CS计算机代考程序代写 AI Mathematical Methods Lebesgue Measure

Mathematical Methods Lebesgue Measure

Mathematical Methods
Lebesgue Measure

Evan Sadler
Columbia University

November 3, 2021

Evan Sadler Math Methods 1/20

Review

Last time:
• Measure spaces (X,F , µ), σ-algebra F closed under
complements and countable unions, measure µ countably
additive

• Any measure satisfies monotonicity, subadditivity, continuity
from above and below

• Any measure on R that generalizes the length of an interval
cannot be defined on all sets

• Borel σ-algebra B(X) generated from open sets

Evan Sadler Math Methods 2/20

Today

Does a measure that extends the length function exist?
• Can we define a countably additive measure µ on B(R) that
µ([a, b]) = b− a?

Yes! Construction of Lebesgue measure

Properties of Lebesgue measure

Evan Sadler Math Methods 3/20

Today

Does a measure that extends the length function exist?
• Can we define a countably additive measure µ on B(R) that
µ([a, b]) = b− a?

Yes! Construction of Lebesgue measure

Properties of Lebesgue measure

Evan Sadler Math Methods 3/20

Algebras

Construction starts with a few more definitions…

Definition
A family of sets A ⊆ 2X is an algebra if it is closed under
complements and finite unions.

Example: Finite subsets of X and their complements form an
algebra
• This is not a σ-algebra if X is uncountable
• For instance, if X = R, the set A = Q is not in this family,
but it must be in any σ-algebra containing all singletons

Evan Sadler Math Methods 4/20

Algebras

Construction starts with a few more definitions…

Definition
A family of sets A ⊆ 2X is an algebra if it is closed under
complements and finite unions.

Example: Finite subsets of X and their complements form an
algebra
• This is not a σ-algebra if X is uncountable
• For instance, if X = R, the set A = Q is not in this family,
but it must be in any σ-algebra containing all singletons

Evan Sadler Math Methods 4/20

Premeasures

Definition
Let A be an algebra. A function µ0 : A → [0,∞] is a
premeasure if
• µ0(∅) = 0, and
• If {Ai}∞i=1 are disjoint sets in A such that ∪∞i=1Ai ∈ A, then

µ0(∪∞i=1Ai) =
∞∑

i=1
µ0(Ai)

Require countable additivity only when the premeasure is defined
for the union, which need not belong to the algebra A

Evan Sadler Math Methods 5/20

Carathéodory’s Extension Theorem

Any premeasure can be extended to a measure

Theorem (Carathéodory’s Extension Theorem)
Given an algebra A ⊆ 2X and a premeasure µ0 on A, there exists
a measure µ that extends µ0 on the σ-algebraM(A) generated
by A. That is, µ is defined onM(A), and we have

µ(A) = µ0(A)

whenever A ∈ A. Moreover, if µ0 is σ-finite, the extension µ is
unique

Evan Sadler Math Methods 6/20

Proof Sketch
For class, only show existence

Define outer measure µ∗ for all B ⊆ X via

µ∗(B) = inf
{ ∞∑

i=1
µ0(Ai) : B ⊆ ∪iAi, Ai ∈ A

}

Non-negative function defined on 2X , clearly monotone, and
µ∗(∪∞i=1Bi) ≤

∑∞
i=1 µ

∗(Bi)
• Note µ∗(A) ≤ µ0(A) for A ∈ A by definition
• Given a cover {Ai}∞i=1 ⊆ A of A, countable additivity of µ0
gives µ0(A) ≤

∑∞
i=1 µ0(Ai), so µ∗(A) = µ0(A) for A ∈ A

Coincides with µ0 on A

Evan Sadler Math Methods 7/20

Proof Sketch
For class, only show existence

Define outer measure µ∗ for all B ⊆ X via

µ∗(B) = inf
{ ∞∑

i=1
µ0(Ai) : B ⊆ ∪iAi, Ai ∈ A

}

Non-negative function defined on 2X , clearly monotone, and
µ∗(∪∞i=1Bi) ≤

∑∞
i=1 µ

∗(Bi)
• Note µ∗(A) ≤ µ0(A) for A ∈ A by definition
• Given a cover {Ai}∞i=1 ⊆ A of A, countable additivity of µ0
gives µ0(A) ≤

