Hausdorff dimension of limsup random fractals

In this paper we find a critical condition for nonempty intersection of a limsup random fractal and an independent fractal percolation set defined on the boundary of a spherically symmetric tree. We then use a codimension argument to derive a formula for the Hausdorff dimension of limsup random fractals.


Introduction
A limsup random fractal on R N can be constructed as follows: (i) for each n ≥ 1, let D n denote the collection of all N -dimensional dyadic hyper-cubes of the form where k ∈ Z N + is an N -dimensional positive integer; (ii) for each n ≥ 1, let {Z n (I) : I ∈ D n } denote a collection of Bernoulli random variables with distribution P(Z n = 1) = q n ; (iii) a limsup random fractal, denoted by A, is then defined by A := where I • denotes the interior of I.
Limsup random fractals arise in the study of stochastic processes.Many interesting random sets are limsup random fractals.For example, the fast points of Brownian motion considered by Orey and Taylor [16], the thick points of Brownian motion investigated by Dembo, Peres, Rosen, and Zeitouni [1], and the Dvoretzky covering set on the unit circle studied by Li, Shieh and Xiao [10], to name only a few.
Limsup random fractals have intimate connection to packing dimension.In particular, if we assume that t = − lim n→∞ n −1 log 2 q n exists and call it the index of A, it is shown by Khoshnevisan, Peres, and Xiao [9] that for all Borel set F ⊂ R N , dim P (F ∩A) = dim P (F ) almost surely, provided that dim P (F ) > t and certain correlation bounds on the Hausdorff dimension of limsup random fractals random variables {Z n (I) : I ∈ D n } hold.Here dim P denotes packing dimension.However, the Hausdorff dimension of a limsup random fractal is unknown in general.It is shown in [9] that for all Borel sets F ⊂ R N with dim H (F ) > t, where dim H denotes Hausdorff dimension.For sets with equal Hausdorff and packing dimension, (1.2) gives the Hausdorff dimension of a limsup random fractal.However, it is well known that there are sets whose Hausdorff dimension is strictly less than its packing dimension.
In this paper we strive to derive a formula for the Hausdorff dimension of limsup random fractals.We notice that the construction of a limsup random fractal on the unit hypercube [0, 1] N generates a 2 N -nary tree T .More specifically, we can associate each sub-hypercube with a vertex, and connect two sub-hypercubes with an edge if one contains the other and the ratio of their side lengths is 2. The collection of all infinite rays of T is called the boundary of T , denoted by ∂T , and it can be made into a nice metric space.Under the assumption that the Bernoulli random variables Z n 's are independent, we have succeeded in obtaining a formula for the Hausdorff dimension of limsup random fractals defined on the boundary of a spherically symmetric tree, where a tree is said to be spherically symmetric if and only if all the vertices at the same generation have same number of children.The boundaries of these trees include many sets whose Hausdorff and packing dimension are different.Thus our result improves (1.2).
For each s ≥ 0, we define a new index on the boundary ∂T of a spherically symmetric tree T via the prescription δ s (∂T ) := lim where d is the tree metric on the boundary and P(∂T ) denotes the collection of all Borel probability measures supported on ∂T .By using this index, we are able to obtain a formula of the Hausdorff dimension of limsup random fractals.
Theorem 1.1.Let T be a spherically symmetric tree and A be a limsup random fractal with parameters {q n } n≥1 .Assume that t = − lim n→∞ n −1 ln q n exists and 0 < t < dim P (∂T ).Furthermore, assume the Bernoulli random variable Z n 's in the definition of the limsup random fractal are independent.Then ||dim H (A)|| L ∞ (P) = sup{s ≥ 0 : δ s (∂T ) > t}, (1.4) with the convention that sup ∅ := 0.
We use a codimension argument developed in Lyons [11] and Peres [17] to prove the above theorem.We consider a fractal percolation set defined on the same boundary of a spherically symmetric tree and independently with respect to the limsup random fractal.A fractal percolation set is defined in a similar manner as limsup random fractals, except that the Bernoulli random variables are independent and identically distributed and this random fractal consists of rays along which the Bernoulli random variables all equal to 1.The codimension argument (see Corollary 3.2) tells us that a nonrandom set will hit a fractal percolation set with positive probability if the Hausdorff dimension of this nonrandom set is large enough and the set will not hit the fractal percolation set almost surely if its Hausdorff dimension is too small.In Theorem 5.6, we strive to find a critical condition for the limsup random fractal hitting an independent fractal percolation set with positive probability.This critical condition will give us estimates of the Hausdorff dimension of the limsup random fractal.
The result of Khoshnevisan, Peres, and Xiao [9] also yields a codimension argument (see Theorem 3.5): a nonrandom set will hit a limsup random fractal almost surely if the packing dimension of this nonrandom set is large enough and it will not hit the limsup random fractal if its packing dimension is too small.As a result of the critical condition derived in Theorem 5.6, we also obtain the packing dimension of a fractal percolation set defined on a spherically symmetric tree in Corollary 5.7.
The remainder of the article is organized as follows.In Section 2, we give some background materials, including definitions for trees, Riesz capacity and fractal dimensions as well as their properties.Then we define random fractals on the boundary of a tree and review some known results in Section 3. In Section 4, we define two new indices and discuss their properties.Finally in Section 5, we estimates the hitting probability of a limsup random fractal and an independent fractal percolation set, and use codimension arguments and the new indices to compute the Hausdorff dimension of limsup random fractals and packing dimension of fractal percolation sets.We also give an example in which we explicitly calculate the dimension of the two random fractals.

