Mathematical Methods Measure Theory
Mathematical Methods
Measure Theory
Evan Sadler
Columbia University
October 27, 2021
Evan Sadler Math Methods 1/21
Today
Motivation for measure theory
Definitions and examples
Evan Sadler Math Methods 2/21
Review of Countable and Uncountable Sets
Set A is countable if its elements can be exhaustively listed
x1, x2, …
• A is either finite, or can be bijectively mapped to N
Set of integers Z and set of rational numbers Q are countable
• We saw that R is uncountable
Similarly, any open interval (a, b) is uncountable
One criterion for evaluating the size of a set
Evan Sadler Math Methods 3/21
Review of Countable and Uncountable Sets
Set A is countable if its elements can be exhaustively listed
x1, x2, …
• A is either finite, or can be bijectively mapped to N
Set of integers Z and set of rational numbers Q are countable
• We saw that R is uncountable
Similarly, any open interval (a, b) is uncountable
One criterion for evaluating the size of a set
Evan Sadler Math Methods 3/21
Review of Countable and Uncountable Sets
Set A is countable if its elements can be exhaustively listed
x1, x2, …
• A is either finite, or can be bijectively mapped to N
Set of integers Z and set of rational numbers Q are countable
• We saw that R is uncountable
Similarly, any open interval (a, b) is uncountable
One criterion for evaluating the size of a set
Evan Sadler Math Methods 3/21
Review of Countable and Uncountable Sets
Set A is countable if its elements can be exhaustively listed
x1, x2, …
• A is either finite, or can be bijectively mapped to N
Set of integers Z and set of rational numbers Q are countable
• We saw that R is uncountable
Similarly, any open interval (a, b) is uncountable
One criterion for evaluating the size of a set
Evan Sadler Math Methods 3/21
Motivation for Measure Theory
Measure theory provides an alternative perspective on the size of
a set
Recall, sufficiently well-behaved functions are Riemann integrable
• Functions that are continuous “almost everywhere”
What this really means is that the set of points at which the
function is not continuous has measure zero
• We still need to define the measure of a set A ⊆ R
Measure is fundamentally a geometric notion, extends the idea of
the length of an interval
Evan Sadler Math Methods 4/21
Motivation for Measure Theory
Measure theory provides an alternative perspective on the size of
a set
Recall, sufficiently well-behaved functions are Riemann integrable
• Functions that are continuous “almost everywhere”
What this really means is that the set of points at which the
function is not continuous has measure zero
• We still need to define the measure of a set A ⊆ R
Measure is fundamentally a geometric notion, extends the idea of
the length of an interval
Evan Sadler Math Methods 4/21
Motivation for Measure Theory
Measure theory provides an alternative perspective on the size of
a set
Recall, sufficiently well-behaved functions are Riemann integrable
• Functions that are continuous “almost everywhere”
What this really means is that the set of points at which the
function is not continuous has measure zero
• We still need to define the measure of a set A ⊆ R
Measure is fundamentally a geometric notion, extends the idea of
the length of an interval
Evan Sadler Math Methods 4/21
Motivation for Measure Theory
Measure theory provides an alternative perspective on the size of
a set
Recall, sufficiently well-behaved functions are Riemann integrable
• Functions that are continuous “almost everywhere”
What this really means is that the set of points at which the
function is not continuous has measure zero
• We still need to define the measure of a set A ⊆ R
Measure is fundamentally a geometric notion, extends the idea of
the length of an interval
Evan Sadler Math Methods 4/21
A Loose Definition
A measure is a function µ defined on subsets of R
Some properties we want µ to have:
• Non-negativity: µ(A) ≥ 0 wherever µ is defined, we do allow
µ(A) =∞
• Identity: µ([0, 1]) = 1
• Translation invariance: µ(A+ t) = µ(A) for any t ∈ R, with
A+ t = {x+ t : x ∈ A}
• Countable additivity: µ (∪∞i=1Ai) =
∑∞
i=1 µ(Ai) whenever
A1, A2, … are disjoint
Evan Sadler Math Methods 5/21
A Loose Definition
A measure is a function µ defined on subsets of R
Some properties we want µ to have:
• Non-negativity: µ(A) ≥ 0 wherever µ is defined, we do allow
µ(A) =∞
• Identity: µ([0, 1]) = 1
• Translation invariance: µ(A+ t) = µ(A) for any t ∈ R, with
A+ t = {x+ t : x ∈ A}
• Countable additivity: µ (∪∞i=1Ai) =
∑∞
i=1 µ(Ai) whenever
A1, A2, … are disjoint
Evan Sadler Math Methods 5/21
Finite Additivity?