∑∞
i=1 µ0(Ai), so µ∗(A) = µ0(A) for A ∈ A

Coincides with µ0 on A

Evan Sadler Math Methods 7/20

Proof Sketch
For class, only show existence

Define outer measure µ∗ for all B ⊆ X via

µ∗(B) = inf
{ ∞∑

i=1
µ0(Ai) : B ⊆ ∪iAi, Ai ∈ A

}

Non-negative function defined on 2X , clearly monotone, and
µ∗(∪∞i=1Bi) ≤

∑∞
i=1 µ

∗(Bi)
• Note µ∗(A) ≤ µ0(A) for A ∈ A by definition
• Given a cover {Ai}∞i=1 ⊆ A of A, countable additivity of µ0
gives µ0(A) ≤

∑∞
i=1 µ0(Ai), so µ∗(A) = µ0(A) for A ∈ A

Coincides with µ0 on A

Evan Sadler Math Methods 7/20

Proof Sketch
For class, only show existence

Define outer measure µ∗ for all B ⊆ X via

µ∗(B) = inf
{ ∞∑

i=1
µ0(Ai) : B ⊆ ∪iAi, Ai ∈ A

}

Non-negative function defined on 2X , clearly monotone, and
µ∗(∪∞i=1Bi) ≤

∑∞
i=1 µ

∗(Bi)
• Note µ∗(A) ≤ µ0(A) for A ∈ A by definition
• Given a cover {Ai}∞i=1 ⊆ A of A, countable additivity of µ0
gives µ0(A) ≤

∑∞
i=1 µ0(Ai), so µ∗(A) = µ0(A) for A ∈ A

Coincides with µ0 on A

Evan Sadler Math Methods 7/20

Proof Continued

We will call a set S ⊆ X µ∗-measurable if

µ∗(A) = µ∗(A ∩ S) + µ∗(A ∩ Sc)

for all A ∈ A

Need to show three things:
• The set F of µ∗-measurable S ⊆ X is a σ-algebra
• µ∗ is countably additive on F (so it is a measure on F)
• Each A ∈ A is µ∗-measurable, so σ(A) ⊆ F

You will do these on the homework

Evan Sadler Math Methods 8/20

Proof Continued

We will call a set S ⊆ X µ∗-measurable if

µ∗(A) = µ∗(A ∩ S) + µ∗(A ∩ Sc)

for all A ∈ A

Need to show three things:
• The set F of µ∗-measurable S ⊆ X is a σ-algebra
• µ∗ is countably additive on F (so it is a measure on F)
• Each A ∈ A is µ∗-measurable, so σ(A) ⊆ F

You will do these on the homework

Evan Sadler Math Methods 8/20

Proof Continued

We will call a set S ⊆ X µ∗-measurable if

µ∗(A) = µ∗(A ∩ S) + µ∗(A ∩ Sc)

for all A ∈ A

Need to show three things:
• The set F of µ∗-measurable S ⊆ X is a σ-algebra
• µ∗ is countably additive on F (so it is a measure on F)
• Each A ∈ A is µ∗-measurable, so σ(A) ⊆ F

You will do these on the homework

Evan Sadler Math Methods 8/20

Specializing to R

Suppose X = R; let’s choose an algebra A that contains the
intervals and a corresponding premeasure µ0

Let A be the algebra generated by the half open intervals (a, b]
with −∞ ≤ a ≤ b ≤ ∞:
• Includes ∅ taking a = b
• Includes (a,∞) and (−∞, b] taking a = −∞ and b =∞

Exercise: the algebra A consists of finite disjoint unions of half
open intervals

Evan Sadler Math Methods 9/20

Specializing to R

Suppose X = R; let’s choose an algebra A that contains the
intervals and a corresponding premeasure µ0

