Porosity and regularity in metric measure spaces

0.1 MB·

by  Antti Käenmäki

POROSITY AND REGULARITY IN METRIC MEASURE
SPACES
ANTTI KA¨ENMA¨KI
7
Abstract. ThisisareportofajointworkwithE.J¨arvenp¨aa¨,M.J¨arvenp¨aa¨,
1
0 T. Rajala, S. Rogovin, and V. Suomala. In [3], we characterized uniformly
2 porous sets in s-regular metric spaces in terms of regular sets by verifying
that a set A is uniformly porous if and only if there is t < s and a t-regular
n
a set F ⊃ A. Here we outline the main idea of the proof and also present an
J alternative proof for the crucial lemma needed in the proof of the result.

0
3

]
A 1. Setting and result
C
We assume that X = (X,d) is a complete separable metric space. If s > 0
.
h then a measure µ on X is called s-regular on a set A ⊂ X provided that there
t
a are constants 0 < a ≤ b < ∞ and r > 0 such that
µ µ µ
m
a rs ≤ µ B(x,r) ≤ b rs
[ µ µ

1 for every x ∈ A and 0 < r < rµ. A s(cid:0)et A ⊂(cid:1)X is s-regular if there is a measure
v µ on X which is s-regular on A and µ(X \A) = 0. We call a measure µ on X
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doubling if there are constants c ≥ 1 and r > 0 such that
9 µ µ
5
0 < µ B(x,2r) ≤ c µ B(x,r) < ∞
8 µ
0 (cid:0) (cid:1) (cid:0) (cid:1)
for every x ∈ X and 0 < r < r . Clearly, if µ is s-regular on X then it is also
. µ
1 doubling. Finally, a set A ⊂ X is called uniformly (lower) ̺-porous provided
0
that there is a constant r > 0 such that
7 p
1
por(A,x,r) > ̺
:
v
i for all x ∈ A and 0 < r < r , where
X p
r por(A,x,r) = sup{̺ ≥ 0 : B(y,̺r) ⊂ B(x,r)\A for some y ∈ X}.
a
It is easy to see that if µ is a doubling measure on X and 0 < α < 1 then
µ B(x,αr) ≥ c⌊log2α⌋µ B(x,r) (1.1)
µ
whenever x ∈ X and 0 <(cid:0) r < r . (cid:1)Here ⌊a⌋ =(cid:0)max{n (cid:1)∈ Z : n ≤ a} for a ∈ R.
µ
It is also straightforward to show that if A ⊂ X is uniformly ̺-porous for some
Date: January 31, 2017.
2000 Mathematics Subject Classification. Primary 28A80; Secondary 51F99.
Key words and phrases. Metric measure space, porosity, regularity.
The author acknowledges the support of the Academy of Finland (project #114821).
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2 ANTTI KA¨ENMA¨KI
̺ > 0 then µ(A) = 0 for any doubling measure µ on X. This is true essentially
because the points of A are nondensity points.
Observe that, by [5, Theorem 5.7] and [1, Theorem 3.16], both the Hausdorff
dimension and the packing dimension of A equal to s provided that A ⊂ X is
s-regular. An unpublished result of Saaranen shows that if X is s-regular then
for each 0 < t < s there exists a t-regular set A ⊂ X. This is the only place
where we need the completeness of X. The separability assumption is natural
since the Hausdorff dimension of any nonseparable metric space is infinity. Also,
it can be shown that no σ-finite doubling measures exist in nonseparable spaces.
Our result under consideration is the following. A complete proof with a
slightly better result can be found from [3].
Theorem 1.1. Suppose X is s-regular. Then A is uniformly ̺-porous for some
̺ > 0 if and only if there is t < s and a t-regular set F ⊃ A.
2. Alternative approach
We begin with the following lemma that gives an alternative proof for [3,
Corollaries 4.5 and 4.6]. The proof was found while preparing the article [3]
and is also credited by the authors of that article. Observe, however, that in [3,
