Proof. By definition of the neighbour set, if $\Delta $ is any net interval, we have

$$ \begin{align*} \Delta\cap K = \bigcup_{f\in{\mathcal{V}}(\Delta)}(T_{\Delta}\circ f(K))\cap\Delta. \end{align*} $$

Set $g=T_{\Delta _{2}}\circ T_{\Delta _{1}}^{-1}$ so that g is clearly a similarity, and applying this observation to $\Delta _{1}$ and $\Delta _{2}$ , we have

$$ \begin{align*} g(\Delta_{1}\cap K) &= \bigcup_{f\in{\mathcal{V}}(\Delta_{1})}g(T_{\Delta_{1}}\circ f(K)\cap\Delta_{1}) = \bigcup_{f\in{\mathcal{V}}(\Delta_{1})}(g\circ T_{\Delta_{1}}\circ f(K))\cap g(\Delta_{1})\\[3pt] &= \bigcup_{f\in{\mathcal{V}}(\Delta_{2})}(T_{\Delta_{2}}\circ f(K))\cap\Delta_{2} = \Delta_{2}\cap K. \end{align*} $$

Thus g is surjective with the correct image.

We now verify the measure property. By the invariant property of the self-similar measure in equation (2.2), if $\Delta \in \mathcal {F}_{t}$ is any net interval and $E\subseteq \Delta $ is any Borel set,

$$ \begin{align*} {\mu_{\boldsymbol{p}}}(E) &= \sum_{\sigma\in\Lambda_{t}}p_{\sigma{\mu_{\boldsymbol{p}}}}\circ S_{\sigma}^{-1}(E)= \sum_{f\in{\mathcal{V}}(\Delta)}{\mu_{\boldsymbol{p}}}(f^{-1}\circ T_{\Delta}^{-1}(E))\sum_{\substack{\sigma\in\Lambda_{t}\\ \sigma\text{ generates }f}}p_{\sigma}. \end{align*} $$

Since f is a neighbour of $\Delta $ , there is at least one $\sigma $ generating f. In particular, say $\Delta _{1}\in \mathcal {F}_{t_{1}}$ and $\Delta _{2}\in \mathcal {F}_{t_{2}}$ , write ${\mathcal {V}}(\Delta _{1})={\mathcal {V}}(\Delta _{2})=\{f_{1},\ldots ,f_{m}\}$ , and set for each $1\leq i\leq m$ and $j=1,2$

$$ \begin{align*} q_{i,j} := \sum_{\substack{\sigma\in\Lambda_{t_{j}}\\ \sigma\text{ generates }f_{i}}}p_{\sigma}>0. \end{align*} $$

Set $c_{1} = \min \{q_{i,2}/q_{i,1}:1\leq i\leq m\}$ . We then have for $E\subseteq \Delta _{1}$ that $g(E)\subseteq \Delta _{2}$ so that

$$ \begin{align*} {\mu_{\boldsymbol{p}}}(g(E)) &= \sum_{i=1}^{m}{\mu_{\boldsymbol{p}}}(f_{i}^{-1}\circ T_{\Delta_{2}}^{-1}\circ g(E))q_{i,2}\\[3pt] &\geq c_{1}\sum_{i=1}^{m} {\mu_{\boldsymbol{p}}}(f_{i}^{-1}\circ T_{\Delta_{1}}^{-1}(E)) q_{i,1}= c_{1}{\mu_{\boldsymbol{p}}}(E). \end{align*} $$

Similarly, we have ${\mu _{\boldsymbol {p}}}(g(E))\leq c_{2}{\mu _{\boldsymbol {p}}}(E)$ , where $c_{2}=\min \{q_{i,1}/q_{i,2}:1\leq i\leq m\}$ .

Proof. Given a map $f(x)=rx+d$ , we set $R(f)=|r|$ .

Write ${\mathcal {V}}(\Delta ^{\prime })={\mathcal {V}}(\Delta )=\{f_{1},\ldots ,f_{m}\}$ , and let

$$ \begin{align*} \mathcal{W}^{\prime} &= \{T_{\Delta^{\prime}}\circ f_{i}:R(f_{i})={{R_{\max}}}(\Delta^{\prime}),1\leq i\leq m\},\\ \mathcal{W} &= \{T_{\Delta}\circ f_{i}:R(f_{i})={{R_{\max}}}(\Delta),1\leq i\leq m\} \end{align*} $$

denote the corresponding sets of neighbours corresponding to functions with maximal contraction factor, where ${{R_{\max }}}(\Delta ^{\prime })={{R_{\max }}}(\Delta )$ . Then let

$$ \begin{align*} \mathcal{C}^{\prime} &= \{S_{\tau}:\tau\in\Lambda_{\operatorname{\mathrm{tg}}(\Delta^{\prime})},S_{\tau}(K)\cap (\Delta^{\prime})^{\circ}\neq\emptyset\},\\ \mathcal{C} &= \{S_{\tau}:\tau\in\Lambda_{\operatorname{\mathrm{tg}}(\Delta)},S_{\tau}(K)\cap\Delta^{\circ}\neq\emptyset\}. \end{align*} $$

In other words, $\mathcal {C}$ is the set of words of generation $\operatorname {\mathrm {tg}}(\Delta )$ which contribute to some child of $\Delta $ , and similarly for $\Delta ^{\prime }$ . Using the observation that the only new words are those which are one-level descendants of those which generate neighbours of maximal length, we have

(2.3) $$ \begin{align} \mathcal{C} &= \{f\circ S_{j}:f\in \mathcal{W},f\circ S_{j}(K)\cap\Delta^{\circ}\neq\emptyset\}\cup\{T_{\Delta}^{-1}\circ f_{i}:R(f_{i})\neq {{R_{\max}}}(\Delta)\}\nonumber\\ &= \{T_{\Delta}\circ T_{\Delta^{\prime}}^{-1}\circ f:f\in \mathcal{C}^{\prime}\}. \end{align} $$

Note that, in the above set of equalities, we use the fact that for $f\in \mathcal {W}$ ,

$$ \begin{align*} f\circ S_{j}(K)\cap\Delta^{\circ}\neq\emptyset &\iff T_{\Delta}^{-1}\circ f\circ S_{j}(K)\cap (0,1)\neq\emptyset\\ &\iff T_{\Delta^{\prime}}\circ T_{\Delta}^{-1}\circ f\circ S_{j}(K)\cap(\Delta^{\prime})^{\circ}\neq\emptyset, \end{align*} $$

where $T_{\Delta ^{\prime }}\circ T_{\Delta }^{-1}\circ f\in \mathcal {W}^{\prime }$ .

Write $\Delta =[a,b]$ and $\Delta ^{\prime }=[a^{\prime },b^{\prime }]$ . Now consider the set $H=\{a,b\}\cup \{f(0),f(1):f\in \mathcal {C}\}\cap \Delta $ so that H is the set of all endpoints of generation $\operatorname {\mathrm {tg}}(\Delta )$ contained in $\Delta $ . Then if $H^{\prime }=\{a^{\prime },b^{\prime }\}\cup \{f(0),f(1):f\in \mathcal {C}^{\prime }\}\cap \Delta ^{\prime }$ , we observe by equation (2.3) that $T_{\Delta ^{\prime }}^{-1}(H^{\prime })=T_{\Delta }^{-1}(H)$ . Let $a=h_{1}<\cdots <h_{k+1}=b$ denote the ordered elements of H and $a^{\prime }=h_{1}^{\prime }<\cdots <h_{k+1}^{\prime }=b^{\prime }$ the ordered elements of $H^{\prime }$ , where $k=|H|-1=|H^{\prime }|-1$ . By Lemma 2.3, $(h_{i},h_{i+1})\cap K\neq \emptyset $ if and only if $(h_{i}^{\prime },h_{i+1}^{\prime })\cap K\neq \emptyset $ . Thus the children of $\Delta $ are $\{[h_{i},h_{i+1}]:(h_{i},h_{i+1})\cap K\neq \emptyset \}$ and the children of $\Delta ^{\prime }$ are $\{T_{\Delta ^{\prime }}\circ T_{\Delta }^{-1}([h_{i},h_{i+1}]):(h_{i},h_{i+1})\cap K\neq \emptyset \}$ , so $k=n=n^{\prime }$ .

Now fix some $1\leq i\leq n$ . Note that $T_{\Delta }\circ T_{\Delta ^{\prime }}^{-1}(\Delta _{i}^{\prime })=\Delta _{i}$ so that $T_{\Delta _{i}}^{-1}\circ T_{\Delta }\circ T_{\Delta ^{\prime }}^{-1}=T_{\Delta _{i}^{\prime }}^{-1}$ .

