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Topology
Notes 32.1.3 Besicovitch Covering Lemma
There exists N = N(d)³1 such that for any cover by balls we can choose a sub-cover { }, such that
i
any point x lies in at most N balls.
Thus we can bound
1 N
H d + (X) å diam ( ) d + å (B ) .
i i
i C i C
N
Letting 0 we have that H d + (X) . Since C > 0 can be chosen arbitrarily large we deduce
C
that H d + (X) = 0. In particular, dim (X) d +for all> 0. Since > 0 is arbitrary, we deduce
H
that dim (X)d.
H
The proof of the first statement is similar, except that a replace X by a set of full measure for
which d (x) d .
Example 1: If L : X X is a surjective Lipschitz map i.e. C > 0 such that
1 2
|L(x) – L(y)| C|x – y|,
then dim (X )dim (X ).
H 1 H 2
Example 2: If L : X X is a bijective bi-Lipschitz map i.e.C > 0 such that
1 2
( ) |x – y| |L(x) – L(y)|C|x – y|,
1
C
then dim (X ) = dim (X ).
H 1 H 2
Solution: For part 1, consider an open cover for X with dim ( )for all . Then the
1 i i
images = {L( ) : } are a cover for X with dim (L( ))L for all . Thus, from the
2 i
definitions, H (X ) H (X ) .³ In particular, letting0 we see that H (X )³H (X ). Finally,
L 2 1 1 2
from the definitions dim (X )dim (X ).
H 1 H 2
–1
For part 2, we can apply the first part a second time with replaced by L .
32.1.4 Bernoulli’s Measures
Example 3: For an iterated function scheme T , ……, T : we can denote as before
1 k
{
= x = (x ) ¥ : x {1, ,k }}
m m = 0
m
with the Tychonoff product topology. The shift map : is a local homeomorphism defined
by (x) = x . The kth level cylinder is defined by,
m m+1
[x , ,x k–1 ] = { (i ) ¥ : i = x for 0 m k - 1 }
m
m
0
m m = 0
(i.e., all sequence which begin with x , …, x ). We denote by W = {(x , …, x ]} the set of all kth
0 k–1 k 0 k–1
level cylinders (of which there are precisely k ).
n
Notation: For a sequence i and a symbol r{1, …, k} we denote by k ( i ) = card{0mk–1 :
r
i = r} the number of occurrences of r in the first k terms of i .
m
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