Chapter 3: Lebesgue Measure
Written by
MenGen Tsai
email: [email protected]
1. Let
m
be a countably additive measure defined for all sets in a
σ
algebra
M
.
If
A
and
B
are two sets in
M
with
A
⊂
B
, then
mA
≤
mB
.
This
property is called
monotonicity
.
Proof:
B
=
A
(
A

B
).
A
and
A

B
are disjoint.
Since
m
is a
countably additive measure,
mB
=
mA
+
m
(
A

B
). Note that
m
is
nonnegative, and
m
(
A

B
)
≥
0. Hence
mA
≤
mB
.
2. Let
< E
n
>
be any sequence of sets in
M
. Then
m
(
E
n
)
≤
∑
mE
n
.
[Hint: Use Proposition 1.2.] This property of a measure is called
count
able subadditivity
.
Proof:
By Proposition 1.2 on page 17, since
M
is a
σ
algebra, there
is a sequence
< B
n
>
of sets in
M
such that
B
n
B
m
=
φ
for
n
=
m
and
∞
i
=1
B
i
=
∞
i
=1
E
i
.
Since
m
is a countably additive measure and
B
i
⊂
E
i
for all
i
, by
Problem 3.1 I have that
m
(
E
n
) =
m
(
B
n
) =
mB
n
≤
mE
n
.
3. If there is a set
A
in
M
such that
mA <
∞
, then
mφ
= 0.
Proof:
Note that
A
=
A
φ
and
A
and
φ
are disjoint, and thus
mA
=
mA
+
mφ.
Since
mA <
∞
,
mφ
= 0 precisely.
1
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4.
5. Let
A
be the set of rational numbers between 0 and 1, and let
{
I
n
}
be
a finite collection of open intervals covering
A
. Then
∑
l
(
I
n
)
≥
1.
Proof 1:
(due to MengGen Tsai) Since 0
∈
A
, there is an open interval
J
1
in
{
I
n
}
such that 0
∈
J
1
. Let
J
1
= (
a
1
, b
1
). Note that
a
1
<
0 and
b
1
>
0. If
b
1
≥
1, then
l
(
I
n
)
≥
l
(
J
1
) =
b
1

a
1
≥
1
.
Suppose not. If
a
1
is rational, then I can find an open interval
J
2
∈ {
I
n
}
such that
a
1
∈
J
2
. If
a
1
is irrational, I consider the following cases.
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 open interval, Topological space, measure, Lebesgue measure, countably additive measure

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