E
c
∈
S
, it suffices to prove the measurability of sets
E
such that
E
⊂
(
∞
,
0].
Assume thus
E
⊂
(
∞
,
0]. For every subset
A
of
R
, we’ll have sup(
A
∩
E
)
≤
0, thus
μ
*
(
A
∩
E
) = 0. Assume
first that
A
contains some
x
,
x >
0. Then it is a simple exercise to prove that sup
A
= sup(
A
∩
(0
,
∞
))
>
0, thus
μ
*
(
A
) = sup(
A
∩
(0
,
∞
)). We also have
A
∩
(0
,
∞
) =
A
∩
E
c
∩
(0
,
∞
) and
0
<
sup(
A
∩
(0
,
∞
)) = sup(
A
∩
E
c
∩
(0
,
∞
)) = sup(
A
∩
E
c
) =
μ
*
(
A
∩
E
c
);
thus
μ
*
(
A
) =
μ
*
(
A
∩
E
c
) if
a
∩
(0
,
∞
)
6
=
∅
. If
A
∩
(0
,
∞
) =
∅
, then
A
⊂
(
∞
,
0] and
μ
*
(
A
) = 0,
μ
*
(
A
∩
E
c
) = 0
≤
μ
*
(
A
) = 0 , so in all cases we have
μ
*
(
A
) =
μ
*
(
A
∩
E
c
) Thus
μ
*
(
A
) = 0 +
μ
*
(
A
) =
μ
*
(
A
∩
E
) +
μ
*
(
A
∩
E
c
)
proving
E
is measurable.
To conclude, we have to see that if
E /
∈
S
, then
E
is not measurable. Assume
E /
∈
S
. Then
E
6⊂
(
∞
,
0], thus
there is
x
∈
E
,
x >
0. But
E
6
=
F
∪
(0
,
∞
) for a set
F
⊂
(
∞
,
0]; i.e.,
E
6⊃
(0
,
∞
), hence there is
y >
0,
y /
∈
E
.
Let
A
=
{
x, y
}
. Then
A
∩
E
=
{
x
}
, hence
μ
*
(
A
∩
E
) = sup(
A
∩
E
) =
x
,
μ
*
(
A
∩
E
c
) = sup(
A
∩
E
)
=
y
, and
μ
*
(
A
∩
E
) +
μ
*
(
A
∩
E
c
) =
x
+
y >
max(
x, y
) =
μ
*
(
A
). Thus
E /
∈
S
.
(c)
μ
*
(
A
) =
1
if
A
is uncountable,
0
if
A
is countable (or finite).
Solution.
The empty set is finite thus
μ
*
(
∅
) = 0;
OM1
holds. If
A
⊂
B
, if
B
is uncountable, then
μ
*
(
A
)
≤
1 =
μ
*
(
B
); if
B
is countable, then so is
A
and
μ
*
(
A
) =
μ
*
(
B
) = 0.
We see
OM2
holds.
If
A
n
⊂
R
for
all
n
∈
N
, if
A
n
is uncountable for some
n
∈
N
, then
∑
∞
n
=1
μ
*
(
A
n
)
≥
μ
*
(
A
n
) = 1
≥
μ
*
(
S
∞
n
=1
A
n
).
On the
other hand, if all
A
n
’s are countable, so is their union (a countable union of countable sets is countable) and
∑
∞
n
=1
μ
*
(
A
n
)
≥
μ
*
(
A
n
) = 0 =
μ
*
(
S
∞
n
=1
A
n
).
Let
S
consist of all sets that are either countable or that have countable complement. We
claim
this is precisely
the
σ
algebra of measurable sets. Let
E
∈
S
. Suppose
A
⊂
R
. Assume first
A
is countable. Then both
A
∩
E
and
A
∩
E
c
are countable and
μ
*
(
A
) = 0 =
μ
*
(
A
∩
E
) +
μ
*
(
A
∩
E
c
)
.
On the other hand if
A
is uncountable then precisely one of
A
∩
E, A
∩
E
c
is uncountable, the other one countable.
In fact, one has to be uncountable, because
A
is uncountable. But one has to be countable, because one of
E, E
c
is countable. Thus
μ
*
(
A
) = 1 = 1 + 0 =
μ
*
(
A
∩
E
) +
μ
*
(
A
∩
E
c
)
.
This proves that all sets of
S
are measurable. Assume now that
E /
∈
S
. Since
R
is uncountable, the only way this
is possible is if both
E
and its complement are uncountable. Then
1 =
μ
*
(
R
)
<
1 + 1 =
μ
*
(
R
∩
E
) +
μ
*
(
R
∩
E
c
)
,
showing that
E
is not measurable.
2. Prove: If
E, F
are measurable, if
E
⊂
F
and if
m
(
E
)
<
∞
, then
m
(
F
\
E
) =
m
(
F
)

m
(
E
). Why do we need to assume
m
(
E
)
<
∞
?
Proof.
If
E, F
are measurable, so is
E
\
F
and (in case
E
⊂
F
)
F
=
E
∪
(
F
\
E
).
Since this is a disjoint union,
m
(
F
) =
m
(
E
) +
m
(
F
\
E
). If
m
(
E
)
<
∞
we can subtract
m
(
E
) to get
m
(
F
\
E
) =
m
(
F
)

m
(
E
). If
m
(
E
) =
∞
, so is
m
(
F
) and
m
(
F
)

m
(
E
) is undefined. The set
F
\
E
could be a set having finite or infinite measure.
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 Spring '11
 Speinklo
 Sets, Empty set, measure, Basic concepts in set theory, Lebesgue measure

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