Proof. Let $ y\in \mathcal U^\kappa (x)\setminus \{Tx\} $ and let $ i $ be the smallest integer satisfying $ y\notin B(Tx, i^{-\kappa }) $ . By the definition of $ \mathcal U^\kappa (x) $ , for any integer $ N> i $ large enough, there exists $ 1\le n\le N $ such that $ y\in B(T^nx, N^{-\kappa }) $ . Moreover, the condition $ N> i $ implies that $ n\ne 1 $ , and hence $ y\in B(T^{n-1}(Tx),N^{-\kappa }) $ . This gives that $ y\in \mathcal U^\kappa (Tx) $ ; therefore, $ \mathcal U^\kappa (x)\setminus \{Tx\}\subset \mathcal U^{\kappa }(Tx) $ .

Proof of Theorem 1.1: Lebesgue measure part

For any $ \kappa>0 $ , define the function $ g_\kappa :x\mapsto \unicode{x3bb} (\mathcal U^\kappa (x)) $ . We claim that $ g_\kappa $ is measurable. In fact, it suffices to observe that

$$ \begin{align*}g_\kappa(x)=\lim_{i\to \infty}\lim_{m\to\infty}\unicode{x3bb}\bigg(\bigcap_{N=i}^m\bigcup_{n=1}^NB(T^nx,N^{-\kappa})\bigg)\end{align*} $$

and that, by the piecewise continuity of $ T^n $ ( $ n\ge 1 $ ), the set

$$ \begin{align*}\bigg\{x\in[0,1]:\unicode{x3bb}\bigg(\bigcap_{N=i}^m\bigcup_{n=1}^NB(T^nx,N^{-\kappa})\bigg)>t\bigg\}\end{align*} $$

is measurable for any $ t\in \mathbb R $ .

By Lemma 2.1, we see that $ \unicode{x3bb} (\mathcal U^\kappa (Tx))\ge \unicode{x3bb} (\mathcal U^\kappa (x)) $ , or equivalently, $ g_\kappa (Tx)\ge g_\kappa (x) $ . Since $ g_\kappa $ is measurable, by the fact $ 0\le g_\kappa \le 1 $ and the invariance of $ \nu $ , we have that $ g_\kappa $ is invariant with respect to $ \nu $ , that is, $ g_\kappa (Tx)=g_\kappa (x) $ for $ \nu $ -a.e. $ x $ . In the presence of ergodicity of $ \nu $ , $ g_\kappa $ is constant almost surely.

Proof of Theorem 1.1: Hausdorff dimension part

Fix $ \kappa>0 $ and define the function $ f_\kappa :x\mapsto \operatorname {\mathrm {\dim _H}}\mathcal U^\kappa (x) $ . Again by Lemma 2.1, we see that $ f_\kappa (Tx)\ge f_\kappa (x) $ . Proceeding in the same way as the Lebesgue measure part, we get that $ f_\kappa $ is constant almost surely provided that $ f_\kappa $ is measurable.

To show that $ f_\kappa $ is measurable, it suffices to prove that for any $ t>0 $ , the set

$$ \begin{align*}\begin{split} A(t):=\{x\in[0,1]:f_\kappa(x)<t\}=\bigg\{x\in[0,1]:\operatorname{\mathrm{\dim_H}}\!\bigg(\bigcup_{i=1}^\infty\bigcap_{N= i}^\infty\bigcup_{n=1}^NB(T^nx, N^{-\kappa})\!\bigg)\!<t\bigg\} \end{split}\end{align*} $$

is measurable. Throughout the proof of this part, we will assume that the ball $ B(T^nx, N^{-\kappa }) $ is closed. This makes the proof achievable and it does not change the Hausdorff dimension of $ \mathcal U^\kappa (x) $ .

By the definition of Hausdorff dimension, a point $ x\in A(t) $ if and only if there exists $ h\in \mathbb N $ such that

$$ \begin{align*}\mathcal H^{t-{1}/{h}}\bigg(\bigcup_{i=1}^\infty\bigcap_{N= i}^\infty\bigcup_{n=1}^NB(T^nx, N^{-\kappa})\bigg)=0,\end{align*} $$

or equivalently for all $ i\ge 1 $ ,

(2.3) $$ \begin{align} \mathcal H^{t-{1}/{h}}\bigg(\bigcap_{N= i}^\infty\bigcup_{n=1}^NB(T^nx, N^{-\kappa})\bigg)=0. \end{align} $$

By the definition of Hausdorff measure, equation (2.3) holds if and only if for any $ j,k\in \mathbb N $ ,

$$ \begin{align*}\mathcal H_{{1}/{j}}^{t-{1}/{h}}\bigg(\bigcap_{N= i}^\infty\bigcup_{n=1}^NB(T^nx, N^{-\kappa})\bigg)<\frac{1}{k}.\end{align*} $$

Hence, we see that

$$ \begin{align*} A(t)&=\bigcup_{h=1}^\infty\bigcap_{i=1}^\infty\bigcap_{j=1}^\infty\bigcap_{k=1}^\infty\bigg\{ x\in [0,1]:\mathcal H_{{1}/{j}}^{t-{1}/{h}}\bigg(\bigcap_{N= i}^\infty\bigcup_{n=1}^NB(T^nx, N^{-\kappa})\bigg)<\frac{1}{k}\bigg\}\\&=:\bigcup_{h=1}^\infty\bigcap_{i=1}^\infty\bigcap_{j=1}^\infty\bigcap_{k=1}^\infty B_{h,i,j,k}. \end{align*} $$

