6. Let
{
f
n
}
∞
n
=1
be a sequence of real valued functions on a subset
E
of
R
.
Define the following sets: For each triple
(
n, m, k
)
∈
N
3
define
F
n,m,k
=
{
x
∈
E
:

f
n
(
x
)

f
m
(
x
)

<
1
k
}
.
For each pair (
N, k
)
∈
N
2
define
G
N,k
=
\
n
≥
N,m
≥
N
F
m,n,k
.
For each
k
∈
N
, define
H
k
=
∞
[
N
=1
G
N,k
. Finally, define
Z
=
∞
\
k
=1
H
k
.
Prove:
x
∈
Z
if and only if the sequence of real numbers
{
f
n
(
x
)
}
converges.
Solution.
Assume first
x
∈
Z
. We prove
{
f
n
(
x
)
}
is a Cauchy sequence of real numbers. For this let
² >
0 be given.
Let
k
∈
N
be such that 1
/k < ²
. Since
x
∈
Z
,
x
∈
H
k
by the definition of
Z
. By the definition of
H
k
, there is
N
∈
N
such that
x
∈
G
N,k
. By the definition of
G
N,k
,
x
∈
F
n,m,k
for all
n, m
≥
N
, thus

f
n
(
x
)

f
m
(
x
)

<
1
/k < ²
if
n, m
≥
N
.
It follows the sequence is Cauchy, hence converges. Conversely, let
x
∈
R
be such that
{
f
n
(
x
)
}
converges. Then it is
Cauchy and for every
k
∈
N
there is
N
∈
N
such that

f
n
(
x
)

f
m
(
x
)

<
1
/k
if
n, m
≥
N
, thus there is
N
∈
N
such
that
x
∈
F
n,m,k
for
n, m
≥
N
, hence
x
∈
G
N,k
. This proves that for every
k
∈
N
, there is
N
∈
N
such that
x
∈
G
N,k
,
thus for every
k
∈
N
,
x
∈
S
∞
N
=1
G
N,k
=
H
k
. Thus
x
∈
H
k
for all
k
, hence
x
∈
Z
.
2
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7. Let
E
be a measurable subset of
R
and let
f
:
E
→
[0
,
∞
] be measurable. Prove that the function 1
/f
defined by
(1
/f
)(
x
) =
1
/f
(
x
)
if 0
< f
(
x
)
<
∞
∞
if
f
(
x
) = 0
0
if
f
(
x
) =
∞
for
x
∈
E
, is measurable.
Solution.
I’ll present two solutions.
First solution,
the obvious one.
Let
α
∈
R
.
It suffices to prove that
{
x
∈
E
: 1
/f
(
x
)
> α
}
is measurable.
We
have to distinguish three cases:
α >
0,
α
= 0,
α <
0. If
α >
0, then 1
/f
(
x
)
> α
if and only if either
f
(
x
) = 0 (then
1
/f
(
x
) =
∞
) or if
f
(
x
)
<
1
/α
. But since 1
/α >
0, the set
{
f
= 0
} ⊂ {
f <
1
/α
}
. Thus
{
x
∈
E
: 1
/f
(
x
)
> α
}
=
{
x
∈
E
:
f
(
x
)
<
1
/α
}
,
the latter set being measurable because
f
is measurable, it follows that
{
x
∈
E
: 1
/f
(
x
)
> α
}
is measurable if
α >
0.
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
(
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 Spring '11
 Speinklo
 Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Nonmeasurable set

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