(d) Chapter 3, # 8 (p. 58).
Solution.
Since
A
⊂
A
∪
B
, we have
m
*
(
A
)
≤
m
*
(
A
∪
B
), by monotonicity. By
σ
subadditivity (which implies
finite subadditivity in conjunction with
m
*
(
∅
) = 0),
m
*
(
A
∪
B
)
≤
m
*
(
A
) +
m
*
(
B
) =
m
*
(
A
).
(e) Chapter 3, # 9 (p. 64). This exercise is an immediate consequence of Royden’s Exercise 7.
Solution.
Assume
E
⊂
R
is measurable and let
x
∈
R
.
It is easy (basically trivial) to verify that the map
A
7→
A
+
x
from subsets of
R
to subsets of
R
is a Boolean isomorphism; i.e., it preserves all set operations so that
for all
A
⊂
R
A
∩
(
E
+
x
) = [(
A

x
)
∩
E
] +
x,
A
∩
(
E
+
x
)
c
=
A
∩
(
E
c
+
x
) = [(
A

x
)
∩
E
c
] +
x,
thus, because
m
*
is translation invariant,
m
*
(
A
)
=
m
*
(
A

x
) =
m
*
((
A

x
)
∩
E
) +
m
*
((
A

x
)
∩
E
c
) =
m
*
([(
A

x
)
∩
E
] +
x
) +
m
*
([(
A

x
)
∩
E
c
] +
x
)
=
m
*
(
A
∩
(
E
+
x
) +
m
*
(
A
∩
(
E
+
x
)
c
)
proving
E
+
x
measurable.
3
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(f) Chapter 3, # 10 (p. 64). (Not graded)
Solution.
Since
E
=
E
\
F
∪
(
E
∩
F
) and (
E
\
F
)
∩
(
E
∩
F
=
∅
,
m
(
E
) =
m
(
E
\
F
) +
m
(
E
∩
F
)
.
Similarly
m
(
F
) =
m
(
F
\
E
) +
m
(
E
∩
F
)
.
Since the sets
E
\
F, F
\
E, E
∩
F
are pairwise disjoint,
m
(
E
)+
m
(
F
) =
m
(
F
\
E
)+
m
(
E
\
F
)+2
m
(
E
∩
F
) =
m
(
F
\
E
∪
E
\
F
∪
(
E
∩
F
))+
m
(
E
∩
F
) =
m
(
E
∪
F
)+
m
(
E
∩
F
)
.
(g) Chapter 3, # 11 (p. 64).
Solution.
Let (for example)
E
n
= (
n,
∞
) for
n
= 1
,
2
, . . .
.
Then
E
1
⊃
E
2
⊃ · · ·
,
T
∞
n
=1
E
n
=
∅
, but
lim
n
→∞
m
(
E
n
) =
∞ 6
= 0 =
m
(
∅
).
(h) Chapter 3, # 14 a (p. 64). There are several ways of doing this. If you look at the construction I gave in the notes
on the Cantor set, it should be easy to determine
m
(
C
n
) for all
n
. Use all properties of Lebesgue measurable sets
that have been proved so far (in Royden). That is, use that intervals, open, closed, semiopen, semiclosed, are
measurable and that a measure is a measure; that is, it is
σ
additive, hence also additive. Then the result can be
obtained as a consequence of Proposition 14 in Royden.
Solution.
This is essentially done in my notes on the Cantor set.
(i) Chapter 3, # 14 b (p. 64). This is a bit harder. Let’s call it optional, for now.
4.
??
Let
A
be the family of all subsets of
R
that can be obtained as a finite union of sets of one of the following forms:
I.
(
a, b
]
for some
a, b,
∞ ≤
a < b <
∞
,
II.
(
a,
∞
)
for some
a,
∞ ≤
a.
The sets in the list above include
R
, which is of the second form with
a
=
∞
. “Finite unions” include unions with
no sets at all, putting the empty set into the mix, and unions involving only one set. So
A
contains
R
,
∅
, all intervals
of the form (
a, b
], all intervals of the form (
a,
∞
), all unions of two sets of the type just mentioned, all unions of three
such sets, etc.
Prove
A
is an algebra.
Hint:
The only difficult part (i.e., nontrivial part) is to see that
A
∈ A
implies
A
c
∈ A
. For this purpose you should
prove first: If
A, B
∈ A
, then
A
∩
B
∈ A
. Once you have this, you can proceed by induction. Let us call the sets listed
in
I, II
basic sets. Go by induction on
n
, proving: If
A
is a union of
n
basic sets, then
A
c
∈ A
. But if you find a better
way, by all means use it.
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 Spring '11
 Speinklo
 Sets, Empty set, measure, Basic concepts in set theory, Lebesgue measure

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