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Mathematical Methods Lebesgue Integration

Mathematical Methods
Lebesgue Integration

Evan Sadler
Columbia University

November 8, 2021

Evan Sadler Math Methods 1/21

Review

Lebesgue measure λ on R has the following properties:
• λ(A) is non-negative and countably additive
• λ(A) is translation invariant and homogeneous
• Any interval (a, b) has measure b− a
• Any countable set has measure zero
• Defined on the complete Lebesgue σ-algebra; every subset of a
measure-zero set is measurable

• Not all sets are Lebesgue measurable

Evan Sadler Math Methods 2/21

Today

Measurable functions

Integration with respect to a measure

Absolute continuity and density

Evan Sadler Math Methods 3/21

Measurable Functions
Definition
Suppose (X,F) is a measurable space. The function
f : X → R is measurable if for any open U ⊆ R, the inverse
image f−1(U) is a measurable set in F .

Inverse image of open sets measurable =⇒ continuous functions
are measurable
• Should say “F -measurable,” but F typically clear from context

Simpler way to check measurability

Proposition
f is measurable iff f−1((a,∞)) is measurable for any a ∈ R. The
same is true replacing (a,∞) with [a,∞) or (−∞, a) or (−∞, a].

Suffices to check inverse image of a set of generators of B(R)

Evan Sadler Math Methods 4/21

Measurable Functions
Definition
Suppose (X,F) is a measurable space. The function
f : X → R is measurable if for any open U ⊆ R, the inverse
image f−1(U) is a measurable set in F .

Inverse image of open sets measurable =⇒ continuous functions
are measurable
• Should say “F -measurable,” but F typically clear from context

Simpler way to check measurability

Proposition
f is measurable iff f−1((a,∞)) is measurable for any a ∈ R. The
same is true replacing (a,∞) with [a,∞) or (−∞, a) or (−∞, a].

Suffices to check inverse image of a set of generators of B(R)

Evan Sadler Math Methods 4/21

Measurable Functions
Definition
Suppose (X,F) is a measurable space. The function
f : X → R is measurable if for any open U ⊆ R, the inverse
image f−1(U) is a measurable set in F .

Inverse image of open sets measurable =⇒ continuous functions
are measurable
• Should say “F -measurable,” but F typically clear from context

Simpler way to check measurability

Proposition
f is measurable iff f−1((a,∞)) is measurable for any a ∈ R. The
same is true replacing (a,∞) with [a,∞) or (−∞, a) or (−∞, a].

Suffices to check inverse image of a set of generators of B(R)
Evan Sadler Math Methods 4/21

Indicator Functions
The simplest kind of measurable function

Definition
For any A ⊆ X, the indicator function (or characteristic
function) of A is

χA(x) =


1 if x ∈ A0 if x /∈ A

Defined for any set A, could have A = Q ∩ [0, 1]

Proposition
The indicator function χA is measurable iff A is a measurable set.

Proof: The set χ−1A ((a,∞)) is empty when a ≥ 1, A when
a ∈ [0, 1), and X when a < 0 Evan Sadler Math Methods 5/21 Indicator Functions The simplest kind of measurable function Definition For any A ⊆ X, the indicator function (or characteristic function) of A is χA(x) =  1 if x ∈ A0 if x /∈ A Defined for any set A, could have A = Q ∩ [0, 1] Proposition The indicator function χA is measurable iff A is a measurable set. Proof: The set χ−1A ((a,∞)) is empty when a ≥ 1, A when a ∈ [0, 1), and X when a < 0 Evan Sadler Math Methods 5/21 Indicator Functions The simplest kind of measurable function Definition For any A ⊆ X, the indicator function (or characteristic function) of A is χA(x) =  1 if x ∈ A0 if x /∈ A Defined for any set A, could have A = Q ∩ [0, 1] Proposition The indicator function χA is measurable iff A is a measurable set. Proof: The set χ−1A ((a,∞)) is empty when a ≥ 1, A when a ∈ [0, 1), and X when a < 0 Evan Sadler Math Methods 5/21 Properties of Measurable Functions Proposition If f and g are measurable, then so are f + g and f · g If {fn}n∈N is a sequence of measurable functions, then so are sup n fn(x) inf n fn(x) lim sup n→∞ fn(x) lim inf n→∞ fn(x) Will prove supn fn(x) is measurable, the rest are exercises For any a, we have g(x) := supn fn(x) > a iff fn(x) > a for
some n, so

g−1((a,∞)) = ∪∞n=1f
−1((a,∞))

