Proof. Choose a maximal $({9}/{10})r$ -separated set $\{y_1,\ldots ,y_N\}$ of $\Omega $ .

Claim. There exist a constant $0<c<{1}/{10}$ depending only on G, and $\{g_i\}_{i=1}^N \subset B^G_{{r}/{10}}$ such that for $z_i=g_iy_i$ and for any $0<\delta <cr$ ,

(2.3) $$ \begin{align} {\sum_i(\mu(\partial_\delta (B_r^G\cdot z_i))+\mu(\partial_\delta (B_{{r}/{2}}^G\cdot z_i)))\leq \bigg(\frac{\delta}{r}\bigg)^{{1}/{2}}\mu(B^G_{({12}/{10})r}\Omega).} \end{align} $$

Proof of the claim

To prove this claim, we randomly choose each $g_i$ with the independent uniform distribution on $B^G_{{r}/{10}}$ . For $0<\delta <{r}/{10}$ fixed, we have

For any $y\in B_{({12}/{10})r}^G\Omega $ , the number of $y_i$ terms contained in $B_{({12}/{10})r}^G \cdot y$ is at most $({33}/{9})^{\dim G}$ since $B_{({9}/{20})r}^G \cdot y_i$ terms are disjoint and contained in $B_{({33}/{20})r}^G \cdot y$ . It implies that for any $y\in B_{({12}/{10})r}^G\Omega $ . It follows that

$$ \begin{align*}\mathbb{E}\bigg(\sum_i\mu(\partial_\delta (B_r^G\cdot z_i))\bigg)\ll \frac{\delta}{r}\int_{B_{({12}/{10})r}^G\Omega}4^{\dim G}\,d\mu(y)\ll \frac{\delta}{r}\mu(B_{({12}/{10})r}^G\Omega),\end{align*} $$

where the implied constant depends only on G.

Applying the same argument for $\partial _\delta (B_{{r}/{2}}^G\cdot z_i)$ instead of $\partial _\delta (B_r^G\cdot z_i)$ ,

$$ \begin{align*}\mathbb{E}\bigg(\sum_i(\mu(\partial_\delta (B_r^G\cdot z_i))+\mu(\partial_\delta (B_{{r}/{2}}^G\cdot z_i)))\bigg)\ll\frac{\delta}{r}\mu(B^G_{({12}/{10})r}\Omega).\end{align*} $$

It follows from Chebyshev’s inequality that

$$ \begin{align*}\mathbb{P}\bigg(\sum_i(\mu(\partial_\delta (B_r^G\cdot z_i))+\mu(\partial_\delta (B_{{r}/{2}}^G\cdot z_i)))\ge \frac{1}{2}\bigg(\frac{\delta}{r}\bigg)^{{1}/{2}}\mu(B^G_{({12}/{10})r}\Omega)\bigg)\ll \bigg(\frac{\delta}{r}\bigg)^{{1}/{2}}. \end{align*} $$

Hence, we have

(2.4) $$ \begin{align} \mathbb{P}&\bigg(\bigcap_{k\geq 0}\bigg\{\sum_i(\mu(\partial_{2^{-k}\delta} (B_r^G\cdot z_i))+\mu(\partial_{2^{-k}\delta} (B_{{r}/{2}}^G\cdot z_i)))< \frac{1}{2}\bigg(\frac{2^{-k}\delta}{r}\bigg)^{{1}/{2}}\mu(B^G_{({12}/{10})r}\Omega)\bigg\}\bigg)\nonumber\\&>1-O\bigg( \bigg(\frac{\delta}{r}\bigg)^{{1}/{2}}\bigg). \end{align} $$

Thus, there exists $0<c<{1}/{10}$ so that the right-hand side of equation (2.4) is positive for any $\delta <cr$ . It follows that we can find $\{g_i\}_{i=1}^{N}$ such that the $z_i=g_iy_i$ terms satisfy equation (2.3) for any $0<\delta <cr$ .

Let $c>0$ and $\{g_i\}_{i=1}^N \subset B^G_{{r}/{10}}$ be as in the above claim. The set $\{z_i = g_i y_i\}_{i=1}^N$ is $({7}/{10})r$ -separated since $\{y_i\}_{i=1}^N$ is $({9}/{10})r$ -separated. Let $K\overset {\operatorname {def}}{=}\bigcup _{i=1}^N B_r^G\cdot z_i$ . Since $B_{({9}/{10})r}^G\cdot y_i\subseteq B_r^G\cdot z_i\subseteq B_{({11}/{10})r}^G\cdot y_i$ , we have

$$ \begin{align*}\Omega\subseteq\bigcup_{i=1}^N B_{({9}/{10})r}^G\cdot y_i\subseteq K\subseteq \bigcup_{i=1}^N B_{({11}/{10})r}^G\cdot y_i\subseteq B_{({11}/{10})r}^G \Omega.\end{align*} $$

