2.1 Notation

Given a set S and $f,g:S\to (0,\infty )$ , we write $f\ll g$ if there exists $C>0$ such that $f(s)\leq Cg(s)$ for all $s\in S$ . We write $f\asymp g$ if $f\ll g$ and $g\ll f$ . Given an alphabet $\mathcal {A}$ , we let $\mathcal {A}^*=\bigcup _{n=1}^{\infty }\mathcal {A}^{n}$ denote the set of finite words with entries in $\mathcal {A}$ . Given two finite words $\mathbf {a},\mathbf {b}\in \mathcal {A}^{*}$ , we let $\mathbf {a}\wedge \mathbf {b}$ denote the maximal common prefix of $\mathbf {a}$ and $\mathbf {b}$ . If no such prefix exists we define $\mathbf {a}\wedge \mathbf {b}$ to be the empty word. We denote the length of a finite word $\mathbf {a}\in \mathcal {A}^*$ by $|\mathbf {a}|.$ Given $a\in \mathcal {A}$ and $n\in \mathbb {N}$ , we let $a^{n}=\overbrace {a\cdots a}^{\times n}.$

2.2 Fractal geometry

Let $Y\subset \mathbb {R}^{d}$ be a closed set. A map $\varphi :Y\to Y$ is called a contraction if there exists $r\in (0,1)$ such that $\|\varphi (x)-\varphi (y)\|\leq r\|x-y\|$ for all $x,y\in Y$ . Here $\|\cdot \|$ denotes the Euclidean norm on $\mathbb {R}^{d}$ . An IFS on Y is a finite set of contractions acting on Y. We will often suppress Y from our discussion and simply speak of an iterated function system or IFS. A well-known result due to Hutchinson [Reference Hutchinson30] states that for any IFS $\Phi =\{\varphi _{a}\}_{a\in \mathcal {A}}$ there exists a unique non-empty compact set $X\subset \mathbb {R}^{d}$ satisfying

$$ \begin{align*}X=\bigcup_{a\in \mathcal{A}}\varphi_{a}(X).\end{align*} $$

We call X the invariant set of the IFS. We will always assume that two of the contractions in our IFS have distinct fixed points. This ensures that the invariant set is non-trivial. When an IFS $\Phi $ consists of similarities we say that it is a self-similar IFS and the invariant set is a self-similar set. In the special case where each contraction in our IFS is $C^{1+\alpha }$ and angle preserving we will say that the IFS is self-conformal. By angle preserving, we mean that for all $a\in \mathcal {A}$ , $x\in Y,$ and $y\in \mathbb {R}^{d}$ we have $\|\varphi _{a}'(x)y\|=\|\varphi _{a}'(x)\|\cdot \|y\|$ . We will say that a self-similar IFS is affinely irreducible if there does not exist an affine subspace in $\mathbb {R}^{d}$ that is preserved by each element of the IFS. This is equivalent to the property that the invariant set is not contained in an affine subspace of $\mathbb {R}^d$ . Given $\mathbf {a}=(a_1,\ldots ,a_n)\in \mathcal {A}^{*}$ , we let $\varphi _{\mathbf {a}}=\varphi _{a_1}\circ \cdots \circ \varphi _{a_n}$ .

Given an IFS $\Phi =\{\varphi _{a}\}_{a\in \mathcal {A}}$ and a probability vector $\mathbf {p}=(p_{a})_{a\in \mathcal {A}}$ ( $\mathbf {p} $ is a probability vector if $\sum _{a\in \mathcal {A}} p_a=1$ and $p_{a}>0$ for all $a\in \mathcal {A}$ ), there exists a unique Borel probability measure $\mu $ satisfying $\mu =\sum _{a\in \mathcal {A}}p_{a}\varphi _{a}\mu $ where $\varphi _{a}\mu $ is the pushforward of $\mu $ under $\varphi _{a}$ . We call $\mu $ the stationary measure corresponding to $\Phi $ and $\mathbf {p}$ . It is a consequence of our underlying assumptions that the invariant set is non-trivial that $\mu $ is always non-atomic. When $\Phi $ consists of similarities we will say that $\mu $ is the self-similar measure corresponding to $\Phi $ and $\mathbf {p}$ . Similarly, when $\Phi $ is a self-conformal IFS we will say that $\mu $ is the self-conformal measure corresponding to $\Phi $ and $\mathbf {p}$ . We will say that a self-similar measure is affinely irreducible if the underlying self-similar IFS is affinely irreducible.

We finish this section by stating the following lemma that records some useful properties of self-conformal IFSs. For a proof of this lemma see [Reference Angelevska, Käenmäki and Troscheit7, Lemma 6.1].

Lemma 2.1. Let $\Phi $ be a self-conformal IFS. Then for any $x,y\in Y$ and $\mathbf {a}\in \mathcal {A}^{*}$ we have

$$ \begin{align*} \|\varphi_{\mathbf{a}}(x)-\varphi_{\mathbf{a}}(y)\|\asymp \mathrm{ Diam} (\varphi_{\mathbf{a}}(X))\cdot \|x-y\|. \end{align*} $$

Moreover, for any $\mathbf {a},\mathbf {b}\in \mathcal {A}^{*}$ we have

$$ \begin{align*} \mathrm{Diam}(\varphi_{\mathbf{a}\mathbf{b}}(X))\asymp \mathrm{ Diam}(\varphi_{\mathbf{a}}(X))\mathrm{Diam}(\varphi_{\mathbf{b}}(X)). \end{align*} $$

2.3 A general disintegration framework

In this section we will provide a general framework for disintegrating stationary measures. The ideas appearing in this section have their origins in a paper of Galicer et al. [Reference Galicer, Saglietti, Shmerkin and Yavicoli26]. These ideas have been used to study the absolute continuity of self-similar measures [Reference Käenmäki and Orponen31, Reference Saglietti, Shmerkin and Solomyak46], normal numbers in fractal sets [Reference Algom, Baker and Shmerkin2], and problems related to the Fourier decay of stationary measures [Reference Algom, Rodriguez Hertz and Wang3, Reference Algom, Rodriguez Hertz and Wang4, Reference Baker and Banaji10–Reference Baker and Sahlsten12].

