2
INTRODUCTION TO LEBESGUE INTEGRATION
17
Lecture 11: May 29
Theorem.
If
f
is Riemann integrable over [
a, b
], then
f
is also Lebesgue integrable over [
a, b
], with
R
R
b
a
f
=
R
b
a
f
.
Proof.
Suppose
f
is Riemann integrable. Then
f
is bounded. First, we show that
f
is measurable. Recall that
f
is Riemann integrable i
↵
m
(
E
) = 0 where
E
is the set of discontinuities of
f
. Then
f
=
f
χ
E

{z
}
g
+
f
χ
E
C

{z
}
h
. Notice that
g
= 0 a.e. , since
m
(
E
) = 0, and hence
g
is measurable. Now
h
is a continuous function on
E
C
, so
h
is measurable.
Thus
f
is measurable, so

f

is measurable, so
R
b
a

f

is welldefined. Now since
f
is bounded, there exists
M
such
that

f

M
8
x
. By monotonicity,
R
b
a

f

R
b
a
M
=
M
(
b

a
)
<
1
. Take a partition
P
of [
a, b
]. Then
U
(
f, P
) =
X
M
i
(
x
i

x
i

1
) =
R
Z
b
a
X
M
i
χ
[
x
i

1
,x
i
]
=
Z
b
a
X
M
i
χ
[
x
i

1
,x
i
]
≥
Z
b
a
f
since on [
x
i

1
, x
i
],
f
(
x
)
M
i
. Similarly,
L
(
f, P
) =
X
m
i
(
x
i

x
i

1
) =
R
Z
b
a
X
m
i
χ
[
x
i

1
,x
i
]
=
Z
b
a
X
m
i
χ
[
x
i

1
,x
i
]
Z
b
a
f
since
f
(
x
)
≥
m
i
for
x
2
[
x
i

1
, x
i
].
Thus we have
U
(
f, P
)
≥
R
b
a
f
≥
L
(
f, P
) for any partition.
We also have
U
(
f, P
)
≥
R
R
b
a
f
≥
L
(
f, P
). But inf
U
(
f, P
) = sup
L
(
f, P
) since
f
is Riemann integrable, so
R
R
b
a
f
=
R
b
a
f
.
2.6
L
p
Spaces
We will start with 1
p <
1
.
Definition.
Let
f
:
E
!
C
, with
E
and
f
measurable. We define the
p
norm
as
k
f
k
L
p
(
E
)
=
k
f
k
p
=
✓Z
E

f

p
◆
1
/p
.