Let A be the algebra generated by the half open intervals (a, b]
with −∞ ≤ a ≤ b ≤ ∞:
• Includes ∅ taking a = b
• Includes (a,∞) and (−∞, b] taking a = −∞ and b =∞

Exercise: the algebra A consists of finite disjoint unions of half
open intervals

Evan Sadler Math Methods 9/20

Specializing to R

Suppose X = R; let’s choose an algebra A that contains the
intervals and a corresponding premeasure µ0

Let A be the algebra generated by the half open intervals (a, b]
with −∞ ≤ a ≤ b ≤ ∞:
• Includes ∅ taking a = b
• Includes (a,∞) and (−∞, b] taking a = −∞ and b =∞

Exercise: the algebra A consists of finite disjoint unions of half
open intervals

Evan Sadler Math Methods 9/20

Lebesgue Premeasure
Assign µ0((a, b] = b− a (obviously)
• Element of A takes form ∪ni=1(ai, bi], disjoint intervals,
measure µ0(A) =

∑n
i=1(bi − ai)

Lemma
The function µ0 is a premeasure on the algebra A

Suppose A1, A2, … are disjoint (wlog, assume Ai = (ai, bi]), with
∪∞i=1Ai ∈ A
• Need to show µ0(A) =

∑∞
i=1 µ0(Ai)

Assume also that union A is a single interval (a, b]
• Can iterate argument for finitely many half open intervals that
comprise A

Evan Sadler Math Methods 10/20

Lebesgue Premeasure
Assign µ0((a, b] = b− a (obviously)
• Element of A takes form ∪ni=1(ai, bi], disjoint intervals,
measure µ0(A) =

∑n
i=1(bi − ai)

Lemma
The function µ0 is a premeasure on the algebra A

Suppose A1, A2, … are disjoint (wlog, assume Ai = (ai, bi]), with
∪∞i=1Ai ∈ A
• Need to show µ0(A) =

∑∞
i=1 µ0(Ai)

Assume also that union A is a single interval (a, b]
• Can iterate argument for finitely many half open intervals that
comprise A

Evan Sadler Math Methods 10/20

Lebesgue Premeasure
Assign µ0((a, b] = b− a (obviously)
• Element of A takes form ∪ni=1(ai, bi], disjoint intervals,
measure µ0(A) =

∑n
i=1(bi − ai)

Lemma
The function µ0 is a premeasure on the algebra A

Suppose A1, A2, … are disjoint (wlog, assume Ai = (ai, bi]), with
∪∞i=1Ai ∈ A
• Need to show µ0(A) =

∑∞
i=1 µ0(Ai)

Assume also that union A is a single interval (a, b]
• Can iterate argument for finitely many half open intervals that
comprise A

Evan Sadler Math Methods 10/20

Lebesgue Premeasure
Assign µ0((a, b] = b− a (obviously)
• Element of A takes form ∪ni=1(ai, bi], disjoint intervals,
measure µ0(A) =

∑n
i=1(bi − ai)

Lemma
The function µ0 is a premeasure on the algebra A

Suppose A1, A2, … are disjoint (wlog, assume Ai = (ai, bi]), with
∪∞i=1Ai ∈ A
• Need to show µ0(A) =

∑∞
i=1 µ0(Ai)

Assume also that union A is a single interval (a, b]
• Can iterate argument for finitely many half open intervals that
comprise A

Evan Sadler Math Methods 10/20

Proof
First show b− a ≥

∑∞
i=1(bi − ai]

• Enough to show b− a ≥
∑n

i=1(bi − ai] for all n

Finite disjoint union ∪ni=1(ai, bi] ⊆ (a, b] is a subset, relabel so
a ≤ a1 ≤ b1 ≤ a2 ≤ b2 ≤ … ≤ an ≤ bn ≤ b:

n∑
i=1

(bi − ai) = bn − a1 +
n−1∑
i=1

(bi − ai+1) ≤ b− a

Now show b− a ≥
∑∞

i=1(bi − ai]
• Enough to find n with b− a ≤ �+

∑n
i=1(bi − ai) for all � > 0

Define A′i = (ai, bi + �/2i), slight superset of Ai, open set
• Union ∪iA′i covers compact set [a, b], finite subcover exists
with total length at least b− a:

b− a ≤
n∑

i=1

(
bi − ai +

2i
)