Corollary 4.6] we were able to find a better exponent δ. The lemma generalizes
the argument of [4, Lemma 2.8].
Given a set A ⊂ X, we define A(r) = {x ∈ X : dist(x,A) < r} as r > 0.
With the notation C(·), we mean a constant whose dependence is indicated in
the parentheses.
Lemma 2.1. Suppose that µ is a doubling measure on X. If A ⊂ X is uniformly
̺-porous then
r δ
µ A∩B(x ,r ) (r) ≤ C(c ,̺)µ B(x ,r )
0 0 µ 0 0
r
(cid:16)(cid:0) (cid:1) (cid:17) (cid:0) (cid:1)(cid:16) 0(cid:17)
for every x ∈ X and 0 < r < r ≤ r , where δ = C(c )̺log2cµ/log(1/̺).
0 0 µ µ
Moreover, if µ is s-regular then δ = C(a ,b ,s)̺s/log(1/̺).
µ µ
Proof. We may assume that ̺ ≤ 1. Let us denote for 0 < t ,t < ∞
3 1 2
A(t ,t ) = {x ∈ X : t < dist(x,A) ≤ t }
1 2 1 2
= A(t )\A(t ).
2 1
Furthermore, we denote
α = µ A(r ̺3k,r ̺3(k−1))
k 0 0
as k ∈ N. (cid:0) (cid:1)
We shall first find a number 0 < γ < 1 such that α ≥ γ ∞ α whenever
k i=k+1 i
k ∈ N. To do so, fix k and notice that for every x ∈ A therePis y ∈ B(x,r ̺3k−2)
0
POROSITY AND REGULARITY IN METRIC MEASURE SPACES 3
such that
B(y,r ̺3k−1) ⊂ B(x,r ̺3k−2)\A. (2.1)
0 0
We claim that this implies
A(r ̺3k) ⊂ A(r ̺3k−1,r ̺3k−2) (2r ̺3k−2). (2.2)
0 0 0 0
(cid:0) (cid:1)
To see this, pick z ∈ A(r ̺3k). Since dist(z,A) ≤ r ̺3k, there is x ∈ A such that
0 0
d(z,x) < r ̺3k−1. According to (2.1), there is y ∈ A(r ̺3k−1,r ̺3k−2) for which
0 0 0
d(x,y) < r ̺3k−2. Therefore
0
d(z,y) ≤ d(z,x)+d(x,y) < r ̺3k−1 +r ̺3k−2 < 2r ̺3k−2
0 0 0
and hence (2.2) follows.
Attaching for each z ∈ A(r ̺3k−1,r ̺3k−2) a ball B(z, 1r ̺3k−2), we find, using
0 0 5 0
the 5r-covering theorem (see [2, Theorem 1.2] and [5, Theorem 2.1]), an index
set I and points z ∈ A(r ̺3k−1,r ̺3k−2), i ∈ I, such that
i 0 0
A(r ̺3k−1,r ̺3k−2) ⊂ B(z ,r ̺3k−2) (2.3)
0 0 i 0
[
i∈I
and B(z , 1r ̺3k−2)∩B(z , 1r ̺3k−2) = ∅ for i 6= j. Since for each i ∈ I and for
i 5 0 j 5 0
every x ∈ B(z , 1r ̺3k−1) we have dist(x,A) ≤ d(x,z )+dist(z ,A) ≤ r (1̺3k−1+
i 2 0 i i 0 2
̺3k−2) ≤ r ̺3(k−1) and dist(x,A) ≥ dist(z ,A)−d(z ,x) > r ̺3k−1 − 1r ̺3k−1 ≥
0 i i 0 2 0
r ̺3k, it follows that
0
B(z , 1r ̺3k−1) ⊂ A(r ̺3k,r ̺3(k−1))
i 2 0 0 0
whenever i ∈ I. On the other hand, we have
A(r ̺3k) ⊂ B(z ,r 3̺3k−2)
0 i 0
[
i∈I
by (2.2) and (2.3), and hence, using (1.1),
α ≥ µ B(z , 1r ̺3k−1) ≥ γ µ B(z ,r 3̺3k−2)
k i 2 0 i 0
Xi∈I (cid:0) (cid:1) Xi∈I (cid:0) (cid:1)
(2.4)
∞
≥ γµ A(r ̺3k) = γ α
0 i
(cid:0) (cid:1) i=Xk+1
where γ = c⌊log2(̺/6)⌋. Moreover, if µ is s-regular then we have γ = aµ(̺/6)s.
µ bµ
Next we shall derive a growth inequality for the numbers α . Notice that
k
α ≥ γ ∞ α ≥ γα by (2.4). Assuming inductively
1 i=2 i 2
P
∞
α ≥ γ(γ +1)k−1 α ≥ γ(γ +1)k−1α
1 i k+1
X
i=k+1
4 ANTTI KA¨ENMA¨KI
for k ∈ N, we get
∞ ∞
α ≥ γ(γ +1)k−1 α = γ(γ +1)k−1 α + α
1 i k+1 i
(cid:18) (cid:19)
X X
i=k+1 i=k+2
∞ ∞
≥ γ(γ +1)k−1 γ α + α
i i
(cid:18) (cid:19)
X X
i=k+2 i=k+2
∞
= γ(γ +1)k α ≥ γ(γ +1)kα
i k+2
X
i=k+2
using (2.4) again. Thus we have shown that
α ≤ γ−1(γ +1)2−kα (2.5)
k 1
whenever k ∈ N.
Now let 0 < r < r and choose k ∈ N such that r ̺3(k0+1) ≤ r < r ̺3k0. Then
0 0 0 0
we have
∞
µ A(r) ≤ µ A(r ̺3k0) ≤ µ(A)+ α
0 i
(cid:0) (cid:1) (cid:0) (cid:1) i=Xk0+1
∞
= α ≤ γ−1α ≤ γ−2(γ +1)2−k0α
i k0 1
X
i=k0+1
≤ (γ +1)/γ 2(γ +1)−k0µ A(r )
0
≤ (cid:0)(γ +1)/γ(cid:1)2̺−k0log(γ+1)/(cid:0)log̺µ B(cid:1) (x ,2r )
0 0
(cid:0) (cid:1) (cid:0) γ(cid:1)
≤ (γ +1)/γ 2c ̺−1µ B(x ,r ) (r/r )6log(1/̺)
µ 0 0 0
