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Chapter 2
Daniel H. Luecking
Sept 11, 2020
Daniel H. Luecking Chapter 2 Sept 11, 2020 1 / 13
Recall: The Cantor-Lebesgue function is continuous, nondecreasing, isconstant on each disjoint interval of [0, 1]∼ C and satisfies φ(C ) = [0, 1]
Now consider ψ(x) = x + φ(x) which takes [0, 1] onto [0, 2]. It is strictlyincreasing so it maps open intervals to open intervals (as does its inverse).
If we take one of the removed intervals for the Cantor set, (a,b), we haveψ((a, b)) = φ(a) + (a, b). That is m(ψ(a, b)) = m((a, b)). This impliesm(ψ([0, 1]∼ C )) = 1 so m(ψ(C )) = 1
We will show that any measurable set with positive measure contains anonmeasurable set. Since ψ(C ) has positive measure, it contains anonmeasurable set A. But B = ψ−1(A) ⊆ C so B is measurable. If we letf = ψ−1 we have shown that for the continuous function f there is ameasurable set B such that f −1(B) is not measurable.
We contrast this with
Lemma
If f is continuous then for every Borel set E , f −1(E ) is a Borel set andhence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 2 / 13
Recall: The Cantor-Lebesgue function is continuous, nondecreasing, isconstant on each disjoint interval of [0, 1]∼ C and satisfies φ(C ) = [0, 1]
Now consider ψ(x) = x + φ(x) which takes [0, 1] onto [0, 2]. It is strictlyincreasing so it maps open intervals to open intervals (as does its inverse).If we take one of the removed intervals for the Cantor set, (a,b), we haveψ((a, b)) = φ(a) + (a, b). That is m(ψ(a, b)) = m((a, b)).
This impliesm(ψ([0, 1]∼ C )) = 1 so m(ψ(C )) = 1
We will show that any measurable set with positive measure contains anonmeasurable set. Since ψ(C ) has positive measure, it contains anonmeasurable set A. But B = ψ−1(A) ⊆ C so B is measurable. If we letf = ψ−1 we have shown that for the continuous function f there is ameasurable set B such that f −1(B) is not measurable.
We contrast this with
Lemma
If f is continuous then for every Borel set E , f −1(E ) is a Borel set andhence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 2 / 13
Recall: The Cantor-Lebesgue function is continuous, nondecreasing, isconstant on each disjoint interval of [0, 1]∼ C and satisfies φ(C ) = [0, 1]
Now consider ψ(x) = x + φ(x) which takes [0, 1] onto [0, 2]. It is strictlyincreasing so it maps open intervals to open intervals (as does its inverse).If we take one of the removed intervals for the Cantor set, (a,b), we haveψ((a, b)) = φ(a) + (a, b). That is m(ψ(a, b)) = m((a, b)). This impliesm(ψ([0, 1]∼ C )) = 1 so m(ψ(C )) = 1
We will show that any measurable set with positive measure contains anonmeasurable set. Since ψ(C ) has positive measure, it contains anonmeasurable set A. But B = ψ−1(A) ⊆ C so B is measurable. If we letf = ψ−1 we have shown that for the continuous function f there is ameasurable set B such that f −1(B) is not measurable.
We contrast this with
Lemma
If f is continuous then for every Borel set E , f −1(E ) is a Borel set andhence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 2 / 13
Recall: The Cantor-Lebesgue function is continuous, nondecreasing, isconstant on each disjoint interval of [0, 1]∼ C and satisfies φ(C ) = [0, 1]
Now consider ψ(x) = x + φ(x) which takes [0, 1] onto [0, 2]. It is strictlyincreasing so it maps open intervals to open intervals (as does its inverse).If we take one of the removed intervals for the Cantor set, (a,b), we haveψ((a, b)) = φ(a) + (a, b). That is m(ψ(a, b)) = m((a, b)). This impliesm(ψ([0, 1]∼ C )) = 1 so m(ψ(C )) = 1
We will show that any measurable set with positive measure contains anonmeasurable set. Since ψ(C ) has positive measure, it contains anonmeasurable set A.
But B = ψ−1(A) ⊆ C so B is measurable. If we letf = ψ−1 we have shown that for the continuous function f there is ameasurable set B such that f −1(B) is not measurable.
We contrast this with
Lemma
If f is continuous then for every Borel set E , f −1(E ) is a Borel set andhence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 2 / 13
Recall: The Cantor-Lebesgue function is continuous, nondecreasing, isconstant on each disjoint interval of [0, 1]∼ C and satisfies φ(C ) = [0, 1]
Now consider ψ(x) = x + φ(x) which takes [0, 1] onto [0, 2]. It is strictlyincreasing so it maps open intervals to open intervals (as does its inverse).If we take one of the removed intervals for the Cantor set, (a,b), we haveψ((a, b)) = φ(a) + (a, b). That is m(ψ(a, b)) = m((a, b)). This impliesm(ψ([0, 1]∼ C )) = 1 so m(ψ(C )) = 1
We will show that any measurable set with positive measure contains anonmeasurable set. Since ψ(C ) has positive measure, it contains anonmeasurable set A. But B = ψ−1(A) ⊆ C so B is measurable. If we letf = ψ−1 we have shown that for the continuous function f there is ameasurable set B such that f −1(B) is not measurable.
We contrast this with
Lemma
If f is continuous then for every Borel set E , f −1(E ) is a Borel set andhence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 2 / 13
Recall: The Cantor-Lebesgue function is continuous, nondecreasing, isconstant on each disjoint interval of [0, 1]∼ C and satisfies φ(C ) = [0, 1]
Now consider ψ(x) = x + φ(x) which takes [0, 1] onto [0, 2]. It is strictlyincreasing so it maps open intervals to open intervals (as does its inverse).If we take one of the removed intervals for the Cantor set, (a,b), we haveψ((a, b)) = φ(a) + (a, b). That is m(ψ(a, b)) = m((a, b)). This impliesm(ψ([0, 1]∼ C )) = 1 so m(ψ(C )) = 1
We will show that any measurable set with positive measure contains anonmeasurable set. Since ψ(C ) has positive measure, it contains anonmeasurable set A. But B = ψ−1(A) ⊆ C so B is measurable. If we letf = ψ−1 we have shown that for the continuous function f there is ameasurable set B such that f −1(B) is not measurable.
We contrast this with
Lemma
If f is continuous then for every Borel set E , f −1(E ) is a Borel set andhence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 2 / 13
Proof: Let B be the Borel σ-algebra. Consider the family of setsA = {E : f −1(E ) ∈ B}.
To show that A contains B we need only showthat it is a σ-algebra and that it contains all open sets.
This is a standard argument:
1 Since f −1(∅) = ∅ ∈ B we have ∅ ∈ A.2 If E ∈ A then f −1(E ) ∈ B and so f −1(E c) = f −1(E )c ∈ B. Thus
E c ∈ A.3 Suppose Ei ∈ A, i ∈ N, then f −1(
⋃i Ei ) =
⋃i f−1(Ei ) ∈ B and so⋃
i Ei ∈ A.