If $ x\in B_{h,i,j,k} $ , then there is a countable open cover $ \{U_p\}_{p\ge 1} $ with $ 0<|U_p|<1/j $ satisfying

(2.4) $$ \begin{align} \bigcap_{N= i}^\infty\bigcup_{n=1}^NB(T^nx, N^{-\kappa})\subset \bigcup_{p=1}^\infty U_p\quad \text{and}\quad\sum_{p=1}^\infty |U_p|^{t-{1}/{h}}<\frac{1}{k}. \end{align} $$

The set $ \bigcap _{N= i}^\infty \bigcup _{n=1}^NB(T^nx, N^{-\kappa }) $ can be viewed as the intersection of a family of decreasing compact sets $ \{\bigcap _{N= i}^l\bigcup _{n=1}^NB(T^nx, N^{-\kappa })\}_{l\ge i} $ , and hence there exists $ l_0\ge i $ satisfying

$$ \begin{align*}\bigcap_{N= i}^{l_0}\bigcup_{n=1}^NB(T^nx, N^{-\kappa})\subset\bigcup_{p=1}^\infty U_p,\end{align*} $$

which implies

$$ \begin{align*}\mathcal H_{{1}/{j}}^{t-{1}/{h}}\bigg(\bigcap_{N= i}^{l_0}\bigcup_{n=1}^NB(T^nx, N^{-\kappa})\bigg)<\frac1k.\end{align*} $$

We then deduce that

(2.5) $$ \begin{align} B_{h,i,j,k}\subset\bigcup_{l=i}^{\infty}\bigg\{ x\in [0,1]:\mathcal H_{{1}/{j}}^{t-{1}/{h}}\bigg(\bigcap_{N= i}^l\bigcup_{n=1}^NB(T^nx, N^{-\kappa})\bigg)<\frac1k\bigg\}=:\bigcup_{l=i}^{\infty}C_{h,i,j,k,l}. \end{align} $$

If $ x\in C_{h,i,j,k,l} $ for some $ l\ge 1 $ , then

$$ \begin{align*}\mathcal H_{{1}/{j}}^{t-{1}/{h}}\bigg(\bigcap_{N= i}^\infty\bigcup_{n=1}^NB(T^nx, N^{-\kappa})\bigg)\le\mathcal H_{{1}/{j}}^{t-{1}/{h}}\bigg(\bigcap_{N= i}^l\bigcup_{n=1}^NB(T^nx, N^{-\kappa})\bigg)<\frac{1}{k}.\end{align*} $$

Hence, $ x\in B_{i,j,k,l} $ and the reverse inclusion of equation (2.5) is proved.

Notice that $ T,T^2,\ldots ,T^l $ are continuous on every basic interval of generation greater than $ l $ . For any $ x\in C_{h,i,j,k,l} $ , denote $ \mathcal S(x):=\bigcap _{N= i}^l\bigcup _{n=1}^NB(T^nx, N^{-\kappa }) $ . There is an open cover $ (V_p)_{p\ge 1} $ of $ \mathcal S(x) $ with $ 0<|V_p|<1/j $ and $ \sum _p |V_p|^{t-1/h}<1/k $ . Since $ \mathcal S(x) $ is compact, we see that the distance $ \delta $ between $ \mathcal S(x) $ and the complement of $ \bigcup _{p\ge 1} V_p $ is positive. By the continuities of $ T,T^2,\ldots ,T^l $ , there is some $ l_1:=l_1(\delta )>l $ for which if $ y\in I_{l_1}(x) $ , then $ \mathcal S(y) $ is contained in the $ \delta /2 $ -neighborhood of $ \mathcal S(x) $ . Thus, $ \mathcal S(y) $ can also be covered by $ \bigcup _{p\ge 1} V_p $ and $ I_{l_1}(x)\subset C_{h,i,j,k,l} $ . Finally, $ C_{h,i,j,k,l} $ is a union of some basic intervals, which is measurable.

Now, combining the equalities obtained above, we have

$$ \begin{align*}A(t)=\bigcup_{h=1}^\infty\bigcap_{i=1}^\infty\bigcap_{j=1}^\infty\bigcap_{k=1}^\infty\bigcup_{l=i}^{\infty}C_{h,i,j,k,l},\end{align*} $$

which is a Borel measurable set.

Proof. The top left and bottom right inclusions imply one another. Let us prove the bottom right inclusion. Suppose that $ y\in \mathcal U^\kappa (x) $ . Then for all large enough $ N $ , there is an $ n\le N $ such that $ T^nx \in B(y,N^{-\kappa })$ . Thus, $ \tau _{N^{-\kappa }}(x,y)\le N $ for all $ N $ large enough, which implies $ \overline {R}(x,y)\le 1/\kappa $ .

