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