Math 447
Solutions to Assignment 1
Spring 2009
Exercise 16.4 on page 271
Given any subset of
E
of
R
and any
h
∈
R
, show that
m
*
(
E
+
h
) =
m
*
(
E
), where
E
+
h
=
{
x
+
h
:
x
∈
E
}
.
Solution
The statement, which is part (iii) of Proposition 16.2 on page 269, says that
Lebesgue outer measure is translation invariant. Since outer measure is deﬁned in terms of
coverings by intervals, and since translation preserves the lengths of intervals, the statement
is supposed to be intuitively clear. Here is a formal argument.
If
E
is covered by a sequence of intervals
I
n
, then the intervals
I
n
+
h
have the same
total length and cover the set
E
+
h
. Hence
m
*
(
E
+
h
)
≤
m
*
(
E
). Conversely, if a sequence
of intervals
J
n
covers the set
E
+
h
, then the intervals
J
n

h
have the same total length
and cover the set
E
. Hence
m
*
(
E
)
≤
m
*
(
E
+
h
). Putting the two inequalities together
shows that
m
*
(
E
) =
m
*
(
E
+
h
).
Exercise 16.8 on page 271
Given
δ >
0, show that
m
*
(
E
) = inf
∑
∞
n
=1
‘
(
I
n
) where the
inﬁmum is taken over all coverings of
E
by sequences of intervals (
I
n
), where each
I
n
has
diameter less than
δ
.
Solution
Notice that the indicated property diﬀers from the deﬁnition of outer measure
at the top of page 269 only in the requirement that the intervals
I
n
have uniformly small
length. Therefore it suﬃces to show that every countable collection of intervals (
I
n
) can
be replaced by another countable collection of intervals (
J
n
) such that
S
n
I
n
=
S
n
J
n
and
∑
n
‘
(
I
n
) =
∑
n
‘
(
J
n
) and
‘
(
J
n
)
< δ
for every
n
.
Since a countable collection of ﬁnite sets is countable (a special case of Theorem 2.6 on
page 21), it suﬃces to show that a single interval can be partitioned into a ﬁnite union of
subintervals each of length less than
δ
. That such a partition exists should be obvious, but
here is a formal argument.
One can construct the partition concretely through repeated bisection (for example). If
the interval has length
‘
, then dividing the interval at the midpoint gives 2 subintervals
each of length
‘/
2; bisecting again gives 2
2
subintervals each of length
‘/
2
2
; and so on.
After
k
steps there are 2
k
subintervals each of length
‘/
2
k
. The construction can be stopped
as soon as
‘/
2
k
< δ
, that is,
k >
log
2
(
‘/δ
).
Notice that since the length of an interval, the length of its closure, and the length of its
interior are all equal, it is not necessary to worry in the preceding argument about which
subintervals contain the partition points.
Exercise 16.9 on page 271
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 Spring '08
 Staff
 Math, dx, Dr. Boas

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