The top right and bottom left inclusions imply one another. So, it remains to prove the bottom left inclusion. Consider $ y $ such that $ \overline {R}(x,y)<1/\kappa $ . If $ y\in \mathcal O^+(x) $ with $ y=T^{n_0}x $ for some $ n_0\ge 1 $ , then the system

$$ \begin{align*}|T^nx-y|=|T^nx-T^{n_0}x|<N^{-\kappa}\quad\text{and}\quad 1\le n\le N\end{align*} $$

always has a trivial solution $ n=n_0 $ for all $ N\ge n_0 $ . Therefore, $ y\in \mathcal U^\kappa (x) $ . Now assume that $ y\notin \mathcal O^+(x) $ . By the definition of $ \overline {R}(x,y) $ , there is a positive real number $ r_0<1 $ such that

$$ \begin{align*}\tau_r(x,y)<r^{-1/\kappa} \quad\text{for all }0<r<r_0.\end{align*} $$

Denote $ n_r:=\tau _r(x,y) $ for all $ 0<r<r_0 $ . Since $ y\notin \mathcal O^+(x) $ , the family of positive integers $ \{n_r:0<r<r_0\} $ is unbounded. For each $ N> r_0^{-1/\kappa } $ , denote $ t:=N^{-\kappa } $ . The definition of $ n_{t} $ implies that

$$ \begin{align*}T^{n_{t}}x\in B(y,t)=B(y,N^{-\kappa}).\end{align*} $$

We conclude $ y\in \mathcal U^\kappa (x) $ by noting that $ n_{t}<t^{-1/\kappa }=N $ .

Proof. (a) Let $ 1/\kappa \in (0,\alpha _{\max }) $ . As already observed in §3, the Gibbs measure $ \mu _0 $ associated with $ \log |T'| $ is strongly equivalent to the Lebesgue measure $ \unicode{x3bb} $ . Thus, a set $ F $ has full $ \mu _0 $ -measure if and only if $ F $ has full $ \unicode{x3bb} $ -measure. Corollary 4.4 implies that

$$ \begin{align*}\text{for }\mu_\phi\times\mu_0\text{-}\text{a.e.}\,(x,y),\quad R(x,y)=d_{\mu_\phi}(y)=\frac{-\int\phi\, d\mu_0}{\int\log|T'|\, d\mu_0}=\alpha_{\max}.\end{align*} $$

By Fubini’s theorem, for $ \mu _\phi $ -a.e. $ x $ , the set $ \{y:R(x,y)=d_{\mu _\phi }(y)=\alpha _{\max }\} $ has full $ \mu _0 $ -measure. Then for $ \mu _\phi $ -a.e. $ x $ , we have

$$ \begin{align*}\mu_0(\{ y:\overline{R}(x,y)>1/\kappa \})\ge \mu_0(\{y:R(x,y)=d_{\mu_\phi}(y)=\alpha_{\max}\})=1.\end{align*} $$

By Lemma 4.1, we arrive at the conclusion.

(b) By Lemma 3.5, the level set $ \{y:d_{\mu _\phi }(y)=1/\kappa \} $ is empty if $ 1/\kappa>\alpha _+ $ . Thus, $ D_{\mu _\phi }(1/\kappa )=0 $ and therefore $ \operatorname {\mathrm {\dim _H}}\mathcal B^\kappa (x)\ge 0=D_{\mu _\phi }(1/\kappa ) $ trivially holds for all $ 1/\kappa>\alpha _+ $ .

Let $ 1/\kappa \in [\alpha _{\max },\alpha _+) $ . We can suppose that $ \alpha _{\max }\ne \alpha _+ $ , since otherwise $ [\alpha _{\max },\alpha _+)=\emptyset $ and there is nothing to prove. For any $ s\in (1/\kappa ,\alpha _+) $ , by Proposition 3.4, there exists a real number $ q_s:=q(s) $ such that

$$ \begin{align*}s=\frac{-\int\phi\, d\mu_{q_s}}{\int\log|T'|\, d\mu_{q_s}}\quad\text{and}\quad\mu_{q_s}(\{y:d_{\mu_\phi}(y)=s\})=1.\end{align*} $$

Applying Corollary 4.4, we obtain

$$ \begin{align*}\text{for }\mu_\phi\times\mu_{q_s}\text{-}\text{a.e.}\,(x,y),\quad R(x,y)=d_{\mu_\phi}(y)=\frac{-\int\phi\, d\mu_{q_s}}{\int\log|T'|\, d\mu_{q_s}}=s.\end{align*} $$

It follows from Fubini’s theorem that, for $ \mu _\phi $ -a.e. $ x $ , the set $ \{y:R(x,y)=d_{\mu _\phi }(y)=s\} $ has full $ \mu _{q_s} $ -measure. Consequently, for $ \mu _\phi $ -a.e. $ x $ ,

$$ \begin{align*}\begin{split} \operatorname{\mathrm{\dim_H}}\{y:\overline{R}(x,y)>1/\kappa\}&\ge \operatorname{\mathrm{\dim_H}}\{y:R(x,y)=d_{\mu_\phi}(y)=s\}\\ &\ge \operatorname{\mathrm{\dim_H}} \mu_{q_s}=D_{\mu_\phi}(s). \end{split}\end{align*} $$

We conclude by noting that $ s\in (1/\kappa ,\alpha _+) $ is arbitrary and $ D_{\mu _\phi } $ is continuous on $ [\alpha _{\max },\alpha _+) $ .