Finally if U is open the f −1(U) is open, which is Borel so U ∈ A.
Corollary
There is a measurable set that is not a Borel set.
Proof: We have seen there is a measurable subset B of C and acontinuous function f such that f −1(B) is not measurable. Then Bcannot be a Borel set because if it were then f −1(B) would be a Borel set,hence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 3 / 13
Proof: Let B be the Borel σ-algebra. Consider the family of setsA = {E : f −1(E ) ∈ B}. To show that A contains B we need only showthat it is a σ-algebra and that it contains all open sets.
This is a standard argument:
1 Since f −1(∅) = ∅ ∈ B we have ∅ ∈ A.
2 If E ∈ A then f −1(E ) ∈ B and so f −1(E c) = f −1(E )c ∈ B. ThusE c ∈ A.
3 Suppose Ei ∈ A, i ∈ N, then f −1(⋃
i Ei ) =⋃
i f−1(Ei ) ∈ B and so⋃
i Ei ∈ A.
Finally if U is open the f −1(U) is open, which is Borel so U ∈ A.
Corollary
There is a measurable set that is not a Borel set.
Proof: We have seen there is a measurable subset B of C and acontinuous function f such that f −1(B) is not measurable. Then Bcannot be a Borel set because if it were then f −1(B) would be a Borel set,hence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 3 / 13
Proof: Let B be the Borel σ-algebra. Consider the family of setsA = {E : f −1(E ) ∈ B}. To show that A contains B we need only showthat it is a σ-algebra and that it contains all open sets.
This is a standard argument:
1 Since f −1(∅) = ∅ ∈ B we have ∅ ∈ A.2 If E ∈ A then f −1(E ) ∈ B and so f −1(E c) = f −1(E )c ∈ B. Thus
E c ∈ A.
3 Suppose Ei ∈ A, i ∈ N, then f −1(⋃
i Ei ) =⋃
i f−1(Ei ) ∈ B and so⋃
i Ei ∈ A.
Finally if U is open the f −1(U) is open, which is Borel so U ∈ A.
Corollary
There is a measurable set that is not a Borel set.
Proof: We have seen there is a measurable subset B of C and acontinuous function f such that f −1(B) is not measurable. Then Bcannot be a Borel set because if it were then f −1(B) would be a Borel set,hence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 3 / 13
Proof: Let B be the Borel σ-algebra. Consider the family of setsA = {E : f −1(E ) ∈ B}. To show that A contains B we need only showthat it is a σ-algebra and that it contains all open sets.
This is a standard argument:
1 Since f −1(∅) = ∅ ∈ B we have ∅ ∈ A.2 If E ∈ A then f −1(E ) ∈ B and so f −1(E c) = f −1(E )c ∈ B. Thus
E c ∈ A.3 Suppose Ei ∈ A, i ∈ N, then f −1(
⋃i Ei ) =
⋃i f−1(Ei ) ∈ B and so⋃
i Ei ∈ A.
Finally if U is open the f −1(U) is open, which is Borel so U ∈ A.
Corollary
There is a measurable set that is not a Borel set.
Proof: We have seen there is a measurable subset B of C and acontinuous function f such that f −1(B) is not measurable. Then Bcannot be a Borel set because if it were then f −1(B) would be a Borel set,hence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 3 / 13
Proof: Let B be the Borel σ-algebra. Consider the family of setsA = {E : f −1(E ) ∈ B}. To show that A contains B we need only showthat it is a σ-algebra and that it contains all open sets.
This is a standard argument:
1 Since f −1(∅) = ∅ ∈ B we have ∅ ∈ A.2 If E ∈ A then f −1(E ) ∈ B and so f −1(E c) = f −1(E )c ∈ B. Thus
E c ∈ A.3 Suppose Ei ∈ A, i ∈ N, then f −1(
⋃i Ei ) =
⋃i f−1(Ei ) ∈ B and so⋃
i Ei ∈ A.
Finally if U is open the f −1(U) is open, which is Borel so U ∈ A.
Corollary
There is a measurable set that is not a Borel set.
Proof: We have seen there is a measurable subset B of C and acontinuous function f such that f −1(B) is not measurable. Then Bcannot be a Borel set because if it were then f −1(B) would be a Borel set,hence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 3 / 13
Proof: Let B be the Borel σ-algebra. Consider the family of setsA = {E : f −1(E ) ∈ B}. To show that A contains B we need only showthat it is a σ-algebra and that it contains all open sets.
This is a standard argument:
1 Since f −1(∅) = ∅ ∈ B we have ∅ ∈ A.2 If E ∈ A then f −1(E ) ∈ B and so f −1(E c) = f −1(E )c ∈ B. Thus
E c ∈ A.3 Suppose Ei ∈ A, i ∈ N, then f −1(
⋃i Ei ) =
⋃i f−1(Ei ) ∈ B and so⋃
i Ei ∈ A.
Finally if U is open the f −1(U) is open, which is Borel so U ∈ A.
Corollary
There is a measurable set that is not a Borel set.
Proof: We have seen there is a measurable subset B of C and acontinuous function f such that f −1(B) is not measurable. Then Bcannot be a Borel set because if it were then f −1(B) would be a Borel set,hence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 3 / 13
Proof: Let B be the Borel σ-algebra. Consider the family of setsA = {E : f −1(E ) ∈ B}. To show that A contains B we need only showthat it is a σ-algebra and that it contains all open sets.
This is a standard argument:
1 Since f −1(∅) = ∅ ∈ B we have ∅ ∈ A.2 If E ∈ A then f −1(E ) ∈ B and so f −1(E c) = f −1(E )c ∈ B. Thus
E c ∈ A.3 Suppose Ei ∈ A, i ∈ N, then f −1(
⋃i Ei ) =
⋃i f−1(Ei ) ∈ B and so⋃
i Ei ∈ A.
Finally if U is open the f −1(U) is open, which is Borel so U ∈ A.
Corollary
There is a measurable set that is not a Borel set.
Proof: We have seen there is a measurable subset B of C and acontinuous function f such that f −1(B) is not measurable.
Then Bcannot be a Borel set because if it were then f −1(B) would be a Borel set,hence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 3 / 13
Proof: Let B be the Borel σ-algebra. Consider the family of setsA = {E : f −1(E ) ∈ B}. To show that A contains B we need only showthat it is a σ-algebra and that it contains all open sets.
This is a standard argument:
1 Since f −1(∅) = ∅ ∈ B we have ∅ ∈ A.2 If E ∈ A then f −1(E ) ∈ B and so f −1(E c) = f −1(E )c ∈ B. Thus
E c ∈ A.3 Suppose Ei ∈ A, i ∈ N, then f −1(
⋃i Ei ) =
⋃i f−1(Ei ) ∈ B and so⋃
i Ei ∈ A.