Suppose we are given an IFS $\Phi =\{\varphi _{a}\}_{a\in \mathcal {A}}$ and a probability vector $\mathbf {p}$ . Let $\mathcal {A}_{1},\ldots , \mathcal {A}_{k}\subset \mathcal {A}$ be a collection of non-empty subsets of $\mathcal {A}$ satisfying $\bigcup _{i=1}^{k}\mathcal {A}_{i}=\mathcal {A}$ . We emphasize that we do not assume that $\mathcal {A}_{i}\cap \mathcal {A}_{j}=\emptyset $ for $i\neq j$ . We let $I=\{1,\ldots , k\}$ and $\Omega =I^{\mathbb {N}}$ . We let $\sigma :\Omega \to \Omega $ denote the usual left shift map, that is, $\sigma ((i_n))=(i_{n+1})$ for all $(i_n)\in \Omega $ . We define a probability vector $\mathbf {q}=(q_i)_{i=1}^{k}$ according to the rule

$$ \begin{align*} q_{i}=\sum_{a\in \mathcal{A}_{i}}\frac{p_{a}}{\#\{j\in I: a\in \mathcal{A}_{j}\}} \end{align*} $$

for each $i\in I$ . We let $\mathbb {P}$ denote the infinite product measure on $\Omega $ corresponding to $\mathbf {q}$ . Note that $\mathbb {P}$ is $\sigma $ -invariant. For each $i\in I$ we define a probability vector $\mathbf {q}_{i}=(q_{a}^{i})_{a\in \mathcal {A}_{i}}$ according to the rule

$$ \begin{align*} q_{a}^{i}=\frac{1}{q_{i}}\frac{p_{a}}{\#\{j\in I: a\in \mathcal{A}_{j}\}} \end{align*} $$

for each $i\in \mathcal {A}_{i}$ . Given $\omega =(i_n)\in \Omega $ , we let $\Sigma _{\omega }=\prod _{n=1}^{\infty }\mathcal {A}_{i_n}$ and $m_{\omega }=\prod _{n=1}^{\infty }\mathbf {q}_{i_n}.$ We emphasize that for all $\omega \in \Omega $ the support of $m_{\omega }$ is $\Sigma _{\omega }$ . Given $\omega \in \Omega $ , we also let ${\Pi _{\omega }:\Sigma _{\omega }\to \mathbb {R}^{d}}$ be given by

$$ \begin{align*} \Pi_{\omega}((a_n)_{n=1}^{\infty})= \lim_{n\to\infty}\varphi_{a_1\cdots a_n}(\mathbf{x}). \end{align*} $$

Here $\mathbf {x}$ is any vector in the domain of our IFS. For each $\omega \in \Omega $ we let $X_{\omega }=\Pi _{\omega }(\Sigma _{\omega })$ . It follows from the definition that $X_{\omega }$ satisfies a form of dynamical invariance which resembles that satisfied by invariant sets:

(2.1) $$ \begin{align} X_{\omega}=\bigcup_{a\in \mathcal{A}_{i_1}}\varphi_{a}(X_{\sigma(\omega)}). \end{align} $$

Moreover, iterating (2.1), we have the following relation for all $\omega \in \Omega $ and $n\in \mathbb {N}$ :

$$ \begin{align*} X_{\omega}=\bigcup_{\mathbf{a}\in \prod_{j=1}^{n}\mathcal{A}_{i_j}}\varphi_{\mathbf{a}}(X_{\sigma^{n}\omega}). \end{align*} $$

Last of all, we define $\mu _{\omega }=\Pi _{\omega }m_{\omega }$ . It follows immediately from the definition of $\mu _{\omega }$ that we have the following relation which resembles that satisfied by stationary measures:

(2.2) $$ \begin{align} \mu_{\omega}=\sum_{a\in \mathcal{A}_{i_1}}q_{a}^{i_1}\varphi_{a}\mu_{\sigma(\omega)}. \end{align} $$

Iterating (2.2), we have the following relation for all $\omega \in \Omega $ and $n\in \mathbb {N}$ :

$$ \begin{align*}\mu_{\omega}=\sum_{\mathbf{a}\in \prod_{j=1}^{n}\mathcal{A}_{i_j}}\prod_{j=1}^{n}q_{a_j}^{i_j}\cdot \varphi_{\mathbf{a}}\mu_{\sigma^{n}\omega}.\end{align*} $$

The following result shows that the above framework always gives a disintegration of a stationary measure.

Proposition 2.2. Let $\mu $ be the stationary measure for an IFS $\{\varphi _{a}\}_{a\in \mathcal {A}}$ and a probability vector $\mathbf {p}$ . Let $\mathcal {A}_{1},\ldots ,\mathcal {A}_{k}\subset \mathcal {A}$ satisfy $\bigcup _{i=1}^{k}\mathcal {A}_{i}=\mathcal {A}$ . Then for the family of measures $\{\mu _{\omega }\}_{\omega \in \Omega }$ defined above, we have the following disintegration of $\mu $ :

$$ \begin{align*}\mu=\int \mu_{\omega}\, d\mathbb{P}(\omega).\end{align*} $$

Various versions of Proposition 2.2 appear in the literature (see [Reference Algom, Baker and Shmerkin2, Reference Baker and Banaji10, Reference Galicer, Saglietti, Shmerkin and Yavicoli26, Reference Saglietti, Shmerkin and Solomyak46], for instance). The proof of this proposition is a minor adaptation of the arguments appearing in these papers and is therefore omitted. Our proof of Theorem 1.1 will rely upon a careful choice of $\mathcal {A}_{1},\ldots ,\mathcal {A}_{k}.$

3.1 Proof of Theorem 1.1 for self-conformal measures

Let us fix a self-conformal IFS $\Phi $ and a probability vector $\mathbf {p}$ . We let X denote the invariant set of $\Phi $ . Appealing to our underlying assumption that the IFS is non-trivial, we can assert that there exist $N\in \mathbb {N}$ and $a_1,a_2\in \mathcal {A}$ such that $\varphi _{a_1^{N}}(X)\cap \varphi _{a_{2}^{N}}(X)=\emptyset .$ Replacing N with a potentially larger integer, we can guarantee that the following property also holds. For any $\mathbf {a}\in \mathcal {A}^{N}$ , either ${\varphi _{\mathbf {a}}(X)\cap \varphi _{a_{1}^{N}}(X)=\emptyset }$ or $\varphi _{\mathbf {a}}(X)\cap \varphi _{a_{2}^{N}}(X)=\emptyset .$ It follows from the above, and the well-known fact that any stationary measure for an IFS can be realized as a stationary measure for an iterate of an IFS, that after relabelling digits there is no loss of generality in assuming that the following lemma holds.