Proof. Since each $ A_i $ is a union of at most $ m $ disjoint balls, the exponential mixing property of $ \nu $ gives that, for every $ d\ge 1 $ ,

(5.1) $$ \begin{align} |\nu(A_i\cap T^{-d}B)-\nu(A_i)\nu(B)|\le mC\beta^d\nu(B), \end{align} $$

where $ B $ is a Borel measurable set. In particular, defining

$$ \begin{align*}B_i=A_i\cap T^{-d}A_{i+1}\cap\cdots\cap T^{-d(k-i)}A_k,\end{align*} $$

we get, for any $ i<k $ ,

$$ \begin{align*}|\nu(A_i\cap T^{-d}(B_{i+1}))-\nu(A_i)\nu(B_{i+1})|\le mC\beta^d\nu(B_{i+1}).\end{align*} $$

The above inequality can be written as

$$ \begin{align*} 1-\frac{mC\beta^d}{\nu(A_i)}\le \frac{\nu(A_i\cap T^{-d}B_{i+1})}{\nu(A_i)\nu(B_{i+1})}\le 1+\frac{mC\beta^d}{\nu(A_i)}. \end{align*} $$

Multiplying over all $ i\le k $ and using the identity

$$ \begin{align*}B_{i+1}=A_{i+1}\cap T^{-d}B_{i+2},\end{align*} $$

we have

$$ \begin{align*}\kern-18pt\prod_{i=1}^{k}\bigg(1-\frac{mC\beta^d}{\nu(A_i)}\bigg)\le \frac{\nu(A_1\cap T^{-d}A_2\cap\cdots\cap T^{-d(k-1)}A_k)}{\nu(A_1)\nu(A_2)\cdots\nu(A_k)}\le \prod_{i=1}^{k}\bigg(1+\frac{mC\beta^d}{\nu(A_i)}\bigg). \end{align*} $$

Proof. For each $ i\le N $ , let

$$ \begin{align*}\Delta_{i}:=\{x\in[0,1]:\quad\text{ for all } k\le 2^{nh}, T^kx\notin B_i \}.\end{align*} $$

Obviously, we have $ \mathcal C_{n,N,h}=\bigcup _{i=1}^N\Delta _{i} $ , so it suffices to bound from above each $ \nu (\Delta _{i}) $ . Pick an integer $ \omega $ such that $ \omega>\log _{\beta ^{-1}}2^{h} $ . Let $ k=[2^{nh}/(\omega n)] $ be the integer part of $ 2^{nh}/(\omega n) $ . Then,

$$ \begin{align*}\Delta_{i}\subset\bigcap_{j=1}^{k}\{x\in[0,1]: T^{j\omega n}x\notin B_i\}=\bigcap_{j=1}^{k}T^{-j\omega n}B_i^c.\end{align*} $$

Since $ \omega>\log _{\beta ^{-1}}2^{h}$ , there is an $ n_h $ large enough such that for any $ n\ge n_h $ ,

(5.2) $$ \begin{align} 2C\beta^{\omega n}<2^{-nh-1}\le \nu(B_i)/2 \end{align} $$

and

(5.3) $$ \begin{align} 2^{n+1} \exp\bigg(\frac{-2^{n\epsilon}}{2\omega n}\bigg)\le 2^{-n}. \end{align} $$

Now applying Lemma 5.2 to $ A_l= B_i $ for all $ l\le N $ and to $ m=2 $ , we conclude from equation (5.2) that

$$ \begin{align*} \nu(\Delta_{i})\le\nu(\bigcap_{j=1}^{k}T^{-j\omega n}B_i^c) &\le (\nu(B_i^c)+2C\beta^{\omega n})^{k}\\ &\le (1-\nu(B_i)/2)^{k}\\ &\le (1-2^{-n(h-\epsilon)-1})^{ 2^{nh}/(\omega n)-1}\\ &=(1-2^{-n(h-\epsilon)-1})^{-1} \exp\bigg(\frac{2^{nh}\log(1-2^{-n(h-\epsilon)-1})}{\omega n}\bigg)\\ &\le 2\exp\bigg(\frac{-2^{n\epsilon}}{2\omega n}\bigg). \end{align*} $$

By equation (5.3),

$$ \begin{align*}\nu(C_{n,N,h})\le \sum_{i=1}^{N}\nu(B_i)\le 2^{n+1} \exp\bigg(\frac{-2^{n\epsilon}}{2\omega n}\bigg)\le 2^{-n}.\end{align*} $$

Proof. The case $ s=0 $ is obvious. We therefore assume $ s>0 $ . For any integer $ n\ge 1 $ , let

$$ \begin{align*}\mathcal R_{n,s,\epsilon}(x)&=\{y:\tau(x,B(y,2^{-n+1}))\ge 2^{n(s-\epsilon)}\},\\\mathcal E_{n,s,\epsilon}&=\{y:\nu(B(y,2^{-n}))\le 2^{-n(s-2\epsilon)}\}.\end{align*} $$

By definition, $ y\in \{y:\overline {R}(x,y)\ge s\} $ if and only if for any $ \epsilon>0 $ , there exist infinitely many integers $ n $ such that

$$ \begin{align*}\frac{\log \tau(x,B(y,2^{-n+1}))}{\log 2^n}\ge s-\epsilon.\end{align*} $$

Hence, we have

(5.4) $$ \begin{align} \{y:\overline{R}(x,y)\ge s\}=\bigcap_{\epsilon>0}\limsup_{n\to\infty}\mathcal R_{n,s,\epsilon}(x). \end{align} $$

Similarly,

$$ \begin{align*}\{y:\overline d_{\mu_\phi}(y)\ge s\}=\bigcap_{\epsilon>0}\limsup_{n\to\infty}\mathcal E_{n,s,\epsilon}.\end{align*} $$