Finally if U is open the f −1(U) is open, which is Borel so U ∈ A.
Corollary
There is a measurable set that is not a Borel set.
Proof: We have seen there is a measurable subset B of C and acontinuous function f such that f −1(B) is not measurable. Then Bcannot be a Borel set because if it were then f −1(B) would be a Borel set,hence measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 3 / 13
Existence of a nonmeasurable set
Theorem
For any set E with m∗(E ) > 0 there exists a nonmeasurable subset of E .
Proof: Let E be any set with m∗(E ) > 0. If E is nonmeasurable we aredone, so assume E is measurable. Since [−n, n] ∩ E are ascending withunion E we have m([−n, n] ∩ E )→ m(E ). Thus we may assume E isbounded, say E ⊆ [−N,N].
Define an equivalence relation ≈ on R by x ≈ y if and only if x − y ∈ Q.Use the axiom of choice to select a set L ⊆ E consisting of one point fromeach equivalence class.
Daniel H. Luecking Chapter 2 Sept 11, 2020 4 / 13
Existence of a nonmeasurable set
Theorem
For any set E with m∗(E ) > 0 there exists a nonmeasurable subset of E .
Proof: Let E be any set with m∗(E ) > 0. If E is nonmeasurable we aredone, so assume E is measurable.
Since [−n, n] ∩ E are ascending withunion E we have m([−n, n] ∩ E )→ m(E ). Thus we may assume E isbounded, say E ⊆ [−N,N].
Define an equivalence relation ≈ on R by x ≈ y if and only if x − y ∈ Q.Use the axiom of choice to select a set L ⊆ E consisting of one point fromeach equivalence class.
Daniel H. Luecking Chapter 2 Sept 11, 2020 4 / 13
Existence of a nonmeasurable set
Theorem
For any set E with m∗(E ) > 0 there exists a nonmeasurable subset of E .
Proof: Let E be any set with m∗(E ) > 0. If E is nonmeasurable we aredone, so assume E is measurable. Since [−n, n] ∩ E are ascending withunion E we have m([−n, n] ∩ E )→ m(E ). Thus we may assume E isbounded, say E ⊆ [−N,N].
Define an equivalence relation ≈ on R by x ≈ y if and only if x − y ∈ Q.Use the axiom of choice to select a set L ⊆ E consisting of one point fromeach equivalence class.
Daniel H. Luecking Chapter 2 Sept 11, 2020 4 / 13
Existence of a nonmeasurable set
Theorem
For any set E with m∗(E ) > 0 there exists a nonmeasurable subset of E .
Proof: Let E be any set with m∗(E ) > 0. If E is nonmeasurable we aredone, so assume E is measurable. Since [−n, n] ∩ E are ascending withunion E we have m([−n, n] ∩ E )→ m(E ). Thus we may assume E isbounded, say E ⊆ [−N,N].
Define an equivalence relation ≈ on R by x ≈ y if and only if x − y ∈ Q.
Use the axiom of choice to select a set L ⊆ E consisting of one point fromeach equivalence class.
Daniel H. Luecking Chapter 2 Sept 11, 2020 4 / 13
Existence of a nonmeasurable set
Theorem
For any set E with m∗(E ) > 0 there exists a nonmeasurable subset of E .
Proof: Let E be any set with m∗(E ) > 0. If E is nonmeasurable we aredone, so assume E is measurable. Since [−n, n] ∩ E are ascending withunion E we have m([−n, n] ∩ E )→ m(E ). Thus we may assume E isbounded, say E ⊆ [−N,N].
Define an equivalence relation ≈ on R by x ≈ y if and only if x − y ∈ Q.Use the axiom of choice to select a set L ⊆ E consisting of one point fromeach equivalence class.
Daniel H. Luecking Chapter 2 Sept 11, 2020 4 / 13
Note that any point in E is equivalent to some point in L and soE ⊆
⋃q∈Q(q + L).
If q ∈ Q with |q| > 2N then for each x ∈ E ,|q + x | > N and so (q + L) ∩ E = ∅. Thus
E ⊆ A =⋃
q∈Q∩[−2N,2N]
(q + L).
Now if L is measurable, then so is A since the union is countable.Therefore
m(A) =∑
q∈Q∩[−2N,2N]
m(q + L) =∑
q∈Q∩[−2N,2N]
m(L)
Since A is bounded, its measure is finite and this implies that m(L) = 0and m(A) = 0. Therefore, m(E ) = 0 contradiction the assumption.
Daniel H. Luecking Chapter 2 Sept 11, 2020 5 / 13
Note that any point in E is equivalent to some point in L and soE ⊆
⋃q∈Q(q + L). If q ∈ Q with |q| > 2N then for each x ∈ E ,
|q + x | > N and so (q + L) ∩ E = ∅.
Thus
E ⊆ A =⋃
q∈Q∩[−2N,2N]
(q + L).
Now if L is measurable, then so is A since the union is countable.Therefore
m(A) =∑
q∈Q∩[−2N,2N]
m(q + L) =∑
q∈Q∩[−2N,2N]
m(L)
Since A is bounded, its measure is finite and this implies that m(L) = 0and m(A) = 0. Therefore, m(E ) = 0 contradiction the assumption.
Daniel H. Luecking Chapter 2 Sept 11, 2020 5 / 13
Note that any point in E is equivalent to some point in L and soE ⊆
⋃q∈Q(q + L). If q ∈ Q with |q| > 2N then for each x ∈ E ,
|q + x | > N and so (q + L) ∩ E = ∅. Thus
E ⊆ A =⋃
q∈Q∩[−2N,2N]
(q + L).
Now if L is measurable, then so is A since the union is countable.Therefore
m(A) =∑
q∈Q∩[−2N,2N]
m(q + L) =∑
q∈Q∩[−2N,2N]
m(L)
Since A is bounded, its measure is finite and this implies that m(L) = 0and m(A) = 0. Therefore, m(E ) = 0 contradiction the assumption.
Daniel H. Luecking Chapter 2 Sept 11, 2020 5 / 13
Note that any point in E is equivalent to some point in L and soE ⊆
⋃q∈Q(q + L). If q ∈ Q with |q| > 2N then for each x ∈ E ,
|q + x | > N and so (q + L) ∩ E = ∅. Thus
E ⊆ A =⋃
q∈Q∩[−2N,2N]
(q + L).
Now if L is measurable, then so is A since the union is countable.
Therefore
m(A) =∑
q∈Q∩[−2N,2N]
m(q + L) =∑
q∈Q∩[−2N,2N]
m(L)
Since A is bounded, its measure is finite and this implies that m(L) = 0and m(A) = 0. Therefore, m(E ) = 0 contradiction the assumption.