1. (i) By direct computation,

   $$ \begin{align*} \hspace{-1.5pc} {\mathcal{V}}(\Delta_{i}) &= \{T_{\Delta_{i}}^{-1}\circ f:f\in\mathcal{C},f(K)\cap\Delta_{i}^{\circ}\neq\emptyset\}\\[3pt] &= \{T_{\Delta_{i}}^{-1}\circ T_{\Delta}\circ T_{\Delta^{\prime}}^{-1}\circ f:f\in\mathcal{C}^{\prime},T_{\Delta}\circ T_{\Delta^{\prime}}^{-1}\circ f(K)\cap(T_{\Delta}\circ T_{\Delta^{\prime}}^{-1}(\Delta_{i}^{\prime}))^{\circ}\neq\emptyset\}\\[3pt] &= \{T_{\Delta_{i}^{\prime}}^{-1}\circ f:f\in\mathcal{C}^{\prime},f(K)\cap(\Delta_{i}^{\prime})^{\circ}\neq\emptyset\}={\mathcal{V}}(\Delta_{i}^{\prime}). \end{align*} $$

2. (ii) Since the $T_{\Delta }$ are isometries, $q(\Delta ,\Delta _{i}) = ({h_{i}-h_{1}})/{\operatorname {\mathrm {diam}}(\Delta )}=T_{\Delta }^{-1}(h_{i})$ since $T_{\Delta }^{-1}(h_{1})=0$ . Then the result follows since $T_{\Delta }^{-1}(h_{i})=T_{\Delta ^{\prime }}^{-1}(h_{i}^{\prime })$ .

3. (iii) We have

   $$ \begin{align*} \frac{\operatorname{\mathrm{diam}}(\Delta_{i})}{\operatorname{\mathrm{diam}}(\Delta)}=\operatorname{\mathrm{diam}}(T_{\Delta}^{-1}(\Delta_{i}))=\operatorname{\mathrm{diam}}(T_{\Delta^{\prime}}^{-1}(\Delta_{i}^{\prime}))=\frac{\operatorname{\mathrm{diam}}(\Delta_{i}^{\prime})}{\operatorname{\mathrm{diam}}(\Delta^{\prime})}. \end{align*} $$

4. (iv) Recall that for an arbitrary net interval, $\operatorname {\mathrm {tg}}(\Delta _{0}) = {{R_{\max }}}(\Delta _{0})\cdot \operatorname {\mathrm {diam}}(\Delta _{0})$ where ${{R_{\max }}}(\Delta _{0})$ depends only on ${\mathcal {V}}(\Delta _{0})$ . Apply (i) and (iii).

We thus have the desired result.

Proof. Suppose $\Delta _{0}\in \mathcal {F}_{t}$ and $\Delta _{m}\in \mathcal {F}_{s}$ . Say ${\mathcal {V}}(\Delta _{0})=\{f_{1},\ldots ,f_{\ell }\}$ with $f_{1}<\cdots <f_{\ell }$ and ${\mathcal {V}}(\Delta _{m})=\{g_{1},\ldots ,g_{m}\}$ with $g_{1}<\cdots <g_{m}$ . For each i, assume $\tau _{i}$ generates the neighbour $f_{i}$ , and set $\mathcal {A}_{ij}=\{\omega :\tau _{i}\omega \in \Lambda _{s},\tau _{i}\omega \text { generates }g_{j}\}$ . Then for any $1\leq j\leq m$ , we have

$$ \begin{align*} ({\boldsymbol{Q}_{\boldsymbol{p}}}(\Delta_{0})T(\eta))_{j} &= \sum_{i=1}^{\ell} {\mu_{\boldsymbol{p}}}(f_{i}^{-1}((0,1)))\bigg(\!\sum_{\substack{\sigma\in\Lambda_{t}\\ \sigma\text{ generates }f_{i}}}p_{\sigma}\!\bigg)\cdot\bigg(\sum_{\omega\in\mathcal{A}_{ij}}\frac{{\mu_{\boldsymbol{p}}}(g_{j}^{-1}((0,1))}{{\mu_{\boldsymbol{p}}}(f_{i}^{-1}((0,1))}p_{\omega}\bigg)\\[3pt] &= {\mu_{\boldsymbol{p}}}(g_{j}^{-1}((0,1))) \sum_{i=1}^{\ell}\bigg(\!\sum_{\substack{\sigma\in\Lambda_{t}\\ \sigma\text{ generates }f_{i}}}p_{\sigma}\!\bigg)\cdot\bigg(\sum_{\omega\in\mathcal{A}_{ij}}p_{\omega}\bigg)\\[3pt] &= {\mu_{\boldsymbol{p}}}(g_{j}^{-1}((0,1)))\sum_{\substack{\omega\in\Lambda_{s}\\ \omega\text{ generates }g_{j}}}p_{\omega} \end{align*} $$

so that ${\boldsymbol {Q}_{\boldsymbol {p}}}(\Delta _{0})T(\eta )={\boldsymbol {Q}_{\boldsymbol {p}}}(\Delta _{m})$ .

Proof. It suffices to show that there exists some vertex v such that if w is any other vertex, there exists an admissible path from w to v. Then the essential class is the set of all vertices $v^{\prime }$ for which there is a path from v to $v^{\prime }$ . By Proposition 3.3, there exists some $t>0$ and net interval $\Delta _{0}\in \mathcal {F}_{t}$ such that $\#{\mathcal {V}^{c}}(\Delta _{0})$ is maximal; let $v:= {\mathcal {V}}(\Delta _{0})$ .

Now, let $w\in V(\mathcal {G})$ be arbitrary and $\Delta \in \mathcal {F}$ such that ${\mathcal {V}}(\Delta )=w$ . Since $\Delta ^{\circ }\cap K\neq \emptyset $ , there exists some $\sigma \in \mathcal {I}^{*}$ such that $S_{\sigma }(K)\subseteq \Delta $ and $r_{\sigma }>0$ . Set $\gamma =r_{\sigma }\cdot t$ and let $\Delta _{1}:= S_{\sigma }(\Delta _{0})$ .

Let $\Delta _{0}=[a,b]$ have covering neighbours generated by words $\{\omega _{1},\ldots ,\omega _{m}\}$ with $\omega _{i}\in \Lambda _{t}$ . By definition of $\gamma $ , $\{\sigma \omega _{1},\ldots ,\sigma \omega _{m}\}$ are words of generation $\Lambda _{\gamma }$ . Note that $(\Delta _{1})^{\circ }\cap K\neq \emptyset $ and that the endpoints of $\Delta _{1}$ are of the form $S_{\sigma \zeta }(z)$ , where $z\in \{0,1\}$ and $\zeta \in \Lambda _{t}$ , so that $\sigma \zeta \in \Lambda _{\gamma }$ . In particular, if $\Delta _{1}\notin \mathcal {F}_{\gamma }$ , then there exists some $\tau \in \Lambda _{\gamma }$ such that $S_{\tau }\notin \{S_{\sigma \omega _{1}},\ldots ,S_{\sigma \omega _{m}}\}$ and $S_{\tau }([0,1])\supseteq \Delta _{1}$ . However, then there exists some $\Delta _{2}\in \mathcal {F}_{\gamma }$ with $\Delta _{2}\subseteq \Delta _{1}\cap S_{\tau }([0,1])$ , where $\Delta _{2}$ has distinct covering neighbours generated by $\{\omega _{1},\ldots ,\omega _{m}\}\cup \{\tau \}$ , contradicting the maximality of $\#{\mathcal {V}^{c}}(\Delta _{0})$ .

Thus $\Delta _{1}$ is in fact a net interval of generation $\gamma $ . Moreover, since $r_{\sigma }>0$ , we have $T_{\Delta _{1}}=S_{\sigma }\circ T_{\Delta _{0}}$ , so that

$$ \begin{align*} {\mathcal{V}^{c}}(\Delta_{1})=\{T_{\Delta_{1}}^{-1}\circ S_{\sigma \omega_{i}}\}_{i=1}^{m}=\{T_{\Delta_{0}}^{-1}\circ S_{\sigma }^{-1}\circ S_{\sigma }\circ S_{\omega_{i}}\}_{i=1}^{m}={\mathcal{V}^{c}}(\Delta_{0}). \end{align*} $$

Thus by Remark 3.2, we have ${\mathcal {V}}(\Delta _{1})=v$ and $\Delta _{1}\subseteq \Delta $ , so that there exists a path from ${\mathcal {V}}(\Delta )$ to ${\mathcal {V}}(\Delta _{1})$ , as claimed.

Proof. To see that $S_{\sigma }(U(x_{0},t_{0}))$ also attains the maximal value in equation (3.1), if

$$ \begin{align*} \mathcal{S}_{t_{0}}(U(x_{0},t_{0}))=\{S_{\phi_{1}},\ldots,S_{\phi_{m}}\}, \end{align*} $$

then $S_{\sigma \phi _{i}}\in \mathcal {S}_{|r_{\sigma }|t_{0}}(S_{\sigma }(U(x_{0},t_{0})))$ for each i and $\#\mathcal {S}_{|r_{\sigma }|t_{0}}(S_{\sigma }(U(x_{0},t_{0})))\geq m$ . Then equality holds by maximality of m.

