Math 447
Solutions to Assignment 3
Spring 2009
Exercise 16.58 on page 284
Suppose that
m
*
(
E
)
<
∞
.
Prove that
E
is measurable
if and only if, for every
ε >
0, there is a finite union of bounded intervals
A
such that
m
*
(
E
4
A
)
< ε
(where
E
4
A
is the symmetric difference of
E
and
A
).
Solution
First suppose that
E
is measurable, and fix a positive
ε
.
For each positive
integer
n
, let
E
n
denote the set
E
∩
([
n

1
, n
)
∪
[

n,

n
+ 1)), which is a measurable set
by Lemma 16.14 on page 278. Theorem 16.18 on page 280 implies that
∞
X
n
=1
m
*
(
E
n
) =
m
*
(
E
)
<
∞
.
The convergence of this infinite series of nonnegative terms means that there is a positive
integer
N
such that
∑
∞
n
=
N
m
*
(
E
n
)
< ε/
3.
Let
E
0
denote the bounded measurable set
E
\
S
∞
n
=
N
E
n
. By part (ii) of Theorem 16.21
on page 283, there is an open set
G
containing
E
0
such that
m
*
(
G
\
E
0
)
< ε/
3. There is
no loss of generality in supposing that
G
is a bounded set [simply intersect
G
with the
interval (

N, N
)]. The open set
G
is then a countable union of bounded open intervals of
finite total length. Split this collection of intervals into two subcollections
A
and
B
such
that
A
consists of finitely many intervals, and the intervals in
B
have total length less
than
ε/
3. Points in the symmetric difference
E
4
A
lie either in
S
∞
n
=
N
E
n
or in
B
or in
G
\
E
0
. Since each of these three sets has outer measure less than
ε/
3, the subadditivity
property of outer measure implies that
m
*
(
E
4
A
)
< ε
.
Conversely, fix a positive
ε
, and suppose
A
is a finite union of bounded intervals such
that
m
*
(
E
4
A
)
< ε
. Slightly enlarging each of the intervals produces a finite collection
A
0
of bounded open intervals such that
m
*
(
E
4
A
0
)
<
2
ε
. In particular,
m
*
(
E
\
A
0
)
<
2
ε
.
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 Spring '08
 Staff
 Math, Empty set, Open set, Topological space, measurable set, Dr. Boas

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