Chapter 2

Daniel H. Luecking

Sep 4, 2020

Daniel H. Luecking Chapter 2 Sep 4, 2020 1 / 18

Outline

1 Lebesgue Measurable sets

Daniel H. Luecking Chapter 2 Sep 4, 2020 2 / 18

Measurable sets

We define a set E to be (Lebesgue) measurable iff

For all A ⊆ R, m∗(A ∩ E ) + m∗(A∼ E ) = m∗(A)

The collection of measurable sets is denoted M.

Note that if E and F are disjoint and in M then applying the definition toA ∩ (E ∪ F ) in place of A we get

m∗(A ∩ E ) + m∗(A ∩ F ) = m∗(A ∩ (E ∪ F )).

Applied to A = R we get m∗(E ) + m∗(F ) = m∗(E ∪ F ).

This additivity is part of the reason for this definition. Another reason isthat we automatically have M closed under complementation: if E ismeasurable and A is any set then

m∗(A ∩ E c) + m∗(A∼ E c) = m∗(A∼ E ) + m∗(A ∩ E ) = m∗(A)

Daniel H. Luecking Chapter 2 Sep 4, 2020 3 / 18

Measurable sets

We define a set E to be (Lebesgue) measurable iff

For all A ⊆ R, m∗(A ∩ E ) + m∗(A∼ E ) = m∗(A)

The collection of measurable sets is denoted M.

Note that if E and F are disjoint and in M then applying the definition toA ∩ (E ∪ F ) in place of A we get

m∗(A ∩ E ) + m∗(A ∩ F ) = m∗(A ∩ (E ∪ F )).

Applied to A = R we get m∗(E ) + m∗(F ) = m∗(E ∪ F ).

This additivity is part of the reason for this definition. Another reason isthat we automatically have M closed under complementation: if E ismeasurable and A is any set then

m∗(A ∩ E c) + m∗(A∼ E c) = m∗(A∼ E ) + m∗(A ∩ E ) = m∗(A)

Daniel H. Luecking Chapter 2 Sep 4, 2020 3 / 18

Measurable sets

We define a set E to be (Lebesgue) measurable iff

For all A ⊆ R, m∗(A ∩ E ) + m∗(A∼ E ) = m∗(A)

The collection of measurable sets is denoted M.

Note that if E and F are disjoint and in M then applying the definition toA ∩ (E ∪ F ) in place of A we get

m∗(A ∩ E ) + m∗(A ∩ F ) = m∗(A ∩ (E ∪ F )).

Applied to A = R we get m∗(E ) + m∗(F ) = m∗(E ∪ F ).

This additivity is part of the reason for this definition. Another reason isthat we automatically have M closed under complementation: if E ismeasurable and A is any set then

m∗(A ∩ E c) + m∗(A∼ E c) = m∗(A∼ E ) + m∗(A ∩ E ) = m∗(A)

Daniel H. Luecking Chapter 2 Sep 4, 2020 3 / 18

Measurable sets

We define a set E to be (Lebesgue) measurable iff

For all A ⊆ R, m∗(A ∩ E ) + m∗(A∼ E ) = m∗(A)

The collection of measurable sets is denoted M.

Note that if E and F are disjoint and in M then applying the definition toA ∩ (E ∪ F ) in place of A we get

m∗(A ∩ E ) + m∗(A ∩ F ) = m∗(A ∩ (E ∪ F )).

Applied to A = R we get m∗(E ) + m∗(F ) = m∗(E ∪ F ).

This additivity is part of the reason for this definition. Another reason isthat we automatically have M closed under complementation: if E ismeasurable and A is any set then

m∗(A ∩ E c) + m∗(A∼ E c) = m∗(A∼ E ) + m∗(A ∩ E ) = m∗(A)

Daniel H. Luecking Chapter 2 Sep 4, 2020 3 / 18

We always have m∗(A) ≤ m∗(A ∩ E ) + m∗(A∼ E ) so provingmeasurability reduces to showing m∗(A ∩ E ) + m∗(A∼ E ) ≤ m∗(A).

Our goal now is to show that (1) M contains all intervals, (2) is closedunder countable unions, and (3) m∗ restricted to M is a measure.

To start, let J be the interval (a,∞) and let’s show that J is measurable.We only need to show that for any A we have

m∗(A ∩ (a,∞)) + m∗(A ∩ (−∞, a]) ≤ m∗(A)

Because m∗(A) is an inf, it suffices to show that for any cover of A byintervals In,

m∗(A ∩ J) + m∗(A ∩ Jc) ≤∑n

`(In)

All In ∩ J are intervals that cover A ∩ J and In ∩ Jc are intervals that coverA ∩ Jc .

Daniel H. Luecking Chapter 2 Sep 4, 2020 4 / 18

We always have m∗(A) ≤ m∗(A ∩ E ) + m∗(A∼ E ) so provingmeasurability reduces to showing m∗(A ∩ E ) + m∗(A∼ E ) ≤ m∗(A).

Our goal now is to show that (1) M contains all intervals, (2) is closedunder countable unions, and (3) m∗ restricted to M is a measure.

To start, let J be the interval (a,∞) and let’s show that J is measurable.We only need to show that for any A we have

m∗(A ∩ (a,∞)) + m∗(A ∩ (−∞, a]) ≤ m∗(A)

Because m∗(A) is an inf, it suffices to show that for any cover of A byintervals In,

m∗(A ∩ J) + m∗(A ∩ Jc) ≤∑n

`(In)

All In ∩ J are intervals that cover A ∩ J and In ∩ Jc are intervals that coverA ∩ Jc .