We now see (ii). By definition of net intervals, we know that for any $t>0$ , $U(x_{0},t_{0})\cap K$ is contained in a finite union of net intervals of generation t. In particular, it suffices to show that there is some $t_{1}>0$ such that the set

$$ \begin{align*} \{\Delta\in\mathcal{F}_{t_{1}}:\Delta\cap U(x_{0},t_{0})\neq\emptyset\} \end{align*} $$

is composed only of essential net intervals. Let $\Delta _{0}$ be a fixed essential net interval and let $\sigma _{0}\in \mathcal {I}^{*}$ have $r_{\sigma _{0}}>0$ and $S_{\sigma _{0}}([0,1])\subseteq \Delta _{0}$ . As argued above, $S_{\sigma _{0}}(U(x_{0},t_{0}))$ also attains the maximal value in equation (3.1). Let

$$ \begin{align*} H=\{S_{\sigma}:\sigma\in\Lambda_{r_{\sigma_{0}}t_{0}},S_{\sigma}(K)\cap S_{\sigma_{0}}(U(x_{0},t_{0}))=\emptyset\}. \end{align*} $$

Since $S_{\sigma _{0}}(U(x_{0},t_{0}))$ is open, there exists some $\epsilon _{0}>0$ such that for any $\epsilon $ with $|\epsilon |<\epsilon _{0}$ , $S_{\sigma _{0}}(U(x_{0}+\epsilon ,t_{0}))$ also attains the maximal value in equation (3.1). In particular, if $S_{\sigma }\in H$ is arbitrary, we in fact have $S_{\sigma }(K)\cap S_{\sigma _{0}}(B(x_{0},t_{0}))=\emptyset $ . Since H is a finite set, take

$$ \begin{align*} t_{1} = \min\{\min\{\operatorname{\mathrm{dist}}(f(K),S_{\sigma_{0}}(B(x_{0},t_{0}))) : f\in H\},t_{0}\}>0. \end{align*} $$

It remains to show that such a $t_{1}$ works.

Write $\mathcal {S}_{t_{0}}(U(x_{0},t_{0}))=\{S_{\phi _{1}},\ldots ,S_{\phi _{m}}\}$ and set

$$ \begin{align*} F = \{\Delta\in\mathcal{F}_{t_{1}}:\Delta\cap U(x_{0},t_{0})\neq\emptyset\}. \end{align*} $$

Suppose for contradiction there is some $\Delta \in F$ that is not an essential net interval, and let $\Delta $ have neighbours generated by distinct functions $\{S_{\omega _{1}},\ldots ,S_{\omega _{k}}\}$ with $\omega _{i}\in \Lambda _{t_{1}}$ . As argued in Proposition 3.5, since $\Delta _{1}:= S_{\sigma _{0}}(\Delta )$ is not a net interval with neighbour set ${\mathcal {V}}(\Delta )$ (or $\Delta _{1}$ would be a descendant of $\Delta _{0}$ , and hence essential), there exists some $\tau \in \Lambda _{r_{\sigma _{0}}t_{1}}$ such that $S_{\tau }(K)\cap \Delta _{1}^{\circ }\neq \emptyset $ and $S_{\tau }\neq S_{\sigma _{0}\omega _{i}}$ for each $1\leq i\leq k$ . We also observe that

(3.2) $$ \begin{align} \{S_{\sigma_{0}\omega_{1}},\ldots,S_{\sigma_{0}\omega_{k}}\}=\{S_{\sigma_{0}\xi}:\xi\in\Lambda_{t_{1}},S_{\sigma_{0}\xi}(K)\cap\Delta_{1}^{\circ}\neq\emptyset\}. \end{align} $$

Since $t_{1}\leq t_{0}$ , let $\tau _{1}\preccurlyeq \tau $ be the unique prefix in $\Lambda _{r_{\sigma _{0}} t_{0}}$ . Suppose for contradiction $S_{\tau _{1}}(K)\cap S_{\sigma _{0}}(U(x_{0},t_{0}))\neq \emptyset $ . Since $S_{\sigma _{0}}(U(x_{0},t_{0}))$ attains the maximal value in equation (3.1), we have $S_{\tau _{1}}=S_{\sigma _{0}}\circ S_{\omega }$ for some $S_{\omega }\in \mathcal {S}_{r_{\sigma _{0}} t_{0}}(S_{\sigma }(U(x_{0},t_{0})))$ . Thus there exists some word $\xi $ such that $S_{\tau }=S_{\sigma _{0}}\circ S_{\xi }$ , which contradicts equation (3.2). We thus have that $S_{\tau _{1}}(K)\cap S_{\sigma _{0}}(U(x_{0},t_{0}))=\emptyset $ so that $S_{\tau _{1}}\in H$ .

However, by definition of $\Delta _{1}$ , we have that $\Delta _{1}\cap S_{\sigma _{0}}(U(x_{0},t_{0}))\neq \emptyset $ and $\Delta _{1}^{\circ }\cap S_{\tau _{1}}(K)\neq \emptyset $ , so

$$ \begin{align*} \operatorname{\mathrm{dist}}(S_{\tau_{1}}(K),S_{\sigma_{0}}(U(x_{0},t_{0})))<\operatorname{\mathrm{diam}}(\Delta_{1})\leq t_{1}, \end{align*} $$

contradicting the choice of $t_{1}$ . Thus every $\Delta \in F$ is in fact essential, as claimed.

Proof. Since ${\mu _{\boldsymbol {p}}}(B(x,t))>0$ and ${\mu _{\boldsymbol {p}}}$ is non-atomic, $U(x,t)\cap K\neq \emptyset $ . From the weak separation condition, there exists some $\ell \in {\mathbb {N}}$ such that $\#\mathcal {S}_{t}(B(x,t))\leq \ell $ for any $x\in {\mathbb {R}}$ and $t>0$ . By the invariant property of ${\mu _{\boldsymbol {p}}}$ and since ${\mu _{\boldsymbol {p}}}$ is a probability measure, we have

$$ \begin{align*} {\mu_{\boldsymbol{p}}}(B(x, t)) &= \sum_{\sigma\in\Lambda_{t}(B(x, t))}p_{\sigma{\mu_{\boldsymbol{p}}}}\circ S_{\sigma}^{-1}((B(x, t)))\leq \sum_{\sigma\in\Lambda_{t}(B(x, t))}p_{\sigma}\\[6pt] &= \sum_{S_{\omega}\in\mathcal{S}_{t}(B(x, t))}\sum_{\substack{\sigma\in\Lambda_{t}(B(x, t))\\ S_{\sigma}=S_{\omega}}}p_{\sigma}. \end{align*} $$

In particular, since $\#\mathcal {S}_{t}(B(x, t))\leq \ell $ , get $\omega _{0}$ such that

(3.3) $$ \begin{align} \sum_{\substack{\sigma\in\Lambda_{t}(B(x, t))\\ S_{\sigma}=S_{\omega_{0}}}}p_{\sigma}\geq{\mu_{\boldsymbol{p}}}(B(x, t))/\ell. \end{align} $$

Note that $S_{\omega _{0}}(K)\cap B(x, t)\neq \emptyset $ , so that $S_{\omega _{0}}([0,1])\subseteq B(x,2 t)$ . If $r_{\omega _{0}}<0$ , get $k\in \mathcal {I}$ with $r_{k}<0$ and set $\omega _{1}=\omega _{0}k$ ; otherwise, take $\omega _{1}=\omega _{0}$ . Now, let $\Delta _{0}\in \mathcal {F}_{s_{0}}$ be such that $\#{\mathcal {V}}^{c}(\Delta _{0})$ is maximal. Exactly as argued in Proposition 3.5, $\Delta _{1}:= S_{\omega _{1}}(\Delta _{0})$ is a net interval in generation $r_{\omega _{1}}\cdot s_{0}$ with ${\mathcal {V}}(\Delta _{1})={\mathcal {V}}(\Delta _{0})$ . Moreover, we know that if $\sigma $ generates some neighbour f of $\Delta _{0}$ , then $\omega _{1}\sigma $ generates the same neighbour f of $\Delta _{1}$ . Fix some $1\leq j\leq \#{\mathcal {V}}(\Delta _{1})$ and let $f_{j}$ be the neighbour of $\Delta _{1}$ corresponding to the index j. We then have by using equation (3.3) and the above observation that

$$ \begin{align*} ({\boldsymbol{Q}_{\boldsymbol{p}}}(\Delta_{1}))_{j} &= \mu(f_{j}^{-1}((0,1)))\sum_{\substack{\sigma\in\Lambda_{ s_{0} r_{\omega_{1}}}\\ \sigma\text{ generates }f_{j}}}p_{\sigma}\\[3pt] &\geq p_{k}\bigg(\sum_{\substack{\sigma\in\Lambda_{t}(B(x, t))\\ S_{\sigma}=S_{\omega_{1}}}}p_{\sigma}\bigg)\cdot\mu(f_{j}^{-1}((0,1)))\cdot\sum_{\substack{\sigma\in\Lambda_{ s_{0}}\\ \sigma\text{ generates }f_{j}}}p_{\sigma}\\[3pt] &\geq {\mu_{\boldsymbol{p}}}(B(x, t))\cdot\frac{p_{k}\cdot ({\boldsymbol{Q}_{\boldsymbol{p}}}(\Delta_{0}))_{j}}{\ell}\geq{\mu_{\boldsymbol{p}}}(B(x, t))\cdot C_{1}, \end{align*} $$

where $C_{1}:= p_{k}\cdot \min _{j}({\boldsymbol {Q}_{\boldsymbol {p}}}(\Delta _{0}))_{j}/\ell $ , which depends only on the IFS and choice of probabilities.