Tree Topology
Let T = (V, E) denote a tree with distinct root o, where V is the collection of all vertices and E ⊂ V × V is the collection of all edges.Figure 1 gives an illustration of a typical tree.For x, y ∈ V, if x is on the path from o to y, then we call x an ancestor of y, call y a descendant of x, and use y x to denote this relation.In particluar if (x, y) ∈ E, then we call x the parent of y and y a child of x.For each x ∈ V, let deg(x) denote the number of children x has, that is, If deg(x) is finite for all x ∈ V, then T is called locally finite.We are interested in locally finite trees with infinitely many vertices.Moreover, if deg(x) = deg(y) whenever x and y have the same distance from root o, then the tree is called spherically symmetric.
For a tree T , a ray is an infinite path starting from o, that is, a sequence of vertices {o, v 1 , v 2 , . . .} ⊂ V such that (o, v 1 ) ∈ E and (v n , v n+1 ) ∈ E for all n ≥ 1.The collection of all rays is called the boundary of T and denoted by ∂T .For a ray σ = (o, v 1 , . . .), define σ(n) := v n and denote x ∈ σ if x = v n for some n ≥ 1.For two rays σ = (o, v 1 , . . . ) and γ = (o, u 1 , . . .), σ = γ if and only if v n = u n for all n ≥ 1.Furthermore, let σ γ denote the common vertex on σ and γ which is farthest from o.We can define a metric d on ∂T  The following proposition is well known.

Riesz Energy and Capacity
Let (X, d) be a general metric space.For every Borel subset G ⊂ X, let P(G) denote the collection of all Borel probability measures supported on G.For s ≥ 0 and µ ∈ P(X), the s-dimensional Riesz energy of µ is defined as E s (µ) := d(σ, γ) −s µ(dσ)µ(dγ). (2.4) And the s-dimensional Riesz capacity of G is defined as with the convention Cap s (∅) := 0. When X is the boundary of a tree and d is define in (2.2), we have special forms for the Riesz energy and capacity.For each µ ∈ P(∂T ), we write µ(x) := µ(B(x)).Furthermore, for x, y ∈ V, let σ x and σ y denote two rays such that x ∈ σ x and y ∈ σ y .Then we can define (2.6) Note that x y and x γ do not depend on the choices of σ x and σ y .

