1 µ (B ∩ Ci) + µ (B\ (

⋃∞i=1 Ci))

Using σ-subadditivity of µ we get:µ (B) ≥ µ (

⋃∞i=1 B ∩ Ci) + µ (B\ (

⋃∞i=1 Ci)) = µ (B ∩ (

⋃∞i=1 Ci)) +

µ (B\ (⋃∞

i=1 Ci))

QED

1.1.14 D Measure Spaces

(strw def. 1.1.4)If µ is a measure on X and Σ :=

{A ∈ 2X : A is µ−measurable

}, the triplet

(X, Σ, µ) is called a measure space.

1.1.15 T Properties of µ on Σ

(strw thrm. 1.1.2)Let Ak ∈ Σ ∀ k ∈ N.

1.1.15.1 σ-additivity

• Claim: If Ak ∩Am = ∅∀ k 6= m then µ (⋃∞

k=1 Ak) =∑∞

k=1 µ (Ak).

• P

– ≤Using σ-subadditivity, µ (

⋃∞k=1 Ak) ≤

∑∞k=1 µ (Ak).

– ≥Using 1.1.13 with B = X, we get µ (

⋃ni=1 Ai) =

∑ni=1 µ (Ai) for all

n ∈ N.In the limit where n → ∞:

limn→∞

µ

(n⋃

i=1

Ai

)=

∞∑i=1

µ (Ai)

But using monotonicity, µ (⋃∞

i=1 Ai) ≥ µ (⋃n

i=1 Ai) for all n ∈ N,thus µ (

⋃∞i=1 Ai) ≥ limn→∞ µ (

⋃ni=1 Ai).

QED

15


1.1.15.2 Continuity from inside

• Claim: If Ak ⊂ Ak+1∀k ∈ N then µ(⋃

k∈N Ak)= limk→∞ µ (Ak).

• PDefine a disjoint decomposition of {Ak}:C1 := A1, Ck := Ak\Ak−1∀k ∈ N\ {1}.Note that

⋃∞k=1 Ck =

⋃∞k=1 Ak.

Since {Ck} are disjoint, using the first part of this theorem, we know thatµ (⋃∞

k=1 Ck) =∑∞

k=1 µ (Ck).

Since Ak =⋃k

j=1 Cj by construction, µ (Ak) = µ(⋃k

j=1 Cj

)but again,

since {Ck} are disjoint and using the first part we get µ(⋃k

j=1 Cj

)=∑k

j=1 µ (Cj).

Thus, µ (Ak) =∑k

j=1 µ (Cj).

In the limit where k → ∞, limk→∞ µ (Ak) =∑∞

j=1 µ (Cj) = µ (⋃∞

k=1 Ck) =

µ (⋃∞

k=1 Ak).

QED

1.1.15.3 Continuity from outside

• Claim: If Ak ⊃ Ak+1∀k ∈ N and µ (A1) < ∞ then µ(⋂

k∈N Ak)

=limk→∞ µ (Ak).

• PDefine Ck := A1\Ak∀k ∈ N, where ∅ = C1 ⊂ C2 ⊂ . . . , and observe thatµ (A1) = µ (Ck) + µ (Ak) ∀k ∈ N.Thus in the limit of n → ∞, µ (A1)− limk→∞ µ (Ak) = limk→∞ µ (Ck).But {Ck} satisfies the conditions for 1.1.15.2, so:µ (A1)− limk→∞ µ (Ak) = µ

(⋃k∈N Ck

)= µ

(⋃k∈N (A1\Ak)

)Using de Morgan’s laws we get:

µ(⋃

k∈N (A1\Ak))= µ

(A1\

(⋂k∈N Ak

))But observe that (due to measurability of

⋂k∈N Ak):

µ (A1) = µ((⋂

k∈N Ak))

+ µ(A1\

((⋂k∈N Ak

))).

Thus:

µ(A1\

(⋂k∈N Ak

))= µ (A1)− µ

((⋂k∈N Ak

)).

QED

16

1.2 Construction of Measures 1 σ -ALGEBRAS AND MEASURES

1.1.16 E

(strw ex. 1.1.4)Let X = N and let µ be the counting measure.Define Ak := {n ∈ N |n ≥ k} ∀k ∈ N.Observe that µ (Ak) = ∞∀k ∈ N.Thus, limk→∞ µ (Ak) = ∞.However,

⋂k∈N Ak = ∅, so, µ

(⋂k∈N Ak

)= µ (∅) = 0.

So clearly the assumption that µ (A1) < ∞ is necessary.

1.1.17 D µ-Measure Zero Sets (Null Sets)

(strw def. 1.1.5)N ∈ 2X is called a µ-measure zero set iff µ (N) = 0.

1.1.18 T Sets of Measure Zero are Measurable

(strw thrm. 1.1.3)

• Claim: A µ-null set, N ∈ 2X , is µ-measurable.

• PTake any B ∈ 2X .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 MeasuresLet X 6= ∅ be any set.

1.2.1 D Covering Classes

(strw def. 1.2.1)K ⊂ 2X is called covering class of X in case:

1. ∅ ∈ K

2. ∃ {Kj} ⊂ K : X =⋃∞

j=1 Kj

1.2.2 E Different Covering Classes

(strw ex. 1.2.1)

1.2.2.1 Of Rn The open “n-dimensional-intervals”, I :=∏n

k=1 (ak, bk) suchthat ak ≤ bk and ak, bk ∈ R for all k ∈ N, form a covering class for Rn, as wellas the closed intervals, and the half-open intervals.

17


1.2.2.2 Trivial Covering Class Every algebra A ∈ 22X on a set X is acovering class of X because ∅, X ∈ A.

1.2.3 T Forming Measures from Covering Classes

(strw thrm. 1.2.1)

• Claim: If K is a covering class of X, and ∃λ : K → [0,∞] such thatλ (∅) = 0 then the following is a measure on X:

µ (A) := inf

∞∑j=1

λ (Kj) |Kj ∈ K, A ⊂∞⋃j=1

Kj

∀A ∈ 2X

• P

– First of all, since ∅ ∈ K, and ∅ ⊂ ∅, and λ (∅) = 0, µ (∅) = 0.– Now, take some A,Ak ∈ 2X∀k ∈ N and assume that A ⊂

⋃k∈N Ak.

By the definition of the infimum, ∀k ∈ N, ∀ε > 0 ∃{K

(k),εj

}⊂ K

such that Ak ⊂⋃

j∈N K(k),εj and:

∞∑j=1

λ(K

(k),εj

)< µ (Ak) + ε

Pick εk := 2−kδ where δ > 0.

Thus ∃{K

(k),2−kδj

}⊂ K such that Ak ⊂

⋃j∈N K

(k),2−kδj and

∑∞j=1 K

(k),2−kδj <

µ (Ak) + 2−kδ.

Sum up m equations like that where k ∈ {1, . . . ,m} and take thelimit as m → ∞:

∞∑k=1

∞∑j=1

λ(K

(k),2−kδj

)≤

∞∑k=1

µ (Ak) + δ

But{K

(k),2−kδj

}is a sequence of sets in K whose countable union

covers A, as assumed:

A ⊂⋃k∈N

Ak ⊂⋃k∈N

⋃j∈N

K(k),2−kδj

And so, µ (A), being an infimum over the set of all such countablecovers, must be smaller than

∑k∈N

∑j∈N λ

(K

(k),2−kδj

).

Thus:

18


µ (A) ≤∞∑k=1

µ (Ak) + δ

Since we have not assumed anything about δ other than that it ispositive, we might as well take the limit as δ → 0 and arrive at thedesired result that µ fulfills 1.1.1.

QED

1.2.4 E Generating the Measure defined in 1

(strw ex. 1.2.2)Let X be a nonempty set, and let K = {∅, X}.Let there be a map λ : K → [0, ∞] given by

λ (A) =

{0 A = ∅1 A = X

∀A ∈ K

Then the measure induced by λ according to 1.2.3 is the same as the measuredefined in 1.

• P

– Part 1:µ (∅) = 0Since λ (∅) = 0 and ∅ ⊂ ∅ ∈ K, by definition of the infimumµ (∅) = 0.

– Part 2: µ (A) = 1∀A ∈ 2X\ {∅}Take some A ∈ 2X\ {∅}.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 1.

1.2.5 D Pre-measure

(strw def. 1.2.2)If A ∈ 22X is an algebra, λ : A → [0,∞] is called a pre-measure in case:

1. λ (∅) = 0

2. λ (⋃∞

k=1 Ak) =∑∞

k=1 λ (Ak) where (⋃∞

k=1 Ak) , Ak ∈ A∀k ∈ N and Ak ∩Aj = ∅∀j 6= k.

Note: Thus,⋃∞

k=1 Ak is required to lie in A in addition to all of the Ak. Thisrequirement has to be posed because A is merely an algebra and not a sigma-algebra.

19


1.2.5.1 σ-finite pre-measure λ is called σ-finite in case a cover of X exists(X =

⋃∞k=1 Sk ) such that Sk ∈ A and λ (Sk) < ∞ for all k ∈ N. WLOG we

can take the sets Sk to be disjoint, following the same construction in the proofof 1.1.13.

1.2.6 R

(strw rem. 1.2.1)We may always assume that when taking the infimum we are dealing witha mutually disjoint cover, because if we define the disjoint decomposition tobe as in 1.1.13, then we will always have that λ (Ck) ≤ λ (Ak) due to λ’ssigma-additivity on A, of which Ck are also members of due to A being analgebra.

1.2.7 T Carathéodory-Hahn Extension Theorem

(strw thrm. 1.2.2)Let λ : A → [0, ∞] be a pre-measure on X, where A is an algebra, and let µ bedefined by 1.2.3.

• Claim:

1. 1.2.3 where A is the covering class.2. µ (A) = λ (A)∀A ∈ A.3. A is µ-measurable ∀ A ∈ A.

• P

1. By 1.2.2.2, since every algebra is itself a covering class, we may invoke1.2.3 on λ and A.

2. Take 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’scall it {Aεk}–such that A ⊂

⋃k∈N A

εk and

∑k∈N λ (A

εk) < µ (A)+

ε.By 1.2.6 we may assume that {Aεk} is a mutually disjoint cover.Define a new sequence of sets, Cεk := A

εk ∩A∀ k ∈ N.