We are now in a position to define our subsets of $\mathcal {A}$ so that we can perform our disintegration. For each $a\in \mathcal {A}$ let $\mathcal {A}_{a}=\{a,a_i\}$ where $a_i\in \{a_1,a_2\}$ has been chosen so that

(3.1) $$ \begin{align} \varphi_{a}(X)\cap \varphi_{a_i}(X)=\emptyset. \end{align} $$

This is permissible by Lemma 3.1. We will show that Theorem 1.1 holds for the $\Omega $ , $\mathbb {P}$ , and $\{\mu _{\omega }\}_{\omega \in \Omega }$ corresponding to $\{\mathcal {A}_{a}\}_{a\in \mathcal {A}}$ coming from Proposition 2.2. We emphasize that in this context $\Omega =\mathcal {A}^{\mathbb {N}}$ . At this point it is instructive to highlight an ambiguity in our notation. Elements of $\mathcal {A}$ are being used to encode two objects. They encode contractions in our IFS, as is standard. They also encode the sets with which we perform our disintegration, that is, the sets $\mathcal {A}_1,\ldots , \mathcal {A}_k$ (using the terminology of §2). Thus, to help with our exposition, when an element of $\mathcal {A}$ is being used to encode a contraction we will denote it by a, and when an element of $\mathcal {A}$ is being used to encode a subset of $\mathcal {A}$ we will denote it by b. Elements of $\Omega $ will be denoted by $(b_n)_{n=1}^{\infty }.$

We now make several straightforward observations that follow from the construction of the sets $\{\mathcal {A}_{b}\}_{b\in \mathcal {A}}$ . There exists $c_{1}>0$ such that for any $(b_n)\in \Omega ,$ if $a,a'\in \mathcal {A}_{b_1}$ are distinct then

(3.2) $$ \begin{align} d(\varphi_{a}(X),\varphi_{a'}(X))\geq c_{1}. \end{align} $$

This equation has the important consequence that there exist constants $c_{2},c_{3}>0$ such that for all $\omega \in \Omega $ we have

(3.3) $$ \begin{align} c_{2}\leq \mathrm{Diam}(X_{\omega})\leq c_{3}. \end{align} $$

Combining (3.3) with Lemma 2.1 implies that for any $\omega \in \Omega $ and $\mathbf {a}\in \mathcal {A}^{*}$ we have

(3.4) $$ \begin{align} \mathrm{Diam}(\varphi_{\mathbf{a}}(X_{\omega}))\asymp \mathrm{Diam}(\varphi_{\mathbf{a}}(X)). \end{align} $$

It also follows from the definition of $\mu _{\omega }$ and (3.1) that for any $\omega =(b_n)\in \Omega $ and $(a_1,\ldots a_n)\in \prod _{j=1}^{n}\mathcal {A}_{b_j}$ we have

(3.5) $$ \begin{align} \mu_{\omega}(\varphi_{a_1,\ldots, a_n}(X_{\sigma^{n}\omega}))=\prod_{j=1}^{n}q_{a_j}^{b_j}. \end{align} $$

Moreover, since each $\mathcal {A}_{b}$ contains at least two elements and our probability vector $\mathbf {p}$ is strictly positive, it follows that

$$ \begin{align*} q_{\min}:=\min_{b\in \mathcal{A}}\min_{a\in \mathcal{A}_{b}}\{q_{a}^{b}\}\quad\textrm{ and }\quad q_{\max}:=\max_{b\in \mathcal{A}}\max_{a\in \mathcal{A}_{b}}\{q_{a}^{b}\} \end{align*} $$

satisfy

(3.6) $$ \begin{align} 0<q_{\min}\leq q_{\max}<1. \end{align} $$

We will also require the following lemmas.

Lemma 3.2. For all $(b_n)\in \Omega $ and $\mathbf {a},\mathbf {a}^{\prime }\in \bigcup _{n=1}^{\infty }\prod _{j=1}^{n}\mathcal {A}_{b_{j}}$ such that $\mathbf {a}$ is not a prefix of $\mathbf {a}'$ and $\mathbf {a}^{\prime }$ is not a prefix of $\mathbf {a},$ we have

$$ \begin{align*}d(\varphi_{\mathbf{a}}(X),\varphi_{\mathbf{a}}^{\prime}(X))\asymp \mathrm{Diam}(\varphi_{|\mathbf{a}\wedge \mathbf{a}^{\prime}|}(X)).\end{align*} $$

Proof. Since $\varphi _{\mathbf {a}}(X),\varphi _{\mathbf {a}}^{\prime }(X)\subset \varphi _{|\mathbf {a}\wedge \mathbf {a}^{\prime }|}(X)$ for all $\mathbf {a},\mathbf {a}^{\prime }\in \mathcal {A}^*$ we trivially have

(3.7) $$ \begin{align} d(\varphi_{\mathbf{a}}(X),\varphi_{\mathbf{a}^{\prime}}(X))\leq \mathrm{Diam}(\varphi_{|\mathbf{a}\wedge \mathbf{a}^{\prime}|}(X)). \end{align} $$

We now focus on proving an inequality in the opposite direction. Let $\mathbf {a}=(a_n),\mathbf {a}^{\prime }=(a_{n}^{\prime })\in \bigcup _{n=1}^{\infty }\prod _{j=1}^{n}\mathcal {A}_{b_j}$ satisfy the assumptions of the lemma. Let us also assume that they have a common prefix, that is, $\mathbf {a}\wedge \mathbf {a}^{\prime }$ exists. The case where $\mathbf {a}\wedge \mathbf {a}'$ is the empty word follows immediately from (3.2). Let $x_{\mathbf {a}},x_{\mathbf {a}}^{\prime }$ be such that $d(\varphi _{\mathbf {a}}(X),\varphi _{\mathbf {a}^{\prime }}(X))=\|x_{\mathbf {a}}-x_{\mathbf {a}^{\prime }}\|.$ These points have to exist by compactness. There exist $\tilde {x}_{\mathbf {a}}\in \varphi _{a_{|\mathbf {a}\wedge \mathbf {a}^{\prime }|+1}}(X)$ and $\tilde {x}_{\mathbf {a}^{\prime }}\in \varphi _{a_{|\mathbf {a}\wedge \mathbf {a}^{\prime }|+1}^{\prime }}(X)$ such that $\varphi _{\mathbf {a}\wedge \mathbf {a}^{\prime }}(\tilde {x}_{\mathbf {a}})=x_{\mathbf {a}}$ and $\varphi _{\mathbf {a}\wedge \mathbf {a}^{\prime }}(\tilde {x}_{\mathbf {a}^{\prime }})=x_{\mathbf {a}^{\prime }}.$ Since $a_{|\mathbf {a}\wedge \mathbf {a}^{\prime }|+1}\neq a_{|\mathbf {a}\wedge \mathbf {a}^{\prime }|+1}^{\prime },$ it follows from (3.2) that $\|\tilde {x}_{\mathbf {a}}-\tilde {x}_{\mathbf {a}^{\prime }}\|\geq c_{1}$ . Using this inequality together with Lemma 2.1 yields

(3.8) $$ \begin{align} d(\varphi_{\mathbf{a}}(X),\varphi_{\mathbf{a}^{\prime}}(X))=\|x_{\mathbf{a}}-x_{\mathbf{a}^{\prime}}\|=\|\varphi_{\mathbf{a}\wedge \mathbf{a}^{\prime}}(\tilde{x}_{\mathbf{a}})-\varphi_{\mathbf{a}\wedge \mathbf{a}^{\prime}}(\tilde{x}_{\mathbf{a}^{\prime}})\|\gg \mathrm{Diam}(\varphi_{|\mathbf{a}\wedge \mathbf{a}^{\prime}|}(X)). \end{align} $$

Equations (3.7) and (3.8) together yield our result.

