1. Self-Similar Sets, Hausdorff Dimension, and
Peano’s Curve
Chelsey Erway
Western Washington University
Advised by Dr. David Hartenstine
May 12, 2016
2. Introduction
A set is called self-similar if it contains scaled (and possibly
rotated) copies of itself.
3. Introduction
A set is called self-similar if it contains scaled (and possibly
rotated) copies of itself.
Self-similar sets are also known as fractals.
4. Introduction
A set is called self-similar if it contains scaled (and possibly
rotated) copies of itself.
Self-similar sets are also known as fractals.
Here’s an example:
8. Outline
Construction of the von Koch curve
Hausdorff measure and dimension
Dimension of the von Koch curve
Construction of Peano’s space-filling curve
9. Outline
Construction of the von Koch curve
Hausdorff measure and dimension
Dimension of the von Koch curve
Construction of Peano’s space-filling curve
Properties of the Peano curve
10. The Von Koch Curve
Construction of the Curve
Let n denote the “generation” of the construction, and let Kn be
the corresponding curve.
11. The Von Koch Curve
Construction of the Curve
Let n denote the “generation” of the construction, and let Kn be
the corresponding curve.
K0
12. The Von Koch Curve
Construction of the Curve
Let n denote the “generation” of the construction, and let Kn be
the corresponding curve.
K0
K1
13. The Von Koch Curve
Construction of the Curve
Let n denote the “generation” of the construction, and let Kn be
the corresponding curve.
K0
K1
K2
14. The Von Koch Curve
Construction of the Curve
Let n denote the “generation” of the construction, and let Kn be
the corresponding curve.
K0
K1
K2
Then, the von Koch curve, K, will be limit as n → ∞:
K = lim
n→∞
Kn.
15. The Von Koch Curve
We can think of each Kn as the image of a function Kn(t) that
takes t in the unit interval to the a point (on the curve Kn) in R2.
16. The Von Koch Curve
We can think of each Kn as the image of a function Kn(t) that
takes t in the unit interval to the a point (on the curve Kn) in R2.
Each Kn(t) is continous by construction.
17. The Von Koch Curve
We can think of each Kn as the image of a function Kn(t) that
takes t in the unit interval to the a point (on the curve Kn) in R2.
Each Kn(t) is continous by construction.
We can show that Kn(t) converges uniformly to K(t).
18. The Von Koch Curve
We can think of each Kn as the image of a function Kn(t) that
takes t in the unit interval to the a point (on the curve Kn) in R2.
Each Kn(t) is continous by construction.
We can show that Kn(t) converges uniformly to K(t).
Hence, K(t) is continuous and its image K is a compact set.
19. The Von Koch Curve
Question: What’s the length of K?
20. The Von Koch Curve
Question: What’s the length of K?
K0 l(0) = 1
21. The Von Koch Curve
Question: What’s the length of K?
K0 l(0) = 1
K1 l(1) = 4
3
22. The Von Koch Curve
Question: What’s the length of K?
K0 l(0) = 1
K1 l(1) = 4
3
K2 l(2) = 4(4
3)(1
3) = (4
3)2
23. The Von Koch Curve
Question: What’s the length of K?
K0 l(0) = 1
K1 l(1) = 4
3
K2 l(2) = 4(4
3)(1
3) = (4
3)2
...
Kn l(n) = (4
3)n
24. The Von Koch Curve
Question: What’s the length of K?
K0 l(0) = 1
K1 l(1) = 4
3
K2 l(2) = 4(4
3)(1
3) = (4
3)2
...
Kn l(n) = (4
3)n → ∞ as n → ∞.
26. Lebesgue measure of the curve
Question: What is the Lebesgue measure of K?
Question: What is Lebesgue measure?
27. Lebesgue measure of the curve
Question: What is the Lebesgue measure of K?
Question: What is Lebesgue measure?
In R the Lebesgue measure of a set is its length when
approximated by intervals.
28. Lebesgue measure of the curve
Question: What is the Lebesgue measure of K?
Question: What is Lebesgue measure?
In R the Lebesgue measure of a set is its length when
approximated by intervals.
If E is a measurable set,
m(E) = inf
∞
k=1
|Ik| : E ⊂
∞
k=1
Ik .
29. Lebesgue measure of the curve
Question: What is the Lebesgue measure of K?