Thus, it is sufficient to prove that, for $ \nu $ -a.e. $ x $ , there exists some integer $ n(x) $ such that

(5.5) $$ \begin{align} \text{for all }n\ge n(x),\quad\mathcal R_{n,s,\epsilon}(x)\subset \mathcal E_{n,s,\epsilon}, \end{align} $$

or equivalently,

(5.6) $$ \begin{align} \text{for all }n\ge n(x),\quad\mathcal E^c_{n,s,\epsilon}(x)\subset \mathcal R^c_{n,s,\epsilon}. \end{align} $$

Notice that $ \mathcal E_{n,s,\epsilon }^c $ can be covered by $ N\le 2^n $ balls with center in $ \mathcal E_{n,s,\epsilon }^c $ and radius $ 2^{-n} $ . Let $ \mathcal F_{n,s,\epsilon }:=\{B_1,B_2,\ldots ,B_N\} $ be the collection of these balls. By definition, we have $ \nu (B_i)\le 2^{-n(s-2\epsilon )} $ . Applying Lemma 5.3 to the collection $ \mathcal F_{n,s,\epsilon } $ of balls and to $ h=s-\epsilon $ , we see that

$$ \begin{align*}\sum_{n\ge n_h}\nu(\{x:\text{ there exists } B\in\mathcal F_{n,s,\epsilon} \text{ such that }\tau(x,B)\ge 2^{n(s-\epsilon)}\})\le \sum_{n\ge n_h}2^{-n}<\infty.\end{align*} $$

By the Borel–Cantelli lemma, for $ \nu $ -a.e. $ x $ , there exists an integer $ n(x) $ such that

$$ \begin{align*}\text{for all }n\ge n(x),\text{ for all }B\in \mathcal F_{n,s,\epsilon},\quad \tau(x, B)< 2^{n(s-\epsilon)}.\end{align*} $$

If $ y\in B $ for some $ B\in \mathcal F_{n,s,\epsilon } $ and $ n\ge n(x) $ , then $ B\subset B(y,2^{-n+1}) $ , which implies that $ \tau (x, B(y,2^{-n+1}))<\tau (x,B) $ . We then deduce that $ B $ is included in $ \mathcal R_{n,s,\epsilon }^c $ . This yields $ \mathcal E_{n,s,\epsilon }^c\subset \mathcal R_{n,s,\epsilon }^c $ , which is what we want.

Proof. Recall that Lemma 4.1 asserts that

$$ \begin{align*}\mathcal B^\kappa(x)\subset\{y:\overline{R}(x,y)\ge 1/\kappa\}\cup\mathcal O^+(x).\end{align*} $$

A direct application of Proposition 3.6 and Lemma 5.4 yields the first conclusion.

The second conclusion follows from Lemmas 3.5 and 4.1.

Proof of the items (1), (2), and (4) of Theorem 1.2

Combining with Lemmas 5.1 and 5.5, we get the desired result.

Proof. (a) By Lemma 3.6, the Hausdorff dimension of the level set $ \{y:d_{\mu _\phi }(y)=1/\kappa \} $ is zero if $ 1/\kappa <\alpha _- $ . Therefore, $ \operatorname {\mathrm {\dim _H}}\mathcal U^\kappa (x)\ge 0=D_{\mu _\phi }(1/\kappa ) $ .

The remaining case $ 1/\kappa \in (\alpha _-,\alpha _{\max }] $ holds by the same reasoning as proving Lemma 5.1(b).

(b) Observe that the full Lebesgue measure statement implies the full Hausdorff dimension statement. It follows from item (2) of Theorem 1.2 that $ \operatorname {\mathrm {\dim _H}}\mathcal U^\kappa (x)= 1 $ when $ 1/\kappa \in (\alpha _{\max },+\infty ) $ .

Proof. The proof will be divided into two steps.

Step 1. Given any $ a>1/\kappa $ , we are going to prove that

(5.7) $$ \begin{align} \operatorname{\mathrm{\dim_H}} (\mathcal U^\kappa(x)\cap\{y:\underline d_{\mu_\phi}(y)>a\})=0\quad\text{for }\mu_\phi\text{-}\text{a.e.}\,x. \end{align} $$

Suppose now that equation (5.7) is established. Let $ (a_m)_{m\ge 1} $ be a monotonically decreasing sequence of real numbers converging to $ 1/\kappa $ . Applying equation (5.7) to each $ a_m $ yields a full $ \mu _\phi $ -measure set corresponding to $ a_m $ . Then, by taking the intersection of these countable full $ \mu _\phi $ -measure sets, we conclude from the countable stability of Hausdorff dimension that

$$ \begin{align*}\operatorname{\mathrm{\dim_H}} (\mathcal U^\kappa(x)\cap\{y:\underline d_{\mu_\phi}(y)>1/\kappa\})=0\quad\text{for }\mu_\phi\text{-}\text{a.e.}\,x.\end{align*} $$

As a result, by Lemma 3.6, for $ \mu _\phi $ -a.e. $ x $ ,

$$ \begin{align*} \operatorname{\mathrm{\dim_H}} \mathcal U^\kappa(x)&=\operatorname{\mathrm{\dim_H}} (\mathcal U^\kappa(x)\cap(\{y:\underline{d}_{\mu_\phi}(y)\le 1/\kappa\}\cup \{y:\underline d_{\mu_\phi}(y)>1/\kappa\}))\notag\\ &=\operatorname{\mathrm{\dim_H}} (\mathcal U^\kappa(x)\cap\{y:\underline{d}_{\mu_\phi}(y)\le 1/\kappa\})\notag\\ &\le\operatorname{\mathrm{\dim_H}}\{y:\underline{d}_{\mu_\phi}(y)\le 1/\kappa\}=D_{\mu_\phi}(1/\kappa). \end{align*} $$

This clearly yields the lemma.