Proof. Let $\omega =(b_n)$ , r and x be as in the statement of our lemma. Since $x\in X_{\omega }$ there exists $(a_n)\in \Sigma _{\omega }$ such that $\Pi _{\omega }((a_n))=x$ . For the rest of our proof this $(a_n)$ is fixed. We let

$$ \begin{align*}N_{1}=\max\{n\in \mathbb{N}: B(x,2r)\cap X_{\omega}\subsetneq \varphi_{a_1\cdots a_{n}}(X_{\sigma^{n}\omega})\}\end{align*} $$

and

$$ \begin{align*}N_{2}=\min\{n\in \mathbb{N}: \varphi_{a_1\cdots a_{n}}(X_{\sigma^{n}\omega})\subset B(x,r) \}.\end{align*} $$

It follows from (2.1), (3.2) and our assumption $r\in (0,c_1/2)$ that $N_{1}$ is well defined. It is a consequence of (2.1) that $N_{1}<N_{2}$ . The third and fourth properties in the statement of our lemma follow immediately for this choice of $N_1$ and $N_2$ . It remains to show that $N_{2}-N_{1}$ can be uniformly bounded from above and that the second statement holds.

We focus on proving the second property holds, namely,

(3.9) $$ \begin{align} \mathrm{Diam}(\varphi_{a_1\cdots a_{N_1}}(X_{\sigma^{N_1}\omega}))\asymp \mathrm{Diam}(\varphi_{a_1\cdots a_{N_2}}(X_{\sigma^{N_2}\omega}))\asymp r. \end{align} $$

Since $\varphi _{a_1\cdots a_{N_2}}(X_{\sigma ^{N_2}\omega })\subset \varphi _{a_1\cdots a_{N_1}}(X_{\sigma ^{N_1}\omega })$ by (2.1), it will suffice to show that

(3.10) $$ \begin{align} \mathrm{Diam}(\varphi_{a_1\cdots a_{N_2}}(X_{\sigma^{N_2}\omega}))\gg r\quad \textrm{ and } \quad \mathrm{Diam}(\varphi_{a_1\cdots a_{N_1}}(X_{\sigma^{N_1}\omega}))\ll r. \end{align} $$

Let $N_{3}=\min \{n\in \mathbb {N}: \varphi _{a_1\cdots a_{n}}(X)\subset B(x,r) \}.$ Since $X_{\omega }\subset X$ for any $\omega \in \Omega $ it follows from the definition that $N_{2}\geq N_{3}.$ Appealing to well-known properties of self-conformal IFSs, it can be shown that the sequence $(\mathrm {Diam}(\varphi _{a_1\cdots a_{n}}(X)))_{n=1}^{\infty }$ satisfies

$$ \begin{align*} \mathrm{Diam}(\varphi_{a_1\cdots a_{n+1}}(X))\geq \kappa \mathrm{Diam}(\varphi_{a_1\cdots a_{n}}(X)) \end{align*} $$

for all $n\in \mathbb {N}$ for some $\kappa \in (0,1)$ that does not depend upon $(a_n)$ . Using this property together with the fact that $\Pi _{\omega }((a_n))=x$ , it follows that $\mathrm {Diam}(\varphi _{a_1\cdots a_{N_{3}}}(X))\gg r$ . Combining this with (3.4) then implies that

$$ \begin{align*} \mathrm{Diam}(\varphi_{a_1\cdots a_{N_{3}}}(X_{\sigma^{N_{3}}\omega}))\gg r. \end{align*} $$

We have $N_{2}\leq N_{3}$ so $\varphi _{a_1\cdots a_{N_{3}}}(X_{\sigma ^{N_{3}}\omega })\subset \varphi _{a_1\cdots a_{N_{2}}}(X_{\sigma ^{N_{2}}\omega })$ by (2.1). Therefore, $\mathrm {Diam}(\varphi _{a_1\cdots a_{N_{2}}}(X_{\sigma ^{N_{2}}\omega }))\geq \mathrm {Diam}(\varphi _{a_1\cdots a_{N_{3}}}(X_{\sigma ^{N_{3}}\omega })).$ Thus, the above implies that

(3.11) $$ \begin{align} \mathrm{Diam}(\varphi_{a_1\cdots a_{N_{2}}}(X_{\sigma^{N_{2}}\omega}))\gg r \end{align} $$

and we have proved the first part of (3.10).

We now focus on the second inequality in (3.10). It follows from the definition of $N_{1}$ and (2.1) that there exist distinct $a,a'\in \mathcal {A}_{b_{N_1+1}}$ such that

$$ \begin{align*}B(x,2r)\cap \varphi_{a_1\cdots a_{N_{1}}a}(X_{\sigma^{N_{1}+1}\omega})\neq\emptyset \quad \textrm{ and }\quad B(x,2r)\cap \varphi_{a_1\cdots a_{N_{1}}a'}(X_{\sigma^{N_{1}+1}\omega})\neq\emptyset.\end{align*} $$

Therefore,

$$ \begin{align*}d(\varphi_{a_1\cdots a_{N_{1}}a}(X_{\sigma^{N_{1}+1}\omega}),\varphi_{a_1\cdots a_{N_{1}}a'}(X_{\sigma^{N_{1}+1}\omega}))\ll r.\end{align*} $$

Since $X_{\omega }\subset X$ for all $\omega \in \Omega $ , this now implies that

$$ \begin{align*}d(\varphi_{a_1\cdots a_{N_{1}}a}(X),\varphi_{a_1\cdots a_{N_{1}}a'}(X))\ll r.\end{align*} $$

By Lemma 3.2 this implies that

Since for all $\omega \in \Omega ,$ this in turn implies that

(3.12) $$ \begin{align} \mathrm{Diam}(\varphi_{a_1\cdots a_{N_{1}}}(X_{\sigma^{N_{1}}\omega}))\ll r. \end{align} $$

Thus, we have established the second part of (3.10). As previously remarked, this together with (3.11) implies (3.9). Thus, the second property in the statement of our lemma holds.