Daniel H. Luecking Chapter 2 Sept 11, 2020 5 / 13
Note that any point in E is equivalent to some point in L and soE ⊆
⋃q∈Q(q + L). If q ∈ Q with |q| > 2N then for each x ∈ E ,
|q + x | > N and so (q + L) ∩ E = ∅. Thus
E ⊆ A =⋃
q∈Q∩[−2N,2N]
(q + L).
Now if L is measurable, then so is A since the union is countable.Therefore
m(A) =∑
q∈Q∩[−2N,2N]
m(q + L) =∑
q∈Q∩[−2N,2N]
m(L)
Since A is bounded, its measure is finite and this implies that m(L) = 0and m(A) = 0. Therefore, m(E ) = 0 contradiction the assumption.
Daniel H. Luecking Chapter 2 Sept 11, 2020 5 / 13
Note that any point in E is equivalent to some point in L and soE ⊆
⋃q∈Q(q + L). If q ∈ Q with |q| > 2N then for each x ∈ E ,
|q + x | > N and so (q + L) ∩ E = ∅. Thus
E ⊆ A =⋃
q∈Q∩[−2N,2N]
(q + L).
Now if L is measurable, then so is A since the union is countable.Therefore
m(A) =∑
q∈Q∩[−2N,2N]
m(q + L) =∑
q∈Q∩[−2N,2N]
m(L)
Since A is bounded, its measure is finite and this implies that m(L) = 0and m(A) = 0. Therefore, m(E ) = 0 contradiction the assumption.
Daniel H. Luecking Chapter 2 Sept 11, 2020 5 / 13
Measurable functions
I will use R∗ = R ∪ {−∞,∞} = [−∞,∞].
Definition
If f : E → R∗ we say that f is measurable iff E ∈M and for every realnumber a the set
f −1((a,∞]) = {x ∈ E : f (x) > a}
is measurable.
Theorem
If E ∈M and f : E → R∗, then the following are equivalent.
1 f is measurable.
2 ∀a ∈ R, f −1([−∞, a]) = {x ∈ E : f (x) ≤ a} ∈ M.
3 ∀a ∈ R, f −1([−∞, a)) = {x ∈ E : f (x) < a} ∈ M.
4 ∀a ∈ R, f −1([a,∞]) = {x ∈ E : f (x) ≥ a} ∈ M.
Daniel H. Luecking Chapter 2 Sept 11, 2020 6 / 13
Measurable functions
I will use R∗ = R ∪ {−∞,∞} = [−∞,∞].
Definition
If f : E → R∗ we say that f is measurable iff E ∈M and for every realnumber a the set
f −1((a,∞]) = {x ∈ E : f (x) > a}
is measurable.
Theorem
If E ∈M and f : E → R∗, then the following are equivalent.
1 f is measurable.
2 ∀a ∈ R, f −1([−∞, a]) = {x ∈ E : f (x) ≤ a} ∈ M.
3 ∀a ∈ R, f −1([−∞, a)) = {x ∈ E : f (x) < a} ∈ M.
4 ∀a ∈ R, f −1([a,∞]) = {x ∈ E : f (x) ≥ a} ∈ M.
Daniel H. Luecking Chapter 2 Sept 11, 2020 6 / 13
Measurable functions
I will use R∗ = R ∪ {−∞,∞} = [−∞,∞].
Definition
If f : E → R∗ we say that f is measurable iff E ∈M and for every realnumber a the set
f −1((a,∞]) = {x ∈ E : f (x) > a}
is measurable.
Theorem
If E ∈M and f : E → R∗, then the following are equivalent.
1 f is measurable.
2 ∀a ∈ R, f −1([−∞, a]) = {x ∈ E : f (x) ≤ a} ∈ M.
3 ∀a ∈ R, f −1([−∞, a)) = {x ∈ E : f (x) < a} ∈ M.
4 ∀a ∈ R, f −1([a,∞]) = {x ∈ E : f (x) ≥ a} ∈ M.
Daniel H. Luecking Chapter 2 Sept 11, 2020 6 / 13
Measurable functions
I will use R∗ = R ∪ {−∞,∞} = [−∞,∞].
Definition
If f : E → R∗ we say that f is measurable iff E ∈M and for every realnumber a the set
f −1((a,∞]) = {x ∈ E : f (x) > a}
is measurable.
Theorem
If E ∈M and f : E → R∗, then the following are equivalent.
1 f is measurable.
2 ∀a ∈ R, f −1([−∞, a]) = {x ∈ E : f (x) ≤ a} ∈ M.
3 ∀a ∈ R, f −1([−∞, a)) = {x ∈ E : f (x) < a} ∈ M.
4 ∀a ∈ R, f −1([a,∞]) = {x ∈ E : f (x) ≥ a} ∈ M.
Daniel H. Luecking Chapter 2 Sept 11, 2020 6 / 13
Measurable functions
I will use R∗ = R ∪ {−∞,∞} = [−∞,∞].
Definition
If f : E → R∗ we say that f is measurable iff E ∈M and for every realnumber a the set
f −1((a,∞]) = {x ∈ E : f (x) > a}
is measurable.
Theorem
If E ∈M and f : E → R∗, then the following are equivalent.
1 f is measurable.
2 ∀a ∈ R, f −1([−∞, a]) = {x ∈ E : f (x) ≤ a} ∈ M.
3 ∀a ∈ R, f −1([−∞, a)) = {x ∈ E : f (x) < a} ∈ M.
4 ∀a ∈ R, f −1([a,∞]) = {x ∈ E : f (x) ≥ a} ∈ M.
Daniel H. Luecking Chapter 2 Sept 11, 2020 6 / 13
Proof: The sets f −1([−∞, a]) are complements of the sets f −1((a,∞]) so1⇒ 2.
The sets f −1([−∞, a)) are countable unions of the setsf −1([−∞, a− 1/n]) so 2⇒ 3.
The sets f −1([a,∞]) are complements of the sets f −1([−∞, a)) so 3⇒ 4.
The sets f −1((a,∞]) are countable unions of the sets f −1([a + 1/n,∞])so 4⇒ 1.
Theorem
If f is measurable then for every Borel set B ∈ B, f −1(B) is measurable.
Proof: Let f be measurable. The family of sets A = {A : f −1(A) ∈M} isa σ-algebra that contains all open intervals so it contains all Borel sets.
Daniel H. Luecking Chapter 2 Sept 11, 2020 7 / 13
Proof: The sets f −1([−∞, a]) are complements of the sets f −1((a,∞]) so1⇒ 2.
The sets f −1([−∞, a)) are countable unions of the setsf −1([−∞, a− 1/n]) so 2⇒ 3.
The sets f −1([a,∞]) are complements of the sets f −1([−∞, a)) so 3⇒ 4.
The sets f −1((a,∞]) are countable unions of the sets f −1([a + 1/n,∞])so 4⇒ 1.
Theorem
If f is measurable then for every Borel set B ∈ B, f −1(B) is measurable.
