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