δ
(2
N
)
p
+
δ
(
b

a
)
<
2
"
Lecture 16: June 10
Now suppose
f
2
L
p
[
a, b
] is arbitrary.
Define
f
n
(
x
) =
(
f
(
x
)

f
(
x
)

N
0
otherwise
.
k
f
N
(
x
)
k
N
is bounded in
L
p
[
a, b
].
Notice that
f
N
!
f
a.e. pointwise, so

f
(
x
)

f
N
(
x
)

p
!
0 a.e. pointwise and

f
(
x
)

F
N
(
x
)

p

f

p
2
L
1
[
a, b
]. By
DCT,
k
f
N

f
k
p
p
=
Z
b
a

f

f
n

p
!
Z
b
a
0 = 0
Pick
N
such that
k
f
N

f
k
p
<
"
2
, and get
h
continuous such that
k
f
N

h
k
p
<
"
2
(by the first part of the proof). Then
k
f

h
k
p
k
f

f
N
k
p
+
k
f
N

h
k
p
<
"
.
Finally we show that
C
[
a, b
] is not dense in
L
1
[
a, b
]. Find
c
2
(
a, b
). Let
f
[
a,c
)
= 0 and
f
[
c,b
]
= 1. We claim there is
not a continuous
g
such that
f
=
g
a.e. . Suppose there is. For any
δ
>
0, (
c, c
+
δ
) has positive measure, so
g
=
f
(
x
)
at some
x
2
(
c, c
+
δ
). Hence
g
(
x
) = 1 at some
x
2
(
c, c
+
δ
), so by continuity,
g
(0) = 1. Similarly, approaching from
the other side gives
g
(0) = 0, a contradiction. So
f
2
L
1
[
a, b
] and the equivalence class containing
f
contains no
continuous function. Now suppose there is a sequence
f
n
2
C
[
a, b
] such that
f
n
!
f
in
L
1
norm. (
f
n
) is a Cauchy
sequence in
L
1
norm, but the
L
1
norm is the sup norm on the continuous functions, so really (
f
n
) is Cauchy in
C
([
a, b
]) with the usual sup norm.
C
[
a, b
] is complete with its usual sup norm, so
f
n
!
g
in sup norm where
g
is
continuous. So
k
f
n

g
k
sup
!
0, but
f
n
!
f
in
L
1
, so
f
=
g
a.e. , a contradiction. Thus
C
[
a, b
] is a closed subspace
of
L
1
[
a, b
], whereas cl
C
[
a, b
] =
L
p
[
a, b
] for 1
p <
1
.
Corollary.
Polynomials in
L
p
[0
,
1] for
p <
1
.
Proof.
Note that
k
f
k
p
p
=
⇣
R
1
0

f

p
⌘
1
/p
k
f
k
p
1
. Let
"
>
0. Take
f
2
L
p
and get
g
continuous such that
k
g

f
k
p
<
"
2
by Lusin’s theorem. Now by the StoneWeierstrass, there exists a polynomial
Q
(
x
) such that
k
Q
(
x
)

g
(
x
)
k
sup
<
"
2
.
Then
k
Q

g
k
p
k
Q

g
k
1
=
k
Q

g
k
sup
<
"
2
so
k
f

Q
k
p
k
f

g
k
p
+
k
g

Q
k
p
<
"
2
+
"
2
=
"
.
2.7
Interchanging the Order of Integration
Fubini’s Theorem.
Suppose
f
:
R
2
!
R
where
f

1
(
O
) is Borel in
R
2
for any open set
O
✓
R
. If
Z
R
✓Z
R

f
(
x, y
)

dm
(
x
)
◆
dm
(
y
)
<
1