= �+
n∑

i=1
(bi − ai)

Evan Sadler Math Methods 11/20

Proof
First show b− a ≥

∑∞
i=1(bi − ai]

• Enough to show b− a ≥
∑n

i=1(bi − ai] for all n

Finite disjoint union ∪ni=1(ai, bi] ⊆ (a, b] is a subset, relabel so
a ≤ a1 ≤ b1 ≤ a2 ≤ b2 ≤ … ≤ an ≤ bn ≤ b:

n∑
i=1

(bi − ai) = bn − a1 +
n−1∑
i=1

(bi − ai+1) ≤ b− a

Now show b− a ≥
∑∞

i=1(bi − ai]
• Enough to find n with b− a ≤ �+

∑n
i=1(bi − ai) for all � > 0

Define A′i = (ai, bi + �/2i), slight superset of Ai, open set
• Union ∪iA′i covers compact set [a, b], finite subcover exists
with total length at least b− a:

b− a ≤
n∑

i=1

(
bi − ai +

2i
)

= �+
n∑

i=1
(bi − ai)

Evan Sadler Math Methods 11/20

Proof
First show b− a ≥

∑∞
i=1(bi − ai]

• Enough to show b− a ≥
∑n

i=1(bi − ai] for all n

Finite disjoint union ∪ni=1(ai, bi] ⊆ (a, b] is a subset, relabel so
a ≤ a1 ≤ b1 ≤ a2 ≤ b2 ≤ … ≤ an ≤ bn ≤ b:

n∑
i=1

(bi − ai) = bn − a1 +
n−1∑
i=1

(bi − ai+1) ≤ b− a

Now show b− a ≥
∑∞

i=1(bi − ai]
• Enough to find n with b− a ≤ �+

∑n
i=1(bi − ai) for all � > 0

Define A′i = (ai, bi + �/2i), slight superset of Ai, open set
• Union ∪iA′i covers compact set [a, b], finite subcover exists
with total length at least b− a:

b− a ≤
n∑

i=1

(
bi − ai +

2i
)

= �+
n∑

i=1
(bi − ai)

Evan Sadler Math Methods 11/20

Proof
First show b− a ≥

∑∞
i=1(bi − ai]

• Enough to show b− a ≥
∑n

i=1(bi − ai] for all n

Finite disjoint union ∪ni=1(ai, bi] ⊆ (a, b] is a subset, relabel so
a ≤ a1 ≤ b1 ≤ a2 ≤ b2 ≤ … ≤ an ≤ bn ≤ b:

n∑
i=1

(bi − ai) = bn − a1 +
n−1∑
i=1

(bi − ai+1) ≤ b− a

Now show b− a ≥
∑∞

i=1(bi − ai]
• Enough to find n with b− a ≤ �+

∑n
i=1(bi − ai) for all � > 0

Define A′i = (ai, bi + �/2i), slight superset of Ai, open set
• Union ∪iA′i covers compact set [a, b], finite subcover exists
with total length at least b− a:

b− a ≤
n∑

i=1

(
bi − ai +

2i
)

= �+
n∑

i=1
(bi − ai)

Evan Sadler Math Methods 11/20

Lebesgue Measure
Letting A be the algebra of finite unions of half open intervals,
and using the premeasure µ0, apply CET:

Theorem
There exists a unique measure µ on B(R) such that
µ((a, b]) = b− a. We call this the Lebesgue measure on R,
denoted λ(·).