Now we define a partition $\mathcal {P}$ of K inductively as follows:

$$ \begin{align*}P_i\overset{\operatorname{def}}{=} B^G_r\cdot z_i\setminus\bigg(\bigcup_{j=1}^{i-1}P_j\cup\bigcup_{j=i+1}^{N}B^G_{{r}/{2}}\cdot z_j\bigg)\end{align*} $$

for $1\leq i\leq N$ . By definition, we have $B_{{r}/{5}}^G\cdot z_i\subseteq P_i\subseteq B_{r}^G\cdot z_i$ and $z_i\in B^G_{{r}/{10}}\Omega $ for  $1\leq i\leq N$ . We also observe that the $\delta $ -neighborhood of $\mathcal {P}$ is contained in ${\bigcup _{i=1}^N (\partial _\delta (B_r^G\cdot z_i) \cup \partial _\delta (B_{{r}/{2}}^G\cdot z_i))}$ . Hence, it follows from the above claim that

$$ \begin{align*}\mu(\partial_\delta\mathcal{P})\leq \sum_i(\mu(\partial_\delta (B_r^G\cdot z_i))+\mu(\partial_\delta (B_{{r}/{2}}^G\cdot z_i)))\leq \bigg(\frac{\delta}{r}\bigg)^{{1}/{2}}\mu(B^G_{({12}/{10})r}\Omega)\end{align*} $$

for any $0<\delta <cr$ .

Proof. We will construct $\{K_k\}_{k\ge 1}$ and $\{\mathcal {P}_k\}_{k\ge 1}$ using Lemma 2.3 inductively. For each $k\geq 1$ , let us say that $K_k$ and $\mathcal {P}_k$ satisfy  $(\spadesuit _k)$ if they satisfy the three conditions in the statement. We will also need auxiliary bounded sets $K_k'\subset Y$ and corresponding partitions $\mathcal {P}_k'$ of $K_k'$ during the inductive process. Let us say that $K_k'$ and $\mathcal {P}_k'$ satisfy $(\clubsuit _k)$ if they satisfy the following three conditions:

1. (1) $Y(2^{-k+1})\setminus \bigcup _{j=1}^{k-1}K_j\subseteq K_k'\subseteq B^G_{({11}/{10})r_02^{-k-1}}(Y(2^{-k+1}) \setminus \bigcup _{j=1}^{k-1}K_j)$ ;

2. (2) for each $1\leq i\leq N_{k}$ , there exists $z_{ki}\in B^G_{({1}/{10})r_0 2^{-k-1}}K_k'$ such that

   $$ \begin{align*}B_{({1}/{5})r_0 2^{-k-1}}^G\cdot z_{ki}\subseteq P_{ki}'\subseteq B_{r_0 2^{-k-1}}^G\cdot z_{ki}\quad\text{and} \quad K_k'=\bigcup_{i=1}^{N}B_{r_02^{-k-1}}^G\cdot z_{ki};\end{align*} $$

3. (3) $\mu (\partial _\delta \mathcal {P}_k')\leq (r_0^{-1}2^{k+1}\delta )^{1/2}\mu (Y(2^{-k})\setminus Y(2^{-k+3}))$ for any $0<\delta <cr_02^{-k-1}$ .

Here, $\bigcup _{j=1}^{0} K_j \overset {\operatorname {def}}{=} \emptyset $ .

We first choose $\Omega _1 = Y(1)$ and apply Lemma 2.3 with $r=r_02^{-2}$ and $\Omega =\Omega _1 \subset Y({r_0}/{2})$ . Then we have a subset $K_1' \subset Y$ and a partition $\mathcal {P}_1'$ of $K_1'$ satisfying conditions (1), (2) of $(\clubsuit _{1})$ , and

$$ \begin{align*}\mu(\partial_\delta\mathcal{P}_1')\leq (r_0^{-1} 2^2\delta)^{{1}/{2}}\mu(B_{({12}/{10})r_02^{-2}}^G \Omega_1)\end{align*} $$

for any $0<\delta <cr_02^{-2}$ . It follows from equation (2.5) that $B_{({12}/{10})r_02^{-2}}^G Y(1) \subset Y(\tfrac 12)$ , which implies condition (3) of $(\clubsuit _{1})$ since $Y(4)=\varnothing $ . Note that $K_1' \subset B_{({11}/{10})r_02^{-2}}^G Y(1)\subset Y(\tfrac 12)$ .

