econ 508b: lecture 2

Econ 508B: Lecture 2Lebesgue Measure and Lebesgue Measurable Functions

Hongyi Liu

Washington University in St. Louis

July 18, 2017

Hongyi Liu (Washington University in St. Louis)Math Camp 2017 Stats July 18, 2017 1 / 35

Outline

1 Review of Real Analysis

2 Lebesgue Outer Measure

3 Lebesgue Measure

4 Lebesgue Measurable Functions


Outline



3 Lebesgue Measure



Topological Space

Definition 1.1

A topological space is a pair (S, C), where S is a non-empty set and C isa collection of subsets of S such that

∅,S ∈ C,(finite intersection:)C1, C2 ∈ C ⇒ C1 ∩ C2 ∈ C, and

(finite or infinite union:)Ck : k ∈ N+ ⊂⇒ ∪k∈N+ ∈ C

Remark 1.1

The elements of C are called open sets and the collection C iscalled a topology on S.

Using de Morgan’s laws, the above axioms defining open setsbecome axioms defining closed sets.


Union/Intersection

Definition 1.2

A sequence Ck of sets is said to increase to ∪kCk if Ck ⊂ Ck+1 for allk and to decrease to ∩kCk if Ck ⊃ Ck+1 for all k; we use the notationsCk ∪kCk and Ck ∩kCk to denote these two possibilities.If Ck∞k=1 is a sequence of sets, we define

lim supCk =

∞⋂j=1

∞⋃k=j

Ck

, lim infCk =

∞⋃j=1

∞⋂k=j

Ck

noting that

Uj =⋃∞k=j Ck and Vj =

⋂∞k=j Ck satisfy Uj lim supCk and

Vj lim infCk,

lim infCk ⊂ lim supCk.


Norm on RN

(1) |x| ≥ 0 and |x| = 0⇔ x = 0,

(2) |αx| = |α||x|, x ∈ RN , α ∈ R,

(3) |x+ y| ≤ |x|+ |y|,∀x, y ∈ R,

(4) (Cauchy-Schwarz inequality:)|x· y| ≤ |x||y|.

Proof for (4):

∀x, y ∈ R, xy ≤ 1

2x2 +

1

2y2

x · y =∑

xkyk ≤1

2

∑x2k +

1

2

∑y2k =

1

2|x|2 +

1

2|y|2

x′ = λx, y′ =1

λy, λ 6= 0(to be chosen), x′ · y′ = x · y

≤ 1

2|λ|2|x|2 +

1

2|λ|2|y|2 = |x||y| (choose λ =

√|y||x|

)


Metric Space

Definition 1.3

A metric space is a pair (S, d) where S is a nonempty set and d is afunction from S× S to R+ (d is called a metric on S) satisfying

(i) d(x, y) = d(y, x) for all x, y ∈ S,

(ii) d(x, y) = 0 iff x = y,

(iii) (triangle inequality:)d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ S.

A metric space (S, d) is a topological space where a set C is open if forall x ∈ C, ∃ an ε > 0 such that B(x, ε) ≡ y : d(y, x) < ε ⊂ C.


Outline



3 Lebesgue Measure



Lebesgue Outer (Exterior) Measure

Define closed n-dimensional intervals I = x : aj ≤ xj ≤ bj , j = 1, ..., nand their volumes v(I) =

∏nj=1(bj − aj). To define the outer measure

of an arbitrary subset C of Rn, cover C by a countable collection S ofintervals Ik, and let

σ(S) =∑Ik∈S

v(Ik)

The Lebesgue outer measure of C, denoted as µ∗(C), is defined by

µ∗(C) = infσ(S)

where the infimum is taken over all covers S of C. Thus,0 ≤ µ∗(C) ≤ +∞


Properties of Outer Measure

For an interval I, µ∗(I) = v(I).

Monotonicity: If C1 ⊂ C2, then µ∗(C1) ≤ µ∗(C2).

Countable sub-additivity: If C = ∪Ck is a countable union ofsets, then µ∗(C) ≤

∑µ∗(Ck).

Empty set: The empty set has outer measure zero, e.g., Q.

Remark 2.1

In particular, any subset of a set with outer measure zero has outermeasure zero and that the countable union of sets with outer measurezero has outer measure zero as shown by the example of Q.