Some notes:
• Continuity from above/below tells us measure of open/closed
interval also its length

• Translation invariance is clear
• Also: λ(tA) = |t| · λ(A)

Evan Sadler Math Methods 12/20

Borel vs. Lebesgue
Lebesgue measure defined on Borel σ-algebra
• Can extend Lebesgue measure to larger Lebesgue σ-algebra

Definition
Given a measure space (X,F , µ), a set N ∈ F with µ(N) = 0 is
null.

If a subset of a null set is measurable, it must have measure 0

Proposition
Given (X,F , µ), define

F = {A ∪B : A ∈ F , B ⊆ N ∈ F , N null}.

The set F is a σ-algebra, and µ(A ∪B) = µ(A) for B ⊆ N
extends the measure µ to F (“completion” of µ)

Evan Sadler Math Methods 13/20

Borel vs. Lebesgue
Lebesgue measure defined on Borel σ-algebra
• Can extend Lebesgue measure to larger Lebesgue σ-algebra

Definition
Given a measure space (X,F , µ), a set N ∈ F with µ(N) = 0 is
null.

If a subset of a null set is measurable, it must have measure 0

Proposition
Given (X,F , µ), define

F = {A ∪B : A ∈ F , B ⊆ N ∈ F , N null}.

The set F is a σ-algebra, and µ(A ∪B) = µ(A) for B ⊆ N
extends the measure µ to F (“completion” of µ)

Evan Sadler Math Methods 13/20

Borel vs. Lebesgue
Lebesgue measure defined on Borel σ-algebra
• Can extend Lebesgue measure to larger Lebesgue σ-algebra

Definition
Given a measure space (X,F , µ), a set N ∈ F with µ(N) = 0 is
null.

If a subset of a null set is measurable, it must have measure 0

Proposition
Given (X,F , µ), define

F = {A ∪B : A ∈ F , B ⊆ N ∈ F , N null}.

The set F is a σ-algebra, and µ(A ∪B) = µ(A) for B ⊆ N
extends the measure µ to F (“completion” of µ)

Evan Sadler Math Methods 13/20

Borel vs. Lebesgue
Lebesgue measure defined on Borel σ-algebra
• Can extend Lebesgue measure to larger Lebesgue σ-algebra

Definition
Given a measure space (X,F , µ), a set N ∈ F with µ(N) = 0 is
null.

If a subset of a null set is measurable, it must have measure 0

Proposition
Given (X,F , µ), define

F = {A ∪B : A ∈ F , B ⊆ N ∈ F , N null}.

The set F is a σ-algebra, and µ(A ∪B) = µ(A) for B ⊆ N
extends the measure µ to F (“completion” of µ)

Evan Sadler Math Methods 13/20

Proof

Must show F is closed under complements and countable unions

Since B ⊆ N , we have (A ∪N)c ⊆ (A ∪B)c
• Set difference at most N −B
• Write (A ∪B)c = (A ∪N)c ∪B′ for B′ = N −B ⊆ N
• Since (A ∪N)c ∈ F and B′ ⊆ N , we have (A ∪B)c ∈ F

Consider a sequence {Ai ∪Bi}i∈N, with Bi ⊆ Ni
• Union ∪i(Ai ∪Bi) = (∪iAi) ∪ (∪iBi)
• First union in F , second in null set ∪iNi, so this is in F

Evan Sadler Math Methods 14/20

Proof

Must show F is closed under complements and countable unions

Since B ⊆ N , we have (A ∪N)c ⊆ (A ∪B)c
• Set difference at most N −B
• Write (A ∪B)c = (A ∪N)c ∪B′ for B′ = N −B ⊆ N
• Since (A ∪N)c ∈ F and B′ ⊆ N , we have (A ∪B)c ∈ F

Consider a sequence {Ai ∪Bi}i∈N, with Bi ⊆ Ni
• Union ∪i(Ai ∪Bi) = (∪iAi) ∪ (∪iBi)
• First union in F , second in null set ∪iNi, so this is in F

Evan Sadler Math Methods 14/20

Proof

Must show F is closed under complements and countable unions

Since B ⊆ N , we have (A ∪N)c ⊆ (A ∪B)c
• Set difference at most N −B
• Write (A ∪B)c = (A ∪N)c ∪B′ for B′ = N −B ⊆ N
• Since (A ∪N)c ∈ F and B′ ⊆ N , we have (A ∪B)c ∈ F

