The interior of randomly perturbed self-similar sets on the line

Can we find a self-similar set on the line with positive Lebesgue measure and empty interior? Currently, we do not have the answer for this question for deterministic self-similar sets. In this paper we answer this question negatively for random self-similar sets which are defined with the construction introduced in the paper Jordan, Pollicott and Simon (Commun. Math. Phys., 2007). For the same type of random self-similar sets we prove the Palis-Takens conjecture which asserts that at least typically the algebraic difference of dynamically defined Cantor sets is either large in the sense that it contains an interval or small in the sense that it is a set of zero Lebesgue measure.


Introduction
In this paper we consider only Random self-similar Iterated Function Systems (RIFS) which are defined on the line and which can be obtained as a small random perturbation of a deterministic self-similar Iterated Function System (IFS) on the line.First we give a short description of our results for the expert, and then we provide a more detailed introduction.We do not write "self-similar" in the abbreviation since all iterated function systems considered in this paper are self-similar (random or deterministic).
1.1.Informal description of the main result for experts.Using the construction introduced by Jordan, Pollicott and Simon [7, p. 521], we define self-similar Random Iterated Function Sytems (RIFS) F on the line as follows: We start with a self similar IFS S on the line and we add a small random additive error to every map in every step of the iterative construction of the attractor (see Definition 2 for the precise definition of RIFSs).The scaling parts of the similarities of the deterministic IFS S are left unchanged.So, the similarity dimension s(F) of the RIFS F is the same as the similarity dimension of the deterministic self-similar IFS S. Our main result is that where C F is the attractor of the RIFS F. This implies that whenever C 1 , C 2 are two independent copies of the attractor of the RIFS F then the algebraic difference 1.2.A gentle introduction.A deterministic self-similar Iterated Function System (IFS) on R is a finite list of contracting similarities of R: with contractions r i ∈ (−1, 1) \ {0} and translations 1 t i ∈ R for all i ∈ [L] := {1, . . ., L}.The attractor C S of the IFS S is what we are left with after infinitely many iterations of the system.More formally, it is easy to see that we can find a non-degenerate compact interval I such that S i (I) ⊂ I holds for all i ∈ [L].For all i ∈ [L] n we consider the n-fold iterate (1.4) and form the corresponding n-cylinder interval I i := S i (I).Then the union of all n-cylinders is a nested sequence of non-empty compact sets.The attractor is their intersection: (1.5) C S := In this paper, we consider random IFSs (RIFS) on R, which are small (translational) perturbations of a self-similar IFS of the form (1.3).
Informally, the attractor of an RIFS is obtained by a formula similar to (1.5) with the following difference: Instead of the deterministic n-cylinder intervals I i in (1.5), we work with the random intervals , where the random mappings f i k are small translational perturbations of S i k .Namely, The precise description of the distribution of these random translations is given in Definition 2. The attractor C F of the RIFS F := {f i } L i=1 is defined by a formula analogous to (1.5): we just replace I i with I i in (1.5).
Jordan, Pollicott, and Simon [7] studied this kind of RIFSs in the more general self-affine case.As an immediate consequence of the results in [7], we get that the Hausdorff dimension dim H C F is the minimum of 1 and the similarity dimension s F (solution of (2.5)) almost surely.Moreover, if s F > 1 then the Lebesgue measure of C F is positive almost surely.
1.3.Motivation: the interior of the difference of random Cantor sets.In 1987, Palis and Takens [10] studying the dynamical behavior of diffeomorfisms presented a conjecture about the size of the algebraic difference of two Cantor sets.
Informally, if the size of the Cantor sets is large (see equation (1.6)) then the difference contains an interval.More precisely, if C 1 and C 2 are two Cantor sets then the algebraic difference where dim H denotes the Hausdorff dimension.
In 2001, De Moreira and Yoccoz ( [1]) proved the conjecture for generic dynamically generated non-linear Cantor sets.The conjecture has not been proven for generic linear Cantor sets.
In 1990, Per Larsson put the problem into a probabilistic context in [9], (see also [8]).He considered a very special family of two parameter random Cantor sets C a,b and proved the conjecture for a certain subset of a's and b's.Although the main idea of Larsson's argument is brilliant, unfortunately the proof contains significant gaps and incorrect reasonings.In 2011, three out of the four authors of the present paper gave a precise proof for Larsson's family in [2].We briefly recall the Larsson family from [2]: let a > 1 4 and 3a + 2b < 1.
Since the first condition is equivalent to dim H C a,b > 1/2, which is equivalent to equation (1.6).
Larsson's construction is as follows (see also Figure 1): first remove the interval We write C 1 a,b for their union.In both of the two level one intervals we repeat the same construction independently of each other and of the previous step.In this way we obtain four disjoint intervals of length a 2 .We emphasize that, because of independence, the relative positions of these second level intervals in the first level ones are in general completely different.Similarly, we construct the 2 n level n intervals of length a n .We call their union C n a,b .Then Larsson's random Cantor set is defined by As a corollary of the main result of this paper, we prove that the conjecture by Palis and Takens holds for a very broad class of random linear Cantor sets, including the Larsson family.
The following result is a generalization of the result in [2]: Let F be an RIFS (see Definition 2) with similarity dimension larger than 1 2 .Let C 1 and C 2 be two independent copies of the attractor C F .Then C 2 − C 1 contains an interval a.s.
The proof is presented in Section 5.
It is important to note that in our setting the Hausdorff dimension equals the (the minimum of 1 and) the similarity dimension given by the unique solution of equation (2.5).
We remark that if the Hausdorff dimension of C F is smaller than 1 2 then the set C 2 − C 1 has Hausdorff dimension less than 1 so it cannot possibly contain any intervals.
The essential part of the proof of this theorem is completely different of that of the main result in [2].The proof in [2] was tailored for the Larsson's family, and does not have the potential for generalizations.However, a combination of the ideas of the proof in [2] with the method introduced in Rams, Simon [12] and also invoking an observation from Peres, Shmerkin [11] makes it possible to prove a much more general result with a shorter proof.Namely, both in [2] and the present paper we have to verify that the associated multi-type branching processes are uniformly supercritical, where uniformity is meant in the type of the ancestor.
In both papers this is stated as the Main Lemma and their proofs follow the same path.However, the step where using the Main Lemma one proves the existence of intervals in the arithmetic difference of the random Cantor sets, is where we use the method introduced in [12] and this makes our present proof much more efficient.
We remark here that if also the scalings are random, then the problem becomes easier ( [13]), and can be treated by a method introduced by Hochman and Shmerkin ([5]).