We now focus on establishing the uniform upper bound for $N_2 - N_1$ . By (3.4) we have

$$ \begin{align*}\mathrm{Diam}(\varphi_{a_1\cdots a_{N_1}}(X_{\sigma^{N_1}\omega}))\asymp \mathrm{Diam}(\varphi_{a_1\cdots a_{N_1}}(X))\end{align*} $$

and

$$ \begin{align*}\mathrm{Diam}(\varphi_{a_1\cdots a_{N_2}}(X_{\sigma^{N_2}\omega}))\asymp \mathrm{Diam}(\varphi_{a_1\cdots a_{N_2}}(X)).\end{align*} $$

Combining these equations with our second property and Lemma 2.1, we have

$$ \begin{align*} 1\asymp \frac{r}{r}\asymp \frac{\mathrm{Diam}(\varphi_{a_1\cdots a_{N_2}}(X_{\sigma^{N_2}\omega}))}{\mathrm{Diam}(\varphi_{a_1\cdots a_{N_1}}(X_{\sigma^{N_1}\omega}))}\asymp\frac{\mathrm{Diam}(\varphi_{a_1\cdots a_{N_2}}(X))}{\mathrm{Diam}(\varphi_{a_1\cdots a_{N_1}}(X))}\asymp \mathrm{Diam}(\varphi_{a_{N_1+1}\cdots a_{N_2}}(X)).\end{align*} $$

Since $\mathrm {Diam}(\varphi _{a_{N_1+1}\cdots a_{N_2}}(X))\ll \gamma ^{N_2-N_1}$ for some $\gamma \in (0,1)$ it follows that $1\ll \gamma ^{N_2-N_1}$ . Hence $N_{2}-N_{1}$ must be uniformly bounded from above.

We now give our proof of Theorem 1.1 for self-conformal measures.

Proof of Theorem 1.1 for self-conformal measures

Statement $(1)$ of this theorem is the content of Proposition 2.2. Let us now focus on statement $(2)$ . Let $\omega =(b_n)\in \Omega , x\in \mathrm {supp}(\mu _{\omega })$ and $0<r\leq c_{1}/2$ . We emphasize that to establish statement $(2)$ it suffices to consider r in this interval. Applying Lemma 3.3, we can assert that there exist two words $\mathbf {a},\mathbf {a}'$ such that $\mathbf {a}$ is a prefix of $\mathbf {a}', |\mathbf {a}'|-|\mathbf {a}|\ll 1,$ and

$$ \begin{align*} B(x,2r)\cap X_{\omega}\subset \varphi_{\mathbf{a}}(X_{\sigma^{|\mathbf{a}|}\omega})\quad \textrm{and}\quad\varphi_{\mathbf{a}'}(X_{\sigma^{|\mathbf{a}'|}\omega})\subset B(x,r). \end{align*} $$

Therefore, by (3.5) we have

$$ \begin{align*}\mu_{\omega}(B(x,2r))\leq \mu_{\omega}(\varphi_{\mathbf{a}}(X_{\sigma^{|\mathbf{a}|}\omega}))=\prod_{j=1}^{|\mathbf{a}|}q_{a_j}^{b_{j}} \end{align*} $$

and

$$ \begin{align*}\prod_{j=1}^{|\mathbf{a}'|}q_{a_{j}'}^{b_{j}}= \mu_{\omega}(\varphi_{\mathbf{a}'}(X_{\sigma^{|\mathbf{a}'|}\omega}))\leq \mu_{\omega}(B(x,r)). \end{align*} $$

Now using these inequalities together with the facts that $\mathbf {a}$ is a prefix of $\mathbf {a}'$ and ${|\mathbf {a}'|-|\mathbf {a}|\ll 1}$ , we have

$$ \begin{align*} \frac{\mu_{\omega}(B(x,2r))}{\mu_{\omega}(B(x,r))}\ll \frac{\prod_{j=1}^{|\mathbf{a}|}q_{a_{j}}^{b_{j}}}{\prod_{j=1}^{|\mathbf{a}'|}q_{a_{j}'}^{b_{j}}}& =\bigg(\prod_{j=|\mathbf{a}|+1}^{|\mathbf{a}'|}q_{a_{j}'}^{b_{j}}\bigg)^{-1}\leq q_{\min}^{-(|\mathbf{a}'|-|\mathbf{a}|)}\ll 1. \end{align*} $$

In the final inequality we used (3.6). Thus, $\mu (B(x,2r))\ll \mu (B(x,r))$ and statement $(2)$ follows.

We now turn our attention to statement $(3)$ of our theorem. Let $(b_n)\in \Omega $ , $x\in \mathrm {supp}(\mu _{\omega })$ , $y\in \mathbb {R}^{d}$ , $0<r<c_{1}$ and $0<\epsilon <1$ . We emphasize that there is no loss of generality in restricting to $0<r<c_{1}$ and $0<\epsilon <1$ . We may also assume without loss of generality that $y\in \mathrm {supp}(\mu _{\omega })$ and that $B(y,\epsilon r)\subset B(x,r).$ This final reduction is permissible because of the doubling property guaranteed by the already established statement $(2)$ .

By Lemma 3.3 there exist $\mathbf {a},\mathbf {a}'\in \bigcup _{n=1}^{\infty }\prod _{j=1}^{n}\mathcal {A}_{b_j}$ such that:

• • $\mathrm {Diam}(\varphi _{\mathbf {a}}(X_{\sigma ^{|\mathbf {a}|}\omega }))\asymp r$ ;

• • $\mathrm {Diam}(\varphi _{\mathbf {a}^{\prime }}(X_{\sigma ^{|\mathbf {b}|}\omega }))\asymp \epsilon r$ ;

• • $B(x,r)\cap X_{\omega }\subset \varphi _{\mathbf {a}}(X_{\sigma ^{|\mathbf {a}|}\omega })$ ;

• • $\varphi _{\mathbf {a}^{\prime }}(X_{\sigma ^{|\mathbf {a}^{\prime }|}\omega })\subset B(y,\epsilon r)$ .