Proof: Let f be measurable. The family of sets A = {A : f −1(A) ∈M} isa σ-algebra that contains all open intervals so it contains all Borel sets.
Daniel H. Luecking Chapter 2 Sept 11, 2020 7 / 13
Proof: The sets f −1([−∞, a]) are complements of the sets f −1((a,∞]) so1⇒ 2.
The sets f −1([−∞, a)) are countable unions of the setsf −1([−∞, a− 1/n]) so 2⇒ 3.
The sets f −1([a,∞]) are complements of the sets f −1([−∞, a)) so 3⇒ 4.
The sets f −1((a,∞]) are countable unions of the sets f −1([a + 1/n,∞])so 4⇒ 1.
Theorem
If f is measurable then for every Borel set B ∈ B, f −1(B) is measurable.
Proof: Let f be measurable. The family of sets A = {A : f −1(A) ∈M} isa σ-algebra that contains all open intervals so it contains all Borel sets.
Daniel H. Luecking Chapter 2 Sept 11, 2020 7 / 13
Proof: The sets f −1([−∞, a]) are complements of the sets f −1((a,∞]) so1⇒ 2.
The sets f −1([−∞, a)) are countable unions of the setsf −1([−∞, a− 1/n]) so 2⇒ 3.
The sets f −1([a,∞]) are complements of the sets f −1([−∞, a)) so 3⇒ 4.
The sets f −1((a,∞]) are countable unions of the sets f −1([a + 1/n,∞])so 4⇒ 1.
Theorem
If f is measurable then for every Borel set B ∈ B, f −1(B) is measurable.
Proof: Let f be measurable. The family of sets A = {A : f −1(A) ∈M} isa σ-algebra that contains all open intervals so it contains all Borel sets.
Daniel H. Luecking Chapter 2 Sept 11, 2020 7 / 13
Proof: The sets f −1([−∞, a]) are complements of the sets f −1((a,∞]) so1⇒ 2.
The sets f −1([−∞, a)) are countable unions of the setsf −1([−∞, a− 1/n]) so 2⇒ 3.
The sets f −1([a,∞]) are complements of the sets f −1([−∞, a)) so 3⇒ 4.
The sets f −1((a,∞]) are countable unions of the sets f −1([a + 1/n,∞])so 4⇒ 1.
Theorem
If f is measurable then for every Borel set B ∈ B, f −1(B) is measurable.
Proof: Let f be measurable. The family of sets A = {A : f −1(A) ∈M} isa σ-algebra that contains all open intervals so it contains all Borel sets.
Daniel H. Luecking Chapter 2 Sept 11, 2020 7 / 13
Proof: The sets f −1([−∞, a]) are complements of the sets f −1((a,∞]) so1⇒ 2.
The sets f −1([−∞, a)) are countable unions of the setsf −1([−∞, a− 1/n]) so 2⇒ 3.
The sets f −1([a,∞]) are complements of the sets f −1([−∞, a)) so 3⇒ 4.
The sets f −1((a,∞]) are countable unions of the sets f −1([a + 1/n,∞])so 4⇒ 1.
Theorem
If f is measurable then for every Borel set B ∈ B, f −1(B) is measurable.
Proof: Let f be measurable. The family of sets A = {A : f −1(A) ∈M} isa σ-algebra that contains all open intervals so it contains all Borel sets.
Daniel H. Luecking Chapter 2 Sept 11, 2020 7 / 13
Examples of measurable functions
In particular, f is measurable if and only if f −1(U) is measurable wheneverU is open. By taking complements, f is measurable if and only if f −1(F )is measurable whenever F is closed.
Constants functions are measurable: if f (x) ≡ c then {x ∈ E : f (x) > a}is either ∅ (c ≤ a) or all of E (c > a).
Characteristic functions of measurable sets are measurable: For ameasurable set F ⊆ E , χF (x) = 1 if x ∈ F and χF (x) = 0 if x /∈ F . Then{x ∈ E : χF (x) > a} is either ∅, F or E depending on whether a ≥ 1,a ∈ [0, 1) of a < 0.
Continuous functions are measurable: f −1((a,∞]) = f −1((a,∞)) has theform U ∩ E with U open.
Daniel H. Luecking Chapter 2 Sept 11, 2020 8 / 13
Examples of measurable functions
In particular, f is measurable if and only if f −1(U) is measurable wheneverU is open. By taking complements, f is measurable if and only if f −1(F )is measurable whenever F is closed.
Constants functions are measurable: if f (x) ≡ c then {x ∈ E : f (x) > a}is either ∅ (c ≤ a) or all of E (c > a).
Characteristic functions of measurable sets are measurable: For ameasurable set F ⊆ E , χF (x) = 1 if x ∈ F and χF (x) = 0 if x /∈ F . Then{x ∈ E : χF (x) > a} is either ∅, F or E depending on whether a ≥ 1,a ∈ [0, 1) of a < 0.
Continuous functions are measurable: f −1((a,∞]) = f −1((a,∞)) has theform U ∩ E with U open.
Daniel H. Luecking Chapter 2 Sept 11, 2020 8 / 13
Examples of measurable functions
In particular, f is measurable if and only if f −1(U) is measurable wheneverU is open. By taking complements, f is measurable if and only if f −1(F )is measurable whenever F is closed.
Constants functions are measurable: if f (x) ≡ c then {x ∈ E : f (x) > a}is either ∅ (c ≤ a) or all of E (c > a).
Characteristic functions of measurable sets are measurable: For ameasurable set F ⊆ E , χF (x) = 1 if x ∈ F and χF (x) = 0 if x /∈ F .
Then{x ∈ E : χF (x) > a} is either ∅, F or E depending on whether a ≥ 1,a ∈ [0, 1) of a < 0.
Continuous functions are measurable: f −1((a,∞]) = f −1((a,∞)) has theform U ∩ E with U open.
Daniel H. Luecking Chapter 2 Sept 11, 2020 8 / 13
Examples of measurable functions
In particular, f is measurable if and only if f −1(U) is measurable wheneverU is open. By taking complements, f is measurable if and only if f −1(F )is measurable whenever F is closed.
Constants functions are measurable: if f (x) ≡ c then {x ∈ E : f (x) > a}is either ∅ (c ≤ a) or all of E (c > a).
Characteristic functions of measurable sets are measurable: For ameasurable set F ⊆ E , χF (x) = 1 if x ∈ F and χF (x) = 0 if x /∈ F . Then{x ∈ E : χF (x) > a} is either ∅, F or E depending on whether a ≥ 1,a ∈ [0, 1) of a < 0.
Continuous functions are measurable: f −1((a,∞]) = f −1((a,∞)) has theform U ∩ E with U open.
Daniel H. Luecking Chapter 2 Sept 11, 2020 8 / 13
Examples of measurable functions
In particular, f is measurable if and only if f −1(U) is measurable wheneverU is open. By taking complements, f is measurable if and only if f −1(F )is measurable whenever F is closed.