Observe that {Cεk} is mutually disjoint and that⋃

k∈N Cεk = A.

In addition, since Aεk = Cεk ∪ (Aεk\A), which is a disjoint union,

using the additivity of λ we get that λ (Aεk) = λ (Cεk)+λ (A

εk\A)

so that λ (Aεk) ≥ λ (Cεk).Summing up those equations and going to the limit we get:∑

k∈N λ (Cεk) ≤

∑k∈N λ (A

εk).

20


Using the fact again that λ is σ-additive and that⋃

k∈N Cεk = A is

a disjoint union and that A ∈ A, we can write that∑

k∈N λ (Cεk) =

λ(⋃

k∈N Cεk

)= λ (A).

So finally we arrive at λ (A) ≤ µ (A) + ε, ∀ ε > 0.3. Following 1.1.7, we take any B ∈ 2X , 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’scall it {Bεk}–such that B ⊂

⋃k∈N B

εk and

∑k∈N λ (B

εk) < µ (B) + ε.

By 1.2.6 we may assume that {Bεk} is a mutually disjoint cover.But ∀k ∈ N, Bεk = (Bεk ∩A) ∪ (Bεk\A), (a disjoint union) and sinceBεk, A ∈ A, we conclude that λ (Bεk) = λ (Bεk ∩A) + λ (Bεk\A).In addition, observe that B∩A ⊂

⋃k∈N B

εk∩A and B\A ⊂

⋃k∈N B

εk\A,

so that by the definition of µ for these two sets, we get that µ (B ∩A) ≤∑k∈N λ (B

εk ∩A) and µ (B\A) ≤

∑k∈N λ (B

εk\A).

Putting everything together, we conclude that ∀ε > 0:µ (B)+ε >

∑k∈N λ (B

εk) =

∑k∈N (λ (B

εk ∩A) + λ (Bεk\A)) =

∑k∈N λ (B

εk ∩A)+∑

k∈N λ (Bεk\A) ≥ µ (B ∩A) + µ (B\A)

QED

1.2.8 T Uniqueness of Caratheodory Measure

(strw thrm. 1.2.3)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 inducedmeasure is unique for measurable sets (w.r.t. the induced measure).

• PLet µ be the induced measure by λ (accordng to 1) , and let Σ be theσ-algebra of µ-measurable sets in X.

Pick any measure ν : 2X → [0, ∞] such that ν (A) = λ (A) ∀A ∈ A.

1. Claim: ν (A) ≤ µ (A) ∀A ∈ ΣProof:Take some A ∈ 2X .By the definition of the infimum, ∀ε > 0 ∃ a cover of A in A–let’s callit {Aεk}–such that A ⊂

⋃k∈N A

εk and

∑k∈N λ (A

εk) < µ (A) + ε.

But Aεk ∈ A, so λ (Aεk) = ν (Aεk), so∑

k∈N λ (Aεk) =

∑k∈N ν (A

εk).

But ν is a measure and A ⊂⋃

k∈N Aεk, so according to 2,

∑k∈N λ (A

εk) =∑

k∈N ν (Aεk) ≥ ν (A).

2. Claim: ν (A) ≥ µ (A)∀A ∈ ΣProof: Take some A ∈ Σ.

21


– 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 isequal to µ (S) because A ∈ Σ and A ⊂ S.But S ∈ A, so µ (S) = λ (S) = ν (S), which is smaller than orequal to (by sub-additivity) to: ν (A) + ν (S\A).Thus because the quantities are finite, equality must apply onall expressions involved.

Thus

ν (A) + ν (S\A) = µ (A) + µ (S\A)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 know that ∃ {Sk}, a mutually

disjoint sequence in A, such that X ⊂⋃

k∈N Sk, and λ (Sk) <∞∀k ∈ ∞.Define a disjoint decomposition of A =

⋃k∈N Ak where Ak :=

A ∩ Sk∀k ∈ NTake some n ∈ N.Observe that (

⋃nk=1 Ak) ∈ Σ because A ⊂ Σ and A ∈ Σ. In

addition, (⋃n

k=1 Ak) ⊂ (⋃n

k=1 Sk) ∈ A (since finite intersectionsare in A) and λ (

⋃nk=1 Sk) =

∑nk=1 λ (Sk) < ∞ because {Sk} is

disjoint.So we may apply Case 1 on

⋃nk=1 Ak to get:

ν

(n⋃

k=1

Ak

)= µ

(n⋃

k=1

Ak

)Due to monotonicity of ν:

ν (A) = ν

( ∞⋃k=1

Ak

)≥ ν

(n⋃

k=1

Ak

)Thus:

ν (A) ≥ µ

(n⋃

k=1

Ak

)In the limit of n → ∞, we get: ν (A) ≥ limn→∞ µ (

⋃nk=1 Ak).

Observe that⋃n

k=1 Ak ⊂⋃n+1

k=1 Ak for all n ∈ N, so we may in-voke 1.1.15.2 to conclude limn→∞ µ (

⋃nk=1 Ak) = µ (

⋃∞n=1 (

⋃nk=1 Ak)) =

µ (⋃∞

k=1 Ak) = µ (A).Thus finally ν (A) ≥ µ (A).

22

1.3 The Lebesgue Measure 1 σ -ALGEBRAS AND MEASURES

QED

1.2.9 R

(strw rem 1.2.2)

1.2.9.1 Overlapping Σ’s It is not clear in 1.2.8 wether the Σ of µ-measurablesets contains also all the ν-measurable sets.

1.2.9.2 Cannot Generalize Beyond A It is not possible to improve thegenerality of the uniqeness statement in 1.2.8.

1.2.10 E to 1.2.9.2

(strw ex. 1.2.3)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, isΣ = {∅, X} = A.

• Proof: That both these sets are measurable is clear. That no other set ismeasurable is clear because for any non-empty set, 1 6= 2.

The Lebesgue measure will then not agree with µ.

1.3 The Lebesgue Measure

1.3.1 D Elementary Volume Premeasure

(strw def. 1.3.1)

1.3.1.1 n-dimensional Box (k-cell) Take a =

a1a2. . .an

, b =b1b2. . .bn

∈ Rn.Define

I := (a, b) :=

{∏ni=1 (ai, bi) ai < bi∀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 ai = −∞ or bi = ∞.

23


1.3.1.2 The Volume of an n-dimensional Box Define a function vol :2R

n → [0, ∞] by

vol (I) := vol ((a, b)) := vol ([a, b]) := vol ([a, b)) := vol ([a, b)) :=

{∏ni=1 bi − ai ai < bi∀i ∈ {1, . . . , n}

0 otherwise

1.3.1.3 An elementary Set An elementary set is a finite union of disjointk-cells.Extend the definition of vol to such sets:

vol

(m⋃

k=1

Ik

):=

m∑k=1

vol (Ik)

1.3.2 R

1. The set of elementary sets in Rn forms an algebra (but not a σ-algebra)–de-note it by E .

2. vol is additive.

3. vol is σ-finite.

4. vol is non-negative.

5. vol is monotone.

1.3.3 R

vol is a pre-measure on E .TODO

1.3.4 D The Lebesgue Measure Ln

(def. 1.3.2)The Lebesgue-measure Ln is the Caratheodory-Hahn extension of vol.

1.3.5 T Open Sets are Countable Unions of k-cells

(strw thrm. 1.3.1)Let G ∈ Open (Rn).Claim: ∃ {Ik} ⊂ E such that Ik ∩ Ij∀k 6= j and G =

⋃k∈N Ik

Proof: TODOBut according to 3 Ik are Ln-measurable, and so according to 1.1.13 G =⋃

k∈N Ik is Ln-measurable.

24


1.3.6 C Borel σ-Algebra

The σ-algebra of Ln-measurable sets contains the Borel σ-algebra (smallestσ-algebra containing all open sets).

1.3.7 D Borel Measures

A measure µ on Rn is called Borel if every Borel set is µ-measurable. Thus 1.3.6can be restated as “Ln is a Borel measure”.

1.3.8 T Approximation from Outside by Open Sets

(strw thrm. 1.3.2)Let A ∈ 2Rn .

• Claim: Ln (A) = inf ({Ln (G) |G ∈ Open (Rn) ∧ G ⊃ A})

• Proof:

– Step 1: ≤ The result holds for any member of the set: indeed, dueto monotonicity, for any G ∈ Open (Rn) such that G ⊃ A, Ln (A) ≤Ln (G). Thus the result must hold for the infimum of the set as well.

– Step 2: ≥Approximate Ln (A):∀δ > 0∃

{Iδk}∈ E such that

⋃k∈N I

δk ⊃ A and Ln (A)+δ >

∑k∈N vol

(Iδk).

Construct another sequence of elementary sets,{Jδk}

such that Jδk ∈Open (Rn), Jδk ⊃ Iδk and vol

(Jδk)< vol

(Iδk)+ δ · 2−k for all k ∈ N.

Thus∑

k∈N vol(Jδk)≤ δ +

∑k∈N vol

(Iδk).

By subadditivity, Ln(⋃

k∈N Jδk

)≤∑

k∈N vol(Jδk)

Thus, Ln(⋃

k∈N Jδk

)≤ Ln (A) + 2δ.

∗ Claim: inf ({Ln (G) |G ∈ Open (Rn) ∧ G ⊃ A}) ≤ Ln(⋃

k∈N Jδk

)∗ Proof:

⋃k∈N J

δk ∈ Open (Rn) and

⋃k∈N J

δk ⊃

⋃k∈N I

δk ⊃ A.

Thus in the limit of δ → 0, inf ({Ln (G) |G ∈ Open (Rn) ∧ G ⊃ A}) ≤Ln (A).

1.3.9 D Regular Borel Measure

(strw def. 1.3.4)If µ is a Borel measure on Rn that is both inner regular and outer regular,

it is called a regular Borel measure. According to Struwe: A borel measure isregular if ∀A ∈ 2Rn∃ a Borel set B ⊃ A such that µ (A) = µ (B).