Question: What is Lebesgue measure?
In R the Lebesgue measure of a set is its length when
approximated by intervals.
If E is a measurable set,
m(E) = inf
∞
k=1
|Ik| : E ⊂
∞
k=1
Ik .
In R2 the Lebesgue measure of a set is the area when
approximated by rectangles.
30. Lebesgue measure of the curve
Question: What is the Lebesgue measure of K?
Question: What is Lebesgue measure?
In R the Lebesgue measure of a set is its length when
approximated by intervals.
If E is a measurable set,
m(E) = inf
∞
k=1
|Ik| : E ⊂
∞
k=1
Ik .
In R2 the Lebesgue measure of a set is the area when
approximated by rectangles.
The Lebesgue measure of E is defined only when E is a
measurable set. For arbitrary susbsets of Rn, we have, instead of
m, the outer measure, m∗.
32. Lebesgue measure of the curve
Question: Which sets are measurable?
A set E ⊂ Rn is measurable if for any set A ⊂ Rn,
m∗(A) = m∗(E ∩ A) + m∗(Ec ∩ A).
33. Lebesgue measure of the curve
Question: Which sets are measurable?
A set E ⊂ Rn is measurable if for any set A ⊂ Rn,
m∗(A) = m∗(E ∩ A) + m∗(Ec ∩ A).
That is, E is measurable if we can use it to split up any other set
A so that it has outer measure equal to the sum of the outer
measures of its pieces.
34. The Von Koch Curve
Question: What is the Lebesgue measure of K?
35. The Von Koch Curve
Question: What is the Lebesgue measure of K?
It’s zero!
36. The Von Koch Curve
Question: What is the Lebesgue measure of K?
It’s zero!
We can cover K with countably many rectangles whose total area
is arbitrarily small.
37. The Von Koch Curve
Question: What is the dimension of the curve?
38. The Von Koch Curve
Question: What is the dimension of the curve?
Line segments are one dimensional.
39. The Von Koch Curve
Question: What is the dimension of the curve?
Line segments are one dimensional.
Rectangles are two dimensional.
40. The Von Koch Curve
Question: What is the dimension of the curve?
Line segments are one dimensional.
Rectangles are two dimensional.
Perhaps K is somewhere in between?
42. Hausdorff Measure
To define a new notion of dimension, we need a new measure.
Now, instead of thinking about intervals or rectangles, we’ll think
about covering the set we want to measure with any countable
collection of sets of provided their diameters are smaller than some
size δ.
43. Hausdorff Measure
To define a new notion of dimension, we need a new measure.
Now, instead of thinking about intervals or rectangles, we’ll think
about covering the set we want to measure with any countable
collection of sets of provided their diameters are smaller than some
size δ.
The diameter of E ⊂ Rn, is diam E = sup{d(a, b) : a, b ∈ E}
where d is the usual metric on Rn.
45. Hausdorff Measure
We define the α-dimensional Hausdorff outer measure as
follows:
m∗
α(E) = lim
δ→0
inf
∞
k=1
(diam Fk)α
: E ⊂
∞
k=1
Fk, diam Fk ≤ δ ∀k
46. Hausdorff Measure
We define the α-dimensional Hausdorff outer measure as
follows:
m∗
α(E) = lim
δ→0
inf
∞
k=1
(diam Fk)α
: E ⊂
∞
k=1
Fk, diam Fk ≤ δ ∀k
= lim
δ→0
Hδ
α(E)
47. Hausdorff Measure
We define the α-dimensional Hausdorff outer measure as
follows:
m∗
α(E) = lim
δ→0
inf
∞
k=1
(diam Fk)α
: E ⊂
∞
k=1
Fk, diam Fk ≤ δ ∀k
= lim
δ→0
Hδ
α(E)
where
Hδ
α(E) = inf { k(diam Fk)α : E ⊂ ∞
k=1 Fk, diam Fk ≤ δ ∀k} .
48. Hausdorff Measure
We define the α-dimensional Hausdorff outer measure as
follows:
m∗
α(E) = lim
δ→0
inf
∞
k=1
(diam Fk)α
: E ⊂
∞
k=1
Fk, diam Fk ≤ δ ∀k
= lim
δ→0
Hδ
α(E)
where
Hδ
α(E) = inf { k(diam Fk)α : E ⊂ ∞
k=1 Fk, diam Fk ≤ δ ∀k} .