RIFS
2.1.The formal description of our random Cantor set.First, we give the formal definition of the Random Iterated Functions System (RIFS) F, whose attractor C F is the random Cantor set which is the object of our investigation in this paper.It is convenient to identify the collection of all finite words over the alphabet [L] = {1, 2, . . ., L} with the nodes of the L-ary tree T .The empty word is identified with the root of T , and denoted as ∅.For any n ≥ 1 the level n sets L n of T are defined by The contraction ratios r 1 , . . ., r L ∈ (−1, 1)\{0} are deterministic.About the random translations (D 1 , . . ., D L ), of the functions in F in (2.1), we assume the following: (a) (D 1 , . . ., D L ) is an L-dimensional random variable such that for any i = 1, . . ., L, the random variable D i is absolutely continuous w.r.t. the Lebesgue measure, with a density function ϕ i which is strictly positive, bounded and continuous on (t i − θ i , t i + θ i ) and ϕ i is zero outside (t i − θ i , t i + θ i ), where the t i and θ i > 0 are real numbers.
(b) To define the random translations of the iterates of this system we introduce as a set of i.i.d.random vectors having the same distribution as that of (D 1 , . . ., D L ).
The iterates f i for i ∈ L n are defined as follows: (2.2) Using that the random mappings f i , i ∈ [L] are contractions and the supports of the D i are bounded, we immediately obtain that there exists a deterministic interval [α, β] such that We call [α, β] a supporting interval for F.
The attractor C F of the RIFS F is defined by where d(X) denotes the probability distribution of a random vector X.Then the ambient probability space is (Ω, B, P).A realization ω ω ω ∈ Ω is a labelled tree, ω ω ω = {ω i } i∈T , where ω i is defined for an i = i 1 . . .i n as follows: The dimension theory of the RIFS described above is well understood.The following Fact is a direct consequence of the results in [7].
Fact 3 (Dimension of an RIFS).Let F be an RIFS of size L and let s(F) denote the solution to the equation We say that s(F) is the similarity dimension of F. Then we have for almost all realizations: Moreover, if s(F) > 1 then for almost all realizations: where Leb(•) is the 1-dimensional Lebesgue measure.
With these definitions we can state our Main Theorem: Theorem 4 (Main Theorem).Let C F be the attractor of an RIFS F with s(F) > 1. Then The proof is presented in Section 5.

2.3.
The n-th order RIFS.To introduce the notion of an n-th order RIFS we consider the size n subtrees of T defined by for each k = k 1 . . .k .
Definition 5. Let F = {f i } L i=1 be an RIFS.The n-th order of F, written as F n , is defined for n = 1, 2, . . .by Actually F n is itself an RIFS.Lemma 6.Let F be an RIFS and let F n be its n-th order, for some n ≥ 1.Then F n is an RIFS.
Proof.Recall that the elements of F n indeed have the form f i (x) = r i x + T i , and (see (2.2)) that the random translations T i satisfy (2.8) are independent for all k ≥ 1 and j ∈ L (k−1)n , by disjointness of the size n subtrees rooted at the levels that are multiples of n, it follows that we have the required independent iteration scheme for F n .
At the cost of lowering the similarity dimension with an arbitrary small amount, we can assume that all the scalings in an RIFS are positive: see [3, Lemma 2.10] and a remark in the proof of Proposition 6 in [11].For completeness we combine the results of these references in the following proposition and its proof.
i=1 be an RIFS with attractor C F .Then there exists for every ε > 0 an RIFS F with C F ⊂ C F , such that all the contraction ratios of the similarities in F are positive, and s( F) > s(F) − ε.
Proof.In case all r i are positive, there is nothing to prove.Otherwise we may assume w.l.o.g. that r 1 < 0. For a natural number n, which will be chosen conveniently large later in the proof, we define for all i = i 1 . . .i n ∈ L n Here the r i are the contraction ratios of the f i (x) = r i x + T i mappings from F n . Since were we have taken n such that