25


1.3.10 C The Lebesgue Measure is a regular Borel measure

(strw cor. 1.3.2)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.

1.3.11 T Another Characterization of Measurability

(strw thrm. 1.3.3)A set A ∈ 2Rn is Ln-measurable iff ∀ε > 0∃G ∈ Open (Rn) : G ⊃ A :Ln (G\A) < ε.

• P

– ⇒TODO

– ⇐TODO

1.3.12 C Another Characterization of Measurability

(strw cor. 1.3.3)A set A ∈ 2Rn is Ln-measurable iff ∀ε > 0∃G ∈ Open (Rn) , F ∈ Closed (Rn) :G ⊃ A ⊃ F : Ln (G\A) < ε ∧ Ln (A\F ) < ε.

• PTODO

1.3.13 C Another Characterization of Measurability

(strw cor. 1.3.4)A set A ∈ 2Rn is Ln-measurable iff ∀ε > 0∃G ∈ Open (Rn) , F ∈ Closed (Rn) :

G ⊃ A ⊃ F : Ln (G\F ) < ε.

• PTODO

1.3.14 Comparison with the Jordan Measure

1.3.14.1 D The Jordan Measure A bounded set A ∈ Rn is Jordan-measurable in case µ̄J (A) = µJ (A) =: µJ (A), where

µ̄J (A) := sup ({vol (E) | E ⊂ A, A ∈ E})µ̄J (A) := inf ({vol (E) | E ⊃ A, A ∈ E})where E is the algebra of elementary sets as defined in 1.

26

1.4 The Limits of Measurability 1 σ -ALGEBRAS AND MEASURES

1.3.14.2 T Comparison with Lebesgue-measure (strw thrm 1.3.4)Let A ∈ Rn be bounded. Then µ̄J (A) ≥ Ln (A) ≥ µJ (A) and if A is

Jordan-measurable then µJ (A) = Ln (A).

• PTODO

1.3.15 T Invariance for Rigid Transformations

• Claim: The Lebesgue measure is invariant under translations and orthog-onal transformations.

• Proof: Let Φ : Rn → Rn be defined by Φ(x) := x0 + Rx for all x ∈ Rnwhere x0 ∈ Rn 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 setwith open sets.

1.4 The Limits of Measurability

1.4.1 E Vitali Sets

(strw ex 1.4.1)Let ξ ∈ R\Q and define G := {k + lξ | k, 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

1.4.2 Banach-Tarski

1.5 The Lebesgue-Stieltjes MeasureLet F be a monotonically increasing, continuous from the left function.

Define λF ([a, b)) :=

{F (b)− F (a) b ≥ a0 otherwise

∀a, b ∈ R.

Observe that {[ a, b )} is a covering class of R (but not an algebra on R).Define ΛF to be the measure induced by λF and {[ a, b )} according to 1.2.3,that is: ΛF (A) := inf ({

∑∞k=1 λF ([ak, bk)) : A ⊂

⋃∞k=1[ak, bk)}).

1.5.1 D Distance between two sets

dist (A, B) := inf ({|a− b| | a ∈ A, b ∈ B})

27

1.5 The Lebesgue-Stieltjes Measure 1 σ -ALGEBRAS AND MEASURES

1.5.2 D Metric Measure

(strw def. 1.5.1)A measure µ on Rn is metric in case ∀A,B ∈ 2Rn such that dist (A, B) > 0,µ (A ∪B) = µ (A) + µ (B).

1.5.3 T Caratheodory Criterion for Borel Measures

(strw thrm. 1.5.1)A metric measure is a Borel measure.

• PLet µ be a metric measure on Rn.

1. Step 1: If F ∈ Closed (Rn) then F is µ-measurable.Pick any B ∈ 2Rn .According to 1.1.7, we would like to show that µ (B) ≥ µ (B ∩ F ) +µ (B\F ).Assume µ (B) < ∞, otherwise we are done.Define Fk :=

{x ∈ Rn | dist ({x} , F ) ≤ 1k

}∀k ∈ N.

Observe that dist (B ∩ F, B\Fk) ≥ 1k > 0 ∀k ∈ N, so by 1.5.2:µ ((B ∩ F ) ∪ (B\Fk)) = µ (B ∩ F ) + µ (B\Fk).But µ is a measure, so by 1.1.4, Because B ⊃ (B ∩ F ) ∪ (B\Fk),µ ((B ∩ F ) ∪ (B\Fk)) ≤ µ (B).Finally we get that µ (B) ≥ µ (B ∩ F ) + µ (B\Fk)∀k ∈ N.

– Claim: limk→∞ µ (B\Fk) = µ (B\F )– Proof:

Define Rk := (Fk\Fk+1) ∩B ∀k ∈ N.Observe that {Rk} are disjoint, and that B\F = (B\Fk) ∪(⋃∞

l=k Rl)∀k ∈ N.By 1.1.4, µ (B\Fk) ≤ µ (B\F ).By 1.1.3, µ (B\F ) ≤ µ (B\Fk) +

∑∞l=k µ (Rl).

∗ Claim:∑∞

l=1 µ (Rl) < ∞∗ Proof:

Observe that dist (Ri, Rj) > 0 ∀ j ≥ i+ 2.Since µ is metric, µ (Ri ∪Rj) = µ (Ri) + µ (Rj)∀j ≥ i+ 2.Thus µ (R1 ∪R3) = µ (R1) + µ (R3)Thus by induction: µ (

⋃mk=1 R2k) =

∑mk=1 µ (R2k) and µ (

⋃mk=1 R2k+1) =∑m

k=1 µ (R2k+1).But by 1.1.4, µ (B) ≥ µ (

⋃mk=1 R2k) and µ (B) ≥ µ (

⋃mk=1 R2k+1).

So adding these two inequalities we get that 2µ (B) ≥∑m

k=1 µ (Rk).So that in the limit m → ∞, ∞ > 2µ (B) ≥

∑∞k=1 µ (Rk).

Thus limk→∞∑∞

l=k µ (Rl) = 0.Thus limk→∞ µ (B\Fk) ≤ µ (B\F ) ≤ limk→∞ µ (B\Fk) and sofinally µ (B\F ) = limk→∞ µ (B\Fk).

28


2. Step 2: Generalize from closed set to Borel sets.The set of measurable sets is a sigma-algebra, and the sigma-algebraof Borel sets is generated by closed sets.

1.5.4 T The Lebesgue-Stieltjes Measure is a regular Borel measure

(strw thrm. 1.5.2)P

• Claim: The Lebesgue-Stieltjes measure is metric.

• Proof: Take any A,B ∈ 2R such that dist (A, B) > 0.To satisfy 1.5.2, we would like to show that ΛF (A ∪B) = ΛF (A)+ΛF (B).

– ≤ Follows immediately from 1.1.3.

– ≥Define δ := dist (A, B).ΛF (A ∪B) := inf ({

∑∞k=1 λF ([ak, bk)) : A ⊂

⋃∞k=1[ak, bk)})

Approximate ΛF (A ∪B):∀ε > 0∃ a cover {[ aεk, bεk )}k∈N of A ∪ B by half-open intervals suchthat

∑k∈N λF ([ a

εk, b

εk )) < ΛF (A ∪B) + ε.

∗ Claim: WLOG bk − ak < δ∀k ∈ N∗ Proof: If ∃k0 ∈ N such that bεk0 − a

εk0

≥ δ, define a sequenceof points on R, {xi}mi=1, such that x0 := aεk0 , xm := b

εk0

and0 < xi − xi−1 < δ∀i ∈ {1, . . . ,m}.Then λF

([ aεk0 , b

εk0

))= F

(bεk0)− F

(aεk0)= F (xm)− F (x0) =

F (xm)− F (xm−1) + F (xm−1) · · · − F (x1) + F (x1)− F (x0) =λF ([ xm−1, xm ))+ · · ·+λF ([ x0, x1 )) =

∑mi=1 λF ([ xi−1, xi ))

So we might as well redefine our cover of A∪B to be: {[ aεk, bεk )}k∈N →({[ aεk, bεk )}k∈N \

{[ aεk0 , b

εk0

)})

∪ {[ xi−1, xi )}i∈{1,...,m}.Repeat this procedure for every such k0.

But recall what we defined δ to be: δ := dist (A, B).So that necessarily means that ∀k ∈ N, either [ aεk, bεk ) ∩ A = ∅or [ aεk, b

εk ) ∩ B = ∅. Thus {[ aεk, bεk )}k∈N breaks down into two

separate covers, one of A and the other of B. By the definition of theLebesgue-Stieltjes measure as the infimum, it must be smaller thanthe sum on λF on any cover. So we get:{

ΛF (A) ≤∑

k∈N such that [aεk, bεk )∩A 6=∅ λF ([ aεk, b

εk ))

ΛF (B) ≤∑

k∈N such that [aεk, bεk )∩B 6=∅ λF ([ aεk, b

εk ))

Summing up these two inequalities, we get:ΛF (A) + ΛF (B) ≤

∑k∈N λF ([ a

εk, b

εk )) < ΛF (A ∪B) + ε

29


Thus:ΛF (A) + ΛF (B) ≤ ΛF (A ∪B) + ε for all ε > 0.Thus ΛF (A) + ΛF (B) = ΛF (A ∪B) and so the Lebesgue-Stieltjesmeasure is metric.

Thus by 1.5.3, the Lebesgue-Stieltjes measure is Borel.

• Claim: The measure is Borel regular.

• Proof: Following 1.3.9, pick any A ∈ 2R. If ΛF (A) = ∞, then R would bea Borel set that has a Lebesgue-Stieltjes measure equal to that of A.

Approximate the Lebesgue-Stieltjes measure of A:

∀ε > 0∃ a cover {[ aεk, bεk )}k∈N of A by half-open intervals such that∑k∈N λF ([ a

εk, b

εk )) < ΛF (A) + ε.

In particular, we can pick ε := 1j .