Constants functions are measurable: if f (x) ≡ c then {x ∈ E : f (x) > a}is either ∅ (c ≤ a) or all of E (c > a).
Characteristic functions of measurable sets are measurable: For ameasurable set F ⊆ E , χF (x) = 1 if x ∈ F and χF (x) = 0 if x /∈ F . Then{x ∈ E : χF (x) > a} is either ∅, F or E depending on whether a ≥ 1,a ∈ [0, 1) of a < 0.
Continuous functions are measurable: f −1((a,∞]) = f −1((a,∞)) has theform U ∩ E with U open.
Daniel H. Luecking Chapter 2 Sept 11, 2020 8 / 13
New functions from old
Constant multiples and sums of measurable functions are measurable.
Proof: If b = 0 then bf is constant. If b > 0 then{x ∈ E : bf (x) > a} = {x ∈ E : f (x) > a/b} and if b < 0 then{x ∈ E : bf (x) > a} = {x ∈ E : f (x) < a/b}.If f (x) + g(x) > a then f (x) > a− g(x) so there exists a rational numberr with f (x) > r > a− g(x). Thus
{x ∈ E : f (x) + g(x) > a} ⊆⋃r∈Q{x : f (x) > r} ∩ {x : g(x) > a− r}
Exercise: prove the reverse inclusion so we have that{x ∈ E : f (x) + g(x) > a} is a countable union of measurable sets.
Daniel H. Luecking Chapter 2 Sept 11, 2020 9 / 13
New functions from old
Constant multiples and sums of measurable functions are measurable.
Proof: If b = 0 then bf is constant. If b > 0 then{x ∈ E : bf (x) > a} = {x ∈ E : f (x) > a/b} and if b < 0 then{x ∈ E : bf (x) > a} = {x ∈ E : f (x) < a/b}.
If f (x) + g(x) > a then f (x) > a− g(x) so there exists a rational numberr with f (x) > r > a− g(x). Thus
{x ∈ E : f (x) + g(x) > a} ⊆⋃r∈Q{x : f (x) > r} ∩ {x : g(x) > a− r}
Exercise: prove the reverse inclusion so we have that{x ∈ E : f (x) + g(x) > a} is a countable union of measurable sets.
Daniel H. Luecking Chapter 2 Sept 11, 2020 9 / 13
New functions from old
Constant multiples and sums of measurable functions are measurable.
Proof: If b = 0 then bf is constant. If b > 0 then{x ∈ E : bf (x) > a} = {x ∈ E : f (x) > a/b} and if b < 0 then{x ∈ E : bf (x) > a} = {x ∈ E : f (x) < a/b}.If f (x) + g(x) > a then f (x) > a− g(x) so there exists a rational numberr with f (x) > r > a− g(x).
Thus
{x ∈ E : f (x) + g(x) > a} ⊆⋃r∈Q{x : f (x) > r} ∩ {x : g(x) > a− r}
Exercise: prove the reverse inclusion so we have that{x ∈ E : f (x) + g(x) > a} is a countable union of measurable sets.
Daniel H. Luecking Chapter 2 Sept 11, 2020 9 / 13
New functions from old
Constant multiples and sums of measurable functions are measurable.
Proof: If b = 0 then bf is constant. If b > 0 then{x ∈ E : bf (x) > a} = {x ∈ E : f (x) > a/b} and if b < 0 then{x ∈ E : bf (x) > a} = {x ∈ E : f (x) < a/b}.If f (x) + g(x) > a then f (x) > a− g(x) so there exists a rational numberr with f (x) > r > a− g(x). Thus
{x ∈ E : f (x) + g(x) > a} ⊆⋃r∈Q{x : f (x) > r} ∩ {x : g(x) > a− r}
Exercise: prove the reverse inclusion so we have that{x ∈ E : f (x) + g(x) > a} is a countable union of measurable sets.
Daniel H. Luecking Chapter 2 Sept 11, 2020 9 / 13
New functions from old
Constant multiples and sums of measurable functions are measurable.
Proof: If b = 0 then bf is constant. If b > 0 then{x ∈ E : bf (x) > a} = {x ∈ E : f (x) > a/b} and if b < 0 then{x ∈ E : bf (x) > a} = {x ∈ E : f (x) < a/b}.If f (x) + g(x) > a then f (x) > a− g(x) so there exists a rational numberr with f (x) > r > a− g(x). Thus
{x ∈ E : f (x) + g(x) > a} ⊆⋃r∈Q{x : f (x) > r} ∩ {x : g(x) > a− r}
Exercise: prove the reverse inclusion so we have that{x ∈ E : f (x) + g(x) > a} is a countable union of measurable sets.
Daniel H. Luecking Chapter 2 Sept 11, 2020 9 / 13
Suprema and infima of finite or countable collections of measurablefunctions are measurable.
Proof: If fn are measurable, let f (x) = sup{f1(x), f2(x), f3(x), . . . }. Thenf (x) > a if and only if some fn(x) > a. Therefore,
{x ∈ E : f (x) > a} =⋃n
{x ∈ E : fn(x) > a}.
Lim sups and lim infs of a sequence of measurable functions aremeasurable.
Proof: Let fn be measurable on E and letgn(x) = sup{fn(x), fn+1(x), fn+2(x), . . . }. Then each gn is measurable andso is lim sup
n→∞fn(x) = lim
n→∞gn(x) = inf{g1(x), g2(x), g3(x), . . . }.
Daniel H. Luecking Chapter 2 Sept 11, 2020 10 / 13
Suprema and infima of finite or countable collections of measurablefunctions are measurable.
Proof: If fn are measurable, let f (x) = sup{f1(x), f2(x), f3(x), . . . }. Thenf (x) > a if and only if some fn(x) > a.
Therefore,
{x ∈ E : f (x) > a} =⋃n
{x ∈ E : fn(x) > a}.
Lim sups and lim infs of a sequence of measurable functions aremeasurable.
Proof: Let fn be measurable on E and letgn(x) = sup{fn(x), fn+1(x), fn+2(x), . . . }. Then each gn is measurable andso is lim sup
n→∞fn(x) = lim
n→∞gn(x) = inf{g1(x), g2(x), g3(x), . . . }.
Daniel H. Luecking Chapter 2 Sept 11, 2020 10 / 13
Suprema and infima of finite or countable collections of measurablefunctions are measurable.
Proof: If fn are measurable, let f (x) = sup{f1(x), f2(x), f3(x), . . . }. Thenf (x) > a if and only if some fn(x) > a. Therefore,
{x ∈ E : f (x) > a} =⋃n
{x ∈ E : fn(x) > a}.
Lim sups and lim infs of a sequence of measurable functions aremeasurable.
Proof: Let fn be measurable on E and letgn(x) = sup{fn(x), fn+1(x), fn+2(x), . . . }. Then each gn is measurable andso is lim sup
n→∞fn(x) = lim
n→∞gn(x) = inf{g1(x), g2(x), g3(x), . . . }.