Daniel H. Luecking Chapter 2 Sep 4, 2020 4 / 18

We always have m∗(A) ≤ m∗(A ∩ E ) + m∗(A∼ E ) so provingmeasurability reduces to showing m∗(A ∩ E ) + m∗(A∼ E ) ≤ m∗(A).

Our goal now is to show that (1) M contains all intervals, (2) is closedunder countable unions, and (3) m∗ restricted to M is a measure.

To start, let J be the interval (a,∞) and let’s show that J is measurable.We only need to show that for any A we have

m∗(A ∩ (a,∞)) + m∗(A ∩ (−∞, a]) ≤ m∗(A)

Because m∗(A) is an inf, it suffices to show that for any cover of A byintervals In,

m∗(A ∩ J) + m∗(A ∩ Jc) ≤∑n

`(In)

All In ∩ J are intervals that cover A ∩ J and In ∩ Jc are intervals that coverA ∩ Jc .

Daniel H. Luecking Chapter 2 Sep 4, 2020 4 / 18

Therefore

m∗(A ∩ J) + m∗(A ∩ Jc) ≤∑n

`(In ∩ J) +∑n

`(In ∩ Jc)

=∑n

[`(In ∩ J) + `(In ∩ Jc)] =∑n

`(In)

So intervals of the form (a,∞) are measurable. The identical proof worksfor [a,∞) and complements of these.

Daniel H. Luecking Chapter 2 Sep 4, 2020 5 / 18

Therefore

m∗(A ∩ J) + m∗(A ∩ Jc) ≤∑n

`(In ∩ J) +∑n

`(In ∩ Jc)

=∑n

[`(In ∩ J) + `(In ∩ Jc)] =∑n

`(In)

So intervals of the form (a,∞) are measurable. The identical proof worksfor [a,∞) and complements of these.

Daniel H. Luecking Chapter 2 Sep 4, 2020 5 / 18

Our first step en route to proving M is a σ-algebra is

Lemma

If E ,F ∈M then E ∪ F ∈M.

E F

A

A1

A2A3

A4

Proof: We need to show m∗(A) = m∗(A1 ∪ A2 ∪ A3) + m∗(A4).

Applyingthe measurability of E to A, then of F to A3 ∪ A4 and then measurabilityof E to A1 ∪ A2 ∪ A3, we get

m∗(A) = m∗(A1 ∪ A2) + m∗(A3 ∪ A4)

= m∗(A1 ∪ A2) + m∗(A3) + m∗(A4)

= m∗(A1 ∪ A2 ∪ A3) + m∗(A4)

Daniel H. Luecking Chapter 2 Sep 4, 2020 6 / 18

Our first step en route to proving M is a σ-algebra is

Lemma

If E ,F ∈M then E ∪ F ∈M.

E F

A

A1

A2A3

A4

Proof: We need to show m∗(A) = m∗(A1 ∪ A2 ∪ A3) + m∗(A4). Applyingthe measurability of E to A, then of F to A3 ∪ A4 and then measurabilityof E to A1 ∪ A2 ∪ A3, we get

m∗(A) = m∗(A1 ∪ A2) + m∗(A3 ∪ A4)

= m∗(A1 ∪ A2) + m∗(A3) + m∗(A4)

= m∗(A1 ∪ A2 ∪ A3) + m∗(A4)

Daniel H. Luecking Chapter 2 Sep 4, 2020 6 / 18

M is a σ-algebra

To prove the above statement we need

∅ ∈M: for any set A,m∗(A ∩∅) + m∗(A∼∅) = m∗(∅) + m∗(A) = m∗(A).

If E ∈M then E c ∈M:m∗(A ∩ E c) + m∗(A∼ E c) = m∗(A∼ E ) + m∗(A ∩ E ) = m∗(A).

If Ei ∈M for i ∈ N then⋃Ei ∈M.

Since M is closed under union and complement, it is also closed underintersection and set difference. Also by induction it is closed under finiteunions and finite intersections.

Earlier we saw that if E1,E2 ∈M are disjoint, then for any set A

m∗(A ∩ (E1 ∪ E2)) = m∗(A ∩ E1) + m∗(A ∩ E2).

and this extends to any finite number of disjoint sets in M by induction.

Daniel H. Luecking Chapter 2 Sep 4, 2020 7 / 18

M is a σ-algebra

To prove the above statement we need

∅ ∈M: for any set A,m∗(A ∩∅) + m∗(A∼∅) = m∗(∅) + m∗(A) = m∗(A).

If E ∈M then E c ∈M:m∗(A ∩ E c) + m∗(A∼ E c) = m∗(A∼ E ) + m∗(A ∩ E ) = m∗(A).

If Ei ∈M for i ∈ N then⋃Ei ∈M.

Since M is closed under union and complement, it is also closed underintersection and set difference. Also by induction it is closed under finiteunions and finite intersections.

Earlier we saw that if E1,E2 ∈M are disjoint, then for any set A

m∗(A ∩ (E1 ∪ E2)) = m∗(A ∩ E1) + m∗(A ∩ E2).

and this extends to any finite number of disjoint sets in M by induction.

Daniel H. Luecking Chapter 2 Sep 4, 2020 7 / 18

M is a σ-algebra

To prove the above statement we need

∅ ∈M: for any set A,m∗(A ∩∅) + m∗(A∼∅) = m∗(∅) + m∗(A) = m∗(A).

If E ∈M then E c ∈M:m∗(A ∩ E c) + m∗(A∼ E c) = m∗(A∼ E ) + m∗(A ∩ E ) = m∗(A).

If Ei ∈M for i ∈ N then⋃Ei ∈M.

Since M is closed under union and complement, it is also closed underintersection and set difference. Also by induction it is closed under finiteunions and finite intersections.

