Assume next
α
= 0. We have 1
/f
(
x
)
>
0 if and only if
{
f
(
x
)
<
∞}
and this last set is measurable because
f
is
measurable
*
. Finally, assume
α <
0. Then
{
x
∈
E
: 1
/f
(
x
)
> α
}
=
E
is measurable.
Second solution.
Deﬁne
f
n
(
x
) = max(
n,f
(
x
) +
1
n
)
for
x
∈
E
,
n
= 1
,
2
,
3
...
. Because
f
is measurable, all functions
f
n
are measurable. Moreover,
f
n
takes values in
the interval [1
/n,n
], in which
y
7→
1
/y
is continuous. It follows that 1
/f
n
is measurable for every
n
= 1
,
2
,...
, being
the composition of a measurable function with a continuous one. It is clear (or trivial to prove) that
{
f
n
}
converges
pointwise to
f
and that
{
1
/f
n
}
converges pointwise to 1
.f
, hence 1
/f
is measurable.
Comment
For some credit I needed to see at least that if
α >
0 then 1
/f > α
is equivalent to
f <
1
/α
.
8. For this exercise one must, of course, assume (as one usually does) the existence of some nonmeasurable set. It was
proved in class that if
f
n
:
R
→
R
is measurable for all
n
∈
N
, then sup
n
∈
N
f
, the function deﬁned by
(sup
n
∈
N
f
)(
x
) = sup
{
f
n
(
x
) :
n
∈
N
}
is measurable. Assume now that
A
is an arbitrary set and for each
α
∈
A
,
f
α
:
R
→
R
is measurable. Is it always true
that sup
α
∈
A
f
α
, the function deﬁned by
(sup
α
∈
A
f
α
)(
x
) = sup
{
f
α
(
x
) :
α
∈
A
}
is measurable? Justify your answer.
Solution.
The answer is
NO
. For example, let
A
be a nonmeasurable subset of
R
and for
α
∈
A
deﬁne
f
α
:
R
→
R
by
f
α
(
x
) =
‰
1 if
x
=
α
,
0 otherwise.
Then
f
α
= 0 a.e., hence measurable for each
α
∈
A
. However, sup
α
f
α
=
χ
A
is not measurable.
9.
(a) Let
h
:
R
→
R
2
having the following property: If
I,J
are open intervals in
R
, then
h

1
(
I
×
J
) =
{
x
∈
R
:
h
(
x
)
∈
I
×
J
}
is measurable. Prove that the inverse image under
h
of every open subset of
R
2
is measurable. That is, prove that
if
W
is open in
R
2
, then
h

1
(
W
) is measurable in
R
.
Three hints:
Not every open subset of
R
2
is a rectangle.
This is sort of obvious, but people under the pressure of exams sometimes lose their minds and begin to assume
nonsense just because it seems to help. The second hint is that one might consider proving that every open set in
the plane is a countable union of rectangles. A third hint is, do not try to show it is a union of pairwise disjoint
rectangles, that’s clearly impossible.
*
For measurability it suﬃces to have
{
f < α
}
measurable for all
α
∈
R
; one gets
{
f <
∞}
measurable as an easy consequence; for example
because
{
f <
∞}
=
S
n
{
f < n
}
.
3
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View Full Document(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.
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 Spring '11
 Speinklo
 Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Nonmeasurable set

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