Define

B :=⋂j∈N

(⋃k∈N

[ a1j

k , b1j

k )

)

Observe that B is a Borel set such that A ⊂ B ⊂(⋃

k∈N [ a1j

k , b1j

k )

). Thus

by 1.1.4, ΛF (A) ≤ ΛF (B) ≤ ΛF((⋃

k∈N [ a1j

k , b1j

k )

))≤∑

k∈N λF

([ a

1j

k , b1j

k )

)<

ΛF (A) +1j .

In the limit of j → ∞, ΛF (A) = ΛF (B) and thus the Lebesgue-Stieltjesmeasure is Borel regular.

1.5.5 T Lebesgue-Stieltjes measure on half-open intervals

(strw thrm. 1.5.3)Despite the set of half-open intervals not being an algebra:∀a < b, ΛF ([ a, b )) = λF ([ a, b )).

• P

– ≤ This direction stems immediately from the definitions combinedwith the fact that [ a, b ) ⊂ [ a, b ).

– ≥Approximate the Lebesgue-Stieltjes measure of [ a, b ):∀ε > 0∃ a cover {[ aεk, bεk )}k∈N of A by half-open intervals such that∑

k∈N λF ([ aεk, b

εk )) < ΛF ([ a, b )) + ε.

That is:∀ε > 0∃ a cover {[ aεk, bεk )}k∈N of A by half-open intervals such that∑

k∈N F (bεk)− F (aεk) < ΛF ([ a, b )) + ε.

30


F is continuous from the left (in particular, at b and at any ak).Thus:∀α > 0∃δ (α, b) > 0 such that F (b)− F (b− δ (α)) < α.∀β > 0∃γ (β, k) > 0 such that F (aεk) − F (aεk − γ (β, k)) < β · 2−kfor any k ∈ N.WLOG we may assume that aεk − γ (β, k) < bεk−1 for all k ∈ N, for,if that is not the case, we can always choose a finer cover in ourapproximation, and the result for λF should be the same accordingto the calculation of the preceding proof: 1.5.4.Observe that [a, b− δ (α, b)] is compact.Observe that {(aεk − γ (β, k) , bεk)}k∈N is an open cover for [a, b− δ (α, b)].Thus ∃ a finite cover: {(aεk − γ (β, k) , bεk)}k∈{1,..., m} for some m ∈ N.Recall that F is monotonically increasing, so:{F (b− δ (α)) ≤ F (bεm)F (aε1 − γ (β, 1)) ≤ F (a)

Adding up these two inequalities we get:F (b− δ (α))− F (a) ≤ F (bεm)− F (aε1 − γ (β, 1)) =Using the fact that F is monotonically increasing and that we as-sumed aεk − γ (β, k) < bεk−1 for all k ∈ N, we can keep adding andremoving terms from the expression like so:= F (bεm)−F (aεm − γ (β, m))+F (aεm − γ (β, m))−F (aε1 − γ (β, 1)) ≤F (bεm) − F (aεm − γ (β, m)) + F

(bεm−1

)− F (aε1 − γ (β, 1)) ≤ · · · ≤∑m

k=1 F (bεk)− F (aεk − γ (β, k)).

Plugging in the results from F being left continuous we get:F (b)−F (a) < F (b− δ (α))−F (a)+α ≤

∑mk=1 F (b

εk)−F (aεk − γ (β, k))+

α ≤∑m

k=1

(F (bεk)− F (aεk) + β · 2−k

)+α ≤

∑∞k=1

(F (bεk)− F (aεk) + β · 2−k

)+

α =∑∞

k=1 F (bεk)− F (aεk) + β + α =

∑k∈N λF ([ a

εk, b

εk )) + β + α <

ΛF ([ a, b )) + ε+ β + α

Thus finally:λF ([ a, b )) = F (b) − F (a) < ΛF ([ a, b )) + ε + β + α for all ε >0, α > 0, β > 0.

QED

1.5.6 E Familiar Examples

(ex. 1.5.1)

1. If F (x) := x∀x ∈ R the resulting Lebesgue-Stieltjes measure is theLebesgue measure.

2. If F (x) :=

{1 x ∈ (0, ∞)0 x ∈ ( −∞, 0 ]

∀x ∈ R then λF ([a, b)) :=

{1 x ∈ [ a, b )0 otherwise

∀a, b ∈

R and the resulting Lebesgue-Stieltjes measure is the Dirac measure, which

31

1.6 The Hausdorff Measure 1 σ -ALGEBRAS AND MEASURES

would be just the indicator function of a set at 0: ΛF (A) = χA (0) ∀A ∈2R

n

.

1.6 The Hausdorff MeasureLet s > 0, δ > 0, A ∈ 2Rn .

Define

Hsδ (A) := inf

({ ∞∑k=1

rsk | A ⊂∞⋃k=1

Brk (xk) ∧ rk < δ

})

Note that in the definition, we are intetionally unspecific about where theopen balls are located {xk}. They can be somewhere, just as long as the aboveconditions are met. The infimum is taken over all such sets of open balls.

Claim: Hsδ is a measure on Rn.Proof: The open balls in Rn form a covering class: since we can cover Rn

by a countable number of open balls, and we can take our class to include ∅.Brk (xk) 7→ rsk is a non-negative set function, and so by 1.2.3 Hsδ is a measure

on Rn.Claim: Hsδ1 (A) ≤ H

sδ2(A) whenever δ1 ≥ δ2 and for all A ∈ 2R

n

and s > 0.Proof: Hsδ1 (A) ≤

∑∞k=1 r

sk for all {Brk (xk)} such that A ⊂

⋃∞k=1 Brk (xk) ∧

rk < δ1.On the other hand, ∀ε > 0, ∃

{Bρεk (yk)

}such that A ⊂

⋃∞k=1 Bρεk (yk)∧ ρ

εk <

δ2 and∑∞

k=1 (ρεk)

s< Hsδ2 (A) + ε.

Observe that δ2 ≤ δ1, so that actually ρεk < δ2 ≤ δ1, and{Bρεk (yk)

}is also a

cover in the set of covers over which the infimum would be invoked to computeHsδ1 (A).

Thus we must have Hsδ1 (A) ≤∑∞

k=1 (ρεk)

s< Hsδ2 (A) + ε for all ε > 0.

As a result we conclude that Hsδ is monotonically decreasing in δ, and sothe limit limδ→0+ Hsδ (A) = sup ({Hsδ (A) | δ > 0}) must exist (though it maybe ∞) ∀A ∈ 2Rn .

1.6.0.1 D The Hausdorff Measure (def. 1.6.1)For any A ∈ 2Rn and s ≥ 0 define the s-dimensional Hausdorff Measure to

be

Hs (A) :=

{limδ→0+ Hsδ (A) s > 0|A| s = 0

1.6.1 T The Hausdorff Measure is a regular Borel measure on Rn

(thrm 1.6.1)P

32


1.6.1.1 Step 1 Claim: Hs is a measure.Proof:If s = 0 then Hs 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)

Hsδ (∅) = 0 and so we get Hs (∅) = 0. This satisfies 1.Next, assume that {Ak} ⊂ Rn is a countable sequence of sets such that

A ⊂⋃

k∈N Ak. Because Hsδ is a measure, we know that Hsδ (A) ≤∑

k∈N Hsδ (Ak).On the other hand, because Hs is the supremum over all Hsδ with δ > 0,

Hsδ (Ak) ≤ Hs (Ak) for all k ∈ N,∑

k∈N Hsδ (Ak) ≤∑

k∈N Hs (Ak).So that Hsδ (A) ≤

∑k∈N Hs (Ak), ∀δ > 0.

Thus in the limit of δ → 0+ we get 2.

1.6.1.2 Step 2 Claim: Hs is a metric measure.Proof:Take any A,B ∈ 2Rn such that dist (A, B) > 0.If s = 0 then Hs is the counting measure, which is evidently metric.If s > 0:≤ Since we just showed that Hs is a measure, we know by 1.1.3 that

Hs (A ∪B) ≤ Hs (A) +Hs (B).≥

Firstly, we know that ∀δ > 0, Hsδ (A ∪B) ≤ Hs (A ∪B).Next, given any δ > 0, ∀ε > 0∃

{Brεk (x

εk)}

such that A∪B ⊂⋃

k∈N Brεk (xεk),

rεk < δ and∑

k∈N (rεk)

s< Hsδ (A ∪B) + ε.

Thus given any δ > 0, ∀ε > 0∃{Brεk (x

εk)}

such that A∪B ⊂⋃

k∈N Brεk (xεk),

rεk < δ and∑

k∈N (rεk)

s< Hs (A ∪B) + ε.

What if we pick some δ∗ such that 0 < δ∗ < dist (A, B)?Then our cover

{Brεk (x

εk)}

decomposes into two disjoint covers, one whichcovers only A and another which covers only B: A ⊂

⋃k∈NA Brεk (x

εk) and

B ⊂⋃

k∈NB Brεk (xεk) where NA :=

{k ∈ N | Brεk (x

εk) ∩A 6= ∅

}and similarly

for NB .So that Hsδ∗ (A) ≤

∑k∈NA (r

εk)

s and Hsδ∗ (B) ≤∑

k∈NB (rεk)

s.Thus:Hsδ∗ (A) +Hsδ∗ (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 1.5.3 Hs is Borel.

1.6.1.3 Step 3 Claim: Hs is Borel regular.Proof: Take any A ∈ 2Rn . We need to find a Borel set, B, such that B ⊃ A

and Hs (A) = Hs (B).

33


If s = 0, and A is infinite we can just take the whole space, which is a Borelset. If A is finite then it is itself a Borel set.

If s > 0:Assume that Hs (A) < ∞, otherwise we can just take the Borel set to be the

whole space.Given any δ > 0, ∀ε > 0∃

{Brεk (x

εk)}

such that A ⊂⋃

k∈N Brεk (xεk), r

εk < δ

and∑

k∈N (rεk)

s< Hsδ (A) + ε < Hs (A) + ε.

Pick ε := 1j .

Define our Borel set to be B :=⋂

j∈N

(⋃k∈N Br1/jk

(x1/jk

)).