Earlier we saw that if E1,E2 ∈M are disjoint, then for any set A

m∗(A ∩ (E1 ∪ E2)) = m∗(A ∩ E1) + m∗(A ∩ E2).

and this extends to any finite number of disjoint sets in M by induction.

Daniel H. Luecking Chapter 2 Sep 4, 2020 7 / 18

M is a σ-algebra

To prove the above statement we need

∅ ∈M: for any set A,m∗(A ∩∅) + m∗(A∼∅) = m∗(∅) + m∗(A) = m∗(A).

If E ∈M then E c ∈M:m∗(A ∩ E c) + m∗(A∼ E c) = m∗(A∼ E ) + m∗(A ∩ E ) = m∗(A).

If Ei ∈M for i ∈ N then⋃Ei ∈M.

Since M is closed under union and complement, it is also closed underintersection and set difference. Also by induction it is closed under finiteunions and finite intersections.

Earlier we saw that if E1,E2 ∈M are disjoint, then for any set A

m∗(A ∩ (E1 ∪ E2)) = m∗(A ∩ E1) + m∗(A ∩ E2).

and this extends to any finite number of disjoint sets in M by induction.

Daniel H. Luecking Chapter 2 Sep 4, 2020 7 / 18

M is a σ-algebra

To prove the above statement we need

∅ ∈M: for any set A,m∗(A ∩∅) + m∗(A∼∅) = m∗(∅) + m∗(A) = m∗(A).

If E ∈M then E c ∈M:m∗(A ∩ E c) + m∗(A∼ E c) = m∗(A∼ E ) + m∗(A ∩ E ) = m∗(A).

If Ei ∈M for i ∈ N then⋃Ei ∈M.

Since M is closed under union and complement, it is also closed underintersection and set difference. Also by induction it is closed under finiteunions and finite intersections.

Earlier we saw that if E1,E2 ∈M are disjoint, then for any set A

m∗(A ∩ (E1 ∪ E2)) = m∗(A ∩ E1) + m∗(A ∩ E2).

and this extends to any finite number of disjoint sets in M by induction.

Daniel H. Luecking Chapter 2 Sep 4, 2020 7 / 18

M is a σ-algebra

To prove the above statement we need

∅ ∈M: for any set A,m∗(A ∩∅) + m∗(A∼∅) = m∗(∅) + m∗(A) = m∗(A).

If E ∈M then E c ∈M:m∗(A ∩ E c) + m∗(A∼ E c) = m∗(A∼ E ) + m∗(A ∩ E ) = m∗(A).

If Ei ∈M for i ∈ N then⋃Ei ∈M.

Since M is closed under union and complement, it is also closed underintersection and set difference. Also by induction it is closed under finiteunions and finite intersections.

Earlier we saw that if E1,E2 ∈M are disjoint, then for any set A

m∗(A ∩ (E1 ∪ E2)) = m∗(A ∩ E1) + m∗(A ∩ E2).

and this extends to any finite number of disjoint sets in M by induction.

Daniel H. Luecking Chapter 2 Sep 4, 2020 7 / 18

Let us now prove that version of additivity for a countable number of sets.That is, we want to show

Lemma

If Fi ∈M for i ∈ N are disjoint, then for all A ⊆ R

m∗ (A ∩⋃∞

i=1 Fi ) =∑∞

i=1m∗(A ∩ Fi ) (*)

Proof: The left side is less than are equal to the right by countablesubadditivity so we need only prove the other inequality. Take the finitecase:

n∑i=1

m∗(A ∩ Fi ) = m∗(A ∩

⋃nj=1 Fi

)≤ m∗

(A ∩

⋃∞j=1 Fi

)and then let n→∞ to get what we want.

To prove M is a σ-algebra we now need to show it is closed undercountable unions.

Daniel H. Luecking Chapter 2 Sep 4, 2020 8 / 18

Let us now prove that version of additivity for a countable number of sets.That is, we want to show

Lemma

If Fi ∈M for i ∈ N are disjoint, then for all A ⊆ R

m∗ (A ∩⋃∞

i=1 Fi ) =∑∞

i=1m∗(A ∩ Fi ) (*)

Proof: The left side is less than are equal to the right by countablesubadditivity so we need only prove the other inequality. Take the finitecase:

n∑i=1

m∗(A ∩ Fi ) = m∗(A ∩

⋃nj=1 Fi

)≤ m∗

(A ∩

⋃∞j=1 Fi

)and then let n→∞ to get what we want.

To prove M is a σ-algebra we now need to show it is closed undercountable unions.

Daniel H. Luecking Chapter 2 Sep 4, 2020 8 / 18

Let us now prove that version of additivity for a countable number of sets.That is, we want to show

Lemma

If Fi ∈M for i ∈ N are disjoint, then for all A ⊆ R

m∗ (A ∩⋃∞

i=1 Fi ) =∑∞

i=1m∗(A ∩ Fi ) (*)

Proof: The left side is less than are equal to the right by countablesubadditivity so we need only prove the other inequality. Take the finitecase:

n∑i=1

m∗(A ∩ Fi ) = m∗(A ∩

⋃nj=1 Fi

)≤ m∗

(A ∩

⋃∞j=1 Fi

)

and then let n→∞ to get what we want.

To prove M is a σ-algebra we now need to show it is closed undercountable unions.

Daniel H. Luecking Chapter 2 Sep 4, 2020 8 / 18

Let us now prove that version of additivity for a countable number of sets.That is, we want to show

Lemma

If Fi ∈M for i ∈ N are disjoint, then for all A ⊆ R

m∗ (A ∩⋃∞

i=1 Fi ) =∑∞

i=1m∗(A ∩ Fi ) (*)

Proof: The left side is less than are equal to the right by countablesubadditivity so we need only prove the other inequality. Take the finitecase:

n∑i=1

m∗(A ∩ Fi ) = m∗(A ∩

⋃nj=1 Fi

)≤ m∗

(A ∩

⋃∞j=1 Fi

)and then let n→∞ to get what we want.

