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Royden, Real Analysis 3rd ed.
Chapter 3
Problem 34
Assume
E
1
,E
2
,...
are mutually disjointed, it is easy to
check
¯
¯
¯
¯
¯
n
[
i
=1
E
i
¯
¯
¯
¯
¯
=
n
X
i
=1

E
i

(1)
f
rst for
n
=2
and then for
n
=3
,
4
,...
and so on.
We still need to verify
¯
¯
¯
¯
¯
∞
[
i
=1
E
i
¯
¯
¯
¯
¯
=
∞
X
i
=1

E
i

.
(2)
It is easy to see that
(2)
holds when the RHS is
∞
.
Otherwise, all
E
i
=
∅
for
i
larger than some integer, say,
n.
Then
(2)
is reduced to
(1)
and we are
done.
Problem 39
By assumption
m
∗
A
=
m
∗
(
A
∩
E
)+
m
∗
(
A
∩
E
c
)
for any
A.
Us
ingProb
lem37andrep
lac
ing
A
above by
A
−
y,
m
∗
A
=
m
∗
(
A
−
y
)
=
m
∗
((
A
−
y
)
∩
E
)+
m
∗
((
A
−
y
)
∩
E
c
)
=
m
∗
(((
A
−
y
)+
y
)
∩
(
E
+
y
)) +
m
∗
(((
A
−
y
)+
y
)
∩
(
E
c
+
y
))
=
m
∗
(
A
∩
(
E
+
y
)) +
m
∗
(
A
∩
(
E
+
y
)
c
)
.
Hence
E
+
y
is measurable.
Problem 311
Consider
E
n
=(
n,
∞
)
.
Then
∩
∞
n
=1
E
n
=
∅
and
mE
n
=
∞
.
Problem 312
By Proposition 2, it is enough to show
m
∗
Ã
A
∩
∞
[
i
=1
E
i
!
≥
∞
X
i
=1
m
∗
(
A
∩
E
i
)
.
By monotonicity and Lemma 9,
m
∗
Ã
A
∩
∞
[
i
=1
E
i
!
≥
m
∗
Ã
A
∩
n
[
i
=1
E
i
!
≥
n
X
i
=1
m
∗
(
A
∩
E
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View Full DocumentNow let
n
→∞
.
Problem 313
a. (i), (ii) and (iv) are equivalent.
(i)
⇒
(ii). Applying Proposition 5 to
E
n
=
E
∩
[
n,n
+1]
,
we can
f
nd an
open set
O
n
such that
O
n
⊇
E
n
and
m
∗
(
O
n
)
≤
m
∗
(
E
n
)+
ε
3
·
2
n
.
Since
m
∗
E
n
≤
m
∗
[
n,n
+1]=1
and both
O
n
,E
n
are measurable,
m
∗
(
O
n
\
E
n
)=
m
∗
O
n
−
m
∗
E
n
≤
ε
3
·
2
n
,
(1)
by Lemma 9. Consider
O
=
∪
∞
n
=
−∞
O
n
,
wh
ichisanopense
t
. Then
O
⊇
∪
∞
n
=
−∞
E
n
=
E.
By
(1)
and Problem 312,
m
∗
(
O
\
E
)=
m
∗
Ã
∞
[
n
=
−∞
(
O
n
\
E
n
)
!
≤
∞
X
n
=
−∞
m
∗
(
O
n
\
E
n
)
≤
∞
X
n
=
−∞
ε
3
·
2
n
=
ε
.
(2)
(ii)
⇒
(iv). Let
ε
=1
/n
in (2) and we have open set
O
n
⊇
E
and
m
∗
(
O
n
\
E
)
≤
1
/n.
Let
G
=
∩
∞
n
=1
O
n
.
Then
G
⊇
E
and
0
≤
m
∗
(
G
\
E
)
≤
m
∗
(
O
n
\
E
)
≤
1
n
→
0
,
as
n
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