It is enough to consider homogeneous systems
We call an RIFS homogeneous if all contraction ratios are the same.Using a simple combination of [3, Lemma 2.8] and [11, Proposition 6] it appears that any RIFS can be well-approximated by a homogeneous RIFS.This is Proposition 8.
Given an RIFS F of the form (2.1).Let U ⊂ L n , #U ≥ 2 for an n ≥ 1.We define (3.1) Here the random variables T i are defined in (2.2).According to Lemma 6, F n is an RIFS.Then F U is also an RIFS since F U is a subsystem of F n .For all realizations, the random attractor C F U of F U is a subset of the random attractor C F of F, for the same realization.
be an RIFS as in Definition 2. Then there exists for every ε > 0 a number n, a set U ⊂ L n , and an a ∈ (0, 1) such that the RIFS F U has the form and satisfies For the convenience of the reader, we give a detailed proof of this result in the Appendix.

The Main Lemma
A homogeneous system H has the form (4.1) Here we are motivated by Proposition 7: with an arbitrary small loss in similarity dimension we may assume that a ∈ (0, 1) instead of a ∈ (−1, 1) \ {0}.
It is convenient to introduce a slightly unusual notation.The support of a function f : X → R, for an arbitrary set X, is the set-theoretical support.That is This is slightly unusual since supp(f ) most commonly means the closure of the set in (4.2).
Proposition 9. Let H be a homogeneous RIFS and let C 1 and C 2 be two independent copies of the attractor of the RIFS.Then the algebraic difference C 2 − C 1 is the attractor of a homogeneous RIFS H with similarity dimension s(H ) = 2s(H).
Proof.Let the homogeneous RIFS H be given by ) is an L dimensional random vector such that for every i ∈ [L] the random variable D i is absolutely continuous w.r.t. the Lebesgue measure, with a density ϕ i which is bounded, continuous with supp Moreover, we require that D i and D i are independent.We define Then D i,j is absolutely continuous w.r.t. the Lebesgue mesure, with a density function ϕ i,j which is continuous, bounded with That is, ϕ i,j satisfies all the requirements we set for the density function in Part (a) of Definition 2. Using that, we can define the homogeneous RIFS which consists of a number of L 2 functions H := {ax + D i,j } L i,j=1 .It is straightforward to check that the attractor Λ of H is the algebraic difference of two independent copies of the attractor of H.
In the rest of the paper, we always assume that the following assumptions hold: Here A2 means the following: for all i ∈ [L] there exist an interval [−θ i , θ i ] and an absolutely continuous random variable Y i whose probability density function f i is continuous, bounded and has support supp The self-similarity property of an RIFS is expressed by the position of a point and we define the random variable where the symbol Θ represents that x ∈ J i .(b) For an x ∈ int(I) let φ i (x, •) be the density function of Φ i (x).Then an easy calculation yields that where f i is the density of the random variables Y i defined by (4.5).
The following function plays a crucial role in our argument Main Lemma 11.There exists a T (0) ⊂ I (the pre-type space) which is composed of a finite number of disjoint open intervals and there exists a real number ε MAIN > 0 such that for every ε ∈ (0, ε MAIN ), the so-called type space (4.9) The compact set T (ε) consists of as many intervals as T (0). ( Then there is an index N 0 for which the function m ε N 0 is uniformly positive and bounded on T (ε) × T (ε).
(3) The Perron-Frobenius eigenvalue of the operator The proof is given in Section 8.
The contents of the following theorem form the essential part of the proof of our main result Theorem 4.
Theorem 12. Let H be a homogeneous RIFS with s(H) > 1.Then int(C H ) = ∅ almost surely.Theorem 12 will be proved in Section 7, as a consequence of the Main Lemma 11.