To prove M is a σ-algebra we now need to show it is closed undercountable unions.

Daniel H. Luecking Chapter 2 Sep 4, 2020 8 / 18

Let us now prove that version of additivity for a countable number of sets.That is, we want to show

Lemma

If Fi ∈M for i ∈ N are disjoint, then for all A ⊆ R

m∗ (A ∩⋃∞

i=1 Fi ) =∑∞

i=1m∗(A ∩ Fi ) (*)

Proof: The left side is less than are equal to the right by countablesubadditivity so we need only prove the other inequality. Take the finitecase:

n∑i=1

m∗(A ∩ Fi ) = m∗(A ∩

⋃nj=1 Fi

)≤ m∗

(A ∩

⋃∞j=1 Fi

)and then let n→∞ to get what we want.

To prove M is a σ-algebra we now need to show it is closed undercountable unions.

Daniel H. Luecking Chapter 2 Sep 4, 2020 8 / 18

If Ei ∈M for i ∈ N we can rewrite the union as the union of a disjointsequence: F1 = E1, F2 = E2 ∼ E1, F3 = E3 ∼ (E1 ∪ E2), etc. Then eachFi ∈M by what we have previously proved.

We will need to show that

m∗ (A ∩⋃

i Fi ) + m∗ (A∼⋃

i Fi ) ≤ m∗(A).

as the inequality the other way is automatic.

We can suppose m∗(A) < +∞, because otherwise there is nothing toprove.

Daniel H. Luecking Chapter 2 Sep 4, 2020 9 / 18

If Ei ∈M for i ∈ N we can rewrite the union as the union of a disjointsequence: F1 = E1, F2 = E2 ∼ E1, F3 = E3 ∼ (E1 ∪ E2), etc. Then eachFi ∈M by what we have previously proved. We will need to show that

m∗ (A ∩⋃

i Fi ) + m∗ (A∼⋃

i Fi ) ≤ m∗(A).

as the inequality the other way is automatic.

We can suppose m∗(A) < +∞, because otherwise there is nothing toprove.

Daniel H. Luecking Chapter 2 Sep 4, 2020 9 / 18

If Ei ∈M for i ∈ N we can rewrite the union as the union of a disjointsequence: F1 = E1, F2 = E2 ∼ E1, F3 = E3 ∼ (E1 ∪ E2), etc. Then eachFi ∈M by what we have previously proved. We will need to show that

m∗ (A ∩⋃

i Fi ) + m∗ (A∼⋃

i Fi ) ≤ m∗(A).

as the inequality the other way is automatic.

We can suppose m∗(A) < +∞, because otherwise there is nothing toprove.

Daniel H. Luecking Chapter 2 Sep 4, 2020 9 / 18

Then∑n

i=1m∗(A ∩ Fi ) = m∗

(A ∩

⋃nj=1 Fi

)≤ m∗(A) <∞.

So the

infinite sum converges and so, given any ε > 0 there exists a N such that∑∞i=N+1m

∗(A ∩ Fi ) < ε. Therefore

m∗ (A ∩⋃∞

i=1 Fi ) + m∗ (A∼⋃∞

i=1 Fi )

≤∑∞

i=1m∗(A ∩ Fi ) + m∗

(A∼

⋃Ni=1 Fi

)≤∑N

i=1m∗(A ∩ Fi ) + ε+ m∗

(A∼

⋃Ni=1 Fi

)= m∗

(A ∪

⋃Ni=1 Fi

)+ m∗

(A∼

⋃Ni=1 Fi

)+ ε

= m∗(A) + ε.

Let ε→ 0 to get the required inequality.

Daniel H. Luecking Chapter 2 Sep 4, 2020 10 / 18

Then∑n

i=1m∗(A ∩ Fi ) = m∗

(A ∩

⋃nj=1 Fi

)≤ m∗(A) <∞. So the

infinite sum converges

and so, given any ε > 0 there exists a N such that∑∞i=N+1m

∗(A ∩ Fi ) < ε. Therefore

m∗ (A ∩⋃∞

i=1 Fi ) + m∗ (A∼⋃∞

i=1 Fi )

≤∑∞

i=1m∗(A ∩ Fi ) + m∗

(A∼

⋃Ni=1 Fi

)≤∑N

i=1m∗(A ∩ Fi ) + ε+ m∗

(A∼

⋃Ni=1 Fi

)= m∗

(A ∪

⋃Ni=1 Fi

)+ m∗

(A∼

⋃Ni=1 Fi

)+ ε

= m∗(A) + ε.

Let ε→ 0 to get the required inequality.

Daniel H. Luecking Chapter 2 Sep 4, 2020 10 / 18

Then∑n

i=1m∗(A ∩ Fi ) = m∗

(A ∩

⋃nj=1 Fi

)≤ m∗(A) <∞. So the

infinite sum converges and so, given any ε > 0 there exists a N such that∑∞i=N+1m

∗(A ∩ Fi ) < ε.

Therefore

m∗ (A ∩⋃∞

i=1 Fi ) + m∗ (A∼⋃∞

i=1 Fi )

≤∑∞

i=1m∗(A ∩ Fi ) + m∗

(A∼

⋃Ni=1 Fi

)≤∑N

i=1m∗(A ∩ Fi ) + ε+ m∗

(A∼

⋃Ni=1 Fi

)= m∗

(A ∪

⋃Ni=1 Fi

)+ m∗

(A∼

⋃Ni=1 Fi

)+ ε

= m∗(A) + ε.

