Math 245A
Homework 6
Brett Hemenway
November 30, 2005
46. Let
X
=
Y
= [0
,
1],
M
=
N
=
B
[0
,
1]
,
μ
= Lebesgue measure, and
ν
= counting measure. Let
D
=
{
(
x, x
) :
x
∈
[0
,
1]
} ⊂
X
×
Y
. Then
D
x
=
{
(
x, x
)
}
and
D
y
=
{
(
y, y
)
}
, thus
χ
D
dμdν
=
0
dν
= 0
and
χ
D
dνdμ
=
1
dμ
= 1
Let
{
A
i
×
B
i
}
be a sequence of rectangles in
M × N
such that
D
⊂
∞
i
=1
A
i
×
B
i
.
For each
x
∈
[0
,
1],
x
∈
A
j
×
B
j
for some
j
, then
x
∈
A
j
∩
B
j
. So [0
,
1]
⊂
∞
i
=1
(
A
i
∩
B
i
). Since
μ
([0
,
1]) = 1, we have
that
μ
(
A
j
∩
B
j
)
>
0 for some
j
.
Then
μ
(
A
j
)
>
0, and
μ
(
B
j
)
>
0,
so
ν
(
B
j
) =
∞
. Thus
∑
∞
i
=1
μ
×
ν
(
A
i
×
B
i
) =
∞
for any covering set
{
A
i
×
B
i
}
. Thus
μ
×
ν
(
D
) =
∞
, so have
χ
D
d
(
μ
×
ν
) =
μ
×
ν
(
D
) =
∞
49.
a. Let
E
∈ M × N
and
μ
×
ν
(
E
) = 0. Then by Fubini’s theorem
0 =
μ
×
ν
(
E
) =
χ
E
dμ
×
ν
=
χ
E
x
dν
=
χ
E
y
dμ
So
ν
(
E
x
) =
μ
(
E
y
) = 0.
b. Let
f
be
L
measurable and
f
= 0
λ
a.e. Let
A
=
{
(
x, y
)
∈
X
×
Y
:
f
(
x, y
) = 0
}
1
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We have
λ
(
A
) = 0, so there exists a set
E
∈ M × N
with
A
⊂
E
and
μ
×
ν
(
E
) = 0. Then by Fubini’s theorem
0 =
fdμ
×
ν
=
f
x
dν
=
f
y
dμ
Thus
ν
(
E
x
) =
μ
(
E
y
) = 0. But
A
x
⊂
E
x
and
A
y
⊂
E
y
, so since
μ
and
ν
are complete,
A
x
,
A
y
are measurable and have measure
zero.
To prove Theorem 2.39, Let
f
be a
L
measurable function. Then by
Theorem 2.12, there is a
μ
×
ν
measurable function
g
such that
f
=
g
λ
a.e. If (i)
f
≥
0, or (ii)
f
∈
L
1
(
λ
), then applying lemma (b) to
f

g
we have that
f
x
is
N
measurable for a.e.
x
and
f
y
is
M
measurable
for a.e.
y
.
then by Fubini’s theorem
x
→
g
x
dν
and
y
→
f
y
dμ
are measurable, and by lemma (b)
g
x
dν
=
f
x
dν
for a.e.
x
and
g
y
dμ
=
f
y
dμ
for a.e.
y
.
So
x
→
f
x
dν
and
x
→
f
y
dμ
are
measurable. If we are in case (ii), then
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