Moreover, there are sets with outer measure zero that are notcountable, e.g., Cantor Set.


continued

Theorem 2.1 (Outer Approximation)

Let C ⊂ Rn. Then given ε > 0, ∃ an open set G s.t. C ⊂ G andµ∗(G) ≤ µ∗(C) + ε. Hence,

µ∗(C) = infµ∗(G),

where the infimum is taken over all open sets G containing C.

Theorem 2.2

If C ⊂ Rn, ∃ a set H of type Gδ s.t. C ⊂ H and µ∗(C) = µ∗(H)


Outline



3 Lebesgue Measure



‘Measurable’ & ‘Measure’

Definition 3.1 (Lebesgue measurable)

A subset C of Rn is defined to be Lebesgue measurable , ormeasurable , if given ε > 0, ∃ an open set G s.t.

C ⊂ G, and µ∗(G− C) < ε

Definition 3.2 (Measure)

If C is measurable, its outer measure is referred to as its Lebesguemeasure or simply its measure , and denoted by µ(C) as previouslyillustrated:

µ(C) = µ∗(C), for measurable C.

Example 3.1

Every open set is measurable.

Every set of outer measure zero is measurable.


Remark 3.1

Noting that the condition for measurability should not be confusedwith theorem 2.1, which only states that ∃ an open set G containing Csuch that µ∗(G) ≤ µ∗(C) + ε. Since G = C ∪ (G− C) when C ⊂ G,which only implies that µ∗(G) ≤ µ∗(C) + µ∗(G− C). However, wecannot obtain from µ∗(G) ≤ µ∗(C) + ε that µ∗(G− C) < ε.


Properties of Measurable Set

(Countable subaddtivity:) The union C = ∪Ck of a countablemeasurable sets is measurable and µ(C) ≤

∑µ(Ck).

Every closed set F is measurable.

The complement of a measurable set is measurable.

The intersection C = ∩kCk of a countable measurable sets ismeasurable.

If C1 and C2 are measurable, then C1 − C2 is measurable.

The collection of measurable subsets of Rn is σ-algebra.

Every Borel set is measurable.


Properties of Lebesgue Measure

Lemma 3.1

Set C in Rn is measurable if.f. given ε > 0, ∃ a closed set F ⊂ C suchthat µ∗(C − F ) < ε.

Theorem 3.1

If Ck is a countable collection of disjoint measurable sets, thenµ(⋃k Ck) =

∑k µ(Ck).

Corollary 3.1

Suppose C1 and C2 are measurable, C2 ⊂ C1, and µ(C2) < +∞. Thenµ(C1 − C2) = µ(C1)− µ(C2).


continued

Theorem 3.2

Let Ck∞k=1 be a sequence of measurable sets.

(1) If Ck C, then limk→∞ µ(Ck) = µ(C).

(2) If Ck C and µ(Ck) < +∞ for some k, limk→∞ µ(Ck) = µ(C).

Proof: (1) Assume that µ(Ck) < +∞ for all k, otherwise bothlimk→∞ µ(Ck) and µ(C) are infinite and the statement holds. Break Cvia

C = C1 ∪ (C2 − C1) ∪ ... ∪ (Ck − Ck−1) ∪ ...By theorem 3.1,

µ(C) = µ(C1) + µ(C2 − C1) + ...+ µ(Ck − Ck−1) + · · ·.

By corollary 3.1,

µ(C) = µ(C1)+(µ(C2)−µ(C1))+···+(µ(Ck)−µ(Ck−1))+··· = limk→∞

µ(Ck).


Proof: (2) clearly assume that µ(C1) < +∞. Write

C1 = C ∪ (C1 − C2) ∪ · · · ∪ (Ck − Ck+1) ∪ · · ·.

Likewise,

µ(C1) = µ(C) + (µ(C1)− µ(C2)) + · · ·+ (µ(Ck)− µ(Ck+1)) + · · ·= µ(C) + µ(C1)− lim

k→∞µ(Ck).

Hence, µ(C) = limk→∞ µ(Ck).

Noting that the condition µ(Ck) < +∞ for some k is necessary.


Characterizations of Measurability

Theorem 3.3

C is measurable if and only if C = H − Z, where H is of type Gδand µ(Z) = 0.

C is measurable if and only if C = H ∪ Z, where H is of type Fδand µ(Z) = 0.