Let ε→ 0 to get the required inequality.

Daniel H. Luecking Chapter 2 Sep 4, 2020 10 / 18

Then∑n

i=1m∗(A ∩ Fi ) = m∗

(A ∩

⋃nj=1 Fi

)≤ m∗(A) <∞. So the

infinite sum converges and so, given any ε > 0 there exists a N such that∑∞i=N+1m

∗(A ∩ Fi ) < ε. Therefore

m∗ (A ∩⋃∞

i=1 Fi ) + m∗ (A∼⋃∞

i=1 Fi )

≤∑∞

i=1m∗(A ∩ Fi ) + m∗

(A∼

⋃Ni=1 Fi

)≤∑N

i=1m∗(A ∩ Fi ) + ε+ m∗

(A∼

⋃Ni=1 Fi

)= m∗

(A ∪

⋃Ni=1 Fi

)+ m∗

(A∼

⋃Ni=1 Fi

)+ ε

= m∗(A) + ε.

Let ε→ 0 to get the required inequality.

Daniel H. Luecking Chapter 2 Sep 4, 2020 10 / 18

Lebesgue measure m

So we have shown that M is a σ-algebra. We also have from the previouslemma with A = R that, if Fi ∈M are disjoint then

m∗ (⋃∞

i=1 Fi ) =∑∞

i=1m∗(Fi ) (*)

We let m be the restriction of m∗ to M and we have a measure.

Daniel H. Luecking Chapter 2 Sep 4, 2020 11 / 18

Lebesgue measure m

So we have shown that M is a σ-algebra. We also have from the previouslemma with A = R that, if Fi ∈M are disjoint then

m∗ (⋃∞

i=1 Fi ) =∑∞

i=1m∗(Fi ) (*)

We let m be the restriction of m∗ to M and we have a measure.

Daniel H. Luecking Chapter 2 Sep 4, 2020 11 / 18

What sets are measurable

The following are measurable sets

All intervals: e.g., (a, b] = (a,∞) ∩ (−∞, b]

All open sets: countable unions of open intervals.

All Gδ-sets: countable intersections of open sets.

All closed sets: complements of open sets.

All Fσ-sets: Countable unions of closed sets.

Any set with outer measure 0: If m∗(E ) = 0 and A is any set thenm∗(A ∩ E ) + m∗(A ∩ E c) ≤ 0 + m∗(A).

The Cantor set C is closed and has m∗(C ) = 0. To see this, recall thatC =

⋂n Cn where each Cn is a union of 2n closed intervals, each with

length 1/3n. Therefore m∗(C ) ≤ (2/3)n for all n. This means that everysubset of C is measurable. This implies that the cardinality of M is thesame as that of P(R).

Daniel H. Luecking Chapter 2 Sep 4, 2020 12 / 18

What sets are measurable

The following are measurable sets

All intervals: e.g., (a, b] = (a,∞) ∩ (−∞, b]

All open sets: countable unions of open intervals.

All Gδ-sets: countable intersections of open sets.

All closed sets: complements of open sets.

All Fσ-sets: Countable unions of closed sets.

Any set with outer measure 0: If m∗(E ) = 0 and A is any set thenm∗(A ∩ E ) + m∗(A ∩ E c) ≤ 0 + m∗(A).

The Cantor set C is closed and has m∗(C ) = 0. To see this, recall thatC =

⋂n Cn where each Cn is a union of 2n closed intervals, each with

length 1/3n. Therefore m∗(C ) ≤ (2/3)n for all n. This means that everysubset of C is measurable. This implies that the cardinality of M is thesame as that of P(R).

Daniel H. Luecking Chapter 2 Sep 4, 2020 12 / 18

What sets are measurable

The following are measurable sets

All intervals: e.g., (a, b] = (a,∞) ∩ (−∞, b]

All open sets: countable unions of open intervals.

All Gδ-sets: countable intersections of open sets.

All closed sets: complements of open sets.

All Fσ-sets: Countable unions of closed sets.

Any set with outer measure 0: If m∗(E ) = 0 and A is any set thenm∗(A ∩ E ) + m∗(A ∩ E c) ≤ 0 + m∗(A).

The Cantor set C is closed and has m∗(C ) = 0. To see this, recall thatC =

⋂n Cn where each Cn is a union of 2n closed intervals, each with

length 1/3n. Therefore m∗(C ) ≤ (2/3)n for all n. This means that everysubset of C is measurable. This implies that the cardinality of M is thesame as that of P(R).

Daniel H. Luecking Chapter 2 Sep 4, 2020 12 / 18

What sets are measurable

The following are measurable sets

All intervals: e.g., (a, b] = (a,∞) ∩ (−∞, b]

All open sets: countable unions of open intervals.

All Gδ-sets: countable intersections of open sets.

All closed sets: complements of open sets.

All Fσ-sets: Countable unions of closed sets.

Any set with outer measure 0: If m∗(E ) = 0 and A is any set thenm∗(A ∩ E ) + m∗(A ∩ E c) ≤ 0 + m∗(A).

The Cantor set C is closed and has m∗(C ) = 0. To see this, recall thatC =

⋂n Cn where each Cn is a union of 2n closed intervals, each with

length 1/3n. Therefore m∗(C ) ≤ (2/3)n for all n. This means that everysubset of C is measurable. This implies that the cardinality of M is thesame as that of P(R).

Daniel H. Luecking Chapter 2 Sep 4, 2020 12 / 18

What sets are measurable

The following are measurable sets

All intervals: e.g., (a, b] = (a,∞) ∩ (−∞, b]

All open sets: countable unions of open intervals.