On one hand
On the other hand, for every ε > 0, we can find some ν ∈ P(G) such that (2.14) For each x with |x| = n and ν(x) > 0, by the definition of Riesz capacity, we can find some When the tree T is spherically symmetric, Riesz capacities can be obtained by computing energies of the uniform probability measure.
Lemma 2.5.Let T be a spherically symmetric tree and ν be the uniform probability measure on ∂T , that is, ν satisfies ν(x) = ν(y), for all |x| = |y| = m and all m ≥ 1. (2.16) (2.17) Proof.Since T is spherically symmetric, the degrees of the vertices at the same generation are equal.Thus we can let K n denote the degree of a vertex at generation n.If for each vertex we fix an order for its children, then each ray σ ∈ ∂T can be identified by a sequence of integers (l n ) n≥1 such that σ(n) is the l n th child of σ(n − 1).We will simply use σ = (l n ) n≥1 to denote this identification.Define a binary operation "+" on where σ = (l n ) n≥1 and γ = (m n ) n≥1 .It follows that (∂T, +) is a group.Furthermore, (∂T, +) is a topological group under the metric d.The proof of Theorem 3.1 of Khoshenvisan [8] implies that the equilibrium measure that minimizes the Riesz energies on a topological group is the Haar measure.This completes the proof.
For a Borel set G ⊂ ∂T and s ≥ 0, the collection of Borel probability measures supported on G with finite s-dimensional Riesz energy is of special interest.Let P s (G) denote this collection, that is, We have a necessary condition for µ ∈ P(G) to have finite s-dimensional Riesz energy.

Fractal Dimensions
In this part, we recall the definitions and properties of fractal dimensions.We refer to Mattila [14] for more details.Let (X, ρ) be a locally compact metric space and B be its Borel σ-algebra induced by the metric.For every subset F , let |F | denote the diameter of F , that is, |F | := sup{ρ(x, y) : x, y ∈ F }.For fixed s ≥ 0, define the s-dimensional Hausdorff measure H s by (2.26) We can consider packings instead of coverings to derive a packing measure.This was first done by Tricott in 1982 [18].For every F ⊂ X and δ > 0, a δ-packing of F is a countable collection of disjoint closed balls {B(x n , r n )} n≥1 such that x n ∈ F and r n < δ for all n ≥ 1.For each fixed s ≥ 0, define (2.27) The set function P s is a premeasure and we regularize it to obtain the s-dimensional packing measure P s : (2.28) EJP 18 (2012), paper 39.

Hausdorff dimension of limsup random fractals
Then for every F ⊂ X we define the packing dimension of F by dim P (F ) := inf{s ≥ 0 : P s (F ) = 0} = sup{s ≥ 0 : P s (F ) = ∞}. (2.29) Packing dimension can also be defined in terms upper Minkowski dimension.For every bounded set F ⊂ X and for each ε > 0, let N (F, ε) denote the smallest number of closed balls with radius ε that are needed to cover F : respectively.Theorem 5.11 of Matilla [14] states that for all F ⊂ X.Now let T be a tree and consider (X, ρ) = (∂T, d).Since (∂T, d) is an ultrametric space, Theorem 3 of Haase [4] shows that for each s > 0, (2.33) In fact, there exists F ⊂ ∂T such that dim H (F ) < dim P (F ).
Proof.The fact dim H (∂T ) = dim M (∂T ) when T is spherically symmetric is a standard result in Chapter 1 of Lyons and Peres [13].In order to show the other equality, we use a Baire category argument.Let N n denote the total number of vertices at generation n.
Then a monotone argument shows that dim M (∂T ) = lim n→∞ ln N n n . (2.34) For each x ∈ V, the closed ball B(x) can be regarded as the boundary of a subtree ), where V B(x) includes x's ancestors, x, and x's descendants and denote the total number of vertices at generation m of the subtree T B(x) .Since T is spherically symmetric, we have where n = |x|.
Example 2.9.We can use Lemma 2.8 to construct a tree T such that dim H (∂T ) < dim P (∂T ).Let {n m } m≥1 be a sequence of integers so that (2.37) Then the sequence {n m } m≥1 with n m = m i=1 k i satisfies this requirement.Now construct a tree T by the following scheme: if m = 2k − 1, then all vertices between generation n m + 1 and n m+1 have exactly 3 children; if m = 2k, then all vertices between generation n m + 1 and n m+1 have exactly 2 children.It follows that (2.38) Since this tree is spherically symmetric, we can apply Lemma 2.8 to see that dim H (∂T ) < dim P (∂T ).
3 Random Fractals On Trees