RHS is countable union of measurable sets, so measurable

Evan Sadler Math Methods 6/21

Properties of Measurable Functions
Proposition
If f and g are measurable, then so are f + g and f · g

If {fn}n∈N is a sequence of measurable functions, then so are

sup
n
fn(x) inf

n
fn(x)

lim sup
n→∞

fn(x) lim inf
n→∞

fn(x)

Will prove supn fn(x) is measurable, the rest are exercises

For any a, we have g(x) := supn fn(x) > a iff fn(x) > a for
some n, so

g−1((a,∞)) = ∪∞n=1f
−1((a,∞))

RHS is countable union of measurable sets, so measurable

Evan Sadler Math Methods 6/21

Properties of Measurable Functions
Proposition
If f and g are measurable, then so are f + g and f · g

If {fn}n∈N is a sequence of measurable functions, then so are

sup
n
fn(x) inf

n
fn(x)

lim sup
n→∞

fn(x) lim inf
n→∞

fn(x)

Will prove supn fn(x) is measurable, the rest are exercises

For any a, we have g(x) := supn fn(x) > a iff fn(x) > a for
some n, so

g−1((a,∞)) = ∪∞n=1f
−1((a,∞))

RHS is countable union of measurable sets, so measurable
Evan Sadler Math Methods 6/21

Simple Functions
Definition
A simple function is a finite linear combination of measurable
indicator functions. For some {λi} ⊆ R and {Ai} ⊆ F , we have

f(x) =
n∑
i=1

λiχAi(x).

Equivalently, f is measurable and takes finitely many values
• If the Ai are intervals, also called step functions

Evan Sadler Math Methods 7/21

Simple Functions
Definition
A simple function is a finite linear combination of measurable
indicator functions. For some {λi} ⊆ R and {Ai} ⊆ F , we have

f(x) =
n∑
i=1

λiχAi(x).

Equivalently, f is measurable and takes finitely many values
• If the Ai are intervals, also called step functions

Evan Sadler Math Methods 7/21

Approximation by Simple Functions
We can approximation a general non-negative function using
simple functions

Lemma
If f : X → R+ is measurable, there exists an increasing
sequence {φn}n∈N of non-negative simple functions such that for
each x ∈ X, we have φn(x)→ f(x) as n→∞.

For each n, define φn(x) as truncation of f(x) to 2n, then
rounded down to smallest integer multiple of 2−n
• φn(x) takes finitely many values

1 · 2−n, 2 · 2−n, …, 4n · 2−n = 2n
• φn(x) is measurable as φn(x) = a · 2−n iff
f(x) ∈ [a · 2−n, (a+ 1) · 2−n)

• φn(x) is increasing in n as 2−n is a multiple of 2−(n+1)
• Clearly φn(x)→ f(x)

Evan Sadler Math Methods 8/21

Approximation by Simple Functions
We can approximation a general non-negative function using
simple functions

Lemma
If f : X → R+ is measurable, there exists an increasing
sequence {φn}n∈N of non-negative simple functions such that for
each x ∈ X, we have φn(x)→ f(x) as n→∞.

