(a) Chapter 3, # 4 (p. 55) (This is a very easy exercise, don’t go overboard with it. You may use the obvious, namely
that if you have two disjoint sets
A, B
one with
n
elements, the other one with
m
elements, then
A
∪
B
has
n
+
m
elements. You may also use that if you have a sequence
{
A
n
}
of pairwise disjoint sets, then the union is a finite set
if and only if all sets are finite
and
all but a finite number of the sets
A
n
are empty. The main point of assigning
this exercise is so we can talk of the
counting measure
later on.
(This exercise wasn’t graded. Everything is quite basic and I’ll skip a proof.
(b) Chapter 3, # 5 (p. 58).
Solution.
Assume
I
1
, . . . , I
n
are open intervals and
Q
∩
[0
,
1]
⊂
S
n
k
=1
I
k
.
We may assume all these in
tervals as being bounded;otherwise the conclusion holds trivially.
Let
I
k
= (
a
k
, b
k
) for
k
= 1
, . . . , n
.
Let
F
=
{
a
1
, a
2
, . . . , a
n
, b
1
, b
2
, . . . , b
n
}
be the set of all the endpoints of these intervals.
Then
F
is a finite set,
thus measurable and of measure 0. We notice that [0
,
1]
\
F
⊂
S
n
k
=1
I
k
. In fact, let
x
∈
[0
,
1]
\
F
. Since the rational
numbers are dense in [0
,
1], there is a sequence
{
r
m
}
in [0
,
1]
∩
Q
converging to
x
. There being only a finite number
of intervals
I
k
, one of these intervals must contain an infinite number of terms of the sequence; i.e., a subsequence.
But a subsequence in an interval (
a
k
, b
k
) can only converge to an interior point or to an endpoint. Since
x /
∈
F
,
we conclude it is an interior point in one of the intervals. Now
1 =
m
([0
,
1]) = ([0
,
1]
\
F
) +
m
([0
,
1]
∩
F
) = ([0
,
1]
\
F
) + 0
≤
m
n
[
k
=1
I
k
!
≤
n
X
k
=1
m
(
I
k
) =
n
X
k
=1
‘
(
I
k
)
.
(c) Chapter 3, # 7 (p. 58). Translating an open interval results in an open interval of exactly the same length.
Solution.
If
I
is an interval (open or not), it is a triviality to verify that
I
+
x
is an interval and
‘
(
I
+
x
) =
‘
(
I
)
for all
x
∈
R
. It is also trivial to verify that if
I
n
is an interval (or, actually any subset of
R
, and the union doesn’t
even have to be countable), then (
bigcup
n
∈
N
I
n
) +
x
= (
bigcup
n
∈
N
I
n
+
x
)) for all
x
∈
R
. But, trivial or not, these
facts should be at the very least stated.
Assume now
A
⊂
R
. Let
{
I
n
}
be a countable collection of open intervals such that
A
⊂
S
∞
n
=1
I
n
. Then
{
I
n
+
x
}
is a collection of intervals covering
A
+
x
and it follows that
m
*
(
A
+
x
)
≤
∞
X
n
=1
‘
(
I
n
+
x
) =
∞
X
n
=1
‘
(
I
n
)
.
Because
{
I
n
}
was an arbitrary countable collection of open intervals such that
A
⊂
S
∞
n
=1
I
n
, we proved that
m
*
(
A
+
x
) is a lower bound of the set of (extended) numbers of which
m
*
(
A
) is the infimum, thus
m
*
(
A
+
x
)
≤
m
*
(
A
). Since
A
⊂
R
, x
∈
R
were arbitrary, we also have
m
*
(
A
) =
m
*
((
A
+
x
) + (

x
))
≤
m
*
(
A
+
x
)
for all
A
⊂
R
, x
∈
R
. Equality follows.