Definition of Random Fractals
In this section we define the main object we are studying, namely the limsup random fractal .For each x ∈ V, define a random variable Z x with distribution where n = |x| and 0 ≤ q n ≤ 1.Note that q n may vary as n changes.Define the limsup random fractal A with parameters {q n } n≥1 by Thus if σ ∈ A, then Z σ(n) = 1 for infinitely many n.Throughout this paper we will assume that and call t the index of the limsup random fractal A. Moreover, we assume that A satisfies the independence assumption: Let {W i } i∈I be a collection of subsets of V so that x i is the youngest (the root o is the oldest) common ancestor of all x ∈ W i .We assume that the collections of random variables {{Z x } x∈Wi } i∈I are mutually independent if x i x j for all i, j ∈ I.
There is a dual object of limsup random fractal, namely the fractal percolation set .
Define i.i.d.random variables {Y x } x∈V with where 0 ≤ p ≤ 1. Define the fractal percolation set E with parameter p by Thus if σ ∈ E, then Y σ(n) = 1 for all n ≥ 1.We will call s := − ln p the index of the fractal percolation set E.

Fractal Percolation Sets and Hausdorff Dimension
There is a close connection between fractal percolation sets defined in (3.5) and Hausdorff dimension.The following theorem is due to Lyons [11].
Theorem 3.1.(Lyons [11]) Let T be a tree and E be a fractal percolation set defined on ∂T with index s.Then: We can generalize this result to all Borel subsets of ∂T .Corollary 3.2.Let T be a tree and E be a fractal percolation set defined on ∂T with index s.Then for all Borel set F ⊂ ∂T : Proof.(i) First, if F is a closed subset, then we can regard F as the boundary of a subtree T F .In fact let T F := (V F , E F ), where In general, for every Borel subset F with dim H (F ) > s, we can find some t such that s < t < dim H (F ).Then Corollary 7 of Howroyd [5] implies the existence of a closed subset (ii) For every fixed s > dim H (F ), the definition of Hausdorff dimension implies H s (F ) = 0. Then for every ε > 0, we can find a ball covering When (∂T, d) is replaced by the Euclidean space R N equipped with the classic metric, the above corollary is same as Lemma 5.1 of Peres [17].Moreover we can apply the above corollary to an independent fractal percolation set and estimate the Hausdorff dimension of E.
Theorem 3.3.(Falconer [2], Mauldin and Williams [15]) Let T be a tree and E be a fractal percolation set defined on ∂T with index s.Then The fractal percolation set E is also closely related to the Riesz capacity of the boundary ∂T .Theorem 3.4.(Lyons [12]) Let T be a tree and E be a fractal percolation set defined on ∂T with index s.Then 1 2 Cap s (∂T ) ≤ P{E = ∅} ≤ 2Cap s (∂T ). (3.10)

Limsup Random Fractals and Packing Dimension
Limsup random fractals and packing dimension are closely related.The following theorem is due to Khoshnevisan, Peres, and Xiao [9].Theorem 3.5.(Khoshnevisan, Peres, and Xiao [9]) Let T be a tree and A a limsup random fractal defined on ∂T with index t.Then for all Borel set F ⊂ ∂T : (i) If dim P (F ) > t, then P(E ∩ F = ∅) = 1; and (ii) If dim P (F ) < t, then P(E ∩ F = ∅) = 0.
Remark 3.6.The limsup random fractals studied in [9] are defined on Euclidean spaces.The proofs there rely on the existence of closed sets with positive finite packing measure in Euclidean spaces (see Joyce and Preiss [6]).The existence of such closed sets in an ultrametric space is proved by Haase [4].Thus Theorem 3.1 of [9] still holds on the boundary of a tree.
We can compare this theorem to Lyon's Theorem (Corollary 3.2).Similar to Theorem 3.3, we can apply Theorem 3.5 to an independent limsup random fractal and estimate the packing dimension of A.
Theorem 3.7.(Khoshnevisan, Peres, and Xiao [9]) Let T be a tree and A a limsup random fractal defined on ∂T with index t.Assume that 0 < t < dim P (∂T ).Then with probability one, dim P (A) = dim P (∂T ).We may ask the Hausdorff dimension of a limsup random fractal and the packing dimension of a fractal percolation set.This is answered partially in [9].Theorem 3.8.(Khoshnevisan, Peres, and Xiao [9]) Let T be a tree.Define on the boundary ∂T a limsup random fractal A with index t and a fractal percolation set E with index s.Then:

New Indices
In order to compute the Hausdorff dimension of limsup random fractals, we introduce two families of indices on the boundary of a tree.where E s (µ) is the s-dimensional Riesz energy of a probability measure µ defined in (2.4), x denotes any vertex at generation n and µ x ∈ P(B(x) In particular, δ s (∂T ) is non-increasing in s.
Therefore δ s (∂T ) ≥ 0 for s < dim H (∂T ).The monotonicity of δ s (∂T ) completes the proof of the first part.
(ii) For every s > dim H (∂T ), we have s > dim H (B(x)) for all vertices x ∈ V. Then Frostman's lemma implies that E s (µ) = ∞ for all µ ∈ P(B(x)) and x ∈ V. Therefore J s (n; ∂T ) = ∞ for all n ≥ 1.This completes the proof of the second part.
(iii) This follows from the finiteness of δ s (∂T ) for 0 ≤ s < dim H (∂T ) and Lemma 4.3.
For each x ∈ V, we can treat the ball B(x) as the boundary of a subtree just as in the proof of Corollary 3.2.Then we can extend the index δ s to closed balls.Lemma 4.5.If T is a spherically symmetric tree, then δ s (∂T ) = δ s (B(x)) for all 0 ≤ s < dim H (∂T ) and all x ∈ V.
Proof.Since J s (n; ∂T ) is an energy form obtained by optimizing over all Borel probability measures supported on ∂T , we can use a similar argument as in the proof of where N n is the total number of vertices at generation n and x is an arbitrary vertex at generation n.If we fix n ≥ 1 and treat each closed ball as the boundary of a subtree, then for all |x| = n and m ≥ n, J s (m; ∂T ) = N −1 n J s (m; B(x)). (4.9) Since N n is finite, the above equation implies that δ s (∂T ) = δ s (B(x)).Lemma 4.8.For a fixed tree T : Proof.(i) Let s 0 := dim M (∂T ) − t.Then s 0 ≥ 0 and Lemma 4.3 implies that δ s0 (∂T ) + s 0 ≤ δ 0 (∂T ) + 0. According to Lemma 4.4 (i), we have δ 0 (∂T ) = dim M (∂T ).Therefore δ s0 (∂T ) ≤ t.This implies D t (∂T ) ≤ s 0 = dim M (∂T ) − t.
We show that δ s (∂T ) = dim P (∂T ) − s = ln 3 − s,  For each x ∈ V, let T x := (V x , E x ) be the subtree rooted at x such that V x = {x and all the descendents of x} and Note that there is a natural one-to-one correspondence between ∂T x and B(x).Let σ x denote the ray in ∂T x which corresponds to σ ∈ B(x).Then  For each Borel probability measure µ x on ∂T x , it naturally induces a Borel probability measure µ x on B(x), and vice versa.Then (4.15) implies that According to Lemma 2.5, we have where ν x and ν x are the uniform probability measures on ∂T x and B(x), respectively.
Thanks to (4.8) and (4.17), we obtain for any x ∈ V with |x| = n, where the second equality follows from (4.16).Since E s (ν x ) only depends on |x|, we denote it by I |x| .Thus For a k-nary tree (all vertices have degree k), it is well known that its boundary has Hausdorff dimension ln k.The structure of T implies that T x contains a binary tree and T x is contained in a ternary tree for all x ∈ V. Since s < ln 2 = dim H (∂T ), there exist constants c and C such that