Hδ
α(E) is increasing as δ → 0.
49. Hausdorff Measure
m∗
α is an outer measure. If we restrict it to measurable sets, it
becomes the α-dimensional Hausdorff measure mα.
50. Hausdorff Measure
m∗
α is an outer measure. If we restrict it to measurable sets, it
becomes the α-dimensional Hausdorff measure mα.
We can show that Lebesgue measurable sets are also Hausdorff
measurable and, in fact, if α is a postive integer, n, then the
Hausdorff Measure of a set is equal to the n-dimensional Lebesgue
measure up to a constant that depends only on the dimension n.
66. Hausdorff Dimension
Question: What should α be?
It turns out that for any measurable set E there is a unique value
of α with the following property:
mβ(E) =
0 if β > α
∞ if β < α.
67. Hausdorff Dimension
Question: What should α be?
It turns out that for any measurable set E there is a unique value
of α with the following property:
mβ(E) =
0 if β > α
∞ if β < α.
This special α is the Hausdorff Dimension of E.
70. The Von Koch Curve
dim K =
log 4
log 3
Showing this uses the ratios involved in the construction: the
length increases by 4
3 at each generation and the number of
vertices increases 4-fold.
72. The Von Koch Curve
Claim: dim K ≥ log 4
log 3.
Proof.
To show that dim K ≥ log 4
log 3 it suffices to show that mα(K) > 0
where α = log 4
log 3. This follows from the previously stated property.
73. The Von Koch Curve
Claim: dim K ≥ log 4
log 3.
Proof.
To show that dim K ≥ log 4
log 3 it suffices to show that mα(K) > 0
where α = log 4
log 3. This follows from the previously stated property.
If the dimension were smaller than log 4
log 3, then we’d have to have
mlog 4
log 3
(K) = 0.
74. The Von Koch Curve
Claim: dim K ≥ log 4
log 3.
Proof.
To show that dim K ≥ log 4
log 3 it suffices to show that mα(K) > 0
where α = log 4
log 3. This follows from the previously stated property.
If the dimension were smaller than log 4
log 3, then we’d have to have
mlog 4
log 3
(K) = 0.
I want to show that if some collection of sets {Fj}∞
j=1 covers K
and diam Fj < δ ∀j ∈ N, then ∞
j=1(diam Fj)α ≥ c > 0 for some
constant c.
75. The Von Koch Curve
Claim: dim K ≥ log 4
log 3.
Proof.
To show that dim K ≥ log 4
log 3 it suffices to show that mα(K) > 0
where α = log 4
log 3. This follows from the previously stated property.
If the dimension were smaller than log 4
log 3, then we’d have to have
mlog 4
log 3
(K) = 0.
I want to show that if some collection of sets {Fj}∞
j=1 covers K
and diam Fj < δ ∀j ∈ N, then ∞
j=1(diam Fj)α ≥ c > 0 for some
constant c.
We can take {Fj}∞
j=1 to be a collection of open balls (if not we
can just exchange δ for 2δ to fit each Fj inside an open ball.)
76. The Von Koch Curve
Since K is compact, ∃M ∈ N such that {Fj}M
j=1 covers K.
77. The Von Koch Curve
Since K is compact, ∃M ∈ N such that {Fj}M
j=1 covers K.
Now choose the ball in the covering of smallest diameter, δ∗.
We can find n ∈ N so that
1
3n
< δ∗
≤
1
3n−1
.
78. The Von Koch Curve
Since K is compact, ∃M ∈ N such that {Fj}M
j=1 covers K.
Now choose the ball in the covering of smallest diameter, δ∗.
We can find n ∈ N so that
1
3n
< δ∗
≤
1
3n−1
.
We can find a constant c > 0 such that any ball B satisfying
1
3n < diam B ≤ 1
3n−1 can cover at most c vertices of the nth
generation and at most c · 4n−k vertices of the kth generation.
79. The Von Koch Curve
Let Nm denote the number of balls Fj in the covering satisfying
1
3m < diam Fj ≤ 1
3m−1 .
80. The Von Koch Curve
Let Nm denote the number of balls Fj in the covering satisfying
1
3m < diam Fj ≤ 1
3m−1 .
Then,
M
j=1
(diam Fj)α
≥
n
m=1
Nm ·
1
3m
α
.