Proof of Theorem 1 and Theorem 4 assuming Theorem 12
We first prove the Main Theorem: Proof of Theorem 4. Given an RIFS F with similarity dimension s(F) > 1.Then according to Proposition 8, we can find a homogeneous RIFS H := F U , with s(H) > 1.It follows from Theorem 12 that the attractor C H of the homogeneous RIFS H contains an interval almost surely.Then this interval is also contained in the attractor C F of the RIFS F.
Proof of Theorem 1.Given an RIFS F with similarity dimension larger than 1  2 .Let ε > 0 so that ε < s(F) − 1 2 .Using Proposition 8 there exists a homogeneous RIFS Let C 1 , C 2 be two independent copies of the attractor of F. Then we also have two independent copies C H contains an interval almost surely.
6.The multi type branching process Z On the probability space Ω we define a multi type branching process Z = (Z n ) ∞ n=0 .For a fixed a 0 < ε < ε MAIN the type space from Equation (4.9) is denoted by T := T (ε).Actually, the set of types is T ∪ {Θ}.By definition Θ is an absorbing state.That is, all descendants of Θ are equal to Θ.
Let Z 0 = {x}, where x ∈ T , and for an i ∈ [L] let Z i := Φ i (x).Then Note that although we speak of Θ as a type, it is not an element of T ⊂ [α, β].
To define the Z n , we need some preparations.We follow the definition of an RIFS given in Definition 2, but now from the viewpoint of the Y -vectors, instead of the D-vectors.So we are given as a set of i.i.d.random vectors having the same distribution as that of (Y 1 , . . ., Y L ).
For an i = i 1 . . .i n we define (6.1) Clearly, the iterates from (2.2) take the following form for a homogeneous RIFS (6.2) It follows from our assumption on the density f i of Y i that (identifying somewhat carelessly the support of an absolutely continuous random variable with the support of its density function) Let I = [α, β] be the supporting interval of H.We define the level-n (random) cylinder intervals The collection of all of these random level n intervals is denoted by I n .Note that these are intervals of length (β −α)a n .The endpoints of the random interval We get the level n children of an x ∈ T with H i (x) ∈ J i as follows: , where Φ i was defined in Definition 10.In the sequel we will use the notation with the H −1 i and their iterates.
In general, if x ∈ T and Z 0 = {x} and A ⊂ T is a Borel set, then for any n ≥ 1 the set of level n descendants of x contained in A is denoted by D n (x, A).So, (6.4) We remark that the process (Z n ) ∞ n=0 is a Markov chain since an individual in Z n gives birth to descendants independently of the individuals of the same generation if Z n−1 is given.
A major role in our analysis is played by the expectations where φ i (x, y) was defined in part (b) of Definition 10.It follows that Let for n ≥ 1 and A ⊂ T and x ∈ T , We remark that if M 1 has a kernel then M n also has a kernel.Let us write m n (x, •) for the kernel of M n (x, •).That is The branching structure of Z yields (see [4, p.67]) which was already introduced in (4.10),where one has to realize that in the notation we suppressed the dependence on ε of the kernel function m(•, •) in Section 6 and 7.
6.1.Supercritical branching processes with uniformly positive kernel.Part (2) of the Main Lemma asserts that there exists an integer N 0 such that m N 0 is a uniformly positive and uniformly bounded function.That is, there exist a min and a max such that for all x, y ∈ T we have We next consider the following two operators: These operators are closely related to the expectations of the branching process.
Note in particular that (6.10) We cite the following theorem from [4, Theorem 10.1]: Theorem 13 (Harris).It follows from (C1) that the operators in (6.9) have a common dominant eigenvalue ρ.Let f and g be the corresponding eigenfunctions of the first and second operator in (6.9) respectively.Then the functions f and g are bounded and uniformly positive on T .Moreover, apart from a scaling, f and g are the only non-negative eigenfunctions of these operators.Further, if we normalize f and g so that f (x)g(x) dx = 1, which will be henceforth assumed, then for all where the bound ∆ < 1 can be taken independently of x and y, and the constant C 1 is independent of x, y and n.
It follows from part (3) of the Main Lemma that the branching process Z is supercritical.That is the the Perron-Frobenius eigenvalue ρ is greater than one: 7. The proof of Theorem 12 assuming Main Lemma 11 Choose ε MAIN as in (8.26), and fix an ε and an η > 0 such that (7.1) 0 < ε < ε MAIN and 0 < η + ε < ε MAIN .
Lemma 14. Fix an ε ∈ (0, ε MAIN ).Let f be a F -eigenfunction, then there exist 0 < η < ε MAIN − ε and C 0 > 0 such that introducing the notation we have for any x ∈ T = T (ε) Proof.By the fact that f is the right eigenfunction of F corresponding to the Perron-Frobenius eigenvalue ρ we obtain that for every x ∈ T we have .
Hence, for all x ∈ T , The fact that C * > 0 follows from Harris' Theorem (Theorem 13).
Using the definition of m in (6.7),(4.7),and the properties of f i in A2 we obtain that there exists a number U such that (7.5) 0 ≤ m(x, y) ≤ U, for all (x, y) ∈ T × T.
We choose η > 0 so small that the second inequality below holds: Putting together (7.4) and (7.6) we obtain that Hence for all That is, (7.3) holds with By induction we obtain from Lemma 14 that for all x ∈ T and integers n ≥ 1 Dividing both sides by f (x), we therefore obtain the following corollary.
Corollary 15.For any n ≥ 1 we have ∀x ∈ T, Consequently, there exists a positive integer r such that for each n ≥ r Our next lemma is a corollary to the Hoeffding inequality [6]: Lemma 16 (Hoeffding).Assume Y 1 , . . ., Y C are independent random variables such that for any i = 1, . . ., C we have a i ≤ Y i ≤ b i for some real numbers a i , b i .Let S C = C i=1 Y i and let t be a positive real number.Then we have .
The following statement, Lemma 17 is a slight generalization of the easy part of the Cramér theorem.
Lemma 17.Let C ∈ N and Z 1 , . . ., Z C be a sequence of independent random variables such that Z i takes values from [0, L r ] and Let r be as in (7.7).There exists 0 < τ = τ (r) < 1 such that for all C ≥ 1 Proof.Let m i := E Z i and in general m x := EZ x .We have . Now, we want to apply the Hoeffding inequality with First, note that by the definition of r in (7.7) we have m x > 6 for any x ∈ T .
Therefore, t > 0. Further, ES C = 0, where we remind the reader that i=1 is contained in an interval of length L r .Thus, we have where we used m x > 6.Since τ = exp − 32 < 1, this proves the Lemma.
Definition 18.Let us denote by c 1 the length of the smallest interval in T (0).We choose n 1 such that a n 1 Leb(W ) ≈ c 1 , specifically, n 1 := log a c 1