Hitting Probabilities and Proof for the Main Results
Let T be a tree and A be a limsup random fractal defined on the boundary ∂T .According to Corollary 3.2, if E is an independent fractal percolation set defined on ∂T , then we can find estimates of the Hausdorff dimension of A by computing the probability of the event {A ∩ E = ∅}.
Recall the definition of limsup random fractals and fractal percolation sets from (3.2) and (3.5), respectively.We adopt the following notations for certain events: where σ(i) denotes the ith vertex on the ray σ.Then we have and In order to compute P{A ∩ E = ∅}, we estimate the probabilities of events {A n ∩ E = ∅} for all n ≥ 1.According to Theorem 3.4, we have P{B(x) ∩ E = ∅} > 0 if and only if Cap s (B(x)) > 0, where s is the index of the fractal percolation set E. We define for each µ ∈ P(∂T ) where {q n } n≥1 are the parameters associated with the limsup random fractal A and (5.5) Lemma 5.1.For fixed s ≥ 0, where P s (∂T ) denote the collection of Borel probability measures with finite s-dimensional Riesz energy.
Before we proceed to estimate the probability of the event {A n ∩ E = ∅}, we define a new capacity and give some technical lemmas for this capacity.For fixed constants s ≥ 0, K > 0 and n ≥ 1, define a kernel function on the boundary ∂T by (5.7) Then for each µ ∈ P(∂T ) define an energy form E(µ; s, K, n) := f (σ, γ; s, K, n)µ(dσ)µ(dγ). (5.8) where µ x ∈ P(B(x)) satisfies µ x (G) = µ(G ∩ B(x))/µ(x) for all G ∈ T , provided that µ(x) > 0. For each Borel set G ⊂ ∂T , we define a capacity . (5.10) Note that when K = 1, the above energy form and capacity are the same as the sdimensional Riesz energy and capacity, respectively.We will be interested in the capacity Cap(G; s, K, n) with large K and n.
Proof.(i) Let {a n } n≥1 be a sequence of real numbers such that lim n→∞ a n = 0 and q n = e −n(t+an) .We can choose r and r 0 such that δ s (∂T ) < r 0 < r < t and t−r < t−r 0 +a n for all sufficiently large n.Then there exists N > 1 such that J s (n; ∂T ) > e −nr0 and q −1 n J s (n; ∂T ) > e n(t−r) for all n > N .This implies that inf µ∈P(∂T ) for all n > N .Thanks to (5.9), we have Cap(∂T ; s, q −1 n , n) < e −n(t−r) for all n > N .Since t − r > 0, we obtain n≥1 Cap(∂T ; s, q −1 n , n) < ∞. (ii) Choose r and r 0 such that δ s (∂T ) > r 0 > r > t and r − t < r 0 − t − a n for all sufficiently large n.Then we can find a subsequence {n k } k≥1 such that J s (n k ; ∂T ) < e −n k r0 and q −1 n k J s (n k ; ∂T ) < e −n k (r−t) for all large k.This implies that where the first equality follows from Lemma 2.7 and the last one follows from (5.14), (5.15), and (5.16).
With the newly defined capacity, we can estimate the probability of the event {A n ∩ E = ∅}.The following two lemmas generalize the idea of the proof of Theorem 2.1 of Khoshnevisan [7].
Lemma 5.4.Let A be a limsup random fractal with index t and E be an independent fractal percolation set with index s.Assume that 0 < t < dim P (∂T ) and 0 < s < dim H (∂T ).Then for all n ≥ 1, (5.18) Proof.We first estimate P{I n µ > 0} for every fixed µ ∈ P s (∂T ) and then use Lemma Lemma 1.2), if X is a nonnegative random variable with finite second moment, then for all ε ∈ (0, 1), ] . (5.19) Thus in order to estimate P{I n µ > 0}, we will calculate the first two moments of I n µ .
On one hand, from the independence of the random variables {Y x } x∈V and {Z x } x∈V , we have where the second equality follows from (5.2) and the last equality follows from the definition of a µ (see (5.5)).
On the other hand, we apply the independence of the random variables {Y x } x∈V and {Z x } x∈V again to obtain p −|x y| µ(x)µ(y)., .
(5.23) Finally, we take supremum over all µ ∈ P s (B(x)) and apply Lemma 5.1 and Lemma 5.2 to obtain the desired result.
In order to find an upper bound for P{A n ∩ E = ∅} we need to introduce some new notations to describe the structure of the tree.For each n ≥ 1, we label all the vertices at generation n from left to right by x n 1 , x n 2 , . . ., x n Nn , where N n is the total number of vertices at generation n. (We use supscript to denote the generation and subscript to denote the position from left to right.)See Figure 3 for an illustration., . . ., Y x n in ) for 1 ≤ i ≤ N n , where i n = i and x k i k is the parent of x k+1 i k+1 for 1 ≤ k ≤ n − 1.In words, X n i denotes the vector consisting of the random variables {Y v } on the path from root o to the vertex x n i .In particular, Lemma 5.5.Let A be a limsup random fractal with index t and E be an independent fractal percolation set with index s.Assume that 0 < t < dim P (∂T ) and 0 < s < dim H (∂T ).Then for all n ≥ 1, (5.26) For every fixed µ ∈ P s (∂T ) and n ≥ 1, (5.4) implies that I n µ is bounded almost surely.
Thus we can define a bounded martingale {M n µ (k)} 1≤k≤Nn by (5.27) By iterated conditioning and independence, we have ( for k + 1 ≤ i ≤ N .We apply this inequality in (5.28), multiply (5.28) by 1{T n = k} on both sides, and notice the facts ( By the definition of T n and Theorem 3.4, we have ( Let J 1 (k, µ) denote the first summand in the last line of (5.30) and J 2 (k, µ) the second summand.If we use the above µ * n to replace µ, sum over k and take expectations, then we have where we used Theorem 3.4 to derive the last inequality, and  (5.61)If (5.60) holds for some ball B(x 0 ), then the Baire category argument carried at the beginning of (ii) shows that P{A ∩ E ∩ B(x 0 ) = ∅} > 0. In particular, P{A ∩ E = ∅} > 0.
(5.62) Otherwise (5.61) holds for all x ∈ V. Since V is countable, we can first remove a null event and then apply Proposition 3.6 of Falconer [3] to deduce that P{dim P (E) ≥ t } = P{E = ∅}. (5.63) This completes the proof of the theorem.Now we can use Theorem 5.6 to prove Theorem 1.1.In fact we prove a little more.
Corollary 5.7.Let T be a spherically symmetric tree.Then: (i) If A is a limsup random fractal defined on ∂T with index t and 0 < t < dim P (∂T ), then ||dim H (A)|| L ∞ (P) = D t (∂T ).
Proof.(i) For each s > 0, let E(s) be a fractal percolation set with parameter p = e −s so that E(s) is independent of A. Since T is spherically symmetric, Lemma 2.8 implies that dim P (∂T ) = dim M (∂T ).
First, consider D t (∂T ) = dim H (∂T ).On one hand by the monotonicity of Haudorff dimension, we have dim H (A) ≤ D t (∂T ) almost surely.On the other hand, for every 0 < s < D t (∂T ), we have δ s (∂T ) > t.Then Theorem 5.6 (ii) shows that P{A ∩ E(s) = ∅} > 0. Now condition on A and apply Corollary 3.2 to obtain P{dim H (A) > s} > 0. This shows that dim H (A) L ∞ (P) = D t (∂T ) when D t (∂T ) = dim H (∂T ).