Now let $\eta $ be any fixed path from ${\mathcal {V}}(\Delta _{0})$ to v and let $\epsilon $ be the smallest strictly positive entry of $T(\eta )$ . Let $\Delta $ be the unique net interval with symbolic $\gamma \eta $ , where $\gamma $ is the symbolic representation of $\Delta _{0}$ . Since $T(\eta )$ is non-negative and ${\boldsymbol {Q}_{\boldsymbol {p}}}(\Delta )={\boldsymbol {Q}_{\boldsymbol {p}}}(\Delta _{1})T(\eta )$ is a positive vector, we have that $({\boldsymbol {Q}_{\boldsymbol {p}}}(\Delta ))_{j}\geq {\mu _{\boldsymbol {p}}}(B(x, t))\cdot C_{1}\cdot \epsilon $ . Taking $C:= C_{1}\epsilon $ , we see that C satisfies the requirements. Moreover, since $\Delta _{0}\in \mathcal {F}_{r_{\omega _{0}} s_{0}}$ , taking $c= s_{0}L(\eta )\cdot r_{\min }^{2}$ and noting that $ t\cdot r_{\min }\leq |r_{\omega _{0}}|\leq t$ , we have that $\Delta \in \mathcal {F}_{s}$ , where $ s\geq c t$ . Finally, $\Delta \subseteq \Delta _{1}\subseteq S_{\omega _{0}}([0,1])\subseteq B(x,2 t)$ as required.

Proof. By Lemma 3.10, there exist constants $c,C>0$ such that for any $t>0$ and ball $B(x,t)$ with ${\mu _{\boldsymbol {p}}}(B(x,t))>0$ , there exists some net interval $\Delta \in \mathcal {F}$ with $\Delta \subseteq B(x,2t)$ , ${\mathcal {V}}(\Delta )=v$ and ${\mu _{\boldsymbol {p}}}(\Delta )\geq C{\mu _{\boldsymbol {p}}}(B(x,r))$ . We will construct a nested family of sets $E_{1}\supseteq E_{2}\supseteq \cdots \kern-0.1pt$ such that each $E_{n}$ is a finite union of intervals, ${\mu _{\boldsymbol {p}}}(E_{n})\leq (1-C/3)^{n}$ , and $K\setminus E\subseteq \bigcap _{n=1}^{\infty } E_{n}$ . From this, the result clearly follows.

First consider the ball $B_{1}=B(0,1)$ . Get $\Delta _{1}\subseteq B(0,2)$ with ${\mathcal {V}}(\Delta _{1})=v$ , set $E_{1}=[0,1]\setminus \Delta _{1}$ so that ${\mu _{\boldsymbol {p}}}(E_{1})\leq 1-C\leq 1-C/3$ . Since $\Delta _{1}$ is an interval, $E_{1}$ is a finite union of intervals and clearly $K\setminus E\subseteq E_{1}$ . Inductively, suppose $E_{n}$ is a finite union of intervals with ${\mu _{\boldsymbol {p}}}(E_{n})\leq (1-\unicode{x3bb} )^{n}$ . Since each $E_{n}$ is a finite union of intervals, there is a family of balls $\{B(x_{i},t_{i})\}_{i=1}^{m}$ such that the balls only overlap pairwise on endpoints, $E_{n}=\bigcup _{i=1}^{m} B(x_{i},t_{i})$ , and for any distinct $i_{1},i_{2},i_{3}$ ,

(3.4) $$ \begin{align} B(x_{i_{1}},2t_{i_{1}})\cap B(x_{i_{2}},2t_{i_{2}})\cap B(x_{i_{3}},2t_{i_{3}}) \end{align} $$

is either a singleton or the empty set and hence has measure $0$ , as ${\mu _{\boldsymbol {p}}}$ has no atoms. Now for each $1\leq i\leq m$ , apply Lemma 3.10 to get $\Delta _{n}^{i}\subseteq B(x_{i},2t_{i})$ with ${\mu _{\boldsymbol {p}}}(\Delta _{n}^{i})\geq C{\mu _{\boldsymbol {p}}}(B(x_{i},t_{i}))$ . While the $\Delta _{n}^{i}$ need not be disjoint, by equation (3.4), there exists a sub-collection labelled, without loss of generality, $\{\Delta _{n}^{i}\}_{i=1}^{m^{\prime }}$ such that

$$ \begin{align*} \sum_{i=1}^{m^{\prime}} {\mu_{\boldsymbol{p}}}(\Delta_{n}^{i})\geq\frac{1}{3}\sum_{i=1}^{m} {\mu_{\boldsymbol{p}}}(\Delta_{n}^{i}) \end{align*} $$

and $\Delta _{n}^{i}\cap \Delta _{n}^{j}$ is at most a singleton for $i\neq j$ . (To do this, pick the interval $\Delta _{n}^{i}$ with the largest measure and remove any net intervals $\Delta _{n}^{j}$ , where $\Delta _{n}^{j}\cap \Delta _{n}^{i}$ is not a singleton. By equation (3.4) and the geometry in ${\mathbb {R}}$ , there are at most 2 such indices j. Then repeat until the set is exhausted.)

Set $E_{n+1}=E_{n}\setminus \bigcup _{i=1}^{m^{\prime }}\Delta _{n}^{i}$ . Each $\Delta _{n}^{i}$ is an interval with ${\mathcal {V}}(\Delta _{n}^{i})=v$ , so that $E_{n+1}$ is a finite union of intervals with $K\setminus E\subseteq E_{n+1}$ , and

$$ \begin{align*} {\mu_{\boldsymbol{p}}}(E_{n+1}) &= {\mu_{\boldsymbol{p}}}(E_{n})-\sum_{i=1}^{m^{\prime}}{\mu_{\boldsymbol{p}}}(\Delta_{n}^{i})\leq {\mu_{\boldsymbol{p}}}(E_{n})-\frac{C}{3}\sum_{i=1}^{m} {\mu_{\boldsymbol{p}}}(B(x_{i},t))\\[3pt] &\leq (1-C/3){\mu_{\boldsymbol{p}}}(E_{n})\leq (1-C/3)^{n+1} \end{align*} $$

as claimed.

Proof. Suppose the local dimension exists and equals D. Recall that if $\Delta \in \mathcal {F}_{t}$ , then $t\geq \operatorname {\mathrm {tg}}(\Delta )={{R_{\max }}}(\Delta )\operatorname {\mathrm {diam}}(\Delta )$ . Thus there exists some constant $0<\epsilon $ such that for any $t>0$ and $\Delta \in \mathcal {F}_{t}$ with $x\in \Delta $ , $\epsilon t<\operatorname {\mathrm {diam}}(\Delta )$ . Moreover, $\operatorname {\mathrm {diam}}(\Delta )\leq t$ always holds by the net interval construction.

If x is a boundary point, get s such that x is an endpoint of $\Delta _{s}(x)$ and

$$ \begin{align*} B(x,\epsilon s)\subseteq\Delta_{s}^{1}(x)\cup\Delta_{s}^{2}(x)\subseteq B(x,2s), \end{align*} $$

where the notation is as above. Otherwise, if x is not a boundary point, then

$$ \begin{align*} B(x,\epsilon s)\subseteq\Delta^{-}_{s}(x)\cup\Delta_{s}(x)\cup\Delta_{s}^{+}(x)\subseteq B(x,2s). \end{align*} $$

In either case, $B(x,\epsilon s)\subseteq M_{s}(x)\subseteq B(x,2s)$ so that

$$ \begin{align*} &\bigg(\frac{\log\epsilon+\log s}{\log s}\bigg)\bigg(\frac{\log {\mu_{\boldsymbol{p}}}(B(x,\epsilon s))}{\log \epsilon s}\bigg)\\[6pt] &\quad\leq \frac{\log {\mu_{\boldsymbol{p}}}(M_{s}(x))}{\log s}\leq\bigg(\frac{\log s+\log 2}{\log s}\bigg)\bigg(\frac{\log{\mu_{\boldsymbol{p}}}(B(x,2s))}{\log 2s}\bigg). \end{align*} $$

The limit of the left and right both exist and are equal to D; hence, the limit of the middle expression exists and equals D. The arguments for the upper and lower dimension follow similarly.