Leb(W )
. Let 1 be the length of the smallest interval in W .
Lemma 19.For any fixed n ≥ 1 and for every ω ∈ Ω (where Ω is the sample space defined in Section 2.2) there exists a (random) interval J = J(ω) ⊂ T (0) of length Proof.The proof uses the observation that for any bounded integrable function h By the definition of Z n (x, W ) (see (6.4)) we have where we use in the step before the last step that H i (W ) ⊂ T (0).This follows from part (5) of the Main Lemma since W ⊂ T (0).
In this way, for every ω ∈ Ω there exists an We write G l n for the collection of those i ∈ G n for which the center of C i is to the left from H −1 i (x max ).So, for an i ∈ G l n there is an interval of length at least 1 /2, contained in W with right endpoint H −1 i (x max ).Let G r n := G n \ G l n .Then at least one of the sets G l n or G r n (say G l n ) has cardinality at least N (n).This means that for J := (x max − a n 1 /2, x max ) and for every x ∈ J and for every i ∈ G l n we have H −1 i (x) ∈ W . So, by (6.4), we have Z n (x, W ) ≥ #G l n = N (n) for all x ∈ J.To verify that J ⊂ T (0), pick an i ∈ G l n .Then J ⊂ H i (C i ).Using that C i ⊂ W ⊂ T (0) and H i (T (0)) ⊂ T (0) (see Fact 33) yields J ⊂ T (0).
We partition each interval of T (0) into intervals of equal length.If I is a connected component interval of T (0) then we partition it into L := 6 |I| 1 a n subintervals.In this way we obtain a partition of T (0) into the intervals J 1 , . . ., J L (labelled in increasing order) such that for any , Leb(J ) ≤ Leb(J)/3, where J was defined in Lemma 19.
Concerning (7.10).Let X k be a ηa 2n+kr dense set in J l , where η has been set in Lemma 14. X k can be chosen such that Fact 21.It follows from Lemma 17 that for any x ∈ J l (7.12) Before we prove Fact 21 we need to define H i→ij (x), in the following Fact.
Fact 22.Let i ∈ L n and j ∈ L m for some n, m > 0. We define the random function H i→ij on T by Consequently, Proof of Fact 21.Let x ∈ T and U ⊂ T .First we recall that D n (x, U ) was defined in (6.4).Using this for an i ∈ D n (x, T ) and r ≥ 1 now we define where x i := H −1 i (x), and (7.16) D q r,i (x, U ) := {j ∈ D r,i (x, U ) : j 1 = q} .Then by definition (7.17) Observe that for any x ∈ T and i ∈ D n (x, T ) So by Corollary 15 there is a (random) q = q(x, i, ω) ∈ [L] such that (7.19) E #D q r,i (x, W ) > 6.For an x ∈ T and for every i ∈ D n (x, T ), we write Y i := #D q r,i (x, W ). Observe that the random variables (7.20) {Y i } i∈Dn(x,T ) are independent.
Namely, according to (7.14) in Fact 22 for a j ∈ [L] r with j 1 = q the random part of . This implies that for distinct i, i ∈ D n (x, T ) and for j, j ∈ [L] r with j 1 = q = q(i) and j 1 = q = q( i) the random variables H −1 i→ij and H −1 i→ i j are independent.Hence we get that (7.20) holds.Finally, Y i takes values from [0, L r ] and by (7.17) we have (7.21) For an x ∈ J let (7.22) Then by virtue of Lemma 17 and the Markov property there exists a τ ∈ (0, 1) (where τ depends only on r) such that for all x ∈ J l we have: (7.23) To prove equation (7.10) we will use induction.More precisely, we will prove that the inequality holds for any positive integer M .
For M = 1, by (7.12) we obtain: Recall that X 1 was defined as an ηa 2n+r -dense subset of J l .Recall that by (7.11) we have #X 1 ≤ L n+r .Hence, (7.25) Next, our purpose is to extend the inequality (7.25) from all x ∈ X 1 to all x ∈ J l .
Let us fix k ≥ 1.
We will use the following fact.
Fact 23.For any k ≥ 1 and l = 1, . . ., L we have (7.26) and the set Proof of Fact 23.Using the definition of W = T (ε + η) and T = T (ε), if for some x and ω one has Z n+kr (x , W ) > 2 k N (n), then for the same ω and for any x such that |x − x | < ηa n+kr and larger set T we also have Further, since X k is ηa 2n+kr dense, for any x ∈ J l we can find x ∈ X k such that |x − x | < ηa 2n+kr < ηa n+kr .This proves (7.26).
It remains to be proved that the set in (7.27) is measurable which is formally not straightforward since x is running over an interval J l .
First, we note that it is enough to prove that for any fixed x ∈ X k the set We have to take into consideration two facts.T is a union of finite number of intervals and according to (6.5) Z n+kr (x, T ) is a sum of a finite number of indicator functions: Therefore, the function Z n+kr (•, T ) for any ω is a jump function on T with a finite number of jumps.Let {ι i : i ∈ I} denote the partition of T into the intervals on which Z n+kr (•, T ) is constant.So, Z n+kr (x, T ) depends on the interval ι i which x falls into.Therefore, The last set is given by a measurable function of a finite number of random variables hence measurable.
As a consequence of Fact 23 we can exchange X 1 with J l at a price of replacing the set W with the larger set T in (7.25).In this way we obtain For k = 0, using Lemma 19, the definition of Ω l and W ⊂ T , we have Therefore, we have (7.10) for M = 1: Now, assume that we have proved (7.24) for M − 1.We will prove it for M .
The simple fact that for any three events A, B, C of positive probability we have: By induction, it is known that the second term on the right hand side is larger than . Now we use a similar argument, to the one applied as in the case M = 1, for proving that the first term, on the right hand side in the displayed formula above, is larger than 1 − a M (n).Namely, let Ω l,M −1 denote the event in the condition of the first term: As in the proof of (7.12) we use Lemma 17.Let n = n + (M − 1)r, C = Z n+(M −1)r (x, T ), then for any x ∈ J l we have This can be proved in exactly the same way as (7.12) was proved.The continuation is also similar, we first take a dense set X M and prove the counterpart of (7.25), that is, Applying Fact 23 again yields T ).This finishes the proof of Lemma 20.Now, we are ready to present the proof of Theorem 12.
Proof of Theorem 12 assuming the Main Lemma.Using Lemma 20 we have Getting rid of the condition, we obtain that where in the last step we used (7.8).Since ξ can be chosen arbitrarily small this proves Theorem 12.
8. Construction of the pre-typespace T (0) and the type space T (ε) In this section we prove our Main Lemma 11.
We consider the support of m I : It is immediate that (8.1) supp(m where, It is easy to see that for all i ∈ [L], S i is a parallelogram with two horizontal sides: {(x, y) : y = α}, {(x, y) : y = β} and the two other sides are the following two lines of slope 1/a That is Clearly, (8.3) width In general the open filled parallelograms S i are not disjoint.Their union  Observe that the open filled parallelograms S k can be adjacent to each other.
We have (8.4) . The collection of parallelograms having two horizontal sides such that the slopes of the other two sides are equal to 1/a is denoted by P. Particular attention will be given to the parallelograms of the form P k u,v ∈ P, where u ≤ v, and a, b stands for an interval with endpoints a ≤ b about which we do not know if it is closed or open or half-closed and half-open.(8.5) Without loss of generality, we may assume that M k=1 S k is contained in the region between the Western side of S 1 (which is determined by the graph of the function In particular there exists a q such that H ∈ A, and Claim 25.The mapping Ψ satisfies (a) Let y ∈ I. Then Namely, part (a) follows from the second part of (8.2). it is clear that for all and so Ψ(H) consists of finitely many intervals.
The fact that all of these intervals are contained in I follows from the definition of We will prove in Claim 30 that N 0 is finite.
Definition 26.The pre-type space is defined by T (0) := V N 0 .
8.1.Elementary properties of the pre-type space T (0).Using that both V m ∈ A and int( I) \ V m ∈ A for every m we can find an n m , n m and α ≤ α . For an m ≥ 1, the level-m green and red areas are: (8.12) . Hence, In words, we get level m + 1 green area if we take away level m red from level m green area.This implies that the sets {R i } i are pairwise disjoint and where means disjoint union.In words: we get the level-m + 1 green area if we take aways from the level 1 green area all the first m level red areas.That is .
The following Lemma plays an important role in our proofs. .
This follows from elementary geometry by mathematical induction.
Proof of the Lemma.As we have mentioned, it is immediate from the construction that for every the open set π 1 (G ) ⊂ ( α, β) has finitely many components.It follows from (8.6) that if π 1 (G 1 ) \ π 1 (G 2 ) is not empty then it is the disjoint union of finitely many compact intervals.Assume that the same holds for an m ≥ 2 for the non-empty π .
Clearly, this is the case if and only if all endpoints of π 1 (G m−1 ) are also endpoints of π 1 (G m ).That is our induction hypotheses yields that for all i ≤ n m and j ≤ n m ( . Using this, we prove that whenever π 1 (G m ) \ π 1 (G m+1 ) is non-empty then it is the union of finitely many disjoint closed intervals.To verify this we assume that (8.17) Then by (8.12) there is a k ∈ [M ] and i ∈ [n m ], j ∈ [ n m ] and u ∈ G m−1 such that by (8.17) and the induction hypothesis we have Using that u Putting together this, (8.18), (8.17) and (8.12) we obtain that (8.20) Actually we proved a little more.
{a i,k , b i,k } then there is an i(x) ∈ [τ ] and k(x) ∈ [M ] such that Proof.Without loss of generality we may assume that x = a i, k .We know that (ay + t i − θ i , ay Consequently, Proof.The implication in (8.24) follows from (8.9) and from the fact that