All Gδ-sets: countable intersections of open sets.

All closed sets: complements of open sets.

All Fσ-sets: Countable unions of closed sets.

Any set with outer measure 0: If m∗(E ) = 0 and A is any set thenm∗(A ∩ E ) + m∗(A ∩ E c) ≤ 0 + m∗(A).

The Cantor set C is closed and has m∗(C ) = 0. To see this, recall thatC =

⋂n Cn where each Cn is a union of 2n closed intervals, each with

length 1/3n. Therefore m∗(C ) ≤ (2/3)n for all n. This means that everysubset of C is measurable. This implies that the cardinality of M is thesame as that of P(R).

Daniel H. Luecking Chapter 2 Sep 4, 2020 12 / 18

What sets are measurable

The following are measurable sets

All intervals: e.g., (a, b] = (a,∞) ∩ (−∞, b]

All open sets: countable unions of open intervals.

All Gδ-sets: countable intersections of open sets.

All closed sets: complements of open sets.

All Fσ-sets: Countable unions of closed sets.

Any set with outer measure 0: If m∗(E ) = 0 and A is any set thenm∗(A ∩ E ) + m∗(A ∩ E c) ≤ 0 + m∗(A).

The Cantor set C is closed and has m∗(C ) = 0. To see this, recall thatC =

⋂n Cn where each Cn is a union of 2n closed intervals, each with

length 1/3n. Therefore m∗(C ) ≤ (2/3)n for all n. This means that everysubset of C is measurable. This implies that the cardinality of M is thesame as that of P(R).

Daniel H. Luecking Chapter 2 Sep 4, 2020 12 / 18

What sets are measurable

The following are measurable sets

All intervals: e.g., (a, b] = (a,∞) ∩ (−∞, b]

All open sets: countable unions of open intervals.

All Gδ-sets: countable intersections of open sets.

All closed sets: complements of open sets.

All Fσ-sets: Countable unions of closed sets.

Any set with outer measure 0: If m∗(E ) = 0 and A is any set thenm∗(A ∩ E ) + m∗(A ∩ E c) ≤ 0 + m∗(A).

The Cantor set C is closed and has m∗(C ) = 0. To see this, recall thatC =

⋂n Cn where each Cn is a union of 2n closed intervals, each with

length 1/3n. Therefore m∗(C ) ≤ (2/3)n for all n.

This means that everysubset of C is measurable. This implies that the cardinality of M is thesame as that of P(R).

Daniel H. Luecking Chapter 2 Sep 4, 2020 12 / 18

What sets are measurable

The following are measurable sets

All intervals: e.g., (a, b] = (a,∞) ∩ (−∞, b]

All open sets: countable unions of open intervals.

All Gδ-sets: countable intersections of open sets.

All closed sets: complements of open sets.

All Fσ-sets: Countable unions of closed sets.

Any set with outer measure 0: If m∗(E ) = 0 and A is any set thenm∗(A ∩ E ) + m∗(A ∩ E c) ≤ 0 + m∗(A).

The Cantor set C is closed and has m∗(C ) = 0. To see this, recall thatC =

⋂n Cn where each Cn is a union of 2n closed intervals, each with

length 1/3n. Therefore m∗(C ) ≤ (2/3)n for all n. This means that everysubset of C is measurable. This implies that the cardinality of M is thesame as that of P(R).

Daniel H. Luecking Chapter 2 Sep 4, 2020 12 / 18

Properties of measures: ‘continuity’

Theorem

If Ej is an ascending sequence of measurable sets then

m(⋃

j Ej

)= lim

j→∞m(Ej). (∗)

If Ej is a descending sequence of measurable sets and m(E1) <∞ then

m(⋂

j Ej

)= lim

j→∞m(Ej).

Proof: Suppose Ej are ascending. If any m(Ej) =∞, then both sides of(∗) are ∞.

Otherwise, define F1 = E1 and Fj = Ej − Ej−1. The sequence Fj is disjointand has the same union as ths Ej . By additivity m(Fj) = m(Ej)−m(Ej−1).

Daniel H. Luecking Chapter 2 Sep 4, 2020 13 / 18

Properties of measures: ‘continuity’

Theorem

If Ej is an ascending sequence of measurable sets then

m(⋃

j Ej

)= lim

j→∞m(Ej). (∗)

If Ej is a descending sequence of measurable sets and m(E1) <∞ then

m(⋂

j Ej

)= lim

j→∞m(Ej).

Proof: Suppose Ej are ascending. If any m(Ej) =∞, then both sides of(∗) are ∞.

Otherwise, define F1 = E1 and Fj = Ej − Ej−1. The sequence Fj is disjointand has the same union as ths Ej . By additivity m(Fj) = m(Ej)−m(Ej−1).

Daniel H. Luecking Chapter 2 Sep 4, 2020 13 / 18

Then

m(⋃

j Ej

)= m

(⋃j Fj

)=∑

j m(Fj) = m(E1) +∑

j m(Ej)−m(Ej−1)

= limn→∞

m(En)

Now suppose Ej is descending and m(E1) <∞. Let Fj = E1 ∼ Ej so thatFj is ascending. We can write

⋂j Ej = E1 ∼

⋃Fj . Therefore

m(⋂

j Ej

)= m(E1)−m

(⋃j Fj

)= m(E1)− lim

j→∞m(Fj)

= m(E1)− limj→∞

(m(E1)−m(Ej))

= limj→∞

m(Ej)

Daniel H. Luecking Chapter 2 Sep 4, 2020 14 / 18

Then

m(⋃

j Ej

)= m

(⋃j Fj

)=∑

j m(Fj) = m(E1) +∑

j m(Ej)−m(Ej−1)

= limn→∞

m(En)

Now suppose Ej is descending and m(E1) <∞. Let Fj = E1 ∼ Ej so thatFj is ascending. We can write

⋂j Ej = E1 ∼

⋃Fj .