Proof. First, suppose x is a periodic point with two distinct symbolic representations with periods $\theta =(\theta _{1},\ldots ,\theta _{\ell })$ and $\phi =(\phi _{1},\ldots ,\phi _{\ell ^{\prime }})$ , so that x is an endpoint of some net interval $\Delta \in \mathcal {F}$ . We first note that

$$ \begin{align*} \mu_{\boldsymbol{p}}(\Delta_{t}^{1}(x))&=\lVert T(e_{1},\ldots,e_{j},\underbrace{\theta,\ldots,\theta}_{m},\theta_{1},\ldots,\theta_{t})\rVert,\\ \mu_{\boldsymbol{p}}(\Delta_{t}^{2}(x))&=\lVert T(e_{1}^{\prime},\ldots,e^{\prime}_{j^{\prime}},\underbrace{\phi,\ldots,\phi}_{m^{\prime}},\phi_{1},\ldots,\phi_{t^{\prime}})\rVert \end{align*} $$

for t sufficiently small, $t < \ell $ and $t^{\prime }<\ell ^{\prime }$ . Now, get constants $c_{i}$ which do not depend on t such that

(3.5) $$ \begin{align} {\lVert(T(\theta))^{m+1}\rVert} &\leq \lVert T(\underbrace{\theta,\ldots,\theta}_{m},\theta_{1},\ldots,\theta_{t})\rVert\cdot{\lVert T(\theta_{t+1},\ldots,\theta_{\ell})\rVert}\nonumber\\[3pt] &\leq c_{1}\lVert T(e_{1},\ldots,e_{j},\underbrace{\theta,\ldots,\theta}_{m},\theta_{1},\ldots,\theta_{t})\rVert\leq c_{2}{\lVert T(\theta)^{m}\rVert}. \end{align} $$

Moreover, since

$$ \begin{align*} L(e_{1},\ldots,e_{j})L(\theta)^{m} L(\theta_{1},\ldots,\theta_{t}) r_{\min}\leq t\leq L(e_{1},\ldots,e_{j})L(\theta)^{m} L(\theta_{1},\ldots,\theta_{t}), \end{align*} $$

we have $L(\theta )^{m}\asymp t$ with constants of comparability not depending on t. Thus, there exist $k_{i}$ not depending on t so that

$$ \begin{align*} \frac{\log k_{1}{\lVert T(\theta)^{m+1}\rVert}^{1/(m+1)}}{\log k_{3}\cdot L(\theta)}\geq\frac{\log \mu(\Delta_{t}^{1}(x))}{\log t}\geq \frac{\log k_{2}{\lVert(T(\theta))^{m}\rVert}^{1/m}}{\log k_{4}\cdot L(\theta)}, \end{align*} $$

and taking the limit as t goes to 0 yields

$$ \begin{align*} \lim_{t\to 0}\frac{\log \mu_{\boldsymbol{p}}(\Delta_{t}^{1}(x))}{\log t} = \frac{\log \operatorname{\mathrm{sp}}(T(\theta))}{\log L(\theta)}. \end{align*} $$

In the exact same way, we get

$$ \begin{align*} \lim_{t\to 0}\frac{\log \mu_{\boldsymbol{p}}(\Delta_{t}^{2}(x))}{\log t} = \frac{\log \operatorname{\mathrm{sp}}(T(\phi))}{\log L(\phi)}. \end{align*} $$

Now, since x is a periodic point, the set $\{{\mathcal {V}}(\Delta ):x\in \Delta ,\Delta \in \mathcal {F}\kern1.5pt\}$ is finite. Since ${{R_{\max }}}(\Delta )$ depends only on ${\mathcal {V}}(\Delta )$ , $\sup \{{{R_{\max }}}(\Delta ):x\in \Delta ,\Delta \in \mathcal {F}\kern1.5pt\}<\infty $ and the assumptions for Lemma 3.15 hold. Then by the power mean inequality, we have

$$ \begin{align*} {\dim_{\operatorname{\mathrm{loc}}}}\mu_{\boldsymbol{p}}(x) &= \lim_{t\to 0}\frac{\log\mu_{\boldsymbol{p}}(\Delta_{t}^{1}(x))+\mu_{\boldsymbol{p}}(\Delta^{2}_{t}(x))}{\log t}\\[6pt] &= \min\bigg(\lim_{t\to 0}\frac{\log\mu_{\boldsymbol{p}}(\Delta_{t}^{1}(x))}{\log t},\lim_{t\to 0}\frac{\log\mu_{\boldsymbol{p}}(\Delta^{2}_{t}(x))}{\log t}\bigg)\\[6pt] &= \min\bigg(\lim_{t\to 0}\frac{\log\operatorname{\mathrm{sp}} T(\theta)}{\log L(\theta)},\lim_{t\to 0}\frac{\log\operatorname{\mathrm{sp}} T(\phi)}{\log L(\phi)}\bigg), \end{align*} $$

since the final two limits in the maximum exist, as claimed.

If x is an endpoint of some net interval but has only one symbolic representation, then either $\Delta _{t}^{1}(x)$ or $\Delta _{t}^{2}(x)$ is empty for sufficiently small t and the argument is identical, but easier.

Finally, suppose x is not an endpoint of any net interval, and thus has unique symbolic representation $[x] = (e_{1},\ldots ,e_{j},\theta ,\theta ,\ldots )$ , where $\theta =(\theta _{1},\ldots ,\theta _{\ell })$ . In this situation, $\Delta _{1}$ has symbolic representation $(e_{1},\ldots ,e_{j},\theta ^{n})$ and $\Delta _{2}$ has symbolic representation $(e_{1},\ldots ,e_{j},\theta ^{n+1})$ for any $n\in {\mathbb {N}}$ , so we have $\Delta _{2}\subseteq \Delta _{1}^{\circ }$ . Thus for any t sufficiently small, there exists some $m\in {\mathbb {N}}$ , such that $\Delta _{1}\subseteq \Delta _{t}(x)\subseteq M_{t}(x)\subseteq \Delta _{2}$ , where $\Delta _{1}$ has symbolic representation $(e_{1},\ldots ,e_{j},\theta ^{m})$ and $\Delta _{2}$ has symbolic representation $(e_{1},\ldots ,e_{j},\theta ^{m+2})$ . Similarly as argued in equation (3.5), there exist constants $c_{1},c_{2}$ such that ${\lVert T(\theta )^{m+2}\rVert }\leq c_{1}\mu (\Delta _{t}(x))\leq c_{2}{\lVert T(\theta )^{m}\rVert }$ . In addition, since $M_{t}(x)\subseteq \Delta _{1}$ , we have $\mu (M_{t}(x))\leq \mu (\Delta _{1})$ and there exist constants $c_{1}^{\prime },c_{2}^{\prime }$ such that ${\lVert T(\theta )^{m+2}\rVert }\leq c_{1}^{\prime }\mu (M_{t}(x))\leq c_{2}^{\prime }{\lVert T(\theta )^{m}\rVert }$ .

The argument proceeds identically as before.

Proof. Let x be an interior essential point. Either there exists some $s_{0}$ such that there is a unique essential net interval $\Delta _{0}\in \mathcal {F}_{s_{0}}$ containing x, or there exists essential net intervals $\Delta _{0}^{1},\Delta _{0}^{2}$ such that $\{x\}=\Delta _{0}^{1}\cap \Delta _{0}^{1}$ . The cases are similar, but the latter is slightly harder, so we treat that here.