25).
Lemma 34.There is an ε > 0 such that for all 0 < ε < ε the Perron Frobenius eigenvalue of the operator F ε is greater than 1.
The proof can be obtained by obvious modifications from the proof of [2, Lemma 8A].
(a) First we define (b) Fix an arbitrary 0 < ε < ε MAIN .The type space is defined by Claim 36.For an x 0 ∈ T (ε) we define . Then there exists a κ = κ(ε) > 0 such that for all x 0 ∈ T (ε) the set E ε 1 (x 0 ) contains an interval of length κ.
. Now we define the hexagon (8.29)Now we verify that (8.32) Namely, T (ε) ⊂ T (0) and in this way for all x 0 ∈ T (ε) there exists a k ∈ [M ] and an If the condition of (8.33) does not hold then we may assume, without loss of generality that (8.34) a i,k < x 0 < a i,k + ε.
Then we can apply Claim 32 to conclude that there exists an ∈ [M ] and j ∈ [τ ] such that a j, + d < a i,k < b j, − d.Putting together this (8.34) and (8.26) we obtain that a j, + ε < x 0 < b j, − ε.As we have seen above this means that x 0 ∈ π 1 (H j, ).
This proves that (8.32) holds.Putting together (8.32), (8.31) and (8.30) we get that the assertion of the Claim is true.
As a byproduct of the previous proof we obtain that (8.35) Namely, the non-trivial inclusion was verified above.The opposite inclusion is obvious by the definitions.
Our aim is to prove the following proposition which is actually Part (2) of the Main Lemma.
Proposition 37. Fix an 0 < ε < ε MAIN .Then there exists an n such that for every x 0 ∈ T (ε) we have

