Solutions to Homework 1
1. (10 points) Let
Q
be the set of all rationals in the interval [0
,
1]
.
Let
S
=
{
I
1
, I
2
, ..., I
m
}
be a
finite
collection of closed intervals covering
Q.
Show that
m
X
k
=1
v
(
I
k
)
≥
1
.
(0.1)
On the other hand, for any
ε >
0
,
one can
fi
nd
S
=
{
I
1
, I
2
, ..., I
m
,
· · ·}
,
which is a
count
able
collection of closed intervals covering
Q,
such that
∞
X
k
=1
v
(
I
k
)
< ε.
(0.2)
In particular, (0.2) implies that

Q

e
= 0
.
(Now you see the di
ff
erence between the use of
”finite cover”
and
”countable cover”
.)
Solution:
For (0.1), we
fi
rst assume that the intervals
I
1
, I
2
, ... , I
m
are
nonoverlapping
.
In such case we clearly have (0.1).
For arbitrary intervals
I
1
,
I
2
,
... ,
I
m
with overlapping, one can throw away the over
lapping part and the remaining nonoverlapping part, which we denote it as
J
1
, J
2
, ... , J
n
,
satis
fi
es
P
n
k
=1
v
(
J
k
)
≥
1
.
Therefore we have (0.1).
For (0.2), it has been done in class.
¤
2. (10 points) Find a set
E
⊂
R
with outer measure zero and a function
f
:
E
→
R
such that
f
is continuous on
E
and
f
(
E
) = [0
,
1]
.
This exercise says that a