For each n, define φn(x) as truncation of f(x) to 2n, then
rounded down to smallest integer multiple of 2−n
• φn(x) takes finitely many values

1 · 2−n, 2 · 2−n, …, 4n · 2−n = 2n
• φn(x) is measurable as φn(x) = a · 2−n iff
f(x) ∈ [a · 2−n, (a+ 1) · 2−n)

• φn(x) is increasing in n as 2−n is a multiple of 2−(n+1)
• Clearly φn(x)→ f(x)

Evan Sadler Math Methods 8/21

Integral of Indicator Functions
Let (X,F , µ) be a measure space. Over the next several slides,
we gradually define integration w.r.t. the measure µ

Definition
For any measurable set A, the integral of χA is µ(A), written∫

X
χAdµ = µ(A)

When µ is Lebesgue measure, the area under the indicator
function is the “length” of A, multiplied by the height of 1

But we can define this for a larger class of sets than the Riemann
integral
• e.g. if A = Q ∩ [0, 1], then


R χAdλ = 0, but the Riemann

integral does not exist

Evan Sadler Math Methods 9/21

Integral of Indicator Functions
Let (X,F , µ) be a measure space. Over the next several slides,
we gradually define integration w.r.t. the measure µ

Definition
For any measurable set A, the integral of χA is µ(A), written∫

X
χAdµ = µ(A)

When µ is Lebesgue measure, the area under the indicator
function is the “length” of A, multiplied by the height of 1

But we can define this for a larger class of sets than the Riemann
integral
• e.g. if A = Q ∩ [0, 1], then


R χAdλ = 0, but the Riemann

integral does not exist

Evan Sadler Math Methods 9/21

Integral of Indicator Functions
Let (X,F , µ) be a measure space. Over the next several slides,
we gradually define integration w.r.t. the measure µ

Definition
For any measurable set A, the integral of χA is µ(A), written∫

X
χAdµ = µ(A)

When µ is Lebesgue measure, the area under the indicator
function is the “length” of A, multiplied by the height of 1

But we can define this for a larger class of sets than the Riemann
integral
• e.g. if A = Q ∩ [0, 1], then


R χAdλ = 0, but the Riemann

integral does not exist

Evan Sadler Math Methods 9/21

Integral of Indicator Functions
Let (X,F , µ) be a measure space. Over the next several slides,
we gradually define integration w.r.t. the measure µ

Definition
For any measurable set A, the integral of χA is µ(A), written∫

X
χAdµ = µ(A)

When µ is Lebesgue measure, the area under the indicator
function is the “length” of A, multiplied by the height of 1

But we can define this for a larger class of sets than the Riemann
integral
• e.g. if A = Q ∩ [0, 1], then


R χAdλ = 0, but the Riemann

integral does not exist
Evan Sadler Math Methods 9/21

Integral of Simple Functions
Simple functions are finite linear combinations of indicator
functions
• We define the integral as the corresponding linear combination
of the integrals of the indicator functions

Definition
If f =

∑n
i=1 λiχAi for λi ≥ 0 and Ai ∈ F , define∫

X
fdµ =

n∑
i=1

λiµ(Ai)

Homework: show that this gives a unique value for the integral of
a simple function
• That is, if

∑n
i=1 λiχAi(x) =

∑m
i=1 νiχBi(x) for each x, then∑n

i=1 λiµ(Ai) =
∑m
i=1 νiµ(Bi)

Evan Sadler Math Methods 10/21

Integral of Simple Functions
Simple functions are finite linear combinations of indicator
functions
• We define the integral as the corresponding linear combination
of the integrals of the indicator functions

Definition
If f =

∑n
i=1 λiχAi for λi ≥ 0 and Ai ∈ F , define∫

X
fdµ =

n∑
i=1

λiµ(Ai)

Homework: show that this gives a unique value for the integral of
a simple function
• That is, if

∑n
i=1 λiχAi(x) =

∑m
i=1 νiχBi(x) for each x, then∑n

i=1 λiµ(Ai) =
∑m
i=1 νiµ(Bi)

Evan Sadler Math Methods 10/21

Integral of Simple Functions
Simple functions are finite linear combinations of indicator
functions
• We define the integral as the corresponding linear combination
of the integrals of the indicator functions