Choose $ b\in (1/\kappa , a) $ . Put $ A_n:=\{y:\mu _\phi (B(y, r))<r^b\text { for all }r<2^{-n}\} $ . By the definition of $ \underline d_{\mu _\phi }(y) $ , we have

$$ \begin{align*}\{y:\underline d_{\mu_\phi}(y)>a\}\subset\bigcup_{n=1}^\infty A_n.\end{align*} $$

Thus, equation (5.7) is reduced to showing that for any $ n\ge 1 $ ,

(5.8) $$ \begin{align} \operatorname{\mathrm{\dim_H}} (\mathcal U^\kappa(x)\cap A_n)=0\quad\text{for }\mu_\phi\text{-}\text{a.e.}\,x. \end{align} $$

Step 2. The next objective is to prove equation (5.8).

Fix $ n\ge 1 $ . Let $ \epsilon>0 $ be arbitrary. Choose a large integer $ l\ge n $ with

(5.9) $$ \begin{align} 12\times 2^{-\kappa l}<2^{-n}\quad\text{and}\quad \gamma^312^b2^{(1-b\kappa)l}< \epsilon, \end{align} $$

where the constant $ \gamma $ is defined in equation (3.1). Let $ \theta _j=[\kappa j\log _{L_1}2]+1 $ , where $ L_1 $ is given in equation (2.2). Then, by equation (2.2), the length of each basic interval of generation $ \theta _j $ is smaller than $ 2^{-\kappa j} $ . Recall that $ \mathcal E $ is the set of endpoints of basic intervals which is countable. For any $ x\in [0,1]\setminus \mathcal E $ , define

$$ \begin{align*}\mathcal I_{j}(x):=\bigcup_{J\in\Sigma_{\theta_j}\colon d(I_{\theta_j}(x),J)<2^{-\kappa j}}J,\end{align*} $$

where $ d(\cdot ,\cdot ) $ is the Euclidean metric. Clearly, $ \mathcal I_j(x) $ covers the ball $ B(x, 2^{-\kappa j}) $ and is contained in $ B(x, 3\times 2^{-\kappa j}) $ . Moreover, if $ I_{\theta _j}(x)=I_{\theta _j}(y) $ , then $ \mathcal I_j(x)=\mathcal I_j(y) $ . With the notation $ \mathcal I_j(x) $ , we consider the set

$$ \begin{align*}G_{l,i}(x)= A_n\cap\bigg(\bigcap_{j=l}^i\bigcup_{k=1}^{2^j}\mathcal{I}_{j}(T^kx)\bigg).\end{align*} $$

The advantage of using $ \mathcal I_j(x) $ rather than $ B(x, 2^{-\kappa j}) $ is that the map $ x\mapsto G_{l,i}(x) $ is constant on each basic interval of generation $ 2^i+\theta _i $ . We are going to construct inductively a cover of $ G_{l,i}(x) $ by the family $ \{B(T^kx, 3\times 2^{-\kappa i}):k\in S_i(x)\} $ of balls, where $ S_i(x)\subset \{1,2,\ldots ,2^i\} $ .

For $ i=l $ , we let $ S_l(x)\subset \{1,2,\ldots , 2^l\} $ consist of those $ k\le 2^l $ such that $ \mathcal I_l(T^kx) $ intersects $ A_n $ . Suppose now that $ S_i(x) $ has been defined. We define $ S_{i+1}(x) $ to consist of those $ k\le 2^{i+1} $ such that $ \mathcal I_{i+1}(T^kx) $ intersects $ G_{l,i}(x) $ . Then the family $ \{B(T^kx, 3\times 2^{-\kappa (i+1)}):k\in S_{i+1}(x)\} $ of balls forms a cover of $ G_{l,i+1}(x) $ , and the construction is completed.

With the aid of the notation $ \mathcal I_{i+1}(T^kx) $ , one can verify that $ x\mapsto S_{i+1}(x) $ is constant on each basic interval of generation $ 2^{i+1}+\theta _{i+1} $ . Let $ N_{i+1}(x):=\sharp S_{i+1}(x) $ , then $ x\mapsto N_{i+1}(x) $ is also constant on each basic interval of generation $ 2^{i+1}+\theta _{i+1} $ .