81. The Von Koch Curve
Let Nm denote the number of balls Fj in the covering satisfying
1
3m < diam Fj ≤ 1
3m−1 .
Then,
M
j=1
(diam Fj)α
≥
n
m=1
Nm ·
1
3m
α
.
Since each of the Nm balls with diameter between 1
3m and 1
3m−1
can cover at most c · 4k−m vertices of the kth generation we must
have
n
m=1
Nm · c · 4k−m
≥ 4k
82. The Von Koch Curve
Let Nm denote the number of balls Fj in the covering satisfying
1
3m < diam Fj ≤ 1
3m−1 .
Then,
M
j=1
(diam Fj)α
≥
n
m=1
Nm ·
1
3m
α
.
Since each of the Nm balls with diameter between 1
3m and 1
3m−1
can cover at most c · 4k−m vertices of the kth generation we must
have
n
m=1
Nm · c · 4k−m
≥ 4k
or, equivalently,
n
m=1
Nm ·
1
4m
≥
1
c
.
84. The Von Koch Curve
From above we have
M
j=1
(diam Fj)α
≥
n
m=1
Nm ·
1
3m
α
85. The Von Koch Curve
From above we have
M
j=1
(diam Fj)α
≥
n
m=1
Nm ·
1
3m
α
=
n
m=1
Nm ·
1
3m
log 4
log 3
86. The Von Koch Curve
From above we have
M
j=1
(diam Fj)α
≥
n
m=1
Nm ·
1
3m
α
=
n
m=1
Nm ·
1
3m
log 4
log 3
=
n
m=1
Nm ·
1
4m
87. The Von Koch Curve
From above we have
M
j=1
(diam Fj)α
≥
n
m=1
Nm ·
1
3m
α
=
n
m=1
Nm ·
1
3m
log 4
log 3
=
n
m=1
Nm ·
1
4m
≥
1
c
88. The Von Koch Curve
From above we have
M
j=1
(diam Fj)α
≥
n
m=1
Nm ·
1
3m
α
=
n
m=1
Nm ·
1
3m
log 4
log 3
=
n
m=1
Nm ·
1
4m
≥
1
c
>0.
89. The Von Koch Curve
From above we have
M
j=1
(diam Fj)α
≥
n
m=1
Nm ·
1
3m
α
=
n
m=1
Nm ·
1
3m
log 4
log 3
=
n
m=1
Nm ·
1
4m
≥
1
c
>0.
Now, taking the infimum over all coverings and letting δ → 0, we
get mα(K) ≥ 1
c > 0.
90. The Von Koch Curve
To show that dim K ≤ log 4
log 3, we have to use H¨older continuity.
91. The Von Koch Curve
To show that dim K ≤ log 4
log 3, we have to use H¨older continuity.
A function f defined on a subset E of Rn is H¨older continuous
with exponent γ on E if |f(x) − f(y)| ≤ M|x − y|γ for every
x, y ∈ E where M is a constant not dependent on x and y.
92. The Von Koch Curve
To show that dim K ≤ log 4
log 3, we have to use H¨older continuity.
A function f defined on a subset E of Rn is H¨older continuous
with exponent γ on E if |f(x) − f(y)| ≤ M|x − y|γ for every
x, y ∈ E where M is a constant not dependent on x and y.
H¨older continuity is a stronger condition than uniform continuity.
93. The Von Koch Curve
To show that dim K ≤ log 4
log 3, we have to use H¨older continuity.
A function f defined on a subset E of Rn is H¨older continuous
with exponent γ on E if |f(x) − f(y)| ≤ M|x − y|γ for every
x, y ∈ E where M is a constant not dependent on x and y.
H¨older continuity is a stronger condition than uniform continuity.
If we can show that K is H¨older continuous with exponent log 3
log 4 we
can employ the following lemma.
94. The Von Koch Curve
Lemma. If a function f is H¨older continuous with exponent γ on
a compact set E then dim f(E) ≤ 1
γ dim E.
95. The Von Koch Curve
Lemma. If a function f is H¨older continuous with exponent γ on
a compact set E then dim f(E) ≤ 1
γ dim E.
This will give us dim K ≤ log 4
log 3 dim[0, 1] = log 4
log 3.