Definition
If f =

∑n
i=1 λiχAi for λi ≥ 0 and Ai ∈ F , define∫

X
fdµ =

n∑
i=1

λiµ(Ai)

Homework: show that this gives a unique value for the integral of
a simple function
• That is, if

∑n
i=1 λiχAi(x) =

∑m
i=1 νiχBi(x) for each x, then∑n

i=1 λiµ(Ai) =
∑m
i=1 νiµ(Bi)

Evan Sadler Math Methods 10/21

Integral of Non-negative Functions
We then define the integral of any non-negative function via
approximation

Definition
If f ≥ 0 is measurable, we define its integral as∫

X
fdµ = sup

0≤φ≤f
φ simple


X
φdµ

Unlike the Riemann integral, only consider approximation from
below
• Works for unbounded functions (upper sums not defined)

Consider
∫ 1

0
1

2

x
dx, not properly defined

• Could take limit of
∫ 1

1
2

x
dx as �→ 0

• Lebesgue integral justifies direct use of FTC

Evan Sadler Math Methods 11/21

Integral of Non-negative Functions
We then define the integral of any non-negative function via
approximation

Definition
If f ≥ 0 is measurable, we define its integral as∫

X
fdµ = sup

0≤φ≤f
φ simple


X
φdµ

Unlike the Riemann integral, only consider approximation from
below
• Works for unbounded functions (upper sums not defined)

Consider
∫ 1

0
1

2

x
dx, not properly defined

• Could take limit of
∫ 1

1
2

x
dx as �→ 0

• Lebesgue integral justifies direct use of FTC

Evan Sadler Math Methods 11/21

Integral of Non-negative Functions
We then define the integral of any non-negative function via
approximation

Definition
If f ≥ 0 is measurable, we define its integral as∫

X
fdµ = sup

0≤φ≤f
φ simple


X
φdµ

Unlike the Riemann integral, only consider approximation from
below
• Works for unbounded functions (upper sums not defined)

Consider
∫ 1

0
1

2

x
dx, not properly defined

• Could take limit of
∫ 1

1
2

x
dx as �→ 0

• Lebesgue integral justifies direct use of FTC
Evan Sadler Math Methods 11/21

Integral of L1 Functions
Lemma
Any real valued f is a difference of two non-negative functions
f+(x) = max{f(x), 0} and f−(x) = max{−f(x), 0}

For any measurable f , define

f as


f+ −


f−, as long as both

integrals are finite

Definition
A measurable function f is µ-integrable if both


X f

+dµ and∫
X f

−dµ are finite, and define∫
X
fdµ =


X
f+dµ−


X
f−dµ

Write L1(µ) for set of all integrable functions

Evan Sadler Math Methods 12/21

Integral of L1 Functions
Lemma
Any real valued f is a difference of two non-negative functions
f+(x) = max{f(x), 0} and f−(x) = max{−f(x), 0}

For any measurable f , define

f as


f+ −


f−, as long as both

integrals are finite

Definition
A measurable function f is µ-integrable if both


X f

+dµ and∫
X f

−dµ are finite, and define∫
X
fdµ =


X
f+dµ−


X
f−dµ

Write L1(µ) for set of all integrable functions

Evan Sadler Math Methods 12/21

Integral of L1 Functions
Lemma
Any real valued f is a difference of two non-negative functions
f+(x) = max{f(x), 0} and f−(x) = max{−f(x), 0}

For any measurable f , define

f as


f+ −


f−, as long as both

integrals are finite

Definition
A measurable function f is µ-integrable if both


X f

+dµ and∫
X f

−dµ are finite, and define∫
X
fdµ =


X
f+dµ−


X
f−dµ

Write L1(µ) for set of all integrable functions

Evan Sadler Math Methods 12/21

Monotone Convergence Theorem
The integral is linear, meaning


(αf + βg) = α


f + β


g

The integral also preserves limits, under some conditions

Theorem (Monotone Convergence Theorem)
Suppose {fn}n∈N is an increasing sequence of non-negative
measurable functions with fn(x)→ f(x) as n→∞. Then∫

