Math 318 HW #7 Solutions
1. Exercise 20.12. Let
C
be the Cantor set. Let
D
⊂
[0
,
1] be a nowhere dense measurable set
with
m
(
D
)
>
0. Then there is a nonmeasurable set
B
⊂
D
. At each stage of the construction
of
C
and of
D
, a certain finite number of open intervals of [0
,
1] are deleted (put into [0
,
1]
\
C
or [0
,
1]
\
D
).
Let
g
map the intervals put into [0
,
1]
\
D
at the
n
th stage linearly onto the
intervals put into [0
,
1]
\
C
at the
n
th stage, for
n
= 1
,
2
, . . .
. Thus
g
is monotone and defined
at every element of [0
,
1]
\
D
, mapping onto [0
,
1]
\
C
.
(a) Using the fact that
g
is increasing and that [0
,
1]
\
D
is dense in [0
,
1], prove that for every
x
0
∈
D
,
lim
x
→
x

0
g
(
x
) = sup
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x < x
0
}
and
lim
x
→
x
+
0
g
(
x
) = inf
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x > x
0
}
.
Proof.
Since [0
,
1]
\
D
is dense, we know that every
x
0
∈
D
is a limit point of [0
,
1]
\
D
, so
lim
x
→
x

0
g
(
x
) makes sense. Now,
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x < x
0
}
is nonempty (unless
x
0
= 0, in which case lim
x
→
x

0
g
(
x
) is vacuous) and, since
g
is
increasing, bounded above by
g
(
x
1
) for any
x
1
> x
0
such that
x
1
∈
[0
,
1]
\
D
. Therefore,
sup
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x < x
0
}
=
M
exists. But then, for any
>
0, there exists
x
∈
[0
,
1]
\
D
with
x < x
0
such that
g
(
x
)
> M

. Letting
δ
=
x
0

x
, we know that
x < x < x
0
implies that
g
(
x
)
≤
g
(
x
)
≤
M
, and so
g
(
x
)

M
≤
g
(
x
)

M
≤
0
.
Therefore, since
g
(
x
)

M >

, the above implies that

< g
(
x
)

M
≤
0
,
so we can conclude that

x

x
0

< δ
and
x < x
0
implies that

g
(
x
)

M

<
.
Since the choice of
>
0 was arbitrary, we conclude that
lim
x
→
x

0
g
(
x
) =
M
= sup
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x < x
0
}
,
as desired.
An equivalent argument shows that lim
x
→
x
+
0
g
(
x
) = inf
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x >
x
0
}
.
(b) Use (a) to show that for
x
0
∈
D
, lim
x
→
x

0
g
(
x
)
≤
lim
x
→
x
+
0
g
(
x
).
1
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Proof.
Since
g
is monotone increasing, every element of the set
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x >
x
0
}
is an upper bound on the set
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x < x
0
}
. Therefore, by defi
nition of the supremum,
lim
x
→
x

0
g
(
x
) = sup
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x < x
0
}
is less than or equal to every element of
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x > x
0
}
.
In other
words, lim
x
→
x

0
g
(
x
) is a lower bound on the set
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x > x
0
}
, and
so, by the definition of the infimum, is less than or equal to
lim
x
→
x

0
g
(
x
)
≤
inf
{
g
(
x
) :
x
∈
[0
,
1]
\
D
and
x > x
0
}
=
lim
x
→
x
+
0
g
(
x
)
,
as desired.
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 Spring '08
 Staff
 Math, measure, Lebesgue measure, Lebesgue integration, Nonmeasurable set, Fn

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