Solution.
(a) Let
U
be open in
R
2
.
Claim:
There exists a countable family of open rectangles
I
n
×
J
n
such that
U
=
∞
[
n
=1
I
n
×
J
n
.
I consider proving this claim the essential part of this exercise
, the rest is trivial. To prove the claim
we use the fact that the set
D
=
{
(
x,y
)
∈
Q
×
Q
: (
x,y
)
∈
U
}
is countable, thus can be written in the form
of a sequence, namely
D
=
{
(
x
n
,y
n
) :
n
∈
N
}
. Since
U
is open, for each
n
∈
N
we can ﬁnd
r >
0 such that
B
((
x
n
,y
n
)
,r
)
⊂
U
. Let
r
n
be the largest possible such
r
; that is
r
n
=
sup
{
r >
0 :
B
((
x
n
,y
n
)
,r
)
⊂
U
. We
can assume that
U
6
=
R
2
, otherwise
U
=
R
×
R
and
h

1
(
U
) is measurable. We can also assume
U
6
=
∅
, so
D
is
actually inﬁnite countable. The fact that
∅ 6
=
U
6
=
R
2
implies 0
< r
n
<
∞
. One sees that
B
((
x
n
,y
n
)
,r
n
)
⊂
U
(trivial but requiring an argument). Let
I
n
= (
x
n

(
r
n
/
2
,x
n
+ (
r
n
/
2)),
J
n
= (
y
n

(
r
n
/
2
,y
n
+ (
r
n
/
2)), then
(
x
n
,y
n
)
∈
I
n
×
J
n
⊂
U
, hence
[
n
∈
N
(
I
n
×
J
n
)
⊂
U.
Conversely, let (
x,y
)
∈
U
. Then there is
δ >
0 such that
B
((
x,y
)
,δ
)
⊂
U
. Because
D
is dense in
U
, there is
n
∈
N
such that
d
((
x,y
)
,
(
x
n
,y
n
)) =
p
(
x

x
n
)
2
+ (
y

y
n
)
2
< δ/
4
.
Consider
B
((
x
n
,y
n
)
,
3
δ/
4). If (
u,v
)
∈
B
((
x
n
,y
n
)
,
3
δ/
4), then
d
((
u,v
)
,
(
x,y
))
≤
d
((
u,v
)
,
(
x
n
,y
n
)) +
d
((
x
n
,y
n
)
,
(
x,y
))
<
3
δ
4
+
d
4
=
δ
proving (
u,v
)
∈
B
((
x,y
)
,δ
)
⊂
U
. By the deﬁnition of
r
n
,
r
n
≥
3
δ/
4. Thus

x

x
n
 ≤
d
((
x
n
,y
n
)
,
(
x,y
))
<
δ
4
<
3
δ
8
≤
r
n
2
,
proving that
x
n
∈
I
n
. Similarly one sees that
y
n
∈
J
n
. thus (
x,y
)
∈
I
n
×
J
n
. This proves
[
n
∈
N
(
I
n
×
J
n
)
⊃
U,
thus
[
n
∈
N
(
I
n
×
J
n
) =
U.
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
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 Spring '11
 Speinklo
 Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Nonmeasurable set

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