X
fdµ = lim

n→∞


X
fndµ

Reduces to continuity from below if fn and f are indicators
• i.e. fn = χAn and f = χA with A1 ⊆ A2 ⊆ · · ·
• Proof of the full result builds from this, show for simple
functions, then non-negative functions

Evan Sadler Math Methods 13/21

Monotone Convergence Theorem
The integral is linear, meaning


(αf + βg) = α


f + β


g

The integral also preserves limits, under some conditions

Theorem (Monotone Convergence Theorem)
Suppose {fn}n∈N is an increasing sequence of non-negative
measurable functions with fn(x)→ f(x) as n→∞. Then∫

X
fdµ = lim

n→∞


X
fndµ

Reduces to continuity from below if fn and f are indicators
• i.e. fn = χAn and f = χA with A1 ⊆ A2 ⊆ · · ·
• Proof of the full result builds from this, show for simple
functions, then non-negative functions

Evan Sadler Math Methods 13/21

Monotone Convergence Theorem
The integral is linear, meaning


(αf + βg) = α


f + β


g

The integral also preserves limits, under some conditions

Theorem (Monotone Convergence Theorem)
Suppose {fn}n∈N is an increasing sequence of non-negative
measurable functions with fn(x)→ f(x) as n→∞. Then∫

X
fdµ = lim

n→∞


X
fndµ

Reduces to continuity from below if fn and f are indicators
• i.e. fn = χAn and f = χA with A1 ⊆ A2 ⊆ · · ·
• Proof of the full result builds from this, show for simple
functions, then non-negative functions

Evan Sadler Math Methods 13/21

Dominated Convergence Theorem

For a more general sequence of functions, need more conditions

Theorem (Dominated Convergence Theorem)
Suppose {fn}n∈N is a sequence of measurable functions with
fn(x)→ f(x) as n→∞.

Need integrable g that “dominates” each fn
• If fn are indicators, equivalent to continuity from above

Evan Sadler Math Methods 14/21

Dominated Convergence Theorem

For a more general sequence of functions, need more conditions

Theorem (Dominated Convergence Theorem)
Suppose {fn}n∈N is a sequence of measurable functions with
fn(x)→ f(x) as n→∞.

Need integrable g that “dominates” each fn
• If fn are indicators, equivalent to continuity from above

Evan Sadler Math Methods 14/21

Dominated Convergence Theorem

For a more general sequence of functions, need more conditions

Theorem (Dominated Convergence Theorem)
Suppose {fn}n∈N is a sequence of measurable functions with
fn(x)→ f(x) as n→∞.

Need integrable g that “dominates” each fn
• If fn are indicators, equivalent to continuity from above

Evan Sadler Math Methods 14/21

An Example

Consider fn = 1n · χ[0,n], defined on R+


fn = 1n · n = 1 for all n

For each x ∈ R, have fn(x)→ 0 as n→∞
• Integral of limit 6= limit of integrals

Does not contradict DCT: for g to dominate each fn, need
|g| ≥ 1

n
on each interval [n− 1, n]



|g| ≥ 1 + 12 +

1
3 + … =∞

Evan Sadler Math Methods 15/21

An Example

Consider fn = 1n · χ[0,n], defined on R+


fn = 1n · n = 1 for all n

For each x ∈ R, have fn(x)→ 0 as n→∞
• Integral of limit 6= limit of integrals

Does not contradict DCT: for g to dominate each fn, need
|g| ≥ 1

n
on each interval [n− 1, n]



|g| ≥ 1 + 12 +

1
3 + … =∞

Evan Sadler Math Methods 15/21

An Example

Consider fn = 1n · χ[0,n], defined on R+


fn = 1n · n = 1 for all n

For each x ∈ R, have fn(x)→ 0 as n→∞
• Integral of limit 6= limit of integrals

Does not contradict DCT: for g to dominate each fn, need
|g| ≥ 1

n
on each interval [n− 1, n]



