With the claim established, the rest is trivial. Let
U
be open in
R
2
. then
U
=
[
n
∈
N
(
I
n
×
J
n
) for intervals
I
n
, J
n
,
hence
h

1
(
U
) =
[
n
∈
N
h

1
(
I
n
×
J
n
)
is measurable being a countable union of measurable sets.
(b) The easiest way to prove
h
is measurable is to use the characterization by open sets:
h
is measurable if and only
h

1
(
U
) is measurable for every open subset
U
of
R
. Now
h

1
(
U
) =
{
x
∈
R
:
φ
(
f
(
x
)
, g
(
x
))
∈
U
}
=
{
x
∈
R
: (
f
(
x
)
, g
(
x
))
∈
φ

1
(
U
)
}
.
Here we may have to think how to apply part (a) and I admit that the fact that the
h
of this part has nothing to
do with the
h
of part (a) can be confusing. Let us define
H
(
x
) = (
f
(
x
)
, g
(
x
)) so
H
:
R
→
R
2
and
h
=
φ
◦
H
. If
I, J
are open intervals, then
H

1
(
I
×
J
) =
{
x
∈
R
: (
f
(
x
)
, g
(
x
))
∈
I
×
J
}
=
{
x
∈
R
:
f
(
x
)
∈
I
} ∩ {
x
∈
R
:
g
(
x
)
∈
J
}
=
f

1
(
I
)
∩
g

1
(
J
)
so that
H

1
(
I
×
J
) is measurable for all open rectangles
I
×
J
. By part (a),
H

1
(
W
) is measurable for all open
sets
W
⊂
R
2
. Returning now to our open subset
U
of
R
, since
φ
is continuous,
φ

1
(
U
) is open in
R
2
, hence
h

1
(
U
) = (
φ
◦
H
)

1
(
U
) =
H

1
(
φ

1
(
U
)
)
is measurable.
4
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10. Let
{
f
n
}
∞
n
=1
be a sequence of measurable functions on a measurable subset
E
of
R
, converging pointwise to a function
f
. Assume:
Z
E

f
n
 ≤
1
for all
n
∈
N
. Prove:
Z
E

f
 ≤
1
.
A dash of Fatou could help.
Solution.
Because each
f
n
is measurable,

f
n

is nonnegative measurable and Fatou’s lemma applies:
(1)
Z
E
lim inf
n
→∞

f
n
 ≤
lim inf
n
→∞
Z

f
n

.
Since
f
n
→
f
, it follows that lim inf
n
→∞

f
n

= lim
n
→∞

f
n

=

f

.
Since
R

f
n
 ≤
1 for all
n
, we also have
lim inf
n
→∞
R

f
n
 ≤
1. Thus (1) can be expanded to
Z
E

f

=
Z
E
lim inf
n
→∞

f
n
 ≤
lim inf
n
→∞
Z

f
n
 ≤
1
.
We are done.
11. Let
f
n
:
R
→
R
be measurable for all
n
∈
N
. Prove: If
∞
X
n
=1
Z
R

f
n

<
∞
, then
∞
X
n
=1
f
n
converges a.e. In other words, you
have to prove that the set of points
x
∈
R
for which the series of real terms
∞
X
n
=1
f
n
(
x
) does not converge is a null set.
Solution.
By Lebesgue’s Monotone convergence Theorem (series version), since

f
n

is nonnegative, measurable for
all
n
, we have
Z
R
∞
X
n
=1

f
n

=
∞
X
n
=1
Z

f
n

<
∞
.
The function
x
7→
∑
∞
n
=1

f
n
(
x
)

is thus integrable, hence
∑
∞
n
=1

f
n
(
x
)

<
∞
a.e. Since absolute convergence implies
convergence, we are done. (I believe I once mentioned this in class.)
12. Let
f
:
R
→
R
be integrable. Assume
Z
E
f
= 0 for all measurable subsets
E
of
R
. Prove
f
= 0 a.e.
Solution.
We proved in class: If
f
≥
0 is measurable and
R
E
f
= 0, then
f
= 0 a.e. on
E
. Let
A
=
{
x
∈
R
:
f
(
x
)
≥
0
}
,
B
=
{
x
∈
R
:
f
(
x
)
<
0
}
. Then since
A
is measurable, we have
R
A
f
= 0; by the mentioned class result,
f
= 0 a.e.
on
A
since
f
≥
0 on
A
. Similarly
R
b
f
= 0, thus also
R
B
(

f
) = 0; since

f >
0 on
B
we conclude

f
= 0 a.e. on B,
hence
f
= 0 a.e. on
B
. Since
R
=
A
∪
B
and the union of two null sets is a null set,
f
= 0 a.e. on
R
.
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 Spring '11
 Speinklo
 Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Nonmeasurable set

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