Serge Ballif
MATH 501 Homework 5
October 31, 2007
In all problems (
X,
M
, μ
) is a measure space
1.
Let
f
:
X
×
[
a, b
]
→
where
[
a, b
]
is a finite interval. Assume that the function
x
7→
f
(
x, y
)
is integrable for
every
y
∈
[
a, b
]
. Assume also that
∂
y
f
(
x, y
)
exists for every
(
x, y
)
and there exists a function
g
∈
L
1
such that

∂
y
f
(
x, y
)
 ≤
g
(
x
)
for all
(
x, y
)
. Show that the function
x
7→
∂
y
f
(
x, y
)
is measurable for every
y
and
∂
y
Z
f
(
x, y
)
dμ
(
x
) =
Z
∂
y
f
(
x, y
)
dμ
(
x
)
.
For a fixed
y
∈
[
a, b
] define
h
n
(
x, y
) :=
f
(
x,y
+1
/n
)

f
(
x,y
)
1
/n
for
n
∈
.
Then
h
n
is measurable as a multiple of a difference of integrable functions.
Moreover,
lim
n
→∞
h
n
(
x
) exists and is equal to
∂
y
f
(
x, y
). The inequality

h
n
(
x
)

=

∂
y
f
(
x, y
)
 ≤
g
(
x
) allows us to invoke the dominated convergence theorem as follows.
∂
y
Z
f
(
x, y
) = lim
h
→
0
R
f
(
x, y
+
h
)

R
f
(
x, y
)
h
= lim
n
→∞
Z
h
n
=
Z
lim
n
→∞
h
n
=
Z
lim
h
→
0
f
(
x, y
+
h
)

f
(
x, y
)
h
=
Z
∂
y
f
(
x, y
)
.
2.
Prove:
(a)
k
f
n

f
k
∞
→
0
if and only if there exists
E
∈ M
such that
μ
(
E
c
) = 0
and
f
n
→
f
uniformly on
E
.
(b)
(
L
∞
,
k · k
∞
)
is a Banach space.
(a) (
⇒
) Suppose that
k
f
n

f
k
∞
→
0. Then for any
>
0, there exists
N
such
that for
n
≥
N
,
k
f
n

f
k
∞
= inf
{
a
≥
0 :
μ
(
{
x
∈
X
:

f
n
(
x
)

f
(
x
)

> a
}
) =
0
}
<
. Then for each
m
∈
, there exists
N
such that whenever
n > N
the
set
E
m
:=
{
x
∈
X
:

f
n
(
x
)

f
(
x
)

>
1
m
}
has measure 0. Define
E
c
=
S
∞
m
=1
E
m
.
Since
E
m
%
E
c
, we know by continuity from below that
m
(
E
c
) = 0.
To see
that
f
n
→
f
uniformly on
E
we note that for
x
∈
E
, there exists
N
such that
for
n > N
,

f
n
(
x
)

f
(
x
)

<
(else