Consider a sequence {Ai ∪Bi}i∈N, with Bi ⊆ Ni
• Union ∪i(Ai ∪Bi) = (∪iAi) ∪ (∪iBi)
• First union in F , second in null set ∪iNi, so this is in F

Evan Sadler Math Methods 14/20

Almost Everywhere

If a statement about points x ∈ X is true except for x in some
null set, we say it is true almost everywhere or for almost
every x

Function f(x) = 0 for x 6= 0 and f(0) = 1 is continuous almost
everywhere

Function f(x) = |x| is differentiable almost everywhere

Implicitly appealing to Lebesgue measure, not true if our measure
is a point mass at 0

Evan Sadler Math Methods 15/20

Almost Everywhere

If a statement about points x ∈ X is true except for x in some
null set, we say it is true almost everywhere or for almost
every x

Function f(x) = 0 for x 6= 0 and f(0) = 1 is continuous almost
everywhere

Function f(x) = |x| is differentiable almost everywhere

Implicitly appealing to Lebesgue measure, not true if our measure
is a point mass at 0

Evan Sadler Math Methods 15/20

Almost Everywhere

If a statement about points x ∈ X is true except for x in some
null set, we say it is true almost everywhere or for almost
every x

Function f(x) = 0 for x 6= 0 and f(0) = 1 is continuous almost
everywhere

Function f(x) = |x| is differentiable almost everywhere

Implicitly appealing to Lebesgue measure, not true if our measure
is a point mass at 0

Evan Sadler Math Methods 15/20

Almost Everywhere

If a statement about points x ∈ X is true except for x in some
null set, we say it is true almost everywhere or for almost
every x

Function f(x) = 0 for x 6= 0 and f(0) = 1 is continuous almost
everywhere

Function f(x) = |x| is differentiable almost everywhere

Implicitly appealing to Lebesgue measure, not true if our measure
is a point mass at 0

Evan Sadler Math Methods 15/20

Outer and Inner Measure
A useful characterization from the proof of Carathéodory’s
Theorem:

Theorem
For any measurable A ⊆ R, we have

λ(A) = inf
{ ∞∑

i=1
(bi − ai) : A ⊆ ∪∞i=1(ai, bi)

}

Corollary
For any measurable A ⊆ R, we have

λ(A) = inf{λ(U) : A ⊆ U, U is open}

= sup{λ(K) : L ⊆ A, K is compact}

Evan Sadler Math Methods 16/20

Outer and Inner Measure
A useful characterization from the proof of Carathéodory’s
Theorem:

Theorem
For any measurable A ⊆ R, we have

λ(A) = inf
{ ∞∑

i=1
(bi − ai) : A ⊆ ∪∞i=1(ai, bi)

}

Corollary
For any measurable A ⊆ R, we have

λ(A) = inf{λ(U) : A ⊆ U, U is open}

= sup{λ(K) : L ⊆ A, K is compact}

Evan Sadler Math Methods 16/20

The Cantor Set
A countable set must have measure 0
• Can someone tell me why?

Cantor set illustrates an uncountable set with measure 0

Evan Sadler Math Methods 17/20

The Cantor Set
A countable set must have measure 0
• Can someone tell me why?

Cantor set illustrates an uncountable set with measure 0

Evan Sadler Math Methods 17/20

Measure Zero but Uncountable

At each step, we remove 13 of each remaining subinterval, so
measure multiplied by 23
• After n steps, remaining measure is

(
2
3

)n
→ 0

Cantor set includes all points of the form

x =
∞∑

i=1
ai3−i

with ai ∈ {0, 2}—set of infinite sequences of 0s and 2s is
uncountable

Evan Sadler Math Methods 18/20

Measure Zero but Uncountable

At each step, we remove 13 of each remaining subinterval, so
measure multiplied by 23
• After n steps, remaining measure is

