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