Hence we get
Namely, it follows from Fact 38 that every endpoint of a component of V m is an endpoint of a component of T (0).See Figure 6.This implies that Proof.This is immediate from the definitions (8.8) and (8.10).(8.40) Proof.Let z ∈ (α j , β j ) ∩ T (ε).Then it follows from (8.38) that there exists a w Before we start the proof of Proposition 37 we recall some definitions: We defined Furthermore, for n ≥ 1: Similarly, for a k ≥ 1, we defined Proof of Proposition 37. We fix an 0 < ε < ε MAIN and we write T := T (ε), k and κ := κ(ε) (defined in Claim 36).We divide the proof into two steps.First we recall (see part (a) of Definition 31) that the connected components of Step 1 There exists an N such that for all x ∈ T here exists an n(x) ≤ N such that there exists an 1 ≤ i ≤ τ with (8.45) [ Step 2 For every x ∈ T we have Let x ∈ T .Using (8.32) there exists an i ∈ Then either (α i + ε, β i − ε).Without loss of generality, we may assume that U 1 = (α i + ε, w) with w − α i − ε > κ.
Namely, assume that (α , β ) ∩ T (ε) ⊂ E n (x).Then k(j) (y) ∩ T ⊃ (α j , β j ) ∩ T (ε), (8.52)where in the last step we used Claim 42.In this way we have proved that (8.50) holds.Now we fix an x ∈ T and let n = n(x) be defined as in the proof of Step 1.We start with (α i , β i ) obtained in the first step.That is [α i + ε, β i − ε] ⊂ E n (x).By Fact 39 this implies that (α i , β i ) ∩ T ⊂ E n (x).We apply (8.50) N 0 times.The level N 0 -th child of (α i , β i ) ∈ G N 0 is the only element of G N 0 , which is ( α, β).So, we get that E n+N 0 (x) ⊃ ( α, β) ∩ T = T .On the other hand, it is immediate from the definition that E n+N 0 (x) ⊂ T .
The proof of Proposition 37 is immediate if we put together what we obtained in Steps 1, 2.
Proof of the Main Lemma.(1) follows from the definition of ε MAIN .
(2) It is easy to see that m n (x, y) is continuous on T (ε) × T (ε) for n ≥ 2. We apply Proposition 37 to obtain that for an n ≥ 2, m n (x, y) is positive for all (x, y) ∈ T (ε) × T (ε).The uniform positivity follows from the fact, mentioned above, that m n (x, y) is continuous on the compact set T (ε) × T (ε) for n ≥ 2.
(4) This follows easily from the fact that the right eigenfunction f is also an eigenfunction of (F ε ) 2 = F ε • F ε whose kernel m 2 is continuous.
(5) This was proved in Fact 33.Let C > 0 as in Lemma 43.For a large k, which will be conveniently chosen at the end of the proof, we can choose k 1 , . . ., k L ∈ N according to Lemma 43 such that we have the following

2 Figure 1 .
Figure 1.The construction of the Cantor set C a,b .The figure shows the level 1 and level 2 cylinder intervals C 1 a,b and C 2 a,b .