(
2
3

)n
→ 0

Cantor set includes all points of the form

x =
∞∑

i=1
ai3−i

with ai ∈ {0, 2}—set of infinite sequences of 0s and 2s is
uncountable

Evan Sadler Math Methods 18/20

Lebesgue-Stieltjes Measure
Instead of µ((a, b]) = b− a, take µ((a, b]) = F (b)− F (a) as our
premeasure; extends by CET to a measure on the Borel sets

Theorem
Given any increasing and right-continuous F : R→ R, there
exists a unique measure µ on B(R) such that
µ((a, b]) = F (b)− F (a). We call this the Lebesgue-Stieljes
measure.

Right continuity is necessary so µ
(
(a, b+ 1

n
]
)
→ µ((a, b])

Left continuity unnecessary, can capture mass points
• F (x) = 1 for x ≥ 0 and 0 for x < 0 gives point mass at 0 If µ(R) = 1, a probability measure, then F is the CDF Evan Sadler Math Methods 19/20 Lebesgue-Stieltjes Measure Instead of µ((a, b]) = b− a, take µ((a, b]) = F (b)− F (a) as our premeasure; extends by CET to a measure on the Borel sets Theorem Given any increasing and right-continuous F : R→ R, there exists a unique measure µ on B(R) such that µ((a, b]) = F (b)− F (a). We call this the Lebesgue-Stieljes measure. Right continuity is necessary so µ ( (a, b+ 1 n ] ) → µ((a, b]) Left continuity unnecessary, can capture mass points • F (x) = 1 for x ≥ 0 and 0 for x < 0 gives point mass at 0 If µ(R) = 1, a probability measure, then F is the CDF Evan Sadler Math Methods 19/20 Lebesgue-Stieltjes Measure Instead of µ((a, b]) = b− a, take µ((a, b]) = F (b)− F (a) as our premeasure; extends by CET to a measure on the Borel sets Theorem Given any increasing and right-continuous F : R→ R, there exists a unique measure µ on B(R) such that µ((a, b]) = F (b)− F (a). We call this the Lebesgue-Stieljes measure. Right continuity is necessary so µ ( (a, b+ 1 n ] ) → µ((a, b]) Left continuity unnecessary, can capture mass points • F (x) = 1 for x ≥ 0 and 0 for x < 0 gives point mass at 0 If µ(R) = 1, a probability measure, then F is the CDF Evan Sadler Math Methods 19/20 Lebesgue-Stieltjes Measure Instead of µ((a, b]) = b− a, take µ((a, b]) = F (b)− F (a) as our premeasure; extends by CET to a measure on the Borel sets Theorem Given any increasing and right-continuous F : R→ R, there exists a unique measure µ on B(R) such that µ((a, b]) = F (b)− F (a). We call this the Lebesgue-Stieljes measure. Right continuity is necessary so µ ( (a, b+ 1 n ] ) → µ((a, b]) Left continuity unnecessary, can capture mass points • F (x) = 1 for x ≥ 0 and 0 for x < 0 gives point mass at 0 If µ(R) = 1, a probability measure, then F is the CDF Evan Sadler Math Methods 19/20 Lebesgue-Stieltjes Measure Instead of µ((a, b]) = b− a, take µ((a, b]) = F (b)− F (a) as our premeasure; extends by CET to a measure on the Borel sets Theorem Given any increasing and right-continuous F : R→ R, there exists a unique measure µ on B(R) such that µ((a, b]) = F (b)− F (a). We call this the Lebesgue-Stieljes measure. Right continuity is necessary so µ ( (a, b+ 1 n ] ) → µ((a, b]) Left continuity unnecessary, can capture mass points • F (x) = 1 for x ≥ 0 and 0 for x < 0 gives point mass at 0 If µ(R) = 1, a probability measure, then F is the CDF Evan Sadler Math Methods 19/20 Looking Ahead Next time: Lebesgue Integral Measurable functions Integration with respect to a measure Absolute continuity and density Evan Sadler Math Methods 20/20