What if we weaken the last condition to finite additivity, meaning
µ (∪ni=1Ai) =
n∑
i=1
µ(Ai)
for any positive integer n and disjoint sets {Ai}?
Finitely additive measures are used in some applications, but
allowing infinite sums is key for dealing with limits
Evan Sadler Math Methods 6/21
Finite Additivity?
What if we weaken the last condition to finite additivity, meaning
µ (∪ni=1Ai) =
n∑
i=1
µ(Ai)
for any positive integer n and disjoint sets {Ai}?
Finitely additive measures are used in some applications, but
allowing infinite sums is key for dealing with limits
Evan Sadler Math Methods 6/21
Uncountable Additivity?
What if we strengthen the condition to “uncountable additivity,”
meaning
µ (∪i∈IAi) =
∑
i∈I
µ(Ai) := sup
J⊆I
J countable
∑
j∈J
µ(Aj)
for all (potentially uncountable) index sets I with disjoint Ai ⊆ R
An immediate problem with measure of singletons {x}
• A point should have measure 0, but then uncountable
additivity implies
µ([0, 1]) =
∑
x∈[0,1]
µ({x}) = 0
Evan Sadler Math Methods 7/21
Uncountable Additivity?
What if we strengthen the condition to “uncountable additivity,”
meaning
µ (∪i∈IAi) =
∑
i∈I
µ(Ai) := sup
J⊆I
J countable
∑
j∈J
µ(Aj)
for all (potentially uncountable) index sets I with disjoint Ai ⊆ R
An immediate problem with measure of singletons {x}
• A point should have measure 0, but then uncountable
additivity implies
µ([0, 1]) =
∑
x∈[0,1]
µ({x}) = 0
Evan Sadler Math Methods 7/21
Existence of Unmeasurable Sets
Even countable additivity has some similar problems:
Proposition
A function µ defined on all subsets of R cannot simultaneously
satisfy non-negativity, identity, translation invariance, and
countable additivity.
Upshot: we define measure on a family of “measurable” subsets,
some subsets are not measurable
Evan Sadler Math Methods 8/21
Existence of Unmeasurable Sets
Even countable additivity has some similar problems:
Proposition
A function µ defined on all subsets of R cannot simultaneously
satisfy non-negativity, identity, translation invariance, and
countable additivity.
Upshot: we define measure on a family of “measurable” subsets,
some subsets are not measurable
Evan Sadler Math Methods 8/21
Proof
Define the equivalence relation ∼ on the interval [0, 1) via
x ∼ y ⇐⇒ x− y ∈ Q
Consider a set A ⊆ [0, 1) containing exactly one number from
each equivalence class of ∼
• (This relies on axiom of choice)
For each q ∈ [0, 1) ∩Q, define
Aq = {x+q : x ∈ A, x < 1−q}∪{x+q−1 : x ∈ A, x ≥ 1−q} In words, add q to each number in A, and subtract 1 if the sum exceeds 1 • By additivity and translation invariance, must have µ(Aq) = µ(A) Evan Sadler Math Methods 9/21 Proof Define the equivalence relation ∼ on the interval [0, 1) via x ∼ y ⇐⇒ x− y ∈ Q Consider a set A ⊆ [0, 1) containing exactly one number from each equivalence class of ∼ • (This relies on axiom of choice) For each q ∈ [0, 1) ∩Q, define Aq = {x+q : x ∈ A, x < 1−q}∪{x+q−1 : x ∈ A, x ≥ 1−q} In words, add q to each number in A, and subtract 1 if the sum exceeds 1 • By additivity and translation invariance, must have µ(Aq) = µ(A) Evan Sadler Math Methods 9/21 Proof Define the equivalence relation ∼ on the interval [0, 1) via x ∼ y ⇐⇒ x− y ∈ Q Consider a set A ⊆ [0, 1) containing exactly one number from each equivalence class of ∼ • (This relies on axiom of choice) For each q ∈ [0, 1) ∩Q, define Aq = {x+q : x ∈ A, x < 1−q}∪{x+q−1 : x ∈ A, x ≥ 