2
INTRODUCTION TO LEBESGUE INTEGRATION
18
•
k
f
+
g
k
p
k
f
k
p
+
k
g
k
p
.
Note that if
f
= 0 a.e. then
k
f
k
p
= 0, so
k
f
k
p
is not a norm. Moreover if
f
=
g
a.e. , then
k
f
k
p
=
k
g
k
p
. Proof:
Suppose
f
=
g
except on some
A
with
m
(
A
) = 0. Then
k
f
k
p
= (
R
A

f

p
+
R
A
C

g

p
)
1
/p
= (
R
A

g

p
+
R
A
C

g

p
)
1
/p
=
k
g
k
p
.
Definition.
First define the equivalence relation
f
⇠
g
if
f
=
g
a.e. .
Define
L
p
(
E
) to be the set of equivalence
classes under
⇠
where
k
f
k
p
<
1
.
Example.
Suppose
E
= [
a, b
]. Then
C
[
a, b
]
✓
L
p
[
a, b
]. Indeed suppose
f
2
C
[
a, b
], so
9
M
such that

f

M
on
[
a, b
], so
R
[
a,b
]

f

p
M
p
(
b

a
)
<
1
. Moreover, all bounded measurable functions are in
L
P
[
a, b
].
Example.
C
(
R
)
6✓
L
p
(
R
) since, for example,
f
(
x
) = 1
/
2
L
p
(
R
).
Proposition.
L
p
(
E
) is a vector space.
Proof.
Let
f, g
2
L
p
(
E
) and
c
be a scalar. Then
cf
is measurable and
k
cf
k
p
=
✓Z
E

cf

p
◆
1
/p
=
✓Z
E

c

p

f

p
◆
1
/p
=

c

✓Z
E

f

p
◆
1
/p
=

c

k
f
k
p
<
1
.
To check
f
+
g
2
L
p
(
E
), we use the inequality: for
a, b
≥
0,
a
p
+
b
p
2
≥
(
a
+
b
2
)
p
. Then
Z
E
k
f
+
g
k
p
2
p
Z
E
✓

f

+

g

2
◆
p
2
p
Z
E

f

p
+

g

p
2
2
p

1
(
k
f
k
p
p
+
k
g
k
p
p
)
<
1
.
Thus
f
+
g
2
L
p
(
E
), so
L
p
is a vector space.
Proposition.
k
f
k
p
= 0 only if
f
= 0 a.e. , i.e.
f
is the zero element in
L
p
(
E
).
Proof.
Let
E
n
=
{
x
:

f
(
x
)

≥
1
n
}
.
Then
R
E
n

f

p
≥
1
n
m
(
E
n
).
But
R
E
n

f

p
R
E

f

p
= 0, so
m
(
E
n
) = 0.
Let
F
=
S
1
n
=1
E
n
=
{
x
:
f
(
x
)
6
= 0
}
, so
m
(
F
)
P
1
n
=1
E
n
= 0, so
f
= 0 a.e. . Thus
k
[
f
]
k
p
= 0
()
[
f
] = [0].
Definition.
Let
f
be a measurable function defined on
E
. Define the
1
norm
(or
essential supremum norm
)
by
k
f
k
L
1
(
E
)
=
k
f
k
1
=
inf
A
meas
m
(
E
\
A
)=0
{
sup

f
(
x
)

:
x
2
A
}
.
In particular, if
E
=
R
, then
k
f
k
L
1
(
R
)
= inf
m
(
A
C
)=0
{
sup

f
(
x
)

:
x
2
A
}
.
Note.
It is always the case that
k
f
k
L
1
(
E
)
sup
{
f
(
x
)

:
x
2
E
}
, since
E
is one of the sets
A
. It can be less.
Example.
1. If
f
= 0 a.e. take
A
=
{
x
:
f
(
x
) = 0
}
. Then
m
(
A
C
) = 0, and sup
{
f
(
x
)

:
x
2
A
}
= 0, so inf
m
(
A
C
)=0
{
sup

f
(
x
)

:
x
2
A
}
= 0.
2. Suppose
f
is continuous on
R
. We claim that
k
f
k
R
= sup
{
f
(
x
)

:
x
2
R
}
.
Proof.
Suppose
M <
sup

f

. Then there is some
x
0
where

f
(
x
0
)

≥
M
1
> M
. By continuity of
f
, there exists
δ
>
0 such that if
x
2
(
x
0

δ
, x
0
+
δ
), then

f
(
x
)

> M
.
Suppose
k
f
k
L
1
(
R
)
< M
and
m
(
A
C
) = 0.
Then
A
C
6◆
(
x
0

δ
, x
0
+
δ
) because
m
(
x
0

δ
, x
0
+
δ
) = 2
δ
>
0. Thus
A
\
(
x
0

δ
, x
0
+
δ
)
6
=
;
. Pick
y
2
(
x
0

δ
, x
0
+
δ
).
But then

f
(
y
)

> M
since
y
2
(
x
0

δ
, x
0
+
δ
). Thus sup
{
f
(
x
)

:
x
2
A
}
> M
. Since
A
was an arbitrary set
with
m
(
A
C
) = 0,
R
m
(
A
C
)=0
{
sup

f
(
x
)

:
x
2
A
}
≥
M
, a contradiction.