Observe that B is a Borel set such that A ⊂ B ⊂(⋃

k∈N Br1/jk

(x1/jk

)). Thus

by 1.1.4, Hsδ (A) ≤ Hsδ (B) ≤ Hsδ((⋃

k∈N Br1/jk

(x1/jk

)))≤∑

k∈N

(r1/jk

)s<

Hs (A) + 1j .In the limit of j → ∞ and δ → 0, Hs (A) = Hs (B) and thus the Hausdorff

measure is Borel regular.

1.6.2 T Preparation for Hausdroff Dimension

(lemm. 1.6.1)Let A ∈ 2Rn and 0 ≤ s < t < ∞. Then:

1. Hs (A) < ∞ ⇒ Ht (A) = 0

P

Assume that Hs (A) < ∞. Thus Hsδ (A) < ∞ for any δ > 0.Pick some δ > 0.

∀ε > 0∃{Brεk (x

εk)}

such that A ⊂⋃

k∈N Brεk (xεk), r

εk < δ and

∑k∈N (r

εk)

s<

Hsδ (A) + ε < Hs (A) + ε.In addition as

{Brεk (x

εk)}

is a bona fida cover of A with rεk < δ, we knowas well that:

Htδ (A) ≤∑

k∈N (rεk)

t

But observe that (rεk)t= (rεk)

t−s(rεk)

s< δt−s (rεk)

s.

Thus∑

k∈N (rεk)

t ≤ δt−s∑

k∈N (rεk)

s.

Thus Htδ (A) ≤ δt−s (Hs (A) + ε) for all δ > 0, and for all ε > 0.In the limit δ → 0, as Hs (A) < ∞ and ε can be chosen as we wish, weget Ht (A) = 0.

2. Ht (A) > 0 ⇒ Hs (A) = ∞

P

This is the logical negation of the previous claim.

34


Note: This theorem says that the range of values Hs (A) can take for a given Ais basically only three: ∞ for s = 0, some finite value at some s0 ∈ [0, ∞), andthen 0 for all s > s0.

1.6.3 E Compact Sets

(ex. 1.6.1)Let Q := [−1, 1]n ⊂ Rn.

2−nLn (Q) ≤ Hn (Q) ≤ 2−nnn/2Ln (Q)

1 ≤ Hn (Q) ≤ nn/2

P

≤By its definition, Hnδ (Q) ≤

∑k∈N (rk)

n for all {rk} such that rk < δ∀k ∈ Nand Q ⊂

⋃k∈N Brk (xk) for some {xk}.

Define the following cover:Define mδ := dlog2

(√n

2δ

)e+ 1.

Thus mδ > log2(√

n2δ

). Thus 2−mδ−1

√n < δ .

Next define a cover of 2(mδ+1)n balls, each of radius r := 2−mδ−1√n .

If we divided Q into n-dimensional cubes, each of side-length 2−mδ , it isclear that the cover we just defined will have each cube contained in each ball.

ThusHnδ (Q) ≤∑

k∈N (rk)n= 2(mδ+1)nrn = nn/2.

The claim follows from the fact that Ln (Q) = 2+n.≥

Define ωn := Ln (B1 (0)). For example, ω3 = 4π3 . Thus Ln (Br (x)) = ωnr

n.Thus by the σ-subadditivity of Ln, for each cover of B1 (0) with {Brk (xk)}:Ln (B1 (0)) ≤

∑k∈N Ln (Brk (xk)) = ωn

∑k∈N r

nk

Thus we get the relation: 1 ≤∑

k∈N rnk .

Going on to the Hausdorff measure, ∀ε > 0∃{Brεk (x

εk)}

such that B1 (0) ⊂⋃k∈N Brεk (x

εk), r

εk < δ and

∑k∈N (r

εk)

n< Hnδ (B1 (0)) + ε.

Thus we get the relation Hsδ (B1 (0)) ≥ 1.At any rate, since Hnδ is monotone and Q ⊃ B1 (0),Hnδ (Q) ≥ Hnδ (B1 (0)).Thus we get our result in the limit of δ → 0.

1.6.4 R Connection between Lebesgue and Hausdorff measures

(rem. 1.6.1)In fact ∀A ∈ 2Rn which are Lebesgue-measurable, Ln (A) = ωnHn (A).

35


1.6.5 E Different Dimension Hausdorff Measure

(ex. 1.6.2)∀A ∈ 2Rn and s > n, Hs (A) = 0.P

Cover Rn with n-dimensional cubes Ql. By 1.6.3, each cube has Hn (Ql) ≤nn/2 < ∞. Thus for each cube by 1, Hs (Ql) = 0. Thus using 1.1.15.1,Hs (Rn) = 0. Thus by monotonicity, Hs (A) = 0.

1.6.6 D Hausdorff Dimension

(def. 1.6.2)Define dimH (A) := inf ({s ∈ R | s > 0 ∧Hs (A) = 0}) to be the Hausdorff

dimension of A ∈ 2Rn .

1.6.7 E The Cantor “Dust”

(ex. 1.6.3)(Amann-Escher-AnalysisIII-2009-English-Ed-pg. 39)Define A0 := [0, 1]

2 ⊂ R2. Define A1 by dividing A0 into 16 identicalsquares (labelling the top left one 1 all the way to the bottom right one 16), andremoving from it 12 open squares so that in the end only squares 2, 5, 12, 15are left (as closed squares).

Define A2 is a similar fashion within each of the four remaining squares ofA1 (so that it would have in the end 16 smaller squares), and so on.

Define A :=⋂

k∈N Ak.

• Claim: dimH (A) = 1.

• Proof:

– Step 1: H1 (A) ≤√22

Each Ak consists of 4k squares, each of edge-length 4−k. Thus Akcan be covered by 4k balls, each of radius, 4−k

√22 (this is in fact

the smallest radius we could imagine for a ball to cover a square ofedge-length 4−k).

Thus for a given δ > 0, H1δ (Ak) ≤∑4k

j=1 4−k

√22 =

√22 , for all k ∈ N

such that 4−k√22 < δ.

But Ak ⊃ A, so by monotonicity of H1δ (which is a measure), we getthat:H1δ (A) ≤

√22 so that in the limit of δ → ∞, we get the required

result.

– Step 2: H1 (A) ≥ 12Estimate H1 (A):

36

1.7 Radon Measures 1 σ -ALGEBRAS AND MEASURES

Given any δ > 0, ∀ε > 0∃{Brεk

([xεkyεk

])}such that A ⊂

⋃k∈N Brεk

([xεkyεk

]),

rεk < δ and∑

k∈N rεk < H1δ (A) + ε < H1 (A) + ε.

Let π([

xy

]):= x be the projection map of the first coordinate.

Observe that π (A) = [0, 1] (why?).

Observe that since A ⊂⋃

k∈N Brεk

([xεkyεk

]), π (A) ⊂

⋃k∈N π

(Brεk

([xεkyεk

])).

Thus [0, 1] ⊂⋃

k∈N (xεk − rεk, xεk + rεk).

So by monotonicity of the Lebesgue-measure, we get:L1 ([0, 1]) ≤ L1

(⋃k∈N (x

εk − rεk, xεk + rεk)

)≤∑

k∈N L1 ((xεk − rεk, xεk + rεk))But since these are just intervals, the Lebesgue measure of them issimply their length:1 ≤

∑k∈N 2r

εk

So that finally we get the fact that:H1 (A) ≥ 12 + ε for any ε > 0.

Thus 12 ≤ H1 (A) ≤

√22 .

– Step 4: dimH (A) ≤ 1.Take any s > 1. By 1, Hs (A) = 0. Thus Hs (A) = 0∀s > 1. So by1.6.6, dimH (A) ≤ 1.

– Step 5: dimH (A) ≥ 1.Take any s < 1. By 2, Hs (A) = ∞. Thus Hs (A) = ∞∀s < 1. So by1.6.6, dimH (A) ≥ 1.

�

1.7 Radon MeasuresLet µ be a measure on Rn.

1.7.1 D The Radon Measure

(def. 1.7.1)µ is a Radon measure if both the following conditions hold:

1. µ is Borel regular.

2. µ (K) < ∞∀K ∈ 2Rn such that K is compact.

This second condition is equivalent to µ being locally finite.

37


1.7.2 E

(ex. 1.7.1)

1. The Lebesgue and Lebesgue-Stieltjes measures on Rn 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 ∈ 2Rn is µ-measurable, then the measureν, defined by ν (B) := µ (A ∩B)∀B ∈ 2Rn is also a Radon measure.Proof:

• Step 1: ν is a measure.– ν (∅) = µ (A ∩∅) = µ (∅) = 0– Let B ⊂

⋃∞k=1 Bk.

ν (B) = µ (B ∩A)But note that B ∩A ⊂

⋃∞k=1 A ∩Bk.

So that µ (B ∩A) ≤∑

k∈N µ (A ∩Bk) =∑

k∈N ν (Bk).

• Step 2: ν (K) < ∞∀K ∈ 2Rn 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 ∈ 2Rn be a Borel set, and take any B ∈ 2Rn . Then:ν (B) = µ (B ∩A). Since µ is Borel, we know that C is measurable.So we may invoke 1.1.6 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.Take any B ∈ 2Rn . We need to find some C ⊃ B such that C is Boreland 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, invoke 1.1.7 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


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 weget:µ (A ∩B) ≤ µ (C ∩A)To summarise, we achieved at the fact that µ (C\A) ≤ 0 whichmeans that µ (C\A) = 0 (because µ is a non-negative set func-tion).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 equa-tion: 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 invoking 1.1.4 and1.1.3, 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 Rn into a countable union of compact n-dimensional-cubes: Rn = ∪l∈NQl, which only overlap on their “faces”.Thus µ (Ql) < ∞ becuase µ is a Radon measure.By monotonicity then µ (Ql ∩B) < ∞.So we may apply “Case 1” on each Ql ∩ B to get a Borel set Clsuch that Cl ⊃ Ql ∩B and ν (Cl) = ν (Ql ∩B).Thus ∪l∈NCl is a Borel set, ∪l∈NCl ⊃ B.Define {Rl} to be the same n-dimensional-cubes as {Ql}, onlyinstead of being closed they are half-closed, so that they aredisjoint, and their disjoint union now forms Rn. Observe thatν (Ql ∩B) = ν (Rl ∩B).Thus: ν (∪l∈NCl) =

∑l∈N ν (Cl) =

∑l∈N ν (Ql ∩B) and also

that B =⋃

l∈N B∩Rl thus ν (B) = ν(⋃

l∈N B ∩Rl)=∑

l∈N ν (B ∩Rl) =∑l∈N ν (B ∩Ql).