Now let $\Omega _2 = Y(\tfrac 12) \setminus K_1'$ and apply Lemma 2.3 again with $r=r_02^{-3}$ and $\Omega =\Omega _2 \subset Y({r_0}/{4})$ . We have a subset $K_2' \subset Y$ and a partition $\mathcal {P}_2'$ of $K_2'$ satisfying $\Omega _2 \subset K_2' \subset B_{({11}/{10})r_0 2^{-3}}^G \Omega _2$ , (2) of $(\clubsuit _{2})$ , and $\mu (\partial _\delta \mathcal {P}_2')\leq (r_0^{-1} 2^{3}\delta )^{1/2}\mu (B_{({12}/{10})r_0 2^{-3}}^G \Omega _2)$ for any $0<\delta <cr_02^{-3}$ . Setting $K_1 = K_1' \setminus K_2 '$ , condition (1) of $(\clubsuit _{2})$ and condition (1) of $(\spadesuit _1)$ follow since $Y(2)=\varnothing $ . Since $K_1' \supset Y(1)$ , it follows from equation (2.5) that $B_{({12}/{10})r_02^{-3}}^G \Omega _2 \subset Y(\tfrac 14)\setminus Y(2)$ , which implies condition (3) of $(\clubsuit _{2})$ .

Define a partition $\mathcal {P}_1{\kern-1pt} = {\kern-1pt}\{P_{11},\ldots , P_{1N_1}\}$ from $\mathcal {P}_1'{\kern-1pt} ={\kern-1pt} \{ P_{11}',\ldots , P_{1N_1}'\}$ by $P_{1i}{\kern-1pt}={\kern-1pt}P_{1i}' \setminus K_{2}'$ for each $1\leq i\leq N_1$ . For each $1\leq i\leq N_1$ and $y\in B^G_{(1/5)r_02^{-2}}\cdot z_{1i}$ , observe that $y\notin K_{2}'$ since $B^G_{r_02^{-2}}\cdot z_{1i}\subset K_1'$ and $K_{2}'\subset B^G_{({11}/{10})r_02^{-3}}\Omega _2 \subset B^G_{({11}/{10})r_0 2^{-3}}(Y\setminus K_1')$ . Hence, $B^G_{(1/5)r_02^{-2}}\cdot z_{1i}\subset P_{1i}$ holds, so condition (2) of $(\spadesuit _1)$ follows. Since $P_{1i}=P_{1i}' \setminus K_{2}'$ for each $1\leq i\leq N_1$ , we have

$$ \begin{align*} \mu(\partial_\delta \mathcal{P}_1) &\leq \mu(\partial_\delta \mathcal{P}_1') + \mu(\partial_\delta \mathcal{P}_2')\\ &\leq (r_0^{-1}2^2 \delta)^{{1}/{2}}\mu(Y(2^{-1})\setminus Y(2^2)) + (r_0^{-1}2^3 \delta)^{{1}/{2}}\mu(Y(2^{-2})\setminus Y(2)) \\ &\leq (r_0^{-1}2^5 \delta)^{{1}/{2}}\mu(Y(2^{-2})\setminus Y(2^2)) \end{align*} $$

for any $0<\delta <cr_02^{-3}$ . Hence, condition (3) of $(\spadesuit _1)$ follows.

Our desired disjoint sets $\{K_k\}_{k\ge 1}$ and partitions $\{\mathcal {P}_k\}_{k\ge 1}$ will be obtained by applying this process repeatedly.

Proof of the claim

Note that $K_k' \subset B^G_{({11}/{10})r_02^{-k-1}}Y(2^{-k+1})\subset Y(2^{-k})$ and $K_j \subset Y(2^{-j}) \subset Y(2^{-k})$ for each $j=1,\ldots , k-1$ . Let $\Omega _{k+1}=Y(2^{-k})\setminus (\bigcup _{j=1}^{k-1} K_j\cup K_k')$ and apply Lemma 2.3 with $r=r_02^{-k-2}$ and $\Omega =\Omega _{k+1} \subset Y(r_02^{-k-1})$ . There exist $K_{k+1}'\subset Y$ and a partition $\mathcal {P}_{k+1}'=\{P_{(k+1)1}',\ldots ,P_{(k+1)N_{k+1}}'\}$ of $K_{k+1}'$ satisfying $\Omega _{k+1} \subset K_{k+1}' \subset B_{({11}/{10})r_02^{-k-2}}^G \Omega _{k+1}$ , condition (2) of $(\clubsuit _{k+1})$ , and $\mu (\partial _\delta \mathcal {P}_{k+1}')\leq (r_0^{-1}2^{k+2}\delta )^{1/2}\mu (B_{({12}/{10})r_02^{-k-2}}^G \Omega _{k+1})$ for any $0<\delta <cr_02^{-k-2}$ . Setting $K_k=K_k'\setminus K_{k+1}'$ , condition (1) of $(\clubsuit _{k+1})$ follows. Since $\bigcup _{j=1}^{k-1} K_j \supset Y(2^{-k+2})$ and $K_k \subset K_k' \subset Y(2^{-k}) \setminus \bigcup _{j=1}^{k-1}K_j$ , condition (1) of $(\spadesuit _k)$ follows. It follows from $\bigcup _{j=1}^{k-1} K_j \cup K_k' \supset Y(2^{-k+1})$ and equation (2.5) that