Therefore

m(⋂

j Ej

)= m(E1)−m

(⋃j Fj

)= m(E1)− lim

j→∞m(Fj)

= m(E1)− limj→∞

(m(E1)−m(Ej))

= limj→∞

m(Ej)

Daniel H. Luecking Chapter 2 Sep 4, 2020 14 / 18

Then

m(⋃

j Ej

)= m

(⋃j Fj

)=∑

j m(Fj) = m(E1) +∑

j m(Ej)−m(Ej−1)

= limn→∞

m(En)

Now suppose Ej is descending and m(E1) <∞. Let Fj = E1 ∼ Ej so thatFj is ascending. We can write

⋂j Ej = E1 ∼

⋃Fj . Therefore

m(⋂

j Ej

)= m(E1)−m

(⋃j Fj

)= m(E1)− lim

j→∞m(Fj)

= m(E1)− limj→∞

(m(E1)−m(Ej))

= limj→∞

m(Ej)

Daniel H. Luecking Chapter 2 Sep 4, 2020 14 / 18

Some finiteness condition is required in the second part because ifEj = (j ,∞) then

⋂Ej = ∅ while lim

j→∞m((j ,∞)) =∞.

For a descending sequence we have⋂∞

j=1 Ej =⋂∞

j=n Ej for any n, so it isenough that m(En) <∞ for some n.

Daniel H. Luecking Chapter 2 Sep 4, 2020 15 / 18

Some finiteness condition is required in the second part because ifEj = (j ,∞) then

⋂Ej = ∅ while lim

j→∞m((j ,∞)) =∞.

For a descending sequence we have⋂∞

j=1 Ej =⋂∞

j=n Ej for any n, so it isenough that m(En) <∞ for some n.

Daniel H. Luecking Chapter 2 Sep 4, 2020 15 / 18

Theorem

If E ⊆ R then E is measurable if and only if there is a Gδ-set G withE ⊂ G and m∗(G ∼ E ) = 0.

Proof: First suppose that such a set G exists. Then G is measurable andalso G ∼ E is measurable because its outer measure is 0. ThusE = G ∼ (G ∼ E ) is measurable.

Now suppose E is measurable and suppose it has m∗(E ) <∞. Then forany n there exists a family of open intervals Ink , k ∈ N with∑

k `(Ink) < m∗(E ) + 1/n. Let Un =⋃

k Ink so that E ⊂ Un and

m∗(E ) ≤ m∗(Un) ≤∑

k `(Ink) < m∗(E ) + 1/n

Let G =⋂

n Un. Then G is a Gδ-set and m∗(E ) = m∗(G ). Since all setsare measurable with finite measure we havem∗(G ∼ E ) = m(G ∼ E ) = m(G )−m(E ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 16 / 18

Theorem

If E ⊆ R then E is measurable if and only if there is a Gδ-set G withE ⊂ G and m∗(G ∼ E ) = 0.

Proof: First suppose that such a set G exists. Then G is measurable andalso G ∼ E is measurable because its outer measure is 0. ThusE = G ∼ (G ∼ E ) is measurable.

Now suppose E is measurable and suppose it has m∗(E ) <∞. Then forany n there exists a family of open intervals Ink , k ∈ N with∑

k `(Ink) < m∗(E ) + 1/n. Let Un =⋃

k Ink so that E ⊂ Un and

m∗(E ) ≤ m∗(Un) ≤∑

k `(Ink) < m∗(E ) + 1/n

Let G =⋂

n Un. Then G is a Gδ-set and m∗(E ) = m∗(G ). Since all setsare measurable with finite measure we havem∗(G ∼ E ) = m(G ∼ E ) = m(G )−m(E ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 16 / 18

Theorem

If E ⊆ R then E is measurable if and only if there is a Gδ-set G withE ⊂ G and m∗(G ∼ E ) = 0.

Proof: First suppose that such a set G exists. Then G is measurable andalso G ∼ E is measurable because its outer measure is 0. ThusE = G ∼ (G ∼ E ) is measurable.

Now suppose E is measurable and suppose it has m∗(E ) <∞. Then forany n there exists a family of open intervals Ink , k ∈ N with∑

k `(Ink) < m∗(E ) + 1/n. Let Un =⋃

k Ink so that E ⊂ Un and

m∗(E ) ≤ m∗(Un) ≤∑

k `(Ink) < m∗(E ) + 1/n

Let G =⋂

n Un. Then G is a Gδ-set and m∗(E ) = m∗(G ). Since all setsare measurable with finite measure we havem∗(G ∼ E ) = m(G ∼ E ) = m(G )−m(E ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 16 / 18

Theorem

If E ⊆ R then E is measurable if and only if there is a Gδ-set G withE ⊂ G and m∗(G ∼ E ) = 0.

Proof: First suppose that such a set G exists. Then G is measurable andalso G ∼ E is measurable because its outer measure is 0. ThusE = G ∼ (G ∼ E ) is measurable.

Now suppose E is measurable and suppose it has m∗(E ) <∞. Then forany n there exists a family of open intervals Ink , k ∈ N with∑

k `(Ink) < m∗(E ) + 1/n. Let Un =⋃

k Ink so that E ⊂ Un and

m∗(E ) ≤ m∗(Un) ≤∑

k `(Ink) < m∗(E ) + 1/n

Let G =⋂

n Un. Then G is a Gδ-set and m∗(E ) = m∗(G ). Since all setsare measurable with finite measure we havem∗(G ∼ E ) = m(G ∼ E ) = m(G )−m(E ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 16 / 18

Theorem

If E ⊆ R then E is measurable if and only if there is a Gδ-set G withE ⊂ G and m∗(G ∼ E ) = 0.