The analysis given in the proof of statement $(2)$ also yields the following properties of $\mathbf {a}$ and $\mathbf {a}^{\prime }$ :

(3.13) $$ \begin{align} \mu_{\omega}(B(x,r))\asymp \mu_{\omega}(\varphi_{\mathbf{a}}(X_{\sigma^{|\mathbf{a}|}\omega})) \end{align} $$

and

(3.14) $$ \begin{align} \mu_{\omega}(B(y,\epsilon r))\asymp \mu_{\omega}(\varphi_{\mathbf{a}^{\prime}}(X_{\sigma^{|\mathbf{a}^{\prime}|}\omega})). \end{align} $$

It follows from the inclusions $B(y,\epsilon r)\subset B(x,r)$ and $B(x,r)\cap X_{\omega }\subset \varphi _{\mathbf {a}}(X_{\sigma ^{|\mathbf {a}|}\omega })$ that $\mathbf {a}$ must be a prefix of $\mathbf {a}^{\prime }$ . As a consequence of this fact, the above, (3.4), and Lemma 2.1, we have that

(3.15) $$ \begin{align} \epsilon \asymp \frac{\epsilon r}{r}\asymp \frac{\mathrm{Diam}(\varphi_{\mathbf{a}^{\prime}}(X_{\sigma^{|\mathbf{a}^{\prime}|}\omega}))}{\mathrm{Diam}(\varphi_{\mathbf{a}}(X_{\sigma^{|\mathbf{a}|}\omega}))}\asymp \frac{\mathrm{Diam}(\varphi_{\mathbf{a}^{\prime}}(X))}{\mathrm{Diam}(\varphi_{\mathbf{a}}(X))}\asymp \mathrm{Diam}(\varphi_{a_{|\mathbf{a}|+1}^{\prime}\cdots a_{|\mathbf{a}^{\prime}|}^{\prime}}(X)). \end{align} $$

Appealing to well-known properties of self-conformal IFSs, it can be shown that there exist $\gamma ,c_1\in (0,1)$ such that $\mathrm {Diam}(\varphi _{\mathbf {a}}(X))\geq c\gamma ^{|\mathbf {a}|}$ for all $\mathbf {a}\in \mathcal {A}^{*}$ . Substituting this inequality into (3.15) implies that there exist $c_{2},c_{3}>0$ such that

(3.16) $$ \begin{align} |\mathbf{a}'|-|\mathbf{a}|\geq -c_{2}\log \epsilon -c_{3}. \end{align} $$

We now observe that

$$ \begin{align*} \frac{\mu_{\omega}(B(y,\epsilon r)\cap B(x,r))}{\mu_{\omega}(B(x, r))}\stackrel{({3.13}),\, ({3.14})}{\asymp} \frac{\mu_{\omega}(\varphi_{\mathbf{a}^{\prime}}(X_{\sigma^{|\mathbf{a}^{\prime}|}\omega}))}{\mu_{\omega}(\varphi_{\mathbf{a}}(X_{\sigma^{|\mathbf{a}|}\omega}))} & \stackrel{({3.5})}{=}\bigg(\prod_{j=1}^{|\mathbf{a}^{\prime}|}q_{a_{j}^{\prime}}^{b_{j}}\bigg)\cdot \bigg(\prod_{j=1}^{|\mathbf{a}|}q_{a_{j}}^{b_{j}}\bigg)^{-1} \\&\kern5pt =\prod_{j=|\mathbf{a}|+1}^{|\mathbf{a}^{\prime}|}q_{a_{j}'}^{b_{j}} \\&\kern4pt \leq q_{\max}^{|\mathbf{a}^{\prime}|-|\mathbf{a}|} \\& \stackrel{({3.16})}{\ll} q_{\max}^{-c_{2}\log \epsilon} \\&\kern5pt =\epsilon^{\alpha} \end{align*} $$

where $\alpha =-c_{2}\log q_{\max }.$ In the second equality we have used that $\mathbf {a}$ is a prefix of $\mathbf {a}'$ . The fact that $\alpha>0$ follows from (3.6). Thus, statement $(3)$ holds and our proof is complete.

3.2 Proof of Theorem 1.1 for affinely irreducible self-similar measures

Throughout this section we fix an affinely irreducible self-similar IFS $\Phi $ and a probability vector $\mathbf {p}$ . We begin by making some simplifications that are permissible because of our affinely irreducible assumption.

Suppose that for all $n\in \mathbb {N}$ there exists an affine subspace $W_{n}<\mathbb {R}^{d}$ such that our self-similar set X satisfies $X\subset W_{n}^{(1/n)}$ . Then, appealing to a compactness argument, there must exist an affine subspace $W<\mathbb {R}^{d}$ such that $X\subset W$ . This contradicts our affinely irreducible assumption, and so there must exist $\epsilon _{0}>0$ such that for any affine subspace $W<\mathbb {R}^{d}$ we have $X\setminus W^{(\epsilon _0)}\neq \emptyset .$ Replacing $\epsilon _0$ with a potentially smaller constant, it follows that there exist $x_1,\ldots , x_{n}\in X$ such that the following properties are satisfied.

• • $B(x_i,\epsilon _0)\cap B(x_j, \epsilon _0)=\emptyset $ for $i\neq j$ .

• • For any $x\in X$ there exists $1\leq i\leq n$ such that $B(x_i,\epsilon _{0})\subset B(x,3\epsilon _{0})$ .

• • For any affine subspace $W{\kern-1pt}<{\kern-1pt}\mathbb {R}^{d}$ there exists $1{\kern-1pt}\leq{\kern-1pt} i{\kern-1pt}\leq{\kern-1pt} n$ such that ${B(x_i,\epsilon _0){\kern-1pt}\cap{\kern-1pt} W^{(\epsilon _0)}=\emptyset }$ .

Now using the self-similar structure of X, these properties imply that there exist $N\in \mathbb {N}$ and $\mathbf {a}_1,\ldots ,\mathbf {a}_n\in \mathcal {A}^{N}$ such that the following statements hold.

• • $\varphi _{\mathbf {a}_i}(X)\cap \varphi _{\mathbf {a}_j}(X)=\emptyset $ for $i\neq j$ .

• • For any $x\in X$ there exists $1\leq i\leq n$ such that $\varphi _{\mathbf {a}_{i}}(X)\subset B(x,3\epsilon _{0})$ .

• • For any affine subspace $W<\mathbb {R}^{d}$ there exists $1\leq i\leq n$ such that $\varphi _{\mathbf {a}_i}(X)\cap W^{(\epsilon _0)}=\emptyset $ .

Now replacing N with a larger integer if necessary, and $\epsilon _0>0$ with a potentially smaller quantity if necessary, the above implies that there exist $N\in \mathbb {N}$ and $\mathbf {a}_{1},\ldots \mathbf {a}_{n}$ such that the following properties are satisfied.

