Math 598
Feb 22, 2005
1
Geometry and Topology II
Spring 2005, PSU
Lecture Notes 8
2.10
Measure of
C
1
maps
If
X
is a topological space, we say that
A
⊂
X
is
dense
is
X
provided that
A
=
X
, where
A
denotes the closure of
A
. In other words,
A
is dense in
X
if every open subset of
X
intersects
A
.
Theorem 1.
Let
f
:
M
n
→
N
m
be a
C
1
map. Suppose that
m > n
. Then
N
−
f
(
M
)
is dense in
M
.
To prove the above result we need to develop the notion of measure zero,
which is defined as follows. We say that
C
⊂
R
n
is a cube of side length
λ
provided that
C
= [
a
1
, a
1
+
λ
]
× · · · ×
[
a
n
, a
n
+
λ
]
,
for some
a
1
, . . . , a
n
∈
R
n
. We define the mesure or volume of a cube of side
length
λ
by
µ
(
C
) :=
λ
n
.
We say a
X
⊂
R
n
has
measure zero
if for every
>
0, we may cover
X
by a
family of cubes
C
i
,
i
∈
I
, such that
∑
i
∈
I
µ
(
C
i
)
≤
.
Lemma 2.
A countable union of sets of measure zero in
R
n
has measure
zero.
Proof.
Let
X
i
,
i
= 1
,
2
, . . .
be a countable collection of subsets of
R
n
with
measure zero. Then we may cover each
X
i
by a family
C
ij
of cubes such that
∑
j
C
ij
<
/
2
i
. Then
∞
i
=1
j
C
ij
=
∞
i
=1
2
i
=
.
Since
∪
i
X
i
⊂ ∪
ij
C
ij
, it follows then that
∪
i
X
i
has measure zero.
1
Last revised: February 27, 2005
1
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Lemma 3.
If
U
⊂
R
n
is open and nonemepty, then it cannot have measure
zero.
Proof.
By definition, for evey point
p
∈
U
there exists
r >
0 such that
B
r
(
p
)
⊂
U
. Thus, since
U
=
∅
,
U
contains a cube
C
(of side length
λ
≤
2
r/
√
n
). Suppose there is a covering of
U
by a family of cubes. Then, since
C
is compact, there exists a finite subcollection
C
i
,
i
= 1
. . . , m
which cover
C
. Let
N
be the number of integer lattice points (i.e., points with integer
coeﬃcients) which lie in
C
, then
(
max(0
, λ
−
1)
)
n
≤
N
≤
(
λ
+ 1)
n
.
Similarly, if
N
−
i
is the number of integer lattice points in
C
i
and and
C
i
has edge length
λ
i
, then
(
max(0
, λ
i
−
1)
)
n
≤
N
i
≤
(
λ
i
+ 1)
n
.
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 Spring '08
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 Logic, Geometry, Topology, Empty set, Open set, Topological space, measure, rn −1

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