(cid:0) (cid:1) (cid:0) (cid:1)
by using (2.4), (2.5), and the fact that µ(A) = 0. Recalling the definition of γ,
(cid:3)
the claim follows.
3. Main ideas
To show that a t-regular set A is uniformly ̺-porous for some ̺ > 0 whenever
0 < t < s, fix x ∈ A and small r > 0 and consider balls of radius 2−kr with
given k ∈ N. It follows from s-regularity that the number of such balls needed to
cover B(x,r) is at least a constant times 2ks. Taking any sufficiently separated
(by means of the 5r-covering theorem) subcollection from those balls so that
each ball intersects A, it follows from t-regularity that the cardinality of such a
subcollection is at most a constant times 2kt. Choosing k large enough gives the
claim.
To sketch theproofoftheother direction, let δ = c̺s/log(1/̺)beasinLemma
2.1 and take s−δ < t < s. We will next construct the set F ⊃ A. Assume for
simplicity that r and r are large. For all j ∈ N we use the 5r-covering theorem
µ p
POROSITY AND REGULARITY IN METRIC MEASURE SPACES 5
to find a collection of disjoint balls B x ,(1̺)j so that 5 times bigger balls
ji 2 i
cover A. Using the uniform ̺-porosi(cid:8)ty,(cid:0)choose B (cid:1)z(cid:9) ,̺(1̺)j ⊂ B x ,(1̺)j \A
ji 2 ji 2
for each i. Recalling the result of Saaranen, we(cid:0)construct a(cid:1)t-reg(cid:0)ular set F(cid:1) on
ji
each B z ,(1̺)j+1 =: B anddefine F = A∪ F . It suffices to show that ν,
ji 2 ji i,j ji
defined(cid:0)to be the su(cid:1)m of all t-regular measuresSof Fji, is t-regular on F. We will
show that there is a constant c ≥ 1 such that ν B(x,r) ≤ crt for every x ∈ A
and r > 0 small enough. To conclude that ν is t-(cid:0)regular(cid:1)is then straightforward.
Take x ∈ A, fix a scale k ∈ N, and denote N = i : B ∩ B x,(1̺)k 6=
j ji 2
∅ . It is clear that B ⊂ A (1̺)j for every j and(cid:8) i. If j ≤(cid:0)k − 1 t(cid:1)hen
ji 2
B(cid:9) ∩B x,(1̺)k = ∅ for all i.(cid:0)If j ≥(cid:1) k then B ⊂ B x,4(1̺)k+1 as i ∈ N .
ji 2 ji 2 j
Using th(cid:0)ese obse(cid:1)rvations and Lemma 2.1, we have (cid:0) (cid:1)
c#N ̺(j+1)s ≤ µ(B ) ≤ µ A∩B x,4(1̺)k+1 (1̺)j
j ji 2 2
iX∈Nj (cid:16)(cid:0) (cid:0) (cid:1)(cid:1)(cid:0) (cid:1)(cid:17)
(1̺)j δ
≤ cµ B x,4(1̺)k+1 2
2 (cid:18)4(1̺)k+1(cid:19)
(cid:0) (cid:0) (cid:1)(cid:1) 2
≤ c̺(s−δ)(k+1)+jδ
where µ is the s-regular measure, yielding #N ≤ c̺k(s−δ)−j(s−δ) for every j ≥ k.
j
Here c denotes a constant whose value may vary on each instance even within a
line. This implies, for any x ∈ A and (1̺)k+1 ≤ r < (1̺)k,
2 2
∞
ν B(x,r) ≤ ν B ∩B x,(1̺)k ≤ c #N ̺(j+1)t
ji 2 j
(cid:0) (cid:1) Xi,j (cid:0) (cid:0) (cid:1)(cid:1) Xj=k
∞ ̺kt−k(s−δ)
≤ c̺t+k(s−δ) ̺jt−j(s−δ) = c̺t+k(s−δ) ≤ crt
1−̺t−s+δ
X
j=k
which is exactly what we wanted.
References
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metric spaces, Illinois J. Math. 39 (1995), 676–694.
[2] J. Heinonen, Lectures on analysis on metric spaces, Springer-Verlag,New-York, 2001.
[3] E. J¨arvenp¨aa¨,M. J¨arvenp¨aa¨,A. K¨aenm¨aki, T. Rajala, S. Rogovin,and V. Suomala, Small
porosity, dimension and regularity in metric measure spaces, Preprint, 2007.
[4] O.MartioandM.Vuorinen,Whitneycubes, p-capacity, and Minkowski content,Exposition
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Cambridge University Press, Cambridge, 1995.
Department of Mathematics and Statistics, P.O. Box 35 (MaD), FI-40014 Uni-
versity of Jyva¨skyla¨, Finland
E-mail address: [email protected]