Let $t_{0}>0$ be such that $B(x,2t_{0})\subseteq \Delta _{0}^{1}\cup \Delta _{0}^{2}$ . Arguing similarly to Lemma 3.10, there exists constants $c,C>0$ such that for any $0<t\leq t_{0}$ , there exists $\Delta _{t}^{1}\subseteq \Delta _{t}^{2}\subseteq B(x,2t)$ and for each $k=1,2$ , we have $\Delta _{t}^{k}\in \mathcal {F}_{s}$ , where $t\geq s\geq ct$ ,

$$ \begin{align*} \min\{{\boldsymbol{Q}_{\boldsymbol{p}}}(\Delta_{t}^{k})_{j}:1\leq j\leq \# {\mathcal{V}}(\Delta_{t}^{k})\}\geq C{\mu_{\boldsymbol{p}}}(B(x,t)) \end{align*} $$

and ${\mathcal {V}}(\Delta _{i}^{k})={\mathcal {V}}(\Delta _{0}^{k})$ . We may also assume that $\Delta _{t}^{1}$ and $\Delta _{t}^{2}$ do not contain x as an endpoint. In particular, for each $0<t\leq t_{0}$ , there exists some $k\in \{1,2\}$ such that $\Delta _{t}^{k}\subseteq (\Delta _{0}^{k})^{\circ }$ . Set $\Delta _{t}=\Delta _{t}^{k}$ and let $\eta _{t}$ be the path in the transition graph corresponding to $\Delta _{t}^{k}\subseteq \Delta _{0}^{k}$ , which is a cycle since the two net intervals have the same neighbour set. Let $\gamma _{1}$ be the symbolic representation of $\Delta _{0}^{1}$ and $\gamma _{2}$ the symbolic representation of $\gamma _{0}^{2}$ .

For each $0<t\leq t_{0}$ , let $x_{t}$ be any periodic point with period $\eta _{t}$ . We note that since $x_{t}$ is not the boundary point of any net interval, we have by Proposition 3.16

$$ \begin{align*} {\dim_{\operatorname{\mathrm{loc}}}}{\mu_{\boldsymbol{p}}}(x_{t})=\frac{\log\operatorname{\mathrm{sp}} T(\eta_{t})}{\log L(\eta_{t})}. \end{align*} $$

Fix t as above, and let $\Delta _{0}\in \{\Delta _{0}^{1},\Delta _{0}^{2}\}$ be such that $x_{0}\in \Delta _{0}^{\circ }$ . Let $\Delta _{0}$ have symbolic representation $\gamma $ . By definition of c, we observe that $t\geq \operatorname {\mathrm {tg}}(\Delta _{t})\geq c r_{\min }t$ . Since $\operatorname {\mathrm {tg}}(\Delta _{t})=L(\gamma )L(\eta _{t})$ , there exist constants $c_{1},c_{2}>0$ (not depending on t) such that

$$ \begin{align*} c_{1} 2t\leq L(\eta_{t})\leq c_{2} t. \end{align*} $$

We also bound $\operatorname {\mathrm {sp}} T(\eta _{t})$ . Since $\Delta _{t}\subseteq B(x,2t)$ has symbolic representation $\gamma \eta _{t}$ , we have ${\lVert T(\gamma \eta _{t})\rVert }\leq {\mu _{\boldsymbol {p}}}(B(x,2t))$ and since $T(\gamma )$ is a transition matrix, there exists some $C_{1}>0$ such that

$$ \begin{align*} \operatorname{\mathrm{sp}} T(\eta_{t})\leq {\lVert T(\eta_{t})\rVert}\leq C_{1}{\mu_{\boldsymbol{p}}}(B(x,2t)) \end{align*} $$

(just take $C_{1}$ to be the smallest strictly positive entry of $T(\gamma _{1})$ and $T(\gamma _{2})$ ). However, since ${\boldsymbol {Q}_{\boldsymbol {p}}}(\Delta _{t})={\boldsymbol {Q}_{\boldsymbol {p}}}(\Delta _{0})T(\eta _{t})$ , we have

$$ \begin{align*} \operatorname{\mathrm{sp}} T(\eta_{i})&\geq \frac{\min\{{\boldsymbol{Q}_{\boldsymbol{p}}}(\Delta_{t})_{j}:1\leq j\leq \# v\}}{\max\{{\boldsymbol{Q}_{\boldsymbol{p}}}(\Delta_{0})_{j}:1\leq j\leq\# v\}}\geq\frac{C{\mu_{\boldsymbol{p}}}(B(x,t))}{\max\{{\boldsymbol{Q}_{\boldsymbol{p}}}(\Delta_{0}^{k})_{j}:1\leq j\leq\# v,1\leq k\leq 2\}}\\[3pt] &= C_{2} {\mu_{\boldsymbol{p}}}(B(x,t)). \end{align*} $$

To summarize, we have shown that

$$ \begin{align*} \frac{\log C_{2}+\log {\mu_{\boldsymbol{p}}}(B(x,t))}{\log c_{2}+\log t}&\geq {\dim_{\operatorname{\mathrm{loc}}}}{\mu_{\boldsymbol{p}}}(x_{t})=\frac{\log \operatorname{\mathrm{sp}} T(\eta_{t})}{\log L(\eta_{t})}\\[6pt] &\geq\frac{\log C_{1}+\log {\mu_{\boldsymbol{p}}}(B(x,2t))}{\log c_{1}+\log 2t}. \end{align*} $$

Let $\alpha ={\overline {\dim }_{\operatorname {\mathrm {loc}}}}{\mu _{\boldsymbol {p}}}(x)$ and let $\epsilon>0$ be arbitrary. Get some $t_{1}>0$ such that for all $0<t\leq t_{1}$ ,

$$ \begin{align*} \frac{\log C_{2}+\log{\mu_{\boldsymbol{p}}}(B(x,t))}{\log c_{2} +\log t}\leq\alpha+\epsilon \end{align*} $$

and then choose $0<t\leq \min \{t_{0},t_{1}\}$ such that

$$ \begin{align*} \frac{\log C_{1}+\log{\mu_{\boldsymbol{p}}}(B(x,2t))}{\log c_{1}+\log 2t}\geq\alpha-\epsilon. \end{align*} $$

Since $\epsilon>0$ was arbitrary, it follows that the set of local dimensions at periodic points is dense in $\{{\overline {\dim }_{\operatorname {\mathrm {loc}}}}(x):x\in K_{\operatorname {\mathrm {ess}}}\}$ . The result for lower local dimensions holds identically.

Proof. The forward direction is immediate. Conversely, let $0<\epsilon _{0}<1$ be the constant from weak regularity of the boundary of E. Suppose x is in the interior of E relative to K and let $t>0$ . If $B(x,t\epsilon _{0}/4)\cap K=B(x,t\epsilon _{0}/4)\cap E$ , then for $\epsilon <\epsilon _{0}/4$ ,

$$ \begin{align*} \mathcal{H}^{s}(E\cap B(x,t\epsilon_{0}/4)\setminus B(x,\epsilon t))\geq(a(\epsilon_{0}/4)^{s}-\epsilon^{s} b)t^{s}>0 \end{align*} $$

for some $\epsilon>0$ depending only on a, b, s and $\epsilon _{0}$ . Thus $E\cap B(x,t)\setminus B(x,\epsilon t)\neq \emptyset $ . Otherwise, there is some $y\in B(x,t\epsilon _{0}/4)$ in the boundary of E so that

$$ \begin{align*} \emptyset\neq E\cap B(y,t/2)\setminus B(y,t\epsilon_{0}/2)\subset E\cap B(x,t)\setminus B(x,t\epsilon_{0}/4), \end{align*} $$

as required.

Proof. This follows directly for $q\geq 0$ since for all t sufficiently small, $B(x,t)\subset V$ for any $x\in E\cap K$ .

Otherwise, let $q<0$ and let t be sufficiently small such that each interval in E has length at least $2t$ and $B(x,t)\subset V$ for any $x\in E\cap K$ . Let $\{B(x_{i},t)\}_{i}$ be an arbitrary centred packing of $E\cap K$ . By weak regularity, there is some $\epsilon>0$ such that for each i, there is some $y_{i}\in E\cap K$ such that $B(y_{i},\epsilon t)\subseteq B(x_{i},t)\cap E$ . Therefore,

$$ \begin{align*} \sum_{i}\mu|_{E}(B(x_{i},t))^{q}\leq\sum_{i}\mu|_{E}(B(y_{i},\epsilon t))^{q}=\sum_{i} \mu(B(y_{i},\epsilon t))^{q}. \end{align*} $$

However, $\{B(x_{i},t)\}_{i}$ was arbitrary, so the desired result follows.