Daniel H. Luecking Chapter 2 Sept 11, 2020 10 / 13
Suprema and infima of finite or countable collections of measurablefunctions are measurable.
Proof: If fn are measurable, let f (x) = sup{f1(x), f2(x), f3(x), . . . }. Thenf (x) > a if and only if some fn(x) > a. Therefore,
{x ∈ E : f (x) > a} =⋃n
{x ∈ E : fn(x) > a}.
Lim sups and lim infs of a sequence of measurable functions aremeasurable.
Proof: Let fn be measurable on E and letgn(x) = sup{fn(x), fn+1(x), fn+2(x), . . . }. Then each gn is measurable andso is lim sup
n→∞fn(x) = lim
n→∞gn(x) = inf{g1(x), g2(x), g3(x), . . . }.
Daniel H. Luecking Chapter 2 Sept 11, 2020 10 / 13
Suprema and infima of finite or countable collections of measurablefunctions are measurable.
Proof: If fn are measurable, let f (x) = sup{f1(x), f2(x), f3(x), . . . }. Thenf (x) > a if and only if some fn(x) > a. Therefore,
{x ∈ E : f (x) > a} =⋃n
{x ∈ E : fn(x) > a}.
Lim sups and lim infs of a sequence of measurable functions aremeasurable.
Proof: Let fn be measurable on E and letgn(x) = sup{fn(x), fn+1(x), fn+2(x), . . . }. Then each gn is measurable andso is lim sup
n→∞fn(x) = lim
n→∞gn(x) = inf{g1(x), g2(x), g3(x), . . . }.
Daniel H. Luecking Chapter 2 Sept 11, 2020 10 / 13
Composition of a continuous function with a measurable function (in thatorder) is measurable.
Proof: Let f : E → R measurable and φ continuous on an open setcontaining the range of f , then
{x ∈ E : φ(f (x)) > a} = (φ ◦ f )−1((a,∞)) = f −1(φ−1((a,∞))).
Lemma
If E is measurable, then f : E → R∗ is measurable if and only ifA = {x ∈ E : f (x) = +∞} and B = {x ∈ E : f (x) = −∞} aremeasurable and g = f |E∼(A∪B) is measurable.
Proof: Suppose f is measurable. Then A =⋂
n{x ∈ E : f (x) > n} ismeasurable. Similarly, B is measurable. Then g has a measurable domainand
{x ∈ E ∼ (A ∪ B) : g(x) > a} = {x ∈ E : f (x) > a} ∼ (A ∪ B)
is a measurable set.
Daniel H. Luecking Chapter 2 Sept 11, 2020 11 / 13
Composition of a continuous function with a measurable function (in thatorder) is measurable.
Proof: Let f : E → R measurable and φ continuous on an open setcontaining the range of f , then
{x ∈ E : φ(f (x)) > a} = (φ ◦ f )−1((a,∞)) = f −1(φ−1((a,∞))).
Lemma
If E is measurable, then f : E → R∗ is measurable if and only ifA = {x ∈ E : f (x) = +∞} and B = {x ∈ E : f (x) = −∞} aremeasurable and g = f |E∼(A∪B) is measurable.
Proof: Suppose f is measurable. Then A =⋂
n{x ∈ E : f (x) > n} ismeasurable. Similarly, B is measurable. Then g has a measurable domainand
{x ∈ E ∼ (A ∪ B) : g(x) > a} = {x ∈ E : f (x) > a} ∼ (A ∪ B)
is a measurable set.
Daniel H. Luecking Chapter 2 Sept 11, 2020 11 / 13
Composition of a continuous function with a measurable function (in thatorder) is measurable.
Proof: Let f : E → R measurable and φ continuous on an open setcontaining the range of f , then
{x ∈ E : φ(f (x)) > a} = (φ ◦ f )−1((a,∞)) = f −1(φ−1((a,∞))).
Lemma
If E is measurable, then f : E → R∗ is measurable if and only ifA = {x ∈ E : f (x) = +∞} and B = {x ∈ E : f (x) = −∞} aremeasurable and g = f |E∼(A∪B) is measurable.
Proof: Suppose f is measurable. Then A =⋂
n{x ∈ E : f (x) > n} ismeasurable.
Similarly, B is measurable. Then g has a measurable domainand
{x ∈ E ∼ (A ∪ B) : g(x) > a} = {x ∈ E : f (x) > a} ∼ (A ∪ B)
is a measurable set.
Daniel H. Luecking Chapter 2 Sept 11, 2020 11 / 13
Composition of a continuous function with a measurable function (in thatorder) is measurable.
Proof: Let f : E → R measurable and φ continuous on an open setcontaining the range of f , then
{x ∈ E : φ(f (x)) > a} = (φ ◦ f )−1((a,∞)) = f −1(φ−1((a,∞))).
Lemma
If E is measurable, then f : E → R∗ is measurable if and only ifA = {x ∈ E : f (x) = +∞} and B = {x ∈ E : f (x) = −∞} aremeasurable and g = f |E∼(A∪B) is measurable.
Proof: Suppose f is measurable. Then A =⋂
n{x ∈ E : f (x) > n} ismeasurable. Similarly, B is measurable.
Then g has a measurable domainand
{x ∈ E ∼ (A ∪ B) : g(x) > a} = {x ∈ E : f (x) > a} ∼ (A ∪ B)
is a measurable set.
Daniel H. Luecking Chapter 2 Sept 11, 2020 11 / 13
Composition of a continuous function with a measurable function (in thatorder) is measurable.
Proof: Let f : E → R measurable and φ continuous on an open setcontaining the range of f , then
{x ∈ E : φ(f (x)) > a} = (φ ◦ f )−1((a,∞)) = f −1(φ−1((a,∞))).
Lemma
If E is measurable, then f : E → R∗ is measurable if and only ifA = {x ∈ E : f (x) = +∞} and B = {x ∈ E : f (x) = −∞} aremeasurable and g = f |E∼(A∪B) is measurable.
Proof: Suppose f is measurable. Then A =⋂
n{x ∈ E : f (x) > n} ismeasurable. Similarly, B is measurable. Then g has a measurable domainand
{x ∈ E ∼ (A ∪ B) : g(x) > a} = {x ∈ E : f (x) > a} ∼ (A ∪ B)
is a measurable set.
Daniel H. Luecking Chapter 2 Sept 11, 2020 11 / 13
Lemma
If E is measurable and A ⊆ E has m∗(A) = 0 then f : E → R∗ ismeasurable if and only if f |E∼A is measurable.
Proof:{x ∈ E : f (x) > a} = {x ∈ E ∼ A : f (x) > a} ∪ {x ∈ A : f (x) > a}.
Lemma
If Ej , j ∈ N, are disjoint measurable sets with E =⋃
j Ej then f : E → R∗is mearurable if and only if every f |Ej
is measurable.