�

1.7.3 T Approximation of Radon Measures by Open or CompactSets

(thrm. 1.7.1)Let µ be a Radon measure on Rn.Similar to 1.3.8, we have the same theorem for Radon measures. TODO

39

1.8 Topological Definitions 2 MEASURABLE FUNCTIONS

1.8 Topological Definitions

1.8.1 D Locally Compact Space

A topological space X is locally compact if ∀x ∈ X∃V ∈ Open (X) : x ∈V ∧ closureX (V ) is compact.

1.8.2 D Semicontinuous Functions

Let f be a real (or extended-real–in which case the following intervals are half-closed to include the infinities) function on a topological space.

1.8.2.1 Lower Semincontinuous If f−1 ((α, ∞)) ∈ Open (X)∀α ∈ Rthen f is lower semicontinuous.

Equivalently, ∀x0 ∈ X, ∀� > 0∃ a neighborhood U of x0 such that f (x) −f (x0) ≤ � for all x ∈ U .

1.8.2.2 Upper Semicontinuous If f−1 ((−∞, α)) ∈ Open (X)∀α ∈ Rthen f is lower semicontinuous.

Equivalently, ∀x0 ∈ X, ∀� > 0∃ a neighborhood U of x0 such that f (x0) −f (x) ≤ � for all x ∈ U .

1.8.3 D Support

The support of a complex function f on a topological space X is closureX(f−1 (C\ {0})

).

1.8.4 D Continuous Functions with Compact Support

Cc (X) := {f : X → C | f is continuous and support (f) is compact}Observe that Cc (X) is a vector space.

2 Measurable Functions

2.1 Definition and Elementary Characteristics

2.1.1 T Approximation with Step Functions

(strw thrm. 2.1.2)Let f ∈ [0, 1]Ω be µ-measurable.

• Claim: ∃ µ-measurable sets Ak ⊂ Ω ∀k ∈ N such that f =∑

k∈N1kχAk

• PDefine A1 := f−1 ([1, ∞]). A1 is µ-measurable.

Define inductively Ak :={x ∈ Ω : f (x) ≥ 1k +

∑k−1j=1

1jχAj (x)

}∀k ∈

N\ {1}.

40

2.2 Lusin’s Theorem and Egoroff’s Theorem2 MEASURABLE FUNCTIONS

– Claim: f (x) ≥∑

k∈N1kχAk (x) ∀x ∈ Ω

– Proof: Let x ∈ Ω be given.∗ Case 1: sup ({k ∈ N : x ∈ Ak}) = ∞. Then by the definition of

Ak, f (x) = ∞ and we are done.∗ Case 2: ∃k0 ∈ N such that k0 = max ({k ∈ N : x ∈ Ak}). Then

x ∈ Ak0 , so f (x) ≥ 1k0 +∑k0−1

j=11jχAj (x) ≥

∑k0j=1

1jχAj (x) =∑∞

j=11jχAj (x). The last equality is true because χAj (x) =

0∀j ≥ k0 by assumption.– Claim: f (x) ≤

∑k∈N

1kχAk (x) ∀x ∈ Ω

– Proof: Let x ∈ Ω be given.∗ Case 1: f (x) = ∞, then x ∈ Ak∀k ∈ N and so

∑k∈N

1kχAk (x) =∑

k∈N1k = ∞ = f (x).

∗ Case 2: f (x) = 0, then x /∈ Ak∀k ∈ N and so∑

k∈N1kχAk (x) =∑

k∈N1k · 0 = 0 = f (x).

∗ Case 3: f (x) ∈ (0, ∞),· If ∃k0 ∈ N such that x ∈

⋂k≥k0 Ak, then

∑k∈N

1kχAk (x) ≥∑

k≥k01kχAk (x) =

∑k≥k0

1k = ∞ thus

∑k∈N

1kχAk (x) >

f (x) which contradicts the earlier claim.So we conclude that x /∈ Ak for infinitely many k ∈ N.

Let K := {k ∈ N : x /∈ Ak}.Then ∀k ∈ K, by the definition of Ak, we get f (x) < 1k +∑k−1

j=11jχAj (x).

Thus ∀k ∈ K, 0 ≤ f (x)−∑k−1

j=11jχAj (x) <

1k and so in the limit

of k → ∞, k ∈ K we get:f (x) ≤

∑∞j=1

1jχAj (x)

2.1.2 R Monotone Convergence and Real Functions

(strw rem. 2.1.3)

1. Observe that{∑k

j=11jχAj (x)

}k∈N

monotonically converges to f . If f isbounded, the convergence is even uniform.

2. By approximation f+ and f− we could approximate every µ-measurablefunction f : Ω → R by step functions.

2.2 Lusin’s Theorem and Egoroff’s TheoremLet µ be a Radon measure on Rn, let Ω ⊂ Rn such that µ (Ω) < ∞.

41


2.2.1 Egorov’s (Egoroff) Theorem

Let {fk} ∈(RΩ)N

be a sequence of µ-measurable functions, f ∈ RΩ beµ-measurable and µ-almost-everywhere finite. Furthermore, limk→∞ fk (x) =f (x) for µ-almost-all x ∈ Ω.

• Claim: ∀δ > 0∃F ⊂ Ω such that F is compact, µ (Ω\F ) < δ andlimk→∞ sup (|fk (x)− f (x)| : x ∈ F ) = 0.

• P Let δ > 0 be given.Define ∀i, j ∈ N the following set:

Ci,j :=

∞⋃k=j

{x ∈ Ω : |fk (x)− f (x)| > 2−i

}Ci,j is µ-measurable, because f, fk are µ-measurable.

Observe that Ci, (j+1) ⊂ Ci, j ∀i, j ∈ N.Because limk→∞ fk (x) = f (x) for µ-almost-all x ∈ Ω, and µ (Ω) < ∞,µ (Ci, 1) = 0.

So using theorem 1.1.2, we get that limj→∞ µ (Ci, j) = µ(⋂

j∈N Ci, j

).

But again, by assumption that limk→∞ fk (x) = f (x) for µ-almost-allx ∈ Ω, µ

(⋂j∈N Ci, j

).

Finally: limj→∞ µ (Ci, j) = 0∀i ∈ N.Thus ∀i ∈ N∃N (i) ∈ N :

(j ≥ N (i) =⇒ µ (Ci, j) < δ2−i−1

).

Define A := Ω\(⋃

i∈N Ci, N(i)).

Then µ (Ω\A) = µ(⋃

i∈N Ci, N(i))≤∑

i∈N µ(Ci, N(i)

)< δ2 .

Furthermore, by the definition of Ci, j , we see that ∀i ∈ N sup ({|fk (x)− f (x)| : x ∈ A}) ≤2−i∀k ≥ N (i).By theorem 1.7.1, since A is µ-measurable and µ is a Radon measure,µ (A) = sup ({µ (F ) : F ⊂ A ∧ F compact}).So ∀ε > 0∃Fε ⊂ A such that Fε is compact and µ (A)− ε < µ (Fε).In addition, Fε being compact, is closed, and thus, measurable, and so,µ (A) = µ (A ∩ Fε) + µ (A\Fε) = µ (Fε) + µ (A\Fε). Thus µ (A\Fε) =µ (A)− µ (Fε) < ε.Thus µ

(A\Fδ/2

)< δ/2.

Thus µ(Ω\Fδ/2

)≤ µ (Ω\A) + µ

(A\Fδ/2

)< δ.

So Fδ/2 is the compact set we were looking for.

QED

42


2.2.2 Lusin’s Theorem

Let f ∈ RΩ be µ-measurable and µ-almost-everywhere finite.

• Claim: ∀δ > 0∃F ⊂ Ω such that:

1. F is compact2. µ (Ω\F ) < δ3. f |F ∈ C (F, R)

• P

– Case 1: f is a step functionAssume f =

∑Ii=1 biχBi where I ∈ N, Ω =

⋃Ii=1 Bi, Bi∩Bj = ∅∀i 6=

j.By theorem 1.7.1, since Bi are µ-measurable and µ is a Radon mea-sure, µ (Bi) = sup ({µ (F ) : F ⊂ A ∧ F compact}). So we can choosefor each i ∈ {1, . . . , I} a compact set Fi ⊂ Bi such that µ (Bi\Fi) <δ · 2−i ∀i ∈ {1, . . . , I}.Since Bi are disjoint, Fi are also disjoint, and as they are all compact,dist (Fi, Fj) > 0∀i 6= j. Thus f is locally constant, and is thuscontinuous on F :=

⋃Ii=1 Fi. Furthermore, F ⊂ Ω is compact, and:

µ (Ω\F ) =︸︷︷︸⋃Ii=1 Bi=Ω

µ(⋃I

i=1 (Bi\Fi))≤∑I

i=1 µ (Bi\Fi) < δ

– Case 2: f is the limit of step functions, that is, {fk} ∈(RΩ)N is a

sequence of µ-measurable step functions:

f (x) = limk→∞

fk (x)∀x ∈ Ω

where

fk (x) =

k∑j=1

1

jχAj =

Ik∑i=1

bikχBik∀k ∈ N

and

Bi ∩Bj = ∅∀i 6= j ∧ Ω =I⋃

i=1

Bik

and

bik =∑

Bik⊂Aj

1

j∀i ∈ {1, . . . , Ik} , k ∈ N

for each δ > 0 and fk choose a compact set as in “Case 1”, Fk ⊂ Ωwhere

43

2.3 Convergence in Measure 2 MEASURABLE FUNCTIONS

µ (Ω\Fk) < δ · 2−k−1 ∧ fk|Fk ∈ C (Fk, R) ∀k ∈ N

Using Egrovo’s theorem, pick a compact set F0 ⊂ Ω such that:

µ (Ω\F0) <δ

2∧ lim

k→∞sup ({|fk (x)− f (x)| : x ∈ F0}) = 0

Finally define F :=⋂

k∈N∪{0} Fk. Observe that F is compact, andthat

µ (Ω\F ) ≤ µ

⋃k∈N∪{0}

(Ω\Fk)

≤ ∑k∈N∪{0}

µ (Ω\Fk) < δ

and since F ⊂ F0 uniform convergence follows limk→∞ sup ({|fk (x)− f (x)| : x ∈ F}) =0.Thus the continuity of fk|F ∀k ∈ N gives us the continuity of f |F .QED

2.3 Convergence in MeasureLet µ be again any measure on Rn, Ω ∈ 2Rn be a µ-measurable set, and let{fk} ∈

(RΩ)N

be a sequence of µ-measurable functions, f ∈ RΩ be µ-measurableand µ-almost-everywhere finite.