To establish equation (5.8), we need to estimate $ N_{i+1}(x) $ . For those $ k\in S_{i+1}(x)\cap \{1,2,\ldots ,2^i\} $ , since $ \mathcal I_{i+1}(T^kx)\subset \mathcal I_i(T^kx) $ and $ G_{l,i}(x)\subset G_{l,i-1}(x) $ , we must have that $ \mathcal I_i(T^kx) $ intersects $ G_{l,i-1}(x) $ , and hence $ k\in S_i(x) $ . However, since $ \mathcal I_{i+1}(T^kx) $ is contained in $ B(T^kx,3\times 2^{-\kappa (i+1)}) $ , if $ \mathcal I_{i+1}(T^kx) $ has non-empty intersection with $ G_{l,i}(x) $ , then the distance between $ T^kx $ and $ G_{l,i}(x) $ is less than $ 3\times 2^{-\kappa (i+1)} $ . In particular,

(5.10) $$ \begin{align} T^kx \in\{y: d(y, G_{l,i}(x))<3\times2^{-\kappa(i+1)}\}\subset \bigcup_{J\in\Sigma_{\theta_i}\colon d(J,G_{l,i}(x))<3\times2^{-\kappa (i+1)}}J. \end{align} $$

Denote the right-hand side union as $ \hat G_{l,i}(x) $ . The set $ \hat G_{l,i}(x) $ is nothing but the union of cylinders of level $ \theta _i $ whose distance from $ G_{l,i}(x) $ is less than $ 3\times 2^{-\kappa (i+1)} $ . Thus, by the fact that $ x\mapsto G_{l,i}(x) $ is constant on each basic interval of generation $ 2^i+\theta _i $ , we have

(5.11) $$ \begin{align} \hat G_{l,i}(x)=\hat G_{l,i}(y),\quad \text{whenever } I_{2^i+\theta_i}(x)=I_{2^i+\theta_i}(y). \end{align} $$

According to the above discussion, it holds that

$$ \begin{align*}N_{i+1}(x)\le N_i(x)+M_{i+1}(x),\end{align*} $$

where $ M_{i+1}(x) $ is the number of $ 2^i<k\le 2^{i+1} $ for which $ T^k x $ intersects $ \hat G_{l,i}(x) $ .

The function $ M_{i+1}(x) $ can further be written as

$$ \begin{align*} M_{i+1}(x)=\sum_{k=2^i+1}^{2^{i+1}}\chi_{\hat G_{l,i}(x)}(T^kx)=\sum_{k=2^i+1}^{2^i+\theta_i}\chi_{\hat G_{l,i}(x)}(T^kx)+\sum_{k=2^i+\theta_i+1}^{2^{i+1}}\chi_{\hat G_{l,i}(x)}(T^kx). \end{align*} $$

Since the countable set $ \mathcal E $ has zero $\mu _\phi $ -measure, all the functions given above are well defined for $ \mu _\phi $ -a.e. $ x $ . It follows from the locally constant property in equation (5.11) of $ \hat G_{l,i(x)} $ that

(5.12) $$ \begin{align} \notag\int M_{i+1}(x) d\mu_\phi(x)&=\!\sum_{k=2^i+1}^{2^i+\theta_i}\int\! \chi_{\hat G_{l,i}(x)}(T^kx)d\mu_\phi(x)+\!\sum_{k=2^i+\theta_i+1}^{2^{i+1}}\int\! \chi_{\hat G_{l,i}(x)}(T^kx)d\mu_\phi(x)\\ &\le \theta_i+\sum_{J\in\Sigma_{2^i+\theta_i}}\sum_{k=2^i+\theta_i+1}^{2^{i+1}}\int \chi_J(x) \chi_{\hat G_{l,i}(x)}(T^kx)d\mu_\phi(x)\notag\\ &=\theta_i+\sum_{J\in\Sigma_{2^i+\theta_i}}\sum_{k=2^i+\theta_i+1}^{2^{i+1}}\int \chi_J(x) \chi_{\hat G_{l,i}(x_J)}(T^kx)d\mu_\phi(x), \end{align} $$

where $ x_J $ is any fixed point of $ J $ . Now the task is to deal with the right-hand side summation. We deduce from the quasi-Bernoulli property in equation (3.2) of $ \mu _\phi $ that for each $ k>2^i+\theta _i $ ,

(5.13) $$ \begin{align} \int \chi_J(x) \chi_{\hat G_{l,i}(x_J)}(T^kx)d\mu_\phi(x)=\mu_\phi(J\cap T^{-k}(\hat G_{l,i}(x_J)))\le \gamma^3\mu_\phi(J)\mu_\phi(\hat{G}_{l,i}(x_J)). \end{align} $$

Since $ G_{l,i}(x_J) $ can be covered by the family $ \{B(T^kx_J, 3\times 2^{-\kappa i}):k\in S_i(x_J)\} $ of balls, then by equation (5.10), the family $ \mathcal F_i(x_J):=\{B(T^kx_J, 6\times 2^{-\kappa i}):k\in S_i(x_J)\} $ of enlarged balls forms a cover of $ \hat {G}_{l,i}(x_J) $ . Observe that each enlarged ball $ B\in \mathcal F_i(x_J) $ intersects $ A_n $ , and thus $ B\subset B(y,12\times 2^{-\kappa i}) $ for some $ y\in A_n $ . Then, by the definition of $ A_n $ and equation (5.9),

$$ \begin{align*}\mu_\phi(B)\le \mu_\phi(B(y,12\times 2^{-\kappa i}))\le12^b2^{-b\kappa i}.\end{align*} $$

Accordingly,

(5.14) $$ \begin{align} \mu_\phi(\hat{G}_{l,i}(x_J))\le 12^b2^{-b\kappa i}N_i(x_J). \end{align} $$