• • For any $\mathbf {a}\in \mathcal {A}^{2N}$ there exists $1\leq i\leq n$ such that

  $$ \begin{align*} \varphi_{\mathbf{a}}(X)\cap \bigcup_{j=1}^{n}\varphi_{\mathbf{a}_{i}\mathbf{a}_{j}}(X)=\emptyset. \end{align*} $$

• • For any $1\leq i\leq n$ we have $\varphi _{\mathbf {a}_{i}\mathbf {a}_{j}}(X)\cap \varphi _{\mathbf {a}_{i}\mathbf {a}_{j^{\prime }}}(X)=\emptyset $ for $j\neq j^{\prime }$ .

• • For any affine subspace $W<\mathbb {R}^{d}$ and $1\leq i\leq n,$ there exists $1\leq j\leq n$ such that $\varphi _{\mathbf {a}_i\mathbf {a}_j}(X)\cap W^{(\epsilon _0)}=\emptyset $ .

To derive the third property listed above, we have relied upon the fact that our contractions are similarities and therefore map affine subspaces to affine subspaces and contract distances in a uniform way.

It is clear now that after iterating our IFS and relabelling the digits, we can assume without loss of generality that the following lemma holds.

For each $a\in \mathcal {A}$ we let $\mathcal {A}_{a}=\{a,a_{1,i},\ldots a_{n,i}\}$ where $\{a_{1,i},\ldots a_{n,i}\}$ is one of the subsets listed in Lemma 3.4 and has been chosen so that

$$ \begin{align*}\varphi_{a}(X)\cap \bigcup_{j=1}^{n}\varphi_{a_{j,i}}(X)=\emptyset.\end{align*} $$

As in the proof of Theorem 1.1, an ambiguity in our notation arises. Elements of $\mathcal {A}$ encode contractions in our IFS and the sets with which we perform our disintegration. To help with our exposition, when an element of $\mathcal {A}$ is being used to encode a contraction we will denote it by a, and when an element of $\mathcal {A}$ is being used to encode a subset of $\mathcal {A}$ we will denote it by b.

It follows from the above that there exists $c_{1}>0$ such that for any $b\in \mathcal {A},$ if $a,a'\in \mathcal {A}_{b}$ are distinct then

(3.17) $$ \begin{align} d(\varphi_{a}(X),\varphi_{a'}(X))\geq c_{1} \end{align} $$

and

(3.18) $$ \begin{align} \varphi_{a}(X)\cap\varphi_{a'}(X)=\emptyset. \end{align} $$

It follows now from the definition of $\mu _{\omega }$ and (3.18) that for any $\omega =(b_n)\in \Omega $ and $(a_1,\ldots a_n)\in \prod _{j=1}^{n}\mathcal {A}_{b_j},$ we have

(3.19) $$ \begin{align} \mu_{\omega}(\varphi_{a_1,\ldots, a_n}(X_{\sigma^{n}\omega}))=\prod_{j=1}^{n}q_{a_j}^{b_j}. \end{align} $$

As in the proof of Theorem 1.1, since each $\mathcal {A}_{b}$ contains at least two elements and our probability vector $\mathbf {p}$ is strictly positive, it follows that

$$ \begin{align*}q_{\min}:=\min_{b\in \mathcal{A}}\min_{a\in \mathcal{A}_{b}}\{q_{a}^{b}\}\quad\textrm{ and }\quad q_{\max}:=\max_{b\in \mathcal{A}}\max_{a\in \mathcal{A}_{b}}\{q_{a}^{b}\} \end{align*} $$

satisfy

(3.20) $$ \begin{align} 0<q_{\min}\leq q_{\max}<1. \end{align} $$

We now show that the desired properties hold for the $\Omega $ , $\mathbb {P}$ and $\{\mu _{\omega }\}_{\omega }$ coming from Proposition 2.2 for this choice of $\{\mathcal {A}_{b}\}_{b\in \mathcal {A}}.$ The first step is the following proposition which uniformly bounds how much mass a $\mu _{\omega }$ can give to a neighbourhood of an affine subspace.

Proof. We begin by remarking that to prove our proposition it suffices to consider $\epsilon < \epsilon _{0}$ where $\epsilon _{0}$ is as above. Let

$$ \begin{align*} r_{\min}:=\min_{a\in \mathcal{A}}\min_{x,y\in\mathbb{R}^{d},\, x\neq y}\bigg\{\frac{\|\varphi_{a}(x)-\varphi_{a}(y)\|}{\|x-y\|}\bigg\}. \end{align*} $$

Since our IFS consists of similarities we must have $r_{\min }>0$ . Our proof will rely upon showing that for any $\omega \in \Omega $ , $\epsilon <\epsilon _{0}$ and $W<\mathbb {R}^{d}$ there exists an affine subspace $W_{1}<\mathbb {R}^{d}$ such that

(3.21) $$ \begin{align} \mu_{\omega}(W^{(\epsilon)})\leq (1-q_{\min})\mu_{\sigma\omega}(W_{1}^{(\epsilon \cdot r_{\min}^{-1})}). \end{align} $$

As such, let $\omega =(b_n)\in \Omega , W<\mathbb {R}^{d}$ and $\epsilon <\epsilon _{0}$ . Then by (2.2) we have

$$ \begin{align*} \mu_{\omega}(W^{(\epsilon)})=\sum_{a\in \mathcal{A}_{b_1}}q^{b_1}_{a}\varphi_{a}\mu_{\sigma \omega}(W^{(\epsilon)}). \end{align*} $$

By Lemma 3.4 there must exist $a\in \mathcal {A}_{b_1}$ such that $W^{(\epsilon )}\cap \varphi _{a}(X)=\emptyset .$ Since $\mathrm {supp}( \varphi _{a}\mu _{\sigma \omega })\subset \varphi _{a}(X)$ it follows that $\varphi _{a}\mu _{\sigma \omega }(W^{(\epsilon )})=0$ for some $a\in \mathcal {A}_{b_{1}}$ . Consequently, if we choose $a^{*}\in \mathcal {A}_{b_{1}}$ for which $\varphi _{a^*}\mu _{\sigma \omega }(W^{(\epsilon )})=\max _{a\in \mathcal {A}_{b_1}}\{\varphi _{a}\mu _{\sigma \omega }(W^{(\epsilon )})\},$ we have

(3.22) $$ \begin{align} \mu_{\omega}(W^{(\epsilon)})\leq (1-q_{\min})\varphi_{a^*}\mu_{\sigma \omega}(W^{(\epsilon)}). \end{align} $$

Finally, using the fact that our IFS consists of similarities, we have

$$ \begin{align*}\varphi_{a^*}^{-1}(W^{(\epsilon)})\subset W_{1}^{(\epsilon\cdot r_{\min}^{-1})}\end{align*} $$

for some $W_{1}<\mathbb {R}^d$ . Therefore,

(3.23) $$ \begin{align} \varphi_{a^*}\mu_{\sigma \omega}(W^{\epsilon})\leq \mu_{\sigma \omega}(W_{1}^{\epsilon\cdot r_{\min}^{-1}}). \end{align} $$

Combining (3.22) and (3.23) implies (3.21).

