Solution.
We have to see
f
∼
f
,
f
∼
g
⇒
g
∼
f
and
f
∼
g,g
∼
h
⇒
f
∼
h
, for all functions
f,g,h
:
R
→
R
. Since
the empty set is a null set, it is clear that
f
∼
f
for all
f
:
R
→
R
. Assume
f,g
:
R
→
R
and
f
∼
g
. Then the set
E
=
{
x
:
f
(
x
)
6
=
g
(
x
)
}
is a null set; this is of course the same set as
{
x
:
g
(
x
)
6
=
f
(
x
)
}
, thus
g
=
f
a.e. and
g
∼
f
.
Alternatively, the deﬁnition of
f
=
g
a.e. is symmetric in
f,g
, thus
f
=
g
a.e. if and only if
g
=
f
a.e.
. In any case
f
∼
g
⇒
g
∼
f
. Finally assume
f
∼
g,g
∼
h
. The sets
E
=
{
x
:
f
(
x
)
6
=
g
(
x
)
}
,
F
=
{
x
:
g
(
x
)
6
=
h
(
x
)
}
are then null
sets. Clearly
f
(
x
)
6
=
h
(
x
) implies that either
x
∈
E
or
x
∈
F
, thus
{
x
:
f
(
x
)
6
=
h
(
x
)
⊂
E
∪
F
, thus a null set. Thus
f
=
h
a,e. and
f
∼
h
.
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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View Full Document7. 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
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 Spring '11
 Speinklo
 Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Nonmeasurable set

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