|g| ≥ 1 + 12 +

1
3 + … =∞

Evan Sadler Math Methods 15/21

Lebesgue Integral
Consider measure space (R,B, λ),


R fdλ is the Lebesgue

integral

Can integrate over an interval, or any subdomain A ∈ B:∫
A
fdλ =


R
f · χAdλ

More flexible than Riemann integral, if
∫ b
a f(x)dx exists, then∫

[a,b] fdλ is the same value
• Often use same notation

∫ b
a f(x)dx, just expand set of

functions we can integrate
• Note: Lebesgue integral is invariant if we change f on a null
set, e.g., f(x) = 0 for x /∈ Q implies


f = 0

Evan Sadler Math Methods 16/21

Lebesgue Integral
Consider measure space (R,B, λ),


R fdλ is the Lebesgue

integral

Can integrate over an interval, or any subdomain A ∈ B:∫
A
fdλ =


R
f · χAdλ

More flexible than Riemann integral, if
∫ b
a f(x)dx exists, then∫

[a,b] fdλ is the same value
• Often use same notation

∫ b
a f(x)dx, just expand set of

functions we can integrate
• Note: Lebesgue integral is invariant if we change f on a null
set, e.g., f(x) = 0 for x /∈ Q implies


f = 0

Evan Sadler Math Methods 16/21

Lebesgue Integral
Consider measure space (R,B, λ),


R fdλ is the Lebesgue

integral

Can integrate over an interval, or any subdomain A ∈ B:∫
A
fdλ =


R
f · χAdλ

More flexible than Riemann integral, if
∫ b
a f(x)dx exists, then∫

[a,b] fdλ is the same value
• Often use same notation

∫ b
a f(x)dx, just expand set of

functions we can integrate
• Note: Lebesgue integral is invariant if we change f on a null
set, e.g., f(x) = 0 for x /∈ Q implies


f = 0

Evan Sadler Math Methods 16/21

Motivating Absolute Continuity
Recall for Riemann-Stieltjes integral∫
f(x)dg(x) =


f(x)g′(x)dx.

Can do similar exercise with Lebesgue integral, integrate w.r.t.
different measures

Two measures µ and ν on same measurable space (X,F), want
a measurable function h such that∫

X
fdν =


X
f · hdµ, ∀ f measurable

By our construction of the integral, it’s enough to check for
indicator functions, we just need

ν(A) =

A
hdµ, ∀A measurable

Later, h serves role of probability density

Evan Sadler Math Methods 17/21

Motivating Absolute Continuity
Recall for Riemann-Stieltjes integral∫
f(x)dg(x) =


f(x)g′(x)dx.

Can do similar exercise with Lebesgue integral, integrate w.r.t.
different measures

Two measures µ and ν on same measurable space (X,F), want
a measurable function h such that∫

X
fdν =


X
f · hdµ, ∀ f measurable

By our construction of the integral, it’s enough to check for
indicator functions, we just need

ν(A) =

A
hdµ, ∀A measurable

Later, h serves role of probability density

Evan Sadler Math Methods 17/21

Motivating Absolute Continuity
Recall for Riemann-Stieltjes integral∫
f(x)dg(x) =


f(x)g′(x)dx.

Can do similar exercise with Lebesgue integral, integrate w.r.t.
different measures

Two measures µ and ν on same measurable space (X,F), want
a measurable function h such that∫

X
fdν =


X
f · hdµ, ∀ f measurable

By our construction of the integral, it’s enough to check for
indicator functions, we just need

ν(A) =

A
hdµ, ∀A measurable

Later, h serves role of probability density

Evan Sadler Math Methods 17/21

Motivating Absolute Continuity
Recall for Riemann-Stieltjes integral∫
f(x)dg(x) =


f(x)g′(x)dx.