Equipped with (3.21), we can now finish our proof. Let $\omega \in \Omega $ , $W<\mathbb {R}^{d}$ and $\epsilon <\epsilon _{0}$ be arbitrary. Let $M\in \mathbb {N}$ be the unique integer satisfying

(3.24) $$ \begin{align} \epsilon_{0}r_{\min}^{M}\leq \epsilon\leq \epsilon_{0}r_{\min}^{M-1}. \end{align} $$

Repeatedly applying (3.21) yields a sequence of affine subspaces $W_{1},W_{2},\ldots ,W_{M}$ such that

$$ \begin{align*} & \mu_{\omega}(W^{(\epsilon)})\leq (1-q_{\min})\mu_{\sigma \omega}(W_{1}^{(\epsilon r_{\min}^{-1})})\\ & \quad \leq \cdots \leq (1-q_{\min})^{M}\mu_{\sigma^{M} \omega}(W_{M}^{(\epsilon r_{\min}^{-M})})\leq (1-q_{\min})^{M}. \end{align*} $$

Thus, $\mu _{\omega }(W^{(\epsilon )})\leq (1-q_{\min })^{M}$ . Combining this inequality with (3.24) yields

$$ \begin{align*}\mu_{\omega}(W^{(\epsilon)})\ll \epsilon^{\alpha}\end{align*} $$

for $\alpha = ({(\log (1-q_{\min }))}/{(\log r_{\min })}).$ Thus, our result holds.

Duplicating the arguments given in the proof of Lemma 3.3, it is possible to show that the following analogous statement holds for the sets $\{\mathcal {A}_{b}\}_{b\in \mathcal {A}}$ defined in this section. In this lemma $c_{1}$ is as in (3.17).

Equipped with Proposition 3.5 and Lemma 3.6, we can now prove the remaining case of Theorem 1.1.

Proof of Theorem 1.1 for affinely irreducible self-similar measures

Statement $(1)$ of this theorem holds because of Proposition 2.2. Statement $(2)$ follows from an analogous argument to that used in our earlier proof in the self-conformal case where we appeal to (3.19) and Lemma 3.6 instead of (3.5) and Lemma 3.3. As such we omit this argument. We now focus on statement $(4)$ . Notice that statement $(3)$ follows from statement $(4)$ . Fix $\omega =(b_n)\in \Omega $ , $x\in \mathrm {supp}(\mu _{\omega }),$ an affine subspace $W<\mathbb {R}^{d}, r\in (0,c_1/2)$ and $\epsilon \in (0,1)$ . It suffices to consider r and $\epsilon $ contained in this restricted domain.

It follows from an application of Lemma 3.6, and the arguments used in the proof of Theorem 1.1, that there exists $D>0$ depending only on our IFS and $\mathbf {a}\in \bigcup _{n=1}^{\infty }\prod _{j=1}^{n}\mathcal {A}_{b_j}$ such that

(3.25) $$ \begin{align} B(x,r)\cap X_{\omega}\subset \varphi_{\mathbf{a}}(X_{\sigma^{|\mathbf{a}|}\omega}), \end{align} $$

(3.26) $$ \begin{align} \kern-7pt \prod_{j=1}^{|\mathbf{a}|}r_{a_j}\geq \frac{r}{D} \end{align} $$

and

(3.27) $$ \begin{align} \mu_{\omega}(B(x,r))\asymp \mu_{\omega}(\varphi_{\mathbf{a}}(X_{\sigma^{|\mathbf{a}|}\omega})). \end{align} $$

In (3.26) we have adopted the standard notation that for $a\in \mathcal {A}$ we let $r_{a}\in (0,1)$ be such that $\|\varphi _{a}(x)-\varphi _{a}(y)\|=r_{a}\|x-y\|$ for all $x,y\in \mathbb {R}^{d}$ .

We now observe the following:

$$ \begin{align*} \mu_{\omega}(W^{(\epsilon r)}\cap B(x,r))&\stackrel{({2.2})}{=}\sum_{\mathbf{a}'\in \prod_{j=1}^{|\mathbf{a}|}\mathcal{A}_{b_{j}}}\prod_{j=1}^{|\mathbf{a}|}q_{a_{j}}^{b_{j}}\cdot \varphi_{\mathbf{a}'}\mu_{\sigma^{|\mathbf{a}|}\omega}(W^{(\epsilon r)}\cap B(x,r))\\&\kern-2pt\stackrel{({3.25})}{=}\prod_{j=1}^{|\mathbf{a}|}q_{a_{j}}^{b_{j}}\cdot \varphi_{\mathbf{a}}\mu_{\sigma^{|\mathbf{a}|}\omega}(W^{(\epsilon r)}\cap B(x,r))\\&\kern3pt\leq \prod_{j=1}^{|\mathbf{a}|}q_{a_{j}}^{b_{j}}\cdot \varphi_{\mathbf{a}}\mu_{\sigma^{|\mathbf{a}|}\omega}(W^{(\epsilon r)})\\&\kern3pt\leq \prod_{j=1}^{|\mathbf{a}|}q_{a_{j}}^{b_{j}}\cdot \mu_{\sigma^{|\mathbf{a}|}\omega}(W_{1}^{(D\epsilon )}). \end{align*} $$

Here $W_{1}=\varphi _{\mathbf {a}}^{-1}(W)$ , and in the final line we have used (3.26). To conclude the second equality in the above we also used that $\varphi _{\mathbf {a}'}(X_{\sigma ^{|\mathbf {a}|}\omega })\cap \varphi _{\mathbf {a}'}(X_{\sigma ^{|\mathbf {a}|}\omega })=\emptyset $ for distinct $\mathbf {a}',\mathbf {a}"\in \prod _{j=1}^{|\mathbf {a}|}\mathcal {A}_{b_{j}}.$ Now applying Proposition 3.5, (3.19) and (3.27), we have

$$ \begin{align*} \mu_{\omega}(W^{(\epsilon r)}\cap B(x,r))\ll \prod_{j=1}^{|\mathbf{a}|}q_{a_{j}}^{b_{j}}\cdot \epsilon^{\alpha}=\epsilon^{\alpha}\mu_{\omega}(\varphi_{\mathbf{a}}(X_{\sigma^{|\mathbf{a}|}\omega}))\ll \epsilon^{\alpha}\mu_{\omega}(B(x,r)). \end{align*} $$

Thus, statement $(4)$ holds and our proof is complete.