1−q} In words, add q to each number in A, and subtract 1 if the sum exceeds 1 • By additivity and translation invariance, must have µ(Aq) = µ(A) Evan Sadler Math Methods 9/21 Proof Continued Our choice of A ensures the sets {Aq}q∈Q are disjoint, and their union is [0, 1), by additivity: µ([0, 1)) = ∑ q µ(Aq) but the LHS is 1, while the RHS sums infinitely many copies of the same number µ(A) Evan Sadler Math Methods 10/21 σ-algebras Definition Given a set X, a σ-algebra F on X is a non-empty family of subsets of X with two properties: • Closure under complements: if A ∈ F , then Ac = X − A ∈ F • Closure under countable unions: if {Ai}i∈N ⊆ F , then ∪∞i=1Ai ∈ F Proposition If F is a σ-algebra, then X ∈ F and ∅ ∈ F . Moreover, F is closed under countable intersections. First claim follows as X = A ∪ Ac for any A ∈ F , second from De Morgan’s Law Evan Sadler Math Methods 11/21 σ-algebras Definition Given a set X, a σ-algebra F on X is a non-empty family of subsets of X with two properties: • Closure under complements: if A ∈ F , then Ac = X − A ∈ F • Closure under countable unions: if {Ai}i∈N ⊆ F , then ∪∞i=1Ai ∈ F Proposition If F is a σ-algebra, then X ∈ F and ∅ ∈ F . Moreover, F is closed under countable intersections. First claim follows as X = A ∪ Ac for any A ∈ F , second from De Morgan’s Law Evan Sadler Math Methods 11/21 Examples F = 2X , the family of all subsets of X, is a σ-algebra F = {X, ∅} is a σ-algebra F containing all countable subsets of X and their complements • Clearly closed under complements • Closed under countable unions: if {Ai}i∈N all countable, so is their union • If X = R, this F does not contain all subsets (e.g. [0, 1] is neither countable, nor the complement of a countable set) Evan Sadler Math Methods 12/21 Examples F = 2X , the family of all subsets of X, is a σ-algebra F = {X, ∅} is a σ-algebra F containing all countable subsets of X and their complements • Clearly closed under complements • Closed under countable unions: if {Ai}i∈N all countable, so is their union • If X = R, this F does not contain all subsets (e.g. [0, 1] is neither countable, nor the complement of a countable set) Evan Sadler Math Methods 12/21 Examples F = 2X , the family of all subsets of X, is a σ-algebra F = {X, ∅} is a σ-algebra F containing all countable subsets of X and their complements • Clearly closed under complements • Closed under countable unions: if {Ai}i∈N all countable, so is their union • If X = R, this F does not contain all subsets (e.g. [0, 1] is neither countable, nor the complement of a countable set) Evan Sadler Math Methods 12/21 Generation Key observation: intersection of σ-algebras is still a σ-algebra Proposition For any family of subsets E ⊆ 2X , there exists a smallest σ-algebraM(E) that contains E , which is the intersection of all σ-algebras that contain E We callM(E) the σ-algebra generated by E Evan Sadler Math Methods 13/21 Generation Key observation: intersection of σ-algebras is still a σ-algebra Proposition For any family of subsets E ⊆ 2X , there exists a smallest σ-algebraM(E) that contains E , which is the intersection of all σ-algebras that contain E We callM(E) the σ-algebra generated by E Evan Sadler Math Methods 13/21 Generation Key observation: intersection of σ-algebras is still a σ-algebra Proposition For any family of subsets E ⊆ 2X , there exists a smallest σ-algebraM(E) that