$$ \begin{align*}B_{({12}/{10})r_02^{-k-2}}^G \Omega_{k+1} \subset Y(2^{-k-1})\setminus Y(2^{-k+2}),\end{align*} $$

which implies condition (3) of $(\clubsuit _{k+1})$ . Define a partition $\mathcal {P}_k=\{P_{k1},\ldots , P_{kN_k}\}$ from $\mathcal {P}_k'=\{P_{k1}',\ldots ,P_{kN_k}'\}$ by $P_{ki}=P_{ki}'\setminus K_{k+1}'$ for any $1\leq i\leq N_k$ . For each $1\leq i\leq N_k$ and $y\in B^G_{(1/5)r_02^{-k-1}}\cdot z_{ki}$ , observe that $y\notin K_{k+1}'$ since $B^G_{r_02^{-k-1}}\cdot z_{ki}\subseteq K_k'$ and $K_{k+1}'\subseteq B^G_{({11}/{10})r_02^{-k-2}}\Omega _{k+1} \subset B^G_{({11}/{10})r_02^{-k-2}}(Y\setminus K_k')$ . Hence, $B^G_{(1/5)r_02^{-k-1}}\cdot z_{ki}\subset P_{ki}$ holds, so condition (2) of $(\spadesuit _k)$ follows. Since $P_{ki}=P_{ki}' \setminus K_{k+1}'$ for each $1\leq i\leq N_k$ , we have

$$ \begin{align*} \mu(\partial_\delta \mathcal{P}_k) &\leq \mu(\partial_\delta \mathcal{P}_k') + \mu(\partial_\delta \mathcal{P}_{k+1}') \\&\leq (r_0^{-1}2^{k+1}\delta)^{{1}/{2}}\mu(Y(2^{-k})\setminus Y(2^{-k+3})) + (r_0^{-1}2^{k+2}\delta)^{{1}/{2}}\mu(Y(2^{-k-1})\setminus Y(2^{-k+2})) \\&\leq (r_0^{-1}2^{k+4}\delta)^{{1}/{2}}\mu(Y(2^{-k-1})\setminus Y(2^{-k+3})) \end{align*} $$

for any $0<\delta <cr_02^{-k-2}$ . Hence, condition (3) of $(\spadesuit _k)$ follows.

The claim concludes the proof of Lemma 2.5.

Proof. We split $E_\delta $ into two subsets

$$ \begin{align*}&E_\delta'=\bigcup_{k=0}^\infty a^k\bigg(\bigcup_{i= 2+\lceil({\alpha}/{\log 2})k-({\log\delta})/({2\log2})\rceil}^\infty\partial_{d_0e^{-k\alpha}\delta}\mathcal{P}_i\bigg) \quad \mathrm{and} \\ &E_\delta"=\bigcup_{k=0}^\infty a^k\bigg(\bigcup_{i= 1}^{1+\lceil({\alpha}/{\log 2})k-({\log\delta})/({2\log2})\rceil}\partial_{d_0e^{-k\alpha}\delta}\mathcal{P}_i\bigg).\end{align*} $$

We first claim that $E_\delta '\subset Y\setminus Y((d_0+d_0^2)\delta ^{1/2})$ . Let $y\in E_\delta '$ , that is, $y\in a^k\partial _{d_0e^{-k\alpha }\delta }P$ for some $k\geq 0$ and $P\in \mathcal {P}_i$ for some $i \geq 2+\lceil ({\alpha }/{\log 2})k-({\log \delta })/({2\log 2})\rceil .$ By Lemma 2.5,

$$ \begin{align*}P \subset K_i \subset Y(2^{-i})\setminus Y(2^{-i+2}) \subset Y(2^{-i+2})^c.\end{align*} $$

It follows from equation (2.5) that

(2.7) $$ \begin{align} { \partial_{d_0e^{-k\alpha}\delta}P \subset B_{d_0e^{-k\alpha}\delta}^G P \subset B_{d_0e^{-k\alpha}\delta}^G Y(2^{-i+2})^c \subset Y(2^{-i+