Proof. For each $t>0$ , set

$$ \begin{align*} \mathcal{H}_{t}=\{\Delta\in\mathcal{F}_{t}:{\mathcal{V}}(\Delta)\in V(\mathcal{G}_{\operatorname{\mathrm{ess}}})\} \end{align*} $$

and let

$$ \begin{align*} U_{t}=\bigg(\bigcup_{\Delta\in\mathcal{H}_{t}}\Delta\bigg)^{\circ}. \end{align*} $$

It follows directly from the definition that $K_{\operatorname {\mathrm {ess}}}=\bigcup _{t>0}U_{t}$ . We may assume $\delta $ is sufficiently small so that $F^{(\delta )}\subset K_{\operatorname {\mathrm {ess}}}$ . Since $F^{(\delta )}$ is compact, get $t_{0}$ such that $F^{(\delta )}\subset U_{t_{0}}$ , let $t_{1}=\min \{t_{0},\delta /2\}$ , and set

$$ \begin{align*} \mathcal{E} &=\{\Delta\in\mathcal{H}_{t_{1}}:\Delta\cap F\neq\emptyset\},\\ E_{0}&=\bigcup_{\Delta\in\mathcal{E}}\Delta. \end{align*} $$

Note that $F\subset E_{0}\subset F^{(\delta /2)}$ since $\operatorname {\mathrm {diam}}(\Delta )\leq t_{1}$ for any $\Delta \in \mathcal {F}_{t_{1}}$ . Now for each $\Delta \in \mathcal {E}$ , get $0<t\leq t_{1}$ and $\sigma ,\tau \in \Lambda _{t}$ such that $r_{\sigma },r_{\tau }>0$ and $\Delta \subseteq [S_{\sigma }(0),S_{\tau }(1)]\subseteq \Delta ^{(\delta /2)}$ . Finally, set

$$ \begin{align*} \mathcal{E}_{\Delta}=\{\Delta^{\prime}\in\mathcal{F}_{t}:\Delta^{\prime}\subset[S_{\sigma}(0),S_{\tau}(1)]\}. \end{align*} $$

Observe that

$$ \begin{align*} K\cap\bigcup_{\Delta^{\prime}\in\mathcal{E}_{\Delta}}\Delta^{\prime}=K\cap[S_{\sigma}(0),S_{\tau}(1)]\subset F^{(\delta)} \end{align*} $$

is weakly regular by Lemma 4.4. Moreover, since $F^{(\delta )}\subset U_{t_{0}}$ , each $\Delta ^{\prime }\in \mathcal {E}_{\Delta }$ is essential. Thus since a union of weakly regular sets is again weakly regular,

$$ \begin{align*} E=\bigcup_{\Delta\in\mathcal{E}}\bigcup_{\Delta^{\prime}\in\mathcal{E}_{\Delta}}\Delta \end{align*} $$

satisfies the requirements.

Proof. Write $S_{i}(x)=r x+d_{i}$ , where $0<r<1$ . It suffices to prove that $[0,x]\cap K$ and $[x,1]\cap K$ are weakly regular for any $x=S_{\sigma }(z)$ where $\sigma \in \mathcal {I}^{*}$ and $z\in \{0,1\}$ , where x is not an isolated point of $[0,x]\cap K$ or $[x,1]\cap K$ . We will prove the case $[0,S_{\sigma }(0)]\cap K$ ; the remaining cases are either analogous or easier.

Suppose for contradiction $[0,S_{\sigma }(0)]$ is not weakly regular and get indices $(k_{n})_{n=1}^{\infty }$ and a sequence $(\epsilon _{n})_{n=1}^{\infty }$ converging monotonically to zero such that

$$ \begin{align*} [S_{\sigma}(0)-r^{k_{n}},S_{\sigma}(0)-\epsilon_{n} r^{k_{n}})\cap K=\emptyset \end{align*} $$

for each $n\in {\mathbb {N}}$ . Since $S_{\sigma }(0)$ is an accumulation point from the right, there is some $\tau _{n}\in \mathcal {I}^{k_{n}}$ such that $S_{\tau _{n}}([0,1])\supseteq [S_{\sigma }(0)-\delta ,S_{\sigma }(0)]$ for some $\delta>0$ sufficiently small. However, $S_{\tau _{n}}(\{0,1\})\cap [S_{\sigma }(0)-r^{k_{n}},S_{\sigma }(0)-\epsilon _{n} r^{k_{n}})=\emptyset $ , which forces

$$ \begin{align*} |S_{\tau_{n}}(0)-S_{\sigma_{n}}(0)|\leq\epsilon_{n} r^{k_{n}}, \end{align*} $$

where $\sigma _{n}\in \mathcal {I}^{k_{n}}$ is the word with $\sigma $ as a prefix and $S_{\sigma _{n}}(0)=S_{\sigma }(0)$ . This contradicts the weak separation condition by [Reference Zerner40, Theorem 1].

Proof. Statement (i) is [Reference Feng and Lau13, Corollary. 5.6]. Statements (ii) and (iii) are implicit in the usage of [Reference Feng and Lau13, Lemma 2.5].

Proof. We split the proof into two parts for clarity.

Let $U_{0}$ be an open ball which attains the maximal value in equation (3.1).

To verify (i), by Proposition 4.9, it suffices to show that

$$ \begin{align*} \tau_{U_{0}}({\mu_{\boldsymbol{p}}},q)&=\tau(\nu,q) \quad D_{U_{0}}({\mu_{\boldsymbol{p}}})=D(\nu) \quad \operatorname{dim}_{\mathrm{H}} K_{U_{0}}({\mu_{\boldsymbol{p}}},\alpha)=\operatorname{dim}_{\mathrm{H}} K(\nu,\alpha). \end{align*} $$

Let $\sigma $ be such that $S_{\sigma }(U_{0})\subseteq E$ , and we see directly from the definitions and Lemma 4.10 that

$$ \begin{align*} \tau_{U_{0}}({\mu_{\boldsymbol{p}}},q)&=\tau_{S_{\sigma}(U_{0})}({\mu_{\boldsymbol{p}}},q)\geq\tau(\nu,q),\\[3pt] D_{U_{0}}({\mu_{\boldsymbol{p}}})&=D_{S_{\sigma}(U_{0})}({\mu_{\boldsymbol{p}}})\subseteq D(\nu),\\[3pt] \operatorname{dim}_{\mathrm{H}} K_{U_{0}}({\mu_{\boldsymbol{p}}},\alpha)&=\operatorname{dim}_{\mathrm{H}} K_{S_{\sigma}(U_{0})}(\nu,\alpha) \leq \operatorname{dim}_{\mathrm{H}} K(\nu,\alpha). \end{align*} $$

We now establish the reverse inequalities.

That $\tau ({\mu _{\boldsymbol {p}}},q)=\tau _{U_{0}}({\mu _{\boldsymbol {p}}},q)=\tau (\nu ,q)$ for $q\geq 0$ is straightforward; see, for example, [Reference Feng and Lau13, Proposition 3.1].

Otherwise, fix $q<0$ . Since $U_{0}$ is open and the $\Delta _{i}$ are essential, there exist net intervals $\Delta _{1}^{*},\ldots ,\Delta _{n}^{*}$ such that ${\mathcal {V}}(\Delta _{i})={\mathcal {V}}(\Delta _{i}^{*})$ for each $1\leq i\leq n$ and the $\Delta _{i}^{*}$ are pairwise disjoint. By Lemma 4.6, there exist compact intervals $F_{i}\supseteq \Delta _{i}^{*}$ such that the $F_{i}$ are weakly regular, pairwise disjoint and have $F_{i}\subset U_{0}$ . Set $E^{*}:= F_{1}\cup \cdots \cup F_{n}$ and let $\nu ^{*}:= \nu |_{E^{*}}$ . Since $E^{*}$ is weakly regular, it follows that $\tau _{U_{0}}({\mu _{\boldsymbol {p}}},q) \leq \tau (\nu ^{*},q)$ by Lemma 4.5.

It remains to show that $\tau (\nu ^{*},q)\leq \tau (\nu ,q)$ for $q<0$ . By Lemma 2.3, get similarities $g_{i}:\Delta _{i}\cap K\to \Delta _{i}^{*}\cap K$ and some $c_{1},c_{2}>0$ such that if $E\subseteq \Delta _{i}$ is an arbitrary Borel set,

(4.1) $$ \begin{align} c_{1}\nu^{*}(g_{i}(E))\leq \nu(E)\leq c_{2}\nu^{*}(g_{i}(E)). \end{align} $$

Let each $g_{i}$ have contraction ratio $\rho _{i}$ .

Now let $t>0$ be sufficiently small so that $2t\leq \min \{\operatorname {\mathrm {diam}}(\Delta _{i}):1\leq i\leq n\}$ and let $\epsilon _{0}>0$ be the constant from weak regularity of $E\cap K$ . Suppose $\{B(x_{j},t)\}_{j=1}^{m}$ is an arbitrary family of disjoint closed balls, where $x_{j}\in E\cap K$ . For each j, there is some $i(j)$ and $y_{j}$ such that

(4.2) $$ \begin{align} B(y_{j},t\epsilon_{0}/4)\subseteq\Delta_{i(j)}\cap B(x_{j},t) \end{align} $$

(this must hold for either $y_{j}=x_{j}$ or $y_{j}\in E\cap K\cap B(x_{j},t/2)\setminus B(x_{j},t\epsilon _{0}/2)$ ).