(2. 3 )
In words, B(x) is the collection of all rays that pass through the vertex x.By definition, we have B(x) = B(σ, r) with r = e −|x| and any σ such that x ∈ σ.Finally, let T denote the Borel σ-alegbra generated by all closed balls.

Figure 1 :
Figure 1: A General Infinite Tree

Remark 2 . 2 .
In (2.2), if we replace e by any b > 1, the resulting topology does not change.In particular, Proposition 2.1 remains valid.

Figure 2
Figure 2 illustrates part of a limsup random fractal and/or a fractal percolation set.

Figure 2 :
Figure 2: A Graphical Interpretation: Solid Line for 1 and Dashed Line for 0

Figure 3 :
Figure 3: Labeling Vertices from Left to Right

(5. 25 )
Proof.Define T n := inf{k ≥ 1 :Z x n k = 1, o → x n k ,and x n k → ∞} with the convention inf ∅ := ∞.In words, T n is the first time (from left to right) such that A n ∩ E = ∅ at generation n.Then T n is a stopping time with respect to the filtration {F k n } 1≤k≤Nn and {T n < ∞} = {A n ∩ E = ∅}.

n
y| µ * n (x)µ * n (y) equality follows from the symmetry of the summation indices.We also change our notation from x n i back to x.Now combining (5.30), (5.33) and (5.34), we getE[M n µ * n (T n )1{T n < ∞}] (k)} 1≤k≤N isa nonnegative bounded martingale and T n is a stopping time, by Bounded Convergence Theorem and Optional Stopping Theorem, we haveE[M n µ * n (T n )1{T n < ∞}] ≤ E lim K→∞ M n µ * n (T n ∧ K) = lim K→∞ E[M n µ * n (T n ∧ K)]equality follows from (5.20).Combining this inequality with (5.35) and (5.26), we get P{A n ∩ E = ∅} ≤ 2 Finally take sup over µ ∈ P(∂T ) and apply Lemma 5.2 to obtain the desired result.Now we can use Lemma 5.4 and Lemma 5.5 to estimate the probability of the event {A ∩ E = ∅}.