contains E , which is the intersection of all σ-algebras that contain E We callM(E) the σ-algebra generated by E Evan Sadler Math Methods 13/21 The Borel σ-algebra Definition Let X be a metric space, and let E be the set of open sets in X. The Borel σ-algebra B(X) on X is the σ-algebraM(E) generated by the open sets. B(R) contains the open intervals (a, b) as well as • Closed intervals [a, b] (take complements) • The half-open intervals (a, b] and [b, a) (why?) • All singletons, and therefore countable sets (why?) We get any countable union of countable intersection...of the above sets • Hard to describe the σ-algebra explicitly Evan Sadler Math Methods 14/21 The Borel σ-algebra Definition Let X be a metric space, and let E be the set of open sets in X. The Borel σ-algebra B(X) on X is the σ-algebraM(E) generated by the open sets. B(R) contains the open intervals (a, b) as well as • Closed intervals [a, b] (take complements) • The half-open intervals (a, b] and [b, a) (why?) • All singletons, and therefore countable sets (why?) We get any countable union of countable intersection...of the above sets • Hard to describe the σ-algebra explicitly Evan Sadler Math Methods 14/21 The Borel σ-algebra Definition Let X be a metric space, and let E be the set of open sets in X. The Borel σ-algebra B(X) on X is the σ-algebraM(E) generated by the open sets. B(R) contains the open intervals (a, b) as well as • Closed intervals [a, b] (take complements) • The half-open intervals (a, b] and [b, a) (why?) • All singletons, and therefore countable sets (why?) We get any countable union of countable intersection...of the above sets • Hard to describe the σ-algebra explicitly Evan Sadler Math Methods 14/21 Measure Definition Given a set X equipped with a σ-algebra F , a measure on F is a function µ : F → [0,∞] such that µ(∅) = 0 and µ is countably additive: µ (∪∞i=1Ai) = ∞∑ i=1 µ(Ai) whenever the {Ai} are disjoint. The pair (X,F) is a measurable space, each set A ∈ F is a measurable set, the triple (X,F , µ) is a measure space The measure µ is finite if µ(X) <∞, the measure is σ-finite if there exists {Ai}i∈N such that X = ∪∞i=1Ai and µ(Ai) <∞ for each i. Evan Sadler Math Methods 15/21 Measure Definition Given a set X equipped with a σ-algebra F , a measure on F is a function µ : F → [0,∞] such that µ(∅) = 0 and µ is countably additive: µ (∪∞i=1Ai) = ∞∑ i=1 µ(Ai) whenever the {Ai} are disjoint. The pair (X,F) is a measurable space, each set A ∈ F is a measurable set, the triple (X,F , µ) is a measure space The measure µ is finite if µ(X) <∞, the measure is σ-finite if there exists {Ai}i∈N such that X = ∪∞i=1Ai and µ(Ai) <∞ for each i. Evan Sadler Math Methods 15/21 More Examples The counting measure: Take F = 2X and µ(A) = |A|, the number of elements in A (we have µ(A) =∞ if A is infinite). This is countably additive Dirac Measure: F = 2X , and for some x∗ ∈ X, we have µ(A) = 1 iff x∗ ∈ A and µ(A) = 0 otherwise. This is also countably additive F = 2X and µ(A) =∞ for any A 6= ∅ Suppose X is uncountable, and F contains all countable subsets and their complements, µ(A) = 0 if A is countable and µ(A) = 1 otherwise Evan Sadler Math Methods 16/21 More Examples The counting measure: Take F = 2X and µ(A) = |A|, the number of elements in A (we have µ(A) =∞ if A is infinite). This is countably additive Dirac Measure: F = 2X , and for some x∗ ∈ X, we have µ(A) = 1 iff x∗ ∈ A and µ(A) = 0 otherwise. This is also countably additive F = 2X and µ(A) =∞ for any A 6= ∅ Suppose X is uncountable, and F contains all countable subsets and their complements, µ(A) = 0 if A is countable and µ(A) = 1 otherwise Evan Sadler Math Methods 16/21 More Examples The counting measure: Take F = 2X and µ(A) = |A|, the number of elements in A (we have µ(A) =∞ if A is infinite). This is countably additive Dirac Measure: F = 2X , and for some x∗ ∈ X, we have µ(A) = 1 iff x∗ ∈ A and µ(A) = 0 otherwise. This is also countably additive F = 2X and µ(A) =∞ for any A 6= ∅ Suppose X is uncountable, and F contains all countable subsets and their complements, µ(A) = 0 if A is countable and µ(A) = 1 otherwise Evan Sadler Math Methods 16/21 More Examples The counting measure: Take F = 2X and µ(A) = |A|, the number of elements in A (we have µ(A) =∞ if A is infinite). This is countably additive Dirac Measure: F = 2X , and for some x∗ ∈ X, we have µ(A) = 1 iff x∗ ∈ A and µ(A) = 0 otherwise. This is also countably additive F = 2X and µ(A) =∞ for any A 6= ∅ Suppose X is uncountable, and F contains all countable subsets and their complements, µ(A) = 0 if A is countable and µ(A) = 1 otherwise Evan Sadler Math Methods 16/21 Not a Measure Suppose X is infinite, F = 2X , and we have µ(A) = 0 if A is finite and µ(A) =∞ otherwise Similar to previous example but not countably additive • A countably infinite set is a countable union of finite sets, countable additivity implies such sets have measure 0 This function µ is finitely additive, however Evan Sadler Math Methods 17/21 Not a Measure Suppose X is infinite, F = 2X , and we have µ(A) = 0 if A is finite and µ(A) =∞ otherwise Similar to previous example but not countably additive • A countably infinite set is a countable union of finite sets, countable additivity implies such sets have measure 0 This function µ is finitely additive, however Evan Sadler Math Methods 17/21 Not a Measure Suppose X is infinite, F = 2X , and we have µ(A) = 0 if A is finite and µ(A) =∞ otherwise Similar to previous example but not countably additive • A countably infinite set is a countable union of finite sets, countable additivity implies such sets have measure 0 This function µ is finitely additive, however Evan Sadler Math Methods 17/21 Properties of Measure Theorem Suppose (X,F , µ) is a measure space. We have the following properties: • Monotonicity: if A,B ∈ F and B ⊆ A, then µ(B) ≤ µ(A) • Subadditivity: if {Ai} ⊆ F , then µ (∪∞i=1Ai) ≤ ∑∞ i=1 µ(Ai) • Continuity from below: If A1 ⊆ A2 ⊆ · · · ∈ F is an increasing sequence, then µ (∪∞i=1Ai) = limi→∞ µ(Ai) • Continuity from above: If A1 ⊇ A2 ⊇ · · · ∈ F is a decreasing sequence, then µ (∩∞i=1Ai) = limi→∞ µ(Ai), as long as µ(Ai) <∞ for some i Evan Sadler Math Methods 18/21 Proof Monotonicity: B ⊆ A implies µ(A) = µ(B) + µ(A−B) ≥ µ(B) Subadditivity: Let Bi = Ai − ∪i−1j=1Aj, a subset of Ai, and the {Bi} are disjoint • By construction, ∪∞i=1Bi = ∪∞i=1Ai, so: µ(∪∞i=1Ai) = µ(∪ ∞ i=1Bi) = ∞∑ i=1 µ(Bi) ≤ ∞∑ i=1 µ(Ai) Continuity from below: If {Ai} is an increasing sequence, then Bi above is Ai − Ai−1, and µ(∪∞i=1Ai) = ∞∑ i=1 µ(Bi) = lim n→∞ n∑ i=1 µ(Bi) = lim n→∞ µ(∪ni=1Bi) which equals limn→∞ µ(An) as desired Evan Sadler Math Methods 19/21 Proof Monotonicity: B ⊆ A implies µ(A) = µ(B) + µ(A−B) ≥ µ(B) Subadditivity: Let Bi = Ai − ∪i−1j=1Aj, a subset of Ai, and the {Bi} are disjoint • By construction, ∪∞i=1Bi = ∪∞i=1Ai, so: µ(∪∞i=1Ai) = µ(∪ ∞ i=1Bi) = ∞∑ i=1 µ(Bi) ≤ ∞∑ i=1 µ(Ai) Continuity from below: If {Ai} is an increasing sequence, then Bi above is Ai − Ai−1, and µ(∪∞i=1Ai) = ∞∑ i=1 µ(Bi) = lim n→∞ n∑ i=1 µ(Bi) = lim n→∞ µ(∪ni=1Bi) which equals limn→∞ µ(An) as desired Evan Sadler Math Methods 19/21 Proof Monotonicity: B ⊆ A implies µ(A) = µ(B) + µ(A−B) ≥ µ(B) Subadditivity: Let Bi = Ai − ∪i−1j=1Aj, a subset of Ai, and the {Bi} are disjoint • By construction, ∪∞i=1Bi = ∪∞i=1Ai, so: µ(∪∞i=1Ai) = µ(∪ ∞ i=1Bi) = ∞∑ i=1 µ(Bi) ≤ ∞∑ i=1 µ(Ai) Continuity from below: If {Ai} is an increasing sequence, then Bi above is Ai − Ai−1, and µ(∪∞i=1Ai) = ∞∑ i=1 µ(Bi) = lim n→∞ n∑ i=1 µ(Bi) = lim n→∞ µ(∪ni=1Bi) which equals limn→∞ µ(An) as desired Evan Sadler Math Methods 19/21 Proof Continuity from above: Suppose {Ai} is a decreasing sequence and µ(A1) <∞ • This is without loss, could just start analysis from first n such that µ(An) <∞ Define A′i = A1 − Ai, which is an increasing sequence • From last slide, we know µ(∪∞i=1A′i) = limn→∞ µ(A′n) LHS is µ(A1)− µ (∩∞i=1Ai), RHS is limn→∞ µ(A1)− µ(An) • Cancel out µ(A1) on both sides to get the result Note result can be false if each Ai has infinite measure • Take X = N, F = 2N and µ the counting measure with Ai = {x ∈ N : x > i}
• Each Ai is infinite, but countable intersection is empty
Evan Sadler Math Methods 20/21
Proof
Continuity from above: Suppose {Ai} is a decreasing sequence
and µ(A1) <∞
• This is without loss, could just start analysis from first n such
that µ(An) <∞
Define A′i = A1 − Ai, which is an increasing sequence
• From last slide, we know µ(∪∞i=1A′i) = limn→∞ µ(A′n)
LHS is µ(A1)− µ (∩∞i=1Ai), RHS is limn→∞ µ(A1)− µ(An)
• Cancel out µ(A1) on both sides to get the result
Note result can be false if each Ai has infinite measure
• Take X = N, F = 2N and µ the counting measure with
Ai = {x ∈ N : x > i}
• Each Ai is infinite, but countable intersection is empty
Evan Sadler Math Methods 20/21
Proof
Continuity from above: Suppose {Ai} is a decreasing sequence
and µ(A1) <∞
• This is without loss, could just start analysis from first n such
that µ(An) <∞
Define A′i = A1 − Ai, which is an increasing sequence
• From last slide, we know µ(∪∞i=1A′i) = limn→∞ µ(A′n)
LHS is µ(A1)− µ (∩∞i=1Ai), RHS is limn→∞ µ(A1)− µ(An)
• Cancel out µ(A1) on both sides to get the result
Note result can be false if each Ai has infinite measure
• Take X = N, F = 2N and µ the counting measure with
Ai = {x ∈ N : x > i}
• Each Ai is infinite, but countable intersection is empty
Evan Sadler Math Methods 20/21
Proof
Continuity from above: Suppose {Ai} is a decreasing sequence
and µ(A1) <∞
• This is without loss, could just start analysis from first n such
that µ(An) <∞
Define A′i = A1 − Ai, which is an increasing sequence
• From last slide, we know µ(∪∞i=1A′i) = limn→∞ µ(A′n)
LHS is µ(A1)− µ (∩∞i=1Ai), RHS is limn→∞ µ(A1)− µ(An)
• Cancel out µ(A1) on both sides to get the result
Note result can be false if each Ai has infinite measure
• Take X = N, F = 2N and µ the counting measure with
Ai = {x ∈ N : x > i}
• Each Ai is infinite, but countable intersection is empty
Evan Sadler Math Methods 20/21
Looking Ahead
Next time: Lebesgue measure
Extending the notion of interval length
Properties of Lebesgue measure
Evan Sadler Math Methods 21/21