2.3.1 D Convergence In Measure

(strw def. 2.3.1){fk}k∈N converges to f in measure (fk

µ−→ f as k → ∞) iff:

∀ε > 0 limk→∞

µ ({x ∈ Ω : |f (x)− fk (x)| > ε}) = 0

2.3.2 T Convergence Almost Everywhere Implies In Measure

(strw thrm. 2.3.1)If µ (Ω) < ∞ and limk→∞ fk = f µ-almost-everywhere then {fk}k∈N convergesto f in measure (fk

µ−→ f as k → ∞).

• PLet ε > 0 be given.

Define Cj :=⋃∞

k=j {x ∈ Ω : |fk (x)− f (x)| > ε} ∀j ∈ N. Since f andfk are both measurable, so is Cj . In addition, Cj+1 ⊂ Cj ∀j ∈ N andµ (C1) ≤ µ (Ω) < ∞ by assumption.

44

3 INTEGRATION

Thus µ ({x ∈ Ω : |fk (x)− f (x)| > ε}) ≤ µ (Cj)∀j ∈ N and so in the limitj → ∞:

limj→∞ µ ({x ∈ Ω : |fk (x)− f (x)| > ε}) ≤ limj→∞ µ (Cj) = µ(⋂

j∈N Cj

)=︸︷︷︸

limk→∞ fk=f µ−a.e.

0

2.3.3 T Convergence in Measure gives a subsequence that Con-verges Almost Everywhere

(strw thrm 2.3.2)If {fk}k∈N converges to f in measure (fk

µ−→ f as k → ∞), then ∃ a subsequenceΛ ⊂ N such that

limk→∞∧k∈Λ

fk = f µ− a.e.

• PTODO

3 Integration

3.1 Definition and Elementary Characteristics

3.1.1 D σ-Step Function

(strw def. 3.1.1)g ∈ RΩ is called a σ-step function if |g (Ω)| ≤ ℵ0.

3.1.2 D Integral for σ-Step Functions

1. If g ∈ [0, ∞]Ω is µ-measurable σ-step-function, define

ˆΩ

gdµ :=

{∑α∈(0,∞) αµ

(g−1 ({α})

)g < ∞µ− a.e.

∞ otherwise

2. If g ∈ RΩ is µ-measurable σ-step-function such that´Ωg+dµ < ∞ ∨´

Ωg−dµ < ∞, define

ˆΩ

gdµ :=

{∑α∈R\{0} αµ

(g−1 ({α})

)|g| < ∞µ− a.e.

±∞ µ(g−1 ({±∞})

)> 0

45

3.2 Convergence Theorems 3 INTEGRATION

3.2 Convergence Theorems

3.3 Absolute continuity of the Integral

3.4 Vitali’s TheoremVitali’s Theorem is an optimal convergence theorem.Let f, fk ∈ R

Ω ∩ L1 (µ) ∀k ∈ N.

3.4.1 D Uniformly Integrable

(strw def. 3.4.1)The family {fk}k∈N is uniformly integrable if ∀ε > 0∃δ (ε) > 0 such that(´

A|fk| dµ

)< ε whenever µ (A) < δ (ε) and ∀k ∈ N.

3.4.2 T Vitali’s Theorem

(strw thrm. 3.4.1)If µ (Ω) < ∞ then the following two statements are equivalent:

1. {fk}k∈N converges to f in measure (fkµ−→ f as k → ∞) ∧ {fk}k∈N is

uniformly integrable.

2. limk→∞´Ω|fk − f | dµ = 0.

• P

– 1 ⇐= 2∗ limk→∞

´Ω|fk − f | dµ = 0 leads to convergence in measure (...).

∗ For the uniform integrability of {fk}:Let ε > 0 be given, and let k0 (ε) ∈ N be such that

´Ω|fk − f | dµ <

ε∀k ≥ k0 (ε).Next, pick a δ > 0 such that if µ (A) < δ then:·Á|f | dµ < δ

· max({´

A|fk| dµ : k ∈ {1, . . . , k0 (ε)}

})< ε

(This is possible, in part, due to theorem 3.3.1)Thus

Á|fk| dµ ≤

Á|f | dµ +

Á|fk − f | dµ ≤ 2ε ∀k ≥ k0 (ε), if

µ (A) < δ.

– 1 =⇒ 2Assume, contrarily, that:

lim supk→∞

ˆΩ

|fk − f | dµ > 0

WLOG we may even assume (by choice of a suitable subsequence)that:

46

3.5 Lp (Ω, µ) spaces4 COMPLEX MEASURES (RUDIN RCA CHAPTER 6)

lim infk→∞

ˆΩ

|fk − f | dµ > 0

Since {fk} converges in measure to f , using theorem 2.3.2, we mayassume that there exists a subsequence which converges µ-almost-everywhere: ∃Λ ⊂ N such that limk∈Λ∧k→∞ fk = f µ-almost-everywhere.Let ε > 0 be given.We may pick a δ > 0 such that if A ⊂ Ω∧µ (A) < δ then

´A|f | dµ <

ε ∧´A|fk| dµ < ε∀k ∈ N.

Using Egorov’s theorem, for this δ > 0 we may pick a F ⊂ Ω which iscompact, such that µ (Ω\F ) < δ∧limk→∞∧k∈Λ sup ({|fk (x)− f (x)| : x ∈ F}) =0.Pick a k0 ∈ N such that

sup ({|fk (x)− f (x)| : x ∈ F}) <ε

µ (Ω)∀k ∈ Λ ∧ k ≥ k0

Thus for k ≥ k0 and k ∈ Λ:´Ω|fk − f | dµ =

´Ω\F |fk − f | dµ+

´F|fk − f | dµ ≤

´Ω\F (|f |+ |fk|) dµ+´

Fsup ({|fk (x)− f (x)| : x ∈ F}) dµ ≤ 3ε

Since ε > 0 was arbitrary, we arrive at a contradiction to our as-sumption.

QED

3.5 Lp (Ω, µ) spacesEssential supremum: the smallest value of a function, such that the measure ofthe set on which it is higher is zero.

4 Complex Measures (Rudin RCA Chapter 6)

4.1 Total VariationLet M be a σ-algebra in a set X. Let E ∈ M.

4.1.1 D A Partition

A partition on E is a countable collection {Ei} ⊂ M such that:

1. Ei ∩ Ej = ∅∀i 6= j

2. E =⋃

i∈N Ei

47

4.1 Total Variation4 COMPLEX MEASURES (RUDIN RCA CHAPTER 6)

4.1.2 D A Complex Measure

A complex measure µ on M is a function µ : M → C such that for every E ∈ M:

1. ∃∑

i∈N µ (Ei)

2. µ (E) =∑

i∈N µ (Ei)

for any partition {Ei} of E.Observe: The series, if it converges, converges absolutely, since every re-

arrangement of the sequence must alos converge (since union doesn’t care aboutorder).

4.1.3 D Total Variation Measure of µ

If µ is a given complex measure on M, define

|µ| (E) := sup

({∑i∈N

|µ (Ei)| | {Ei} is a partition of E

})∀E ∈ M

Observe: |µ| (E) ≥ |µ (E)|, but in general |µ| (E) 6= |µ (E)|.Note: Sometimes the term “the total variation of µ” denotes |µ| (X).

4.1.4 T |µ| is a positive measure on M

(T 6.2)TODO (in paper)

4.1.5 T

(T 6.4)If µ is a given complex measure on X|µ| (X) < ∞

• P|µ (E)| ≤ |µ| (E)

4.1.6 D

If µ and λ are two complex measures on the same σ-algebra M and c ∈ C,define:

(µ+ λ) (E) := µ (E) + λ (E)(cλ) (E) := cµ (E)for any E ∈ M.

• Claim: µ+ λ and cµ are complex measures.

• Thus the collection of all complex measures on a given M is a vector space.

48

4.2 Absolute Continuity4 COMPLEX MEASURES (RUDIN RCA CHAPTER 6)

• Define a norm on this vector space by ||µ|| := |µ| (X). This is a normedlinear space.

4.1.7 Positive and Negative Variations–Jordan Decomposition of µ

If µ is a real measure on a σ-algebra M, then:µ+ := 12 (|µ|+ µ), the positive variation of µµ− := 12 (|µ| − µ), the negative variation of µµ+ and µ− are both positive measures on M, and they are bounded by T6.4.

4.2 Absolute Continuity(6.7)

Let µ be a positive measure on a σ-algebra M and let λ be an arbitrarymeasure (complex xor positive) on M.

• λ is absolutely continuous with respect to µ, if λ (E) = 0∀E ∈ M suchthat µ (E) = 0.

• If ∃A ∈ M such that λ (E) = λ (A ∩ E) for any E ∈ M then λ is concen-trated on A. This is equivalent to E ∩A = ∅ =⇒ λ (E) = 0.

• Suppose λ1 and λ2 are two measures on M, and suppose that ∃ two disjointsets A and B such that λ1 is concentrated on A and λ2 is concentrated onB. Then λ1 and λ2 are mutually singular .

4.2.1 σ-finite Measures

(6.9 Lemma)If µ is a positive σ-finite measure on a σ-algebra M in a set X, then there

is a function w ∈ L1 (µ) such that 0 < w (x) < 1∀x ∈ X.