Can do similar exercise with Lebesgue integral, integrate w.r.t.
different measures

Two measures µ and ν on same measurable space (X,F), want
a measurable function h such that∫

X
fdν =


X
f · hdµ, ∀ f measurable

By our construction of the integral, it’s enough to check for
indicator functions, we just need

ν(A) =

A
hdµ, ∀A measurable

Later, h serves role of probability density
Evan Sadler Math Methods 17/21

Absolute Continuity

Seek measurable h with ν(A) =

A hdµ for all measurable A

If µ(A) = 0 then

A hdµ = 0, so must have ν(A) = 0 if

µ(A) = 0

Definition
The measure ν is absolutely continuous with respect to µ if
for any measurable set A, we have ν(A) = 0 whenever µ(A) = 0.

Evan Sadler Math Methods 18/21

Absolute Continuity

Seek measurable h with ν(A) =

A hdµ for all measurable A

If µ(A) = 0 then

A hdµ = 0, so must have ν(A) = 0 if

µ(A) = 0

Definition
The measure ν is absolutely continuous with respect to µ if
for any measurable set A, we have ν(A) = 0 whenever µ(A) = 0.

Evan Sadler Math Methods 18/21

Absolute Continuity

Seek measurable h with ν(A) =

A hdµ for all measurable A

If µ(A) = 0 then

A hdµ = 0, so must have ν(A) = 0 if

µ(A) = 0

Definition
The measure ν is absolutely continuous with respect to µ if
for any measurable set A, we have ν(A) = 0 whenever µ(A) = 0.

Evan Sadler Math Methods 18/21

Radon-Nikodym Theorem
Absolute continuity is almost sufficient for the density h to exist

Theorem (Radon-Nikodym Theorem)
Suppose µ and ν are σ finite measures, and ν is absolutely
continuous w.r.t. µ. There exists h such that

ν(A) =

A
hdµ, ∀h measurable

The function h is called the Radon-Nikodym derivative,
denoted dνdµ .

Consequence: ∫
X
fdν =


X
f · hdµ

Evan Sadler Math Methods 19/21

Radon-Nikodym Theorem
Absolute continuity is almost sufficient for the density h to exist

Theorem (Radon-Nikodym Theorem)
Suppose µ and ν are σ finite measures, and ν is absolutely
continuous w.r.t. µ. There exists h such that

ν(A) =

A
hdµ, ∀h measurable

The function h is called the Radon-Nikodym derivative,
denoted dνdµ .

Consequence: ∫
X
fdν =


X
f · hdµ

Evan Sadler Math Methods 19/21

Specializing to R

Take X = R, with µ the Lebesgue measure and ν a general
Lebesgue-Stieltjes measure with ν((a, b]) = F (b)− F (a) for
some F that is increasing and right-continuous

If F is not left-continuous at some x, then
ν((−∞, x]) > ν((−∞, x)), indicating a mass point at x
• Not absolutely continuous w.r.t. Lebesgue measure

F continuous is necessary for absolute continuity, but in general
not sufficient

Evan Sadler Math Methods 20/21

Specializing to R

Take X = R, with µ the Lebesgue measure and ν a general
Lebesgue-Stieltjes measure with ν((a, b]) = F (b)− F (a) for
some F that is increasing and right-continuous

If F is not left-continuous at some x, then
ν((−∞, x]) > ν((−∞, x)), indicating a mass point at x
• Not absolutely continuous w.r.t. Lebesgue measure

F continuous is necessary for absolute continuity, but in general
not sufficient

Evan Sadler Math Methods 20/21

Specializing to R

Take X = R, with µ the Lebesgue measure and ν a general
Lebesgue-Stieltjes measure with ν((a, b]) = F (b)− F (a) for
some F that is increasing and right-continuous

If F is not left-continuous at some x, then
ν((−∞, x]) > ν((−∞, x)), indicating a mass point at x
• Not absolutely continuous w.r.t. Lebesgue measure

F continuous is necessary for absolute continuity, but in general
not sufficient

Evan Sadler Math Methods 20/21

Looking Ahead

Next time: Probability

Basic notation and terminology

Convergence of random variables

Evan Sadler Math Methods 21/21