2 + a 2 from
the middle of [0, 1], then the length b parts from both the beginning and the end of the unit interval.Next, put intervals of length a according to a uniform distribution in the remaining two gaps b, 1 2 − a 2 and 1 2 + a 2 , 1 − b .These two randomly chosen intervals of length a are called the level one intervals of the random Cantor set C a,b .
relative to the endpoints of J i , for i ∈ [L].This corresponds to the position of a point that we call Φ i (x) relative to the endpoints of the supporting interval I = [α, β].This leads to the following definition.Definition 10.(a) For an i ∈ [L], we write where J = J(ω) is the interval defined in Lemma 19.Let Ω l = {ω : k(ω) = l} Note that, by Lemma 19 we have

1 i.
say M connected components {S k } M k=1 .By elementary geometry, for all k ∈ [M ] the connected component S k is also an open filled parallelogram having two horizontal sides and the Western non-horizontal sides is one of the lines from 2 i (x) L i=1 .Let us call it 2 k (x).While the Eastern non-horizontal side of S k is one of the lines from Let us call it 1 k (x).

Figure 3 .
Figure 3. Paralellograms S i on the left and paralellograms S k on the right.
α and β.Part (b) and (c) are obvious from the definition.Set (8.10) V m := Ψ m (int( I)) with V 0 := int( I).It follows from part (b) and (c) of Claim 25, that V m+1 ⊂ V m .Put (8.11) components of V m and V m−1 \ V m respectively.We say that the intervals in the first union are level m green intervals and the intervals in the second union are the level m red intervals.See Figure 4. Their collections are denoted by G m and R m respectively.That is
i +ε Then by simple elementary geometry this implies that the distance between x and the boundary ofπ 1 P k α (m−1) j ,β (m−1) jis at least θ min .Having a look at formula(8.12)we can see that x cannot be an endpoint of a component of π 1 (G m ).Hence, any endpoint of any component of π 1 (G m ) is also an endpoint of a certain component of π 1 (G m+1 ).
So, there is an i(x) ∈ [τ ] and a k(x) ∈ [M ] such that x ∈ (a i(x),k(x) , b i(x),k(x) ).Then by the definition of d, (8.22) holds.

8. 1 . 1 .B=1B
The structure of green and red areas.To prove Proposition 37 we need to verify some auxiliary facts about the structure of the green and red areas and intervals.For an m ≤ N 0 , we write B m and J m for the collection of the left and right endpoints respectively, of the level m red intervals (intervals from R m ), and B = ∪ N 0 i=1 B m and J = ∪ N 0 i=1 J m .It is immediate from the construction that the following statements hold: Fact 38.(a) For all 1 ≤ m ≤ N 0 and for all (α , β ) ∈ G m there exists v(α ) ∈ such that (α , β ) = (v(α ), u(β )).(b) All elements of m are right endpoints of an element of G m .Similarly, all elements of m =1 J are left endpoints of an element of G m .
Hence, when we change from T (0) to T (ε) we can loose in a 1 − 1 way intervals of length ε each at the two ends of the interval of L(z, k).Using that |L(z, k)| = w k 2ε we get that |L(z, k) ∩ T (ε)| > w − 2ε, where w is the minimum width of a stripe S k .

Figure 6 .
Figure 6.The red and green intervals .

Figure 8 .
Figure 8.The definition of the parallelograms P j .

m I in ( 4 . 8 )
and we saw that supp(m I ) = M k=1 S k .Moreover, in the statement of the Main Lemma we defined (8.41) m ε = m I • 1 T (ε)×T (ε) and m ε 1 := m ε .

(a) ( 1 k 2 (E 1
(x), 2 k (x)) ⊂ [α i + ε, β i − ε] or (b) either α i + ε or β i − ε is an endpoint of a component of E 1 (x) of length at least κ (c.f.(8.31)).If (b) holds then E 1 (x) has a component of length at least κ which has at least one endpoint in T .If (a) holds then we consider E (z) is still contained in a component of T then J is an interval of length at least (1/a)κ.We continue this process and we obtain that if we choose N 1 such that κ(1/a) N 1 > β − α then for an n = n(x) < N 1 , the set E n (x) has a component U 1 with length at least κ and with at least one of its endpoints contained in τ i=1

β
j − ε then |U 2 | ⊂ T (ε) and |U 2 | > (1/a)|U 1 |.So, we repeat the same for U 2 insteadof U 1 , and we get U 3 and so on until after uniformly bounded (not more than N 2 ) steps U k contains a component of T .This completes the proof for Step 1 with
Using this, and the definition of H i we get that the imageH i (y) of an y ∈ T (0) by the random mapping H i satisfies H i (y) ∈ (ay + t i − θ i , ay + t i + θ i ) for all i ∈ [L].Successive applications of this inclusion yields(8.