(b) Let
E
be a measurable subset of
R
and let
f, g
:
E
→
R
be measurable. Let
φ
:
R
2
→
R
be continuous. Prove:
the function
h
:
E
→
R
defined by
h
(
x
) =
φ
(
f
(
x
)
, g
(
x
)) is measurable.
Hint:
Use Part (a).
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.