Now set

$$ \begin{align*} \rho_{0}=\dfrac{\epsilon_{0}}{4}\min\{\rho_{i}:1\leq i\leq n\}. \end{align*} $$

For each $1\leq j\leq m$ , by equation (4.2),

$$ \begin{align*} \nu(B(x_{j},t))&\geq\nu(B(y_{j},t\epsilon_{0}/4) \geq c_{1}\nu^{*}(B(g_{i(j)}(y_{j}),\rho_{i(j)}t\epsilon_{0}/4))\\[3pt] &\geq c_{1}\nu^{*}(B(g_{i(j)}(y_{j}),\rho_{0} t)), \end{align*} $$

so that $\nu (B(x_{j},t))^{q}\leq c_{1}^{q} \nu ^{*}(B(x_{j}^{*},\rho _{0} t))^{q}$ , where $x_{j}^{*}=g_{i(j)}(y_{j})$ . Observe also that the $B(x_{j}^{*},\rho _{0} t)$ are pairwise disjoint. However, $\{B(x_{j},t)\}_{j=1}^{m}$ was an arbitrary cover, so that

$$ \begin{align*} \frac{\log \sup\sum_{j}\nu(B(x_{j},t))^{q}}{\log t}\geq \frac{\log c_{1}^{q}+\log\sup\sum_{j} \nu^{*}(B(x_{j}^{*},\rho_{0} t))^{q}}{\log \rho_{0}^{-1}+\log \rho_{0}t}. \end{align*} $$

Taking limits, it follows that $\tau (\nu ,q)\geq \tau (\nu ^{*},q)$ for $q<0$ .

We now see that $D(\nu )\subseteq D_{U_{0}}({\mu _{\boldsymbol {p}}})$ . First note that $D_{U_{0}}({\mu _{\boldsymbol {p}}})=[\alpha _{\min },\alpha _{\max }]$ , where

$$ \begin{align*} \alpha_{\min} &= \lim_{q\to +\infty}\frac{\tau(\nu,q)}{q}=\lim_{q\to +\infty}\frac{\tau_{U_{0}}({\mu_{\boldsymbol{p}}},q)}{q}, \nonumber\\[3pt] \alpha_{\max}&=\lim_{q\to -\infty}\frac{\tau(\nu,q)}{q}=\lim_{q\to -\infty}\frac{\tau_{U_{0}}({\mu_{\boldsymbol{p}}},q)}{q}, \end{align*} $$

since $\tau (\nu ,q)=\tau _{U_{0}}({\mu _{\boldsymbol {p}}},q)$ . Let $x\in \operatorname {\mathrm {supp}} \nu $ be arbitrary with $\alpha ={\dim _{\operatorname {\mathrm {loc}}}}\nu (x)$ . Then for any $q\in {\mathbb {R}}$ and $t>0$ , we have

$$ \begin{align*} \log \sup \sum_{i} \nu(B(x_{i},t))^{q}\geq\log \nu(B(x,t))^{q}, \end{align*} $$

where the supremum is over disjoint balls $B(x_{i},t)$ with $x_{i}\in \operatorname {\mathrm {supp}}\nu $ , and therefore $\tau (\nu ,q)\leq q \alpha $ . Since $\tau (\nu ,q)$ is concave, it follows that $\alpha \in [\alpha _{\min },\alpha _{\max }]=D_{U_{0}}({\mu _{\boldsymbol {p}}})$ .

Finally, we verify that $\operatorname {dim}_{\mathrm {H}} K(\nu ,\alpha )\leq \operatorname {dim}_{\mathrm {H}} K_{U_{0}}({\mu _{\boldsymbol {p}}},\alpha )$ . First note by equation (4.1) that if $x\in \Delta _{i}^{\circ }\cap K$ for some i, then $g_{i}(x)\in (\Delta _{i}^{*})^{\circ }\cap K\subset U_{0}$ has

$$ \begin{align*} {\dim_{\operatorname{\mathrm{loc}}}} \nu(x)=\dim\nu^{*}(g_{i}(x))={\dim_{\operatorname{\mathrm{loc}}}}{\mu_{\boldsymbol{p}}}(g_{i}(x)). \end{align*} $$

Thus $g_{i}(K(\nu ,\alpha )\cap \Delta _{i}^{\circ })\subseteq K_{U_{0}}({\mu _{\boldsymbol {p}}},\alpha )$ and

$$ \begin{align*} \operatorname{dim}_{\mathrm{H}}\bigg(K(\nu,\alpha)\cap\bigcup_{i=1}^{n}\Delta_{i}^{\circ}\bigg)\leq\operatorname{dim}_{\mathrm{H}} K_{U_{0}}({\mu_{\boldsymbol{p}}},\alpha). \end{align*} $$

Since $D(\nu )=D_{U_{0}}({\mu _{\boldsymbol {p}}})$ and $E\setminus \bigcup _{i=1}^{n}\Delta _{i}^{\circ }$ is a finite set (and hence has Hausdorff dimension 0), the result follows.

Thus the complete multifractal formalism holds.

Since $U_{0}$ was fixed, $\tau (\nu ,q)$ does not depend on the choice of $\Delta _{1},\ldots ,\Delta _{n}$ .

We now see that

(4.3) $$ \begin{align} D(\nu) = \{{\dim_{\operatorname{\mathrm{loc}}}}{\mu_{\boldsymbol{p}}}(x):x\in K_{\operatorname{\mathrm{ess}}},{\dim_{\operatorname{\mathrm{loc}}}}{\mu_{\boldsymbol{p}}}(x)\text{ exists}\}. \end{align} $$

If $x\in K_{\operatorname {\mathrm {ess}}}$ , by Lemma 4.6, there is a weakly regular finite union of essential net intervals F such that $x\in (F\cap K)^{\circ }$ , where we take the interior relative to K, and

$$ \begin{align*} {\dim_{\operatorname{\mathrm{loc}}}}{\mu_{\boldsymbol{p}}}(x)={\dim_{\operatorname{\mathrm{loc}}}}{\mu_{\boldsymbol{p}}}|_{F}(x)\in D(\nu), \end{align*} $$

since $D(\nu )=D({\mu _{\boldsymbol {p}}}|_{F})$ as proven above. Conversely, if $\alpha \in D(\nu )$ , then there exists some $y\in U_{0}$ such that ${\dim _{\operatorname {\mathrm {loc}}}}{\mu _{\boldsymbol {p}}}(y)=\alpha $ . However, $U_{0}\subseteq K_{\operatorname {\mathrm {ess}}}$ by Proposition 3.7, so that equation (4.3) follows.

By Theorem 4.1, we have that

$$ \begin{align*} P({\mu_{\boldsymbol{p}}})=\{{\dim_{\operatorname{\mathrm{loc}}}}{\mu_{\boldsymbol{p}}}(x):x\in K_{\operatorname{\mathrm{ess}}},x\text{ periodic}\} \end{align*} $$

is dense in the set of upper and lower local dimensions in $K_{\operatorname {\mathrm {ess}}}$ . Now $P({\mu _{\boldsymbol {p}}})\subseteq D(\nu )$ from equation (4.3) and $D(\nu )=[\alpha _{\min },\alpha _{\max }]$ is a closed set with $D(\nu )\subseteq \{{\overline {\dim }_{\operatorname {\mathrm {loc}}}}{\mu _{\boldsymbol {p}}}(x):x\in K_{\operatorname {\mathrm {ess}}}\}$ . However again, Theorem 4.1 shows that $P({\mu _{\boldsymbol {p}}})$ is a dense subset of $\{{\overline {\dim }_{\operatorname {\mathrm {loc}}}}{\mu _{\boldsymbol {p}}}(x):x\in K_{\operatorname {\mathrm {ess}}}\}$ , forcing

$$ \begin{align*} D(\nu) = \{{\overline{\dim}_{\operatorname{\mathrm{loc}}}}{\mu_{\boldsymbol{p}}}(x):x\in K_{\operatorname{\mathrm{ess}}}\}. \end{align*} $$

Of course, we also have $D(\nu )=\{{\underline {\dim }_{\operatorname {\mathrm {loc}}}}{\mu _{\boldsymbol {p}}}(x):x\in K_{\operatorname {\mathrm {ess}}}\}$ by the same argument, finishing the proof of the theorem.

Proof. Since ${\mu _{\boldsymbol {p}}}$ is Borel and $K_{\operatorname {\mathrm {ess}}}$ is a relatively open subset of K with ${\mu _{\boldsymbol {p}}}(K_{\operatorname {\mathrm {ess}}})=1$ by Theorem 3.11, there exists a nested sequence of compact sets $(F_{m})_{m=1}^{\infty }$ with $F_{m}\subset K_{\operatorname {\mathrm {ess}}}$ such that $\lim _{m\to \infty }\mu (F_{m})=1$ . Let $K_{m}\supseteq F_{m}$ be a finite union of essential net intervals given by Lemma 4.6. Then by Theorem 4.11, each $\mu _{m}:= {\mu _{\boldsymbol {p}}}|_{K_{m}}$ satisfies the complete multifractal formalism and $\tau (\mu _{m},q)$ and $D(\mu _{m})$ do not depend in the index m, as required.