chapter 2 · 2020. 9. 4. · chapter 2 daniel h. luecking sep 4, 2020 daniel h. luecking chapter 2...
TRANSCRIPT
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