96. The Von Koch Curve
Lemma. If a function f is H¨older continuous with exponent γ on
a compact set E then dim f(E) ≤ 1
γ dim E.
This will give us dim K ≤ log 4
log 3 dim[0, 1] = log 4
log 3.
Proof.
97. The Von Koch Curve
Lemma. If a function f is H¨older continuous with exponent γ on
a compact set E then dim f(E) ≤ 1
γ dim E.
This will give us dim K ≤ log 4
log 3 dim[0, 1] = log 4
log 3.
Proof.
Suppose {Fk} is a countable collection of sets that covers E. Then
{f(E ∩ Fk)} covers f(E), and f(E ∩ Fk) has diameter smaller
than M(diam Fk)γ. Thus,
98. The Von Koch Curve
Lemma. If a function f is H¨older continuous with exponent γ on
a compact set E then dim f(E) ≤ 1
γ dim E.
This will give us dim K ≤ log 4
log 3 dim[0, 1] = log 4
log 3.
Proof.
Suppose {Fk} is a countable collection of sets that covers E. Then
{f(E ∩ Fk)} covers f(E), and f(E ∩ Fk) has diameter smaller
than M(diam Fk)γ. Thus,
∞
k=1
(diam f(E ∩ Fk))α/γ
≤ Mα/γ
∞
k=1
(diam Fk)α
.
99. The Von Koch Curve
Lemma. If a function f is H¨older continuous with exponent γ on
a compact set E then dim f(E) ≤ 1
γ dim E.
This will give us dim K ≤ log 4
log 3 dim[0, 1] = log 4
log 3.
Proof.
Suppose {Fk} is a countable collection of sets that covers E. Then
{f(E ∩ Fk)} covers f(E), and f(E ∩ Fk) has diameter smaller
than M(diam Fk)γ. Thus,
∞
k=1
(diam f(E ∩ Fk))α/γ
≤ Mα/γ
∞
k=1
(diam Fk)α
.
Now if we take the infimums of these sums and let δ → 0, then
mα/γ(f(E)) ≤ mα(E). Along with the earlier property, this shows
dim f(E) ≤ 1
γ dim E.
103. Hausdorff Dimension
Hausdorff Dimension for some other fractals:
Cantor set, C dim C = log 2
log 3
Sierpinski triangle, S dim S = log 3
log 2
Apollonian gasket, A dim A = 1.308535???
106. A Space-Filling Curve
What is a space-filling curve?
In two dimensions, a space-filling curve is a curve whose range is
the entire unit square.
107. A Space-Filling Curve
What is a space-filling curve?
In two dimensions, a space-filling curve is a curve whose range is
the entire unit square.
The idea of a space-filling curve was first proposed by Giuseppe
Peano in 1890. He was seeking a surjective, continuous mapping
between the unit interval and the unit square—and he found one.
109. A Space-Fillng Curve
Construction of a space-filling curve
1 Divide the unit interval and unit square each into 4 equal
pieces.
Like this:
110. A Space-Filling Curve
2 Create a mapping φ that takes left/right-most intervals to
lower left/right square, and that preserves adjacency.
111. A Space-Filling Curve
2 Create a mapping φ that takes left/right-most intervals to
lower left/right square, and that preserves adjacency.
3 Connect the centers of the squares with line segments in the
order given by φ to obtain P1(t).
112. A Space-Filling Curve
2 Create a mapping φ that takes left/right-most intervals to
lower left/right square, and that preserves adjacency.
3 Connect the centers of the squares with line segments in the
order given by φ to obtain P1(t).
4 Continue to subdivide intervals and squares by the same
method.
P1(t) P2(t) P3(t)
114. A Space-Filling Curve
Pk(t) takes [0,1] to a curve in [0,1] × [0,1] at a constant speed.
Note that Pk(t) is continuous, and it takes any value t to a
subsquare of the square it was mapped to in the previous
generation Pk−1(t).
115. A Space-Filling Curve
Pk(t) takes [0,1] to a curve in [0,1] × [0,1] at a constant speed.
Note that Pk(t) is continuous, and it takes any value t to a
subsquare of the square it was mapped to in the previous
generation Pk−1(t).
The Peano curve is the image:
P(t) = lim
k→∞
Pk(t).
120. A Space-Filling Curve
Properties of P(t)
Continuity
Each Pk(t) is continuous.
