2
INTRODUCTION TO LEBESGUE INTEGRATION
6
2.1
Construction of Lebesgue Measure
Definition.
Let
A
✓
R
and let
C
(
A
) =
{{
I
n
}
1
n
=1
:
I
n
open interval with
S
1
n
=1
I
n
◆
A
}
.
We define the
outer
Lebesgue measure
as
m
⇤
(
A
) = inf
(
1
X
n
=1
`
(
I
n
) :
{
I
n
}
2
C
(
A
)
)
.
Properties.
(1)
m
⇤
(
;
) = 0.
(2)
m
⇤
(
{
x
}
) = 0 for all
x
2
R
.
(3) If
A
✓
B
then
m
⇤
(
A
)
m
⇤
(
B
).
(4)
m
⇤
is translation invariant.
Proposition.
The outer measure of an interval is its length.
Lecture 4: May 11
Proof.
Case: [
a, b
]. If
"
>
0 then [
a, b
]
✓
(
a

"
, b
+
"
). Thus
m
⇤
([
a, b
])
`
((
a

"
, b
+
"
)) =
b

a
+2
"
. Since this is true for
all
"
>
0, so
m
⇤
([
a, b
])
b

a
. Now we claim that whenever
{
I
n
}
2
C
[
a, b
], we have
P
`
(
I
n
)
≥
b

a
. Let
{
I
n
}
1
n
=1
2
C
[
a, b
], so by compactness of [
a, b
] there exists a finite subcover
{
I
j
1
, . . . , I
j
m
}
.
Then
P
1
i
=1
`
(
I
i
)
≥
P
m
k
=1
`
(
I
j
k
).
So it is enough to show that whenever
I
1
, . . . , I
n
is a cover of [
a, b
] by open intervals, then
P
n
j
=1
`
(
I
j
)
≥
b

a
.
Since
S
n
i
=1
I
j
◆
[
a, b
], pick
j
1
such that
a
2
I
j
1
= (
a
1
, b
1
), so
a
1
< a < b
1
.
If
b < b
1
then [
a, b
]
⇢
(
a
1
, b
1
), so
P
n
j
=1
`
(
I
j
)
≥
`
(
I
j
1
)
≥
b
1

a
1
> b

a
and we are done. Otherwise
a < b
1
b
, so
b
1
2
I
j
2
= (
a
2
, b
2
) for some
j
1
6
=
j
2
.
If
b < b
2
, then [
a, b
]
⇢
(
a
1
, b
1
)
[
(
a
2
, b
2
) and
P
n
i
=1
`
(
I
j
)
≥
b
2

a
2
+
b
1

a
1
≥
b
2

a
1
≥
b

a
. Otherwise we repeat
the argument.
Get intervals
I
j
k
= (
a
k
, b
k
) with
a
k
< b
k

1
< b
k
where the
j
k
are all distinct.
If
b < b
k
we have
[
a, b
]
✓
S
k
i
=1
I
j
i
and then
P
k
i
=1
`
(
I
j
i
) =
P
k
i
=1
b
i

a
i
≥
b

a
. This process must terminate after finitely many steps
because there are only finitely many
I
j
s to pick from.
The remaining cases are left as an exercise.
Proposition.
m
⇤
1
[
k
=1
A
k
!
1
X
k
=1
m
⇤
(
A
k
)
8
A
k
⇢
R
Proof.
Fix
"
>
0. For each
A
k
, choose open intervals
{
I
n,k
}
1
n
=1
that cover
A
k
such that
P
1
n
=1
`
(
I
n,k
)
m
⇤
(
A
n
) +
"
2
k
.
Then
{
I
n,k
}
1
n
=1
1
k
=1
2
C
(
S
1
n
=1
A
k
), so
m
⇤
(
S
1
k
=1
A
k
)
P
1
k
=1
P
1
n
=1
`
(
I
n,k
)
P
1
k
=1
(
m
⇤
(
A
k
) +
"
2
k
) =
P
1
k
=1
m
⇤
(
A
k
) +
"
.
But
"
>
0 was arbitrary, therefore
m
⇤
(
S
1
k
=1
A
k
)
P
1
k
=1
m
⇤
(
A
k
).
Corollary.
Let
A
=
{
x
k
}
1
k
=1
be countable. Then
m
⇤
(
A
) = 0.
Proof.
m
⇤
(
A
) =
m
⇤
(
{
x
k
}
1
k
=1
)
P
1
k
=1
m
⇤
(
{
x
k
}
) = 0.
Corollary.
For all
A
✓
R
and
"
>
0, there exists an open set
O
◆
A
such that
m
⇤
(0)
m
⇤
(
A
) +
"
.
Proof.
Choose intervals
{
I
n
}
covering
A
such that
P
1
n
=1
`
(
I
n
)
m
⇤
(
A
) +
"
. Take
O
=
S
1
n
=1
I
n
◆
A
, which is open.
We have
m
⇤
(
O
) =
m
⇤
(
S
1
n
=1
I
n
)
P
1
n
=1
m
⇤
(
I
n
) =
P
n
=1
`
(
I
n
)
m
⇤
(
A
) +
"
.
Definition.
We say that
A
✓
R
is
Lebesgue measurable
if whenever
E
✓
R
,
m
⇤
(
E
) =
m
⇤
(
E
\
A
) +
m
⇤
(
E
\
A
C
).