Proof: First suppose that such a set G exists. Then G is measurable andalso G ∼ E is measurable because its outer measure is 0. ThusE = G ∼ (G ∼ E ) is measurable.

Now suppose E is measurable and suppose it has m∗(E ) <∞. Then forany n there exists a family of open intervals Ink , k ∈ N with∑

k `(Ink) < m∗(E ) + 1/n. Let Un =⋃

k Ink so that E ⊂ Un and

m∗(E ) ≤ m∗(Un) ≤∑

k `(Ink) < m∗(E ) + 1/n

Let G =⋂

n Un. Then G is a Gδ-set and m∗(E ) = m∗(G ). Since all setsare measurable with finite measure we havem∗(G ∼ E ) = m(G ∼ E ) = m(G )−m(E ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 16 / 18

Now suppose m∗(E ) =∞. We can write Ek = E ∩ [k , k + 1) and thenE =

⋃k∈Z Ek .

Let ε > 0 and apply the first part of the previous argument to find opensets Uk with Ek ⊂ Uk and m(Uk ∼ Ek) < ε/2|k|+2.

Let Uε =⋃

k Uk and then m(Uε ∼ E ) ≤∑

k m(Uk ∼ Ek) < ε. If εk is asequence of positive numbers tending to 0 then let G =

⋂k Uεk and we get

m(G ∼ E ) ≤ m(Uε ∼ E ) ≤ εk for all k

Letting εk → 0 we get m∗(G ∼ E ) = m(G ∼ E ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 17 / 18

Now suppose m∗(E ) =∞. We can write Ek = E ∩ [k , k + 1) and thenE =

⋃k∈Z Ek .

Let ε > 0 and apply the first part of the previous argument to find opensets Uk with Ek ⊂ Uk and m(Uk ∼ Ek) < ε/2|k|+2.

Let Uε =⋃

k Uk and then m(Uε ∼ E ) ≤∑

k m(Uk ∼ Ek) < ε. If εk is asequence of positive numbers tending to 0 then let G =

⋂k Uεk and we get

m(G ∼ E ) ≤ m(Uε ∼ E ) ≤ εk for all k

Letting εk → 0 we get m∗(G ∼ E ) = m(G ∼ E ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 17 / 18

Now suppose m∗(E ) =∞. We can write Ek = E ∩ [k , k + 1) and thenE =

⋃k∈Z Ek .

Let ε > 0 and apply the first part of the previous argument to find opensets Uk with Ek ⊂ Uk and m(Uk ∼ Ek) < ε/2|k|+2.

Let Uε =⋃

k Uk and then m(Uε ∼ E ) ≤∑

k m(Uk ∼ Ek) < ε. If εk is asequence of positive numbers tending to 0 then let G =

⋂k Uεk and we get

m(G ∼ E ) ≤ m(Uε ∼ E ) ≤ εk for all k

Letting εk → 0 we get m∗(G ∼ E ) = m(G ∼ E ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 17 / 18

Now suppose m∗(E ) =∞. We can write Ek = E ∩ [k , k + 1) and thenE =

⋃k∈Z Ek .

Let ε > 0 and apply the first part of the previous argument to find opensets Uk with Ek ⊂ Uk and m(Uk ∼ Ek) < ε/2|k|+2.

Let Uε =⋃

k Uk and then m(Uε ∼ E ) ≤∑

k m(Uk ∼ Ek) < ε. If εk is asequence of positive numbers tending to 0 then let G =

⋂k Uεk and we get

m(G ∼ E ) ≤ m(Uε ∼ E ) ≤ εk for all k

Letting εk → 0 we get m∗(G ∼ E ) = m(G ∼ E ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 17 / 18

By taking complements we get

Theorem

If E ⊆ R then E is measurable if and only if there is an Fσ-set F withF ⊂ E and m∗(E ∼ F ) = 0.

Proof: If such a set F exists then E = F ∪ (E ∼ F ). F is measurable andso is E ∼ F because its outer measure is 0. Thus E is measurable

If E is measureable then so is E c . Find a Gδ-set containing E c withm∗(G ∼ E c) = 0. Let F = G c , then F is an Fσ-set andG ∼ E c = G ∩ E = F c ∩ E = E ∼ F so m∗(E ∼ F ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 18 / 18

By taking complements we get

Theorem

If E ⊆ R then E is measurable if and only if there is an Fσ-set F withF ⊂ E and m∗(E ∼ F ) = 0.

Proof: If such a set F exists then E = F ∪ (E ∼ F ). F is measurable andso is E ∼ F because its outer measure is 0. Thus E is measurable

If E is measureable then so is E c . Find a Gδ-set containing E c withm∗(G ∼ E c) = 0. Let F = G c , then F is an Fσ-set andG ∼ E c = G ∩ E = F c ∩ E = E ∼ F so m∗(E ∼ F ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 18 / 18

By taking complements we get

Theorem

If E ⊆ R then E is measurable if and only if there is an Fσ-set F withF ⊂ E and m∗(E ∼ F ) = 0.

Proof: If such a set F exists then E = F ∪ (E ∼ F ). F is measurable andso is E ∼ F because its outer measure is 0. Thus E is measurable

If E is measureable then so is E c . Find a Gδ-set containing E c withm∗(G ∼ E c) = 0. Let F = G c , then F is an Fσ-set andG ∼ E c = G ∩ E = F c ∩ E = E ∼ F so m∗(E ∼ F ) = 0.

Daniel H. Luecking Chapter 2 Sep 4, 2020 18 / 18