Products of measurable functions are measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 12 / 13
Lemma
If E is measurable and A ⊆ E has m∗(A) = 0 then f : E → R∗ ismeasurable if and only if f |E∼A is measurable.
Proof:{x ∈ E : f (x) > a} = {x ∈ E ∼ A : f (x) > a} ∪ {x ∈ A : f (x) > a}.
Lemma
If Ej , j ∈ N, are disjoint measurable sets with E =⋃
j Ej then f : E → R∗is mearurable if and only if every f |Ej
is measurable.
Products of measurable functions are measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 12 / 13
Lemma
If E is measurable and A ⊆ E has m∗(A) = 0 then f : E → R∗ ismeasurable if and only if f |E∼A is measurable.
Proof:{x ∈ E : f (x) > a} = {x ∈ E ∼ A : f (x) > a} ∪ {x ∈ A : f (x) > a}.
Lemma
If Ej , j ∈ N, are disjoint measurable sets with E =⋃
j Ej then f : E → R∗is mearurable if and only if every f |Ej
is measurable.
Products of measurable functions are measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 12 / 13
Lemma
If E is measurable and A ⊆ E has m∗(A) = 0 then f : E → R∗ ismeasurable if and only if f |E∼A is measurable.
Proof:{x ∈ E : f (x) > a} = {x ∈ E ∼ A : f (x) > a} ∪ {x ∈ A : f (x) > a}.
Lemma
If Ej , j ∈ N, are disjoint measurable sets with E =⋃
j Ej then f : E → R∗is mearurable if and only if every f |Ej
is measurable.
Products of measurable functions are measurable.
Daniel H. Luecking Chapter 2 Sept 11, 2020 12 / 13
Proof: Let f , g be measurable on E . We may assume f , g have values inR. By the lemma we may divide the domain into the following measurablesets and prove fg is measurable on all of them.
E1 = {x ∈ E : f (x) = 0} ∪ {x ∈ E : g(x) = 0} where fg is constant andso measurable.E2 is the set where f and g are both positive.E3 is where f is positive and g is negative.E4 has the signs reversed.E5 is where both are negative.
On E2 we have log f and log g both measurable, hence the sumlog f + log g = log fg is measurable and so fg = exp(log fg) is measurable.
On E3 we need only show (−1)fg = f (−g) is measurable, but that’s thesame as E2. ETC.
Daniel H. Luecking Chapter 2 Sept 11, 2020 13 / 13
Proof: Let f , g be measurable on E . We may assume f , g have values inR. By the lemma we may divide the domain into the following measurablesets and prove fg is measurable on all of them.E1 = {x ∈ E : f (x) = 0} ∪ {x ∈ E : g(x) = 0} where fg is constant andso measurable.
E2 is the set where f and g are both positive.E3 is where f is positive and g is negative.E4 has the signs reversed.E5 is where both are negative.
On E2 we have log f and log g both measurable, hence the sumlog f + log g = log fg is measurable and so fg = exp(log fg) is measurable.
On E3 we need only show (−1)fg = f (−g) is measurable, but that’s thesame as E2. ETC.
Daniel H. Luecking Chapter 2 Sept 11, 2020 13 / 13
Proof: Let f , g be measurable on E . We may assume f , g have values inR. By the lemma we may divide the domain into the following measurablesets and prove fg is measurable on all of them.E1 = {x ∈ E : f (x) = 0} ∪ {x ∈ E : g(x) = 0} where fg is constant andso measurable.E2 is the set where f and g are both positive.
E3 is where f is positive and g is negative.E4 has the signs reversed.E5 is where both are negative.
On E2 we have log f and log g both measurable, hence the sumlog f + log g = log fg is measurable and so fg = exp(log fg) is measurable.
On E3 we need only show (−1)fg = f (−g) is measurable, but that’s thesame as E2. ETC.
Daniel H. Luecking Chapter 2 Sept 11, 2020 13 / 13
Proof: Let f , g be measurable on E . We may assume f , g have values inR. By the lemma we may divide the domain into the following measurablesets and prove fg is measurable on all of them.E1 = {x ∈ E : f (x) = 0} ∪ {x ∈ E : g(x) = 0} where fg is constant andso measurable.E2 is the set where f and g are both positive.E3 is where f is positive and g is negative.
E4 has the signs reversed.E5 is where both are negative.
On E2 we have log f and log g both measurable, hence the sumlog f + log g = log fg is measurable and so fg = exp(log fg) is measurable.
On E3 we need only show (−1)fg = f (−g) is measurable, but that’s thesame as E2. ETC.
Daniel H. Luecking Chapter 2 Sept 11, 2020 13 / 13
Proof: Let f , g be measurable on E . We may assume f , g have values inR. By the lemma we may divide the domain into the following measurablesets and prove fg is measurable on all of them.E1 = {x ∈ E : f (x) = 0} ∪ {x ∈ E : g(x) = 0} where fg is constant andso measurable.E2 is the set where f and g are both positive.E3 is where f is positive and g is negative.E4 has the signs reversed.E5 is where both are negative.
On E2 we have log f and log g both measurable, hence the sumlog f + log g = log fg is measurable and so fg = exp(log fg) is measurable.
On E3 we need only show (−1)fg = f (−g) is measurable, but that’s thesame as E2. ETC.
Daniel H. Luecking Chapter 2 Sept 11, 2020 13 / 13
Proof: Let f , g be measurable on E . We may assume f , g have values inR. By the lemma we may divide the domain into the following measurablesets and prove fg is measurable on all of them.E1 = {x ∈ E : f (x) = 0} ∪ {x ∈ E : g(x) = 0} where fg is constant andso measurable.E2 is the set where f and g are both positive.E3 is where f is positive and g is negative.E4 has the signs reversed.E5 is where both are negative.
On E2 we have log f and log g both measurable, hence the sumlog f + log g = log fg is measurable and so fg = exp(log fg) is measurable.
On E3 we need only show (−1)fg = f (−g) is measurable, but that’s thesame as E2. ETC.
Daniel H. Luecking Chapter 2 Sept 11, 2020 13 / 13
Proof: Let f , g be measurable on E . We may assume f , g have values inR. By the lemma we may divide the domain into the following measurablesets and prove fg is measurable on all of them.E1 = {x ∈ E : f (x) = 0} ∪ {x ∈ E : g(x) = 0} where fg is constant andso measurable.E2 is the set where f and g are both positive.E3 is where f is positive and g is negative.E4 has the signs reversed.E5 is where both are negative.
On E2 we have log f and log g both measurable, hence the sumlog f + log g = log fg is measurable and so fg = exp(log fg) is measurable.
On E3 we need only show (−1)fg = f (−g) is measurable, but that’s thesame as E2. ETC.
Daniel H. Luecking Chapter 2 Sept 11, 2020 13 / 13
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