Recall that $ x\mapsto N_i(x) $ is constant on each basic interval of generation $ 2^i+\theta _i $ . Applying the upper bound of equation (5.14) on $ \mu _\phi (\hat {G}_{l,i}(x_J)) $ to equation (5.13), and then substituting equation (5.13) into equation (5.12), we have

$$ \begin{align*} \notag\int M_{i+1}(x) \,d\mu_\phi(x) &\le \theta_i+\sum_{J\in\Sigma_{2^i+\theta_i}}\sum_{k=2^i+\theta_i+1}^{2^{i+1}}\gamma^3\mu_\phi(\hat{G}_{l,i}(x_J))\mu_\phi(J)\\&\le \theta_i+\sum_{J\in\Sigma_{2^i+\theta_i}}\sum_{k=2^i+\theta_i+1}^{2^{i+1}}\gamma^312^b2^{-b\kappa i}N_i(x_J)\mu_\phi(J)\\&= \theta_i+\gamma^312^b2^{-b\kappa i}(2^i-\theta_i)\int N_i(x)\,d\mu_\phi(x)\\&\le \theta_i+\epsilon\int N_i(x)\,d\mu_\phi(x), \end{align*} $$

where the last inequality follows from equation (5.9). Since $ N_{i+1}(x)\le N_i(x)+M_{i+1}(x) $ , we have

(5.15) $$ \begin{align} \int N_{i+1}(x)\,d\mu_\phi(x)\le \theta_i+(1+\epsilon)\int N_i(x)\,d\mu_\phi(x). \end{align} $$

Note that equation (5.15) holds for all $ i\ge l $ and $ N_l(x)\le 2^l $ . Then,

$$ \begin{align*}\begin{split} \int N_{i+1}(x)\,d\mu_\phi(x)&\le \sum_{k=l}^{i}(1+\epsilon)^{i-k}\theta_k+(1+\epsilon)^{i-l+1}\int N_l(x)\,d\mu_\phi(x)\\ &<(1+\epsilon)^i(i\theta_i+2^l). \end{split}\end{align*} $$

By Markov’s inequality,

$$ \begin{align*} &\mu_\phi(\{x:N_{i+1}(x)\ge (1+\epsilon)^{2i}(i\theta_i+2^l)\})\\&\quad\le \mu_\phi\bigg(\bigg\{x:N_{i+1}(x)\ge (1+\epsilon)^i\int N_{i+1}(x)\,d\mu_\phi(x)\bigg\}\bigg)\\ &\quad\le (1+\epsilon)^{-i}, \end{align*} $$

which is summable over $ i $ . Hence, for $ \mu _\phi $ -a.e. $ x $ , there is an $ i_0(x) $ such that

(5.16) $$ \begin{align} N_{i+1}(x)\le (1+\epsilon)^{2i}(i\theta_i+2^l) \end{align} $$

holds for all $ i\ge i_0(x) $ .

Denote by $ F_{l,\epsilon } $ the full measure set on which equation (5.16) holds. Let $ x\in F_{l,\epsilon } $ . Then, such an $ i_0(x) $ exists. With $ i\ge i_0(x) $ , we may cover the set

$$ \begin{align*}G_l(x)=A_n\cap\bigg(\bigcap_{j=l}^\infty\bigcup_{k=1}^{2^j}\mathcal I_j(T^kx)\bigg)\end{align*} $$

by $ N_i(x) $ balls of radius $ 3\times 2^{-\kappa i} $ . Since $ \theta _i=[\kappa i\log _{L_1}2]+1\le ci $ for some $ c>0 $ , we have

(5.17) $$ \begin{align} \operatorname{\mathrm{\dim_H}} G_l(x)\le\limsup_{i\to\infty}\frac{\log N_i(x)}{\log 2^{\kappa i}}\le\limsup_{i\to\infty}\frac{\log ((1+\epsilon)^{2i}(i\theta_i+2^l))}{\log 2^{\kappa i}}=\frac{2\log (1+\epsilon)}{\kappa\log2}. \end{align} $$

Let $ (\epsilon _m)_{m\ge 1} $ be a monotonically decreasing sequence of real numbers converging to $ 0 $ . For each $ \epsilon _m $ , choose an integer $ l_m $ satisfying equation (5.9), but with $ \epsilon $ replaced by $ \epsilon _m $ . For every $ m\ge 1 $ , by the same reason as equation (5.17), there exists a full $ \mu _\phi $ -measure set $ F_{l_m,\epsilon _m} $ such that

$$ \begin{align*}\text{for all }x\in F_{l_m,\epsilon_m},\quad\operatorname{\mathrm{\dim_H}} G_{l_m}(x)\le\frac{2\log (1+\epsilon_m)}{\kappa\log2}.\end{align*} $$

By taking the intersection of the countable full $ \mu _\phi $ -measure sets $ (F_{l_m,\epsilon _m})_{m\ge 1} $ , and using the fact that $ G_l(x) $ is increasing in $ l $ , we obtain that for $ \mu _\phi $ -a.e. $ x $ ,

$$ \begin{align*}\text{for any }l\ge 1,\quad\operatorname{\mathrm{\dim_H}} G_l(x)\le\lim_{m\to \infty}\operatorname{\mathrm{\dim_H}} G_{l_m}(x)=0.\end{align*} $$

We conclude equation (5.8) by noting that

$$ \begin{align*}\mathcal U^\kappa(x)\cap A_n\subset\bigcup_{l\ge 1}G_l(x).\end{align*} $$