Proof: (sufficiency) It is trivial that C is measurable because H andZ are both measurable sets.(necessity): For the first one, suppose that C is measurable andchoose an open set Gk such that C ⊂ Gk and µ(Gk − C) < 1/k fork = 1, 2, .... Let H =

⋂kGk, which is a set of type Gδ, C ⊂ H and

H − C ⊂ Gk − C for every k. As k →∞, µ(H − C) = 0.Secondly, C is measurable, so is the complement set of C, denoted asCc. Then applying the result of first one, we obtain thatCc =

⋂kGk−Z, and using de Morgan’s laws we have C = (

⋃kG

ck)∪Z.


Nonmeasurable set

Construct a nonmeasurable subset of R1, and the construction inRd, d > 1 is similar. The construction of a nonmeasurable set uses thefollowing axiom of choice and rests on equivalence relation among realnumbers in [0, 1].

Axiom 3.1 (Zermelo’s Axiom:)

A family of arbitrary nonempty disjoint sets indexed by a setA, Cα : α ∈ A, ∃ a set consisting of exactly one element from eachCα, α ∈ A.

Lemma 3.2

Let C be a measurable subset of R1 with µ(C) > 0. Then the set ofdifferences d : d = x− y, x ∈ C, y ∈ C contains an interval centeredat the origin.


Vitali Theorem: There exist nonmeasurable sets.

Proof:An equivalence relation, defined as x ∼ y on the real line if x− y isrational, which can be formulated by Cα = α+ r : r is rational. Itmeans that any two classes Cα and Cβ are either identical or disjoint.

Hence, one equivalence class consists of all the rational numbers, andthe other distinct classes are disjoint sets of irrational numbers.

Using Zemelo’s axiom, construct the set C consisting of exactly oneelement from each distinct equivalence class, therefore, any two pointsof C must differ by an irrational number, which implies that the setd : d = x− y, x ∈ C, y ∈ C cannot contain an interval. According toLemma 3.2 , it suffices that either C is not measurable or µ(C) = 0.Let Cr = C + r, r ∈ Q, then ∪r∈QC representing the union of thetranslation of C by every rational is R1, R1 would have measure zero ifC. Then it completes the proof that C is nonmeasurable.


Outline



3 Lebesgue Measure



Measurable Functions

Definition 4.1 (Measurable function)

In general, let Ω1 be a set with a σ-algebra C1, and Ω2 be a set with asigma-algebra C2, and T be a function from Ω1 to Ω2. Say T is〈C1, C2〉−measurable if the inverse image x ∈ Ω1 : Tx ∈ C2 ∈ C1 foreach C2 ∈ C2.

In particular, if Ω2 = R, C2 becomes the Borel σ-algebra B(R).

Example 4.1

Recall the definition for random variable in the previous slide.


〈C,B(R)〉-measurable

In particular, to establish 〈C,B(R)〉-measurability of a map into thereal line, it is simplified to check the inverse images of intervals of theform (a,∞) as follows.

Definition 4.2

Let C be a measurable set in Rn and f be a real-extended function(i.e., f(x) ∈ [−∞,+∞],x ∈ C) defined on C. f is referred to as aLebesgue measurable function on C or measurable function if forevery finite a, the set

x ∈ C : f(x) > a

is a measurable subset of Rn, which is often simply denoted asf > a. why?

C = f = −∞ ∪ (

∞⋃k=1

f > −k)


Elementary properties of measurable functions

The definition for measurable function is equivalent to any of thefollowing statements hold for finite a:

(i) f ≥ a is measurable.

(ii) f < a is measurable.

(ii) f 6 a is measurable.

Corollary 4.1

f > −∞, f < +∞, f = +∞, a 6 f 6 b, f = a, etc, are allmeasurable if f is measurable.


Properties

Theorem 4.1

The finite-valued function f is measurable if and only if f−1(G) ismeasurable for every open set G of R1, and if and only if f−1(F ) ismeasurable for every closed set F of R1.

Theorem 4.2

Let A be a dense subset of R1. Then f is measurable if f > a ismeasurable for all a ∈ A.

Remark 4.1 (dense)

A set C ⊂ C1 is said to be dense in C1 if ∀x1 ∈ C1 and ε > 0,∃ a pointx ∈ C such that 0 < |x− x1| < ε. Example: Q is dense in R.


almost everywhere or a.e.: A property or a statement holds in Cexcept in some subset of C with measure zero. For instance, thestatement “f = 0 a.e. in C” is abbreviated of f(x) = 0 in C with theexception of those x in some subset Z of C with µ(Z) = 0.