4.2.2 The Lebesgue-Radon-Nikodym Theorem

(theorem 6.10)Let µ be a positive σ-finite measure on a σ-algebra M in a set X, and let λ

be a complex measure on M.

1. ∃ a unique pair of complex measures λa, λs on M such that:

• λ = λa + λs• λa is absolutely continuous with respect to µ.• λs and µ are mutually singular.• If λ is positive and finite, then so are λa and λs.

2. (The Radon-Nikodym theorem) ∃ a unique h ∈ L1 (µ) such thatλa (E) =

´Ehdµ for every E ∈ M

h is called the Radon-Nikodym derivative of λa with respect to µ.

λa, λs is called the Lebesgue decomposition of λ relative to µ.

49

4.2 Absolute Continuity4 COMPLEX MEASURES (RUDIN RCA CHAPTER 6)

4.2.3 Theorem

(theorem 6.11)Let µ be a positive measure and λ be a complex measure on the σ-algebra M.

• Claim: The following statements are equivalent:

– λ is absolutely continuous with respect to µ.

– ∀ε > 0∃δ > 0 such that |λ (E)| < ε∀E ∈ M such that µ (E) < δ.

4.2.4 Theorem

(theorem 6.12)Let µ be a complex measure on a σ-algebra M in X.

• Claim: ∃ a measurable function h such that |h (x)| = 1∀x ∈ X and dµ =hd |µ|. (polar representation or polar decomposition of µ).

4.2.5 Theorem

(theorem 6.13)Let µ be a positive measure on a σ-algebra M, g ∈ L1 (µ) and

λ (E) =

ˆE

gdµ ∀E ∈ M

• Claim: |λ| (E) =´E|g| dµ ∀E ∈ M

4.2.6 The Hahn Decomposition Theorem

(theorem 6.14)Let µ be a real measure on a σ-algebra M in a set X.

• Claim: ∃A,B ∈ M such that:

– A ∪B = X– A ∩B = ∅– µ+ (E) = µ (A ∩ E)∀E ∈ M– µ− (E) = −µ (B ∩ E)∀E ∈ M

(A, B) is called the Hahn decomposition of X, induced by µ.

• Proof: TODO (in paper)

50

5 INTERGRATION ON PRODUCT MEASURES (RUDIN RCACHAPTER 8)

5 Intergration on Product Measures (Rudin RCAChapter 8)

5.1 Measurability on Cartesian Products5.1.1 A Rectangle in X × Y

If A ∈ 2X and B ∈ 2Y then A×B is a rectangle in X × Y .

5.1.2 A Measurable Rectangle

If (X, S ) and (Y, T ) are two measurable spaces, then a masurable rectangleis any set of the form A×B where A ∈ S and B ∈ T .

5.1.3 The Class of all Elementary Sets

Q =⋃n

i=1 Ri where each Ri is a measurable rectangle and Ri ∩ Rj = ∅ for alli 6= j then Q is an elementary set. The class of all elementary sets is denotedby E .

5.1.4 S × T

S × T is defined to be the smallest σ-algebra in X × Y which contains everymeasurable rectangle.

5.1.5 A monotone Class

A monotone class M is a collection of sets with the following property:

• If ∀i ∈ N:

1. Ai ∈ M2. Bi ∈ M3. Ai ⊂ Ai+14. Bi ⊃ Bi+1

• Then:⋃∞

i=1 Ai ∈ M and⋂∞

i=1 Bi ∈ M.

5.1.6 x-section and y-section

If E ⊂ X × Y , x ∈ X and y ∈ Y , define:

1. Ex := {y ∈ Y | (x, y) ∈ E} ⊂ Y , the x-section.

2. Ey := {x ∈ X | (x, y) ∈ E} ⊂ X, the y-section.

51

5.2 Product Measures5 INTERGRATION ON PRODUCT MEASURES (RUDIN RCA

CHAPTER 8)

5.1.7 Theorem

(theorem 8.2)If E ∈ S × T , then ∀x ∈ X, ∀y ∈ Y , Ex ∈ T and Ey ∈ S .

5.1.8 Theorem

(theorem 8.3)S × T is the smallest monotone class which contains E .

5.1.9 Theorem

(theorem 8.5)Let f be a (S × T )-measurable function on X × Y . Then:

• ∀x0 ∈ X, f (x0, y) ∈ CY is a T -measurable function on Y .

• ∀y0 ∈ Y , f (x, y0) ∈ CX is a S -measurable function on X.

5.2 Product Measures5.2.1 Theorem

(theorem 8.6)

• Assumptions:

1. Let (X, S , µ) and (Y, T , λ) be σ-finite measure spaces.2. Suppose that Q ∈ S × T .

• Claim: Then

1. x 7→ λ (Qx) is S -measurable.2. y 7→ µ (Qy) is T -measurable3.´Xdµ (x)

´YχQ (x, y) dλ (y) =

´Ydλ (y)

´XχQ (x, y) dµ (x)

5.2.2 The Product of Measures

(definition 8.7)Let (X, S , µ) and (Y, T , λ) be σ-finite measure spaces. Define ∀Q ∈ S ×

T :

(µ× λ) (Q) :=ˆX

λ (Qx) dµ (x) =

ˆY

µ (Qy) dλ (y)

(=

ˆX

dµ (x)

ˆY

χQ (x, y) dλ (y) =

ˆY

dλ (y)

ˆX

χQ (x, y) dµ (x)

)(the equality of the two integrals is given by theorem 8.6)µ× λ is called the product of the measures µ and λ.

• µ× λ is a σ-finite measure on S × T .

52

5.3 The Fubini Theorem5 INTERGRATION ON PRODUCT MEASURES (RUDIN RCA

CHAPTER 8)

5.3 The Fubini Theorem5.3.1 The Fubini Theorem

(theorem 8.8)Let (X, S , µ) and (Y, T , λ) be σ-finite measure spaces. Let f be a S ×

T -measurable function on X × Y .

1. If f ∈ [0, ∞]X×Y :

• Then x 7→Ýf (x, y) dλ is S -measurable, y 7→

´Xf (x, y) dµ is

T -measurable, and´X

(Ýf (x, y) dλ

)dµ =

´X×Y f (x, y) d (µ× λ) =´

Y

(´Xf (x, y) dµ

)dλ.

2. If f ∈ CX×Y and if´X

(Ý|f (x, y)| dλ

)dµ < ∞ then f ∈ L1 (µ× λ).

3. If f ∈ L1 (µ× λ), then f (x, y) ∈ L1 (λ) (f (x, y) being here a map offixed x) for almost all x ∈ X, f (x, y) ∈ L1 (µ) (similar) for almost ally ∈ Y .

• x 7→Ýf (x, y) dλ and y 7→

´Xf (x, y) dµ are in L1 (µ) and L1 (λ)

and´X

(Ýf (x, y) dλ

)dµ =

´X×Y f (x, y) d (µ× λ) =

Ý

(´Xf (x, y) dµ

)dλ.

5.3.2 Counterexamples

(8.9)Sequences of triangles, the counting measure, and Sierpinski.

5.4 Completion of Product Measures5.4.1 Theorem

(8.11)Let mk denote the Lebesgue measure on Rk. If k = r + s where r ≥ 1 and

s ≥ 1. Then mk is the completion of the product measure mr ×ms.TODO

5.5 Convolutions5.5.1 Theorem–Convolutions

(8.14)Suppose f, g ∈ L1

(R1). Thenˆ ∞−∞

|f (x− y) g (y)| dy < ∞

for almost all x. For such x, define h (x) :=´∞−∞ f (x− y) g (y) dy (the

convolution of f and g, written as h = f ∗ g). Then h ∈ L1(R1)

and ||h||1 ≤||f ||1 ||g||1 where ||f ||1 :=

´∞−∞ |f (x)| dx.

53

7 DIFFERENTIATION (RUDIN RCA)

6 Product Measures, Multiple Integrals

6.1 Fubini’s Theorem

6.2 Convolution

7 Differentiation (Rudin RCA)

7.1 Derivatives of MeasuresLet k ∈ N.Let µ be a complex Borel measure on Rk.Let m be the Lebesgue measure on Rk.

1. Associative to any complex Borel measure µ on Rk the quotients:

(Qrµ) (x) :=µ (B (x, r))

m (B (x, r))

2. Define the symmetric derivative of µ at x to be:

(Dµ) (x) := limr→0

(Qrµ) (x)

for any x such that the above limit exists.

3. Define the maximal function as:

(Mµ) (x) := sup ({(Qr |µ|) (x) | r ∈ (0, ∞)})

7.1.1 Lemma

(lemma 7.3)If W =

⋃ni=1 B (xi, ri) then ∃ S ⊂ {1, . . . , n} such that:

• B (xi, ri) ∩B (xj , rj) = ∅∀i ∈ S, j ∈ S\ {i}

• W ⊂⋃

i∈S B (xi, 3ri)

• m (W ) ≤ 3k∑

i∈S m (B (xi, ri))

7.1.2 Theorem

(theorem 7.4)If µ is a complex Borel measure on Rk and λ is some positive number, then:

m((Mµ)

−1((λ, ∞])

)≤ 3k

|µ|(Rk)

λ

54

7.1 Derivatives of Measures 7 DIFFERENTIATION (RUDIN RCA)

7.1.3 Weak L1

(7.5)Define “weak L1” to be the set of all Lebesgue-measurable functions on Rk

such that λm(|f |−1 ((λ, ∞])

)is a bounded function of λ on (0, ∞) for any f

in “weak L1”.Observe that if f ∈ L1

(Rk)

and λ > 0 then λm(|f |−1 ((λ, ∞])

)≤ ||f ||1. Thus

L1 ⊂”weak L1”.Observe that 1x on (0, 1) is in “weak L

1” but not in L1.

7.1.3.1 The Maximal Function (Hardy-Littlewood) Let f ∈ L1(Rk).

Define the maximal function of f , Mf : Rk → [0, ∞] by:

Measure Theory Summary - ISG D-PHYS Personal web space · 2013. 9. 4. · Measure and Integration...

Documents

Transcript of Measure Theory Summary - ISG D-PHYS Personal web space · 2013. 9. 4. · Measure and Integration...