MATH 467/MAST 669/837 – Measure Theory
Solutions to Assignment #1
1. (
Ex. 6, page 34)
Let
A
be the set of irrational numbers in the interval [0
,
1]. Prove that
m
∗
(
A
) = 1.
Proof.
Let
B
=
Q
∩
[0
,
1], the set of rational numbers in [0
,
1], and note that
B
is countable as a
countable subset of a countable set, namely
Q
. Recall that it was proved in class that the outer
measure of a countable set of real numbers is zero. Hence
m
∗
(
B
) = 0.
Now,
A
∪
B
= [0
,
1] and
A
∩
B
=
∅
. Thus, by the subadditivity property of the outer measure among
other things, 1 = length ([0
,
1]) =
m
∗
([0
,
1])
≤
m
∗
(
A
) +
m
∗
(
B
) =
m
∗
(
A
) + 0, or
m
∗
(
A
)
≥
1. On
the other hand,
A
⊂
[0
,
1], and by the monotonicity property of the outer measure, we also have
m
∗
(
A
)
≥
m
∗
([0
,
1]) = 1. In conclusion,
m
∗
(
A
) = 1.
2. (
Ex. 9, page 34)
Prove that if
m
∗
(
A
) = 0, then
m
∗
(
A
∪
B
) =
m
∗
(
B
).
Proof.
From the subadditivity of
m
∗
,
m
∗
(
A
∪
B
)
≤
m
∗
(
A
) +
m
∗
(
B
) =
m
∗
(
B
). On the other hand,
B
⊂
A
∪
B
and, by the monotonicity of
m
∗
, we also have
m
∗
(
B
)
≤
m
∗
(
A
∪
B
). Hence
m
∗
(
A
∪
B
) =
m
∗
(
B
) is proved.
3. Suppose that
A
⊂
E
⊂
B
, where
A
and
B
are measurable sets of ﬁnite measure. Prove that if
m
(
A
) =
m
(
B
), then
E
is measurable.
Proof.
Since
m
(
B
)
<
∞
, and
A
⊂
B
, we have
m
(
B
\
A
) =
m
(
B
)
−
m
(
A
) = 0
.
Recall that
m
is
m
∗
restricted to
M
, the
σ
-algebra of measurable sets. Thus any subset of
B
\
A
will also have outer
measure zero and, consequently, it will be measurable. In particular,
B
\
E
⊂
B
\
A
, so
B
\
E
∈ M
.
On the other hand, as
B
=
E
∪
(
B
\
E
), we have
E
=
B
\
(
B
\
E
) = [
B
∩
(
B
\
E
)
C
]
∈ M
due to the
fact that
M
is an algebra and that both
B
and
B
\
E
are measurable.
4.
(Ex. 24, page 47)
Show that if
E
1
and
E
2
are measurable, then
m
(
E
1
∪
E
2
) +
m
(
E
1
∩
E
2
) =
m
(
E
1
) +
m
(
E
2
)
Proof.
Suppose that we have two measurable sets
E
1
and
E
2
that they are disjoint. Since
E
1
∩
E
2
=
∅
,
then, by Proposition 13 (page 43) foe example, we have
m
(
E
1
∪
E
2
) +
m
(
E
1
∩
E
2
) =
m
(
E
1
) +
m
(
E
2
) + 0 =