Theorem 4.3

If f is measurable and if g = f a.e., then g = f a.e., then g ismeasurable and µ(g > a) = µ(f > a).


Operations on measurable functions

Theorem 4.4

If f is continuous on Rd, then f is measurable. If f is measurable a.e.in C, and Φ is continuous, then Φ f or Φ(f) is measurable.

Remark 4.2

The cases that arise frequently are

Φ(f) = |f |, |f |p(p > 0), ecf

are measurable if f is measurable.Noting another special case is that of

f+ = maxf, 0, f− = −minf, 0


Operations on measurable functions

Suppose f and g are measurable, and fn∞n=1 is a sequence ofmeasurable functions, and λ is any real number, then so are

f > g.f + λ and λf .

f + g.

fg, and f/g if g 6= 0 a.e.

supn fn(x), infn fn(x), lim supn→∞ fn(x), and lim infn→∞ fn(x).

Plus, if limn→∞

fn(x) = f(x) and fn∞n = 1 is a collection of measurable

functions, then f is measurable.


Characteristic function or indicator function

Our attention to the objects that lie at the heart of integration theory:measurable functions. The starting point is the notion of acharacteristic function or indicator function of a set C, which isdefined by

χC(x) =

1 if x ∈ C0 if x /∈ C

Clearly, χC is measurable if and only if C is measurable. χC is anexample of what is referred to as a simple function on Rn: a simplefunction on a set C ⊂ Rn is one that is defined on C and supposesonly a finite number of finite values on C. If f is a simple function onC taking distinct values a1, ..., aN on disjoint subsets C1, ..., CN of C,and C =

⋃Nk=1Ck, then

f(x) =

N∑k=1

akχCk(x), x ∈ C.


Approximation by simple function

Theorem 4.5

Every function can be represented as the convergence of asequence fk of simple functions.

If f ≥ 0, the sequence can be chosen to increasingly converge to f ,i.e. fk ≤ fk+1,∀k.

If f is measurable, then fk can be chosen to be measurable.


Proof

We begin by approximating pointwise, non-negative measurablefunctions by simple functions.

Firstly, start with a truncation within [0, k]. Suppose f ≥ 0,k = 1, 2, ..., subdivide the values of f by partitioning [0, k] intosubintervals [(j − 1)2−k, j2−k], j = 1, ..., k2k. Then

fk(x) =

j−12k

if j−12k≤ f(x) ≤ j

2k, j = 1, ..., k2k

k if f(x) ≥ k.

Then, fk(x)→ f(x) as k goes to infinity for all x. Clearly, fk ≤ fk+1

and fk → f because of 0 ≤ f − fk ≤ 2−k as k tends to infinity whereverf is finite and fk = k → +∞ wherever f = +∞. It completes the prooffor the second theorem of nonnegative case. In fact,

fk(x) =

k2k∑j=1

j − 1

2kχ j−1

2k≤f(x)≤ j

2k + kχf≥k


Approximation by simple function

For the first theorem, using the decomposition of the function f :f = f+ − f−. Since both f+ and f− are nonnegative, then it triviallyyields to apply the above theorem twice.

Theorem 4.6

Suppose f is measurable on Rd. Then ∃ a sequence of simple functionfk∞k=1 that satisfies

|fk(x)| ≤ |fk+1(x)|, limk→∞

fk(x) = f(x), ∀x

In particular, |fk(x)| ≤ |f(x)|, ∀x, k.


Theorems of Egorov and Lusin

Theorem 4.7 (Egorov’s Theorem)

Suppose fk∞k=1 is a sequence of measurable functions defined on ameasurable set C with µ(C) <∞ and assume that fk → f a.e. on C.Given ε > 0, ∃ a closed set Fε ⊂ C such that µ(C − Fε) ≤ ε and fk → funiformly on Fε.

Theorem 4.8 (Lusin)

Suppose f is measurable and finite valued on C with µ(C) <∞.∀ε > 0,∃ a closed set Fε with Fε ⊂ C and µ(C − Fε) ≤ ε and such thatf |Fε is continuous.


econ 508b: lecture 2

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