The Pk(t) converge uniformly to P(t) since ∀k ∈ N and
∀t ∈ [0, 1], |Pk+1(t) − Pk(t)| ≤
√
2 2−k
.
121. A Space-Filling Curve
Properties of P(t)
Continuity
Each Pk(t) is continuous.
The Pk(t) converge uniformly to P(t) since ∀k ∈ N and
∀t ∈ [0, 1], |Pk+1(t) − Pk(t)| ≤
√
2 2−k
.
This also shows that the set P is compact.
122. A Space-Filling Curve
Properties of P(t)
Continuity
Each Pk(t) is continuous.
The Pk(t) converge uniformly to P(t) since ∀k ∈ N and
∀t ∈ [0, 1], |Pk+1(t) − Pk(t)| ≤
√
2 2−k
.
This also shows that the set P is compact.
Surjectivity
123. A Space-Filling Curve
Properties of P(t)
Continuity
Each Pk(t) is continuous.
The Pk(t) converge uniformly to P(t) since ∀k ∈ N and
∀t ∈ [0, 1], |Pk+1(t) − Pk(t)| ≤
√
2 2−k
.
This also shows that the set P is compact.
Surjectivity
P(t) is continuous.
124. A Space-Filling Curve
Properties of P(t)
Continuity
Each Pk(t) is continuous.
The Pk(t) converge uniformly to P(t) since ∀k ∈ N and
∀t ∈ [0, 1], |Pk+1(t) − Pk(t)| ≤
√
2 2−k
.
This also shows that the set P is compact.
Surjectivity
P(t) is continuous.
P is dense in the unit square.
125. A Space-Filling Curve
Properties of P(t)
Continuity
Each Pk(t) is continuous.
The Pk(t) converge uniformly to P(t) since ∀k ∈ N and
∀t ∈ [0, 1], |Pk+1(t) − Pk(t)| ≤
√
2 2−k
.
This also shows that the set P is compact.
Surjectivity
P(t) is continuous.
P is dense in the unit square.
Pick some point q in [0, 1] × [0, 1]. We can find a nested
sequence of squares—one from each generation—whose
intersection is q.
126. A Space-Filling Curve
Properties of P(t)
Continuity
Each Pk(t) is continuous.
The Pk(t) converge uniformly to P(t) since ∀k ∈ N and
∀t ∈ [0, 1], |Pk+1(t) − Pk(t)| ≤
√
2 2−k
.
This also shows that the set P is compact.
Surjectivity
P(t) is continuous.
P is dense in the unit square.
Pick some point q in [0, 1] × [0, 1]. We can find a nested
sequence of squares—one from each generation—whose
intersection is q.
Since P passes through every square and the squares decrease
to arbitrarily small size we have P arbitrarily close to q.
127. A Space-Filling Curve
Properties of P(t)
Continuity
Each Pk(t) is continuous.
The Pk(t) converge uniformly to P(t) since ∀k ∈ N and
∀t ∈ [0, 1], |Pk+1(t) − Pk(t)| ≤
√
2 2−k
.
This also shows that the set P is compact.
Surjectivity
P(t) is continuous.
P is dense in the unit square.
Pick some point q in [0, 1] × [0, 1]. We can find a nested
sequence of squares—one from each generation—whose
intersection is q.
Since P passes through every square and the squares decrease
to arbitrarily small size we have P arbitrarily close to q.
Since P is compact, it must contain its limit points. We have
shown that q is a limit point of P, so q ∈ P.
129. A Space-Filling Curve
1
2 H¨older Continuity
|P(t) − P(s)| ≤ M|t − s|
1
2 for every t, s ∈ [0, 1] for some
constant M not dependent on t and s.
130. A Space-Filling Curve
1
2 H¨older Continuity
|P(t) − P(s)| ≤ M|t − s|
1
2 for every t, s ∈ [0, 1] for some
constant M not dependent on t and s.
Preservation of Lebesgue Measure
131. A Space-Filling Curve
1
2 H¨older Continuity
|P(t) − P(s)| ≤ M|t − s|
1
2 for every t, s ∈ [0, 1] for some
constant M not dependent on t and s.
Preservation of Lebesgue Measure
A subset, E of [0, 1] is measurable if and only if P(E) is
measurable and m1(E) = m2(P(E)) where m1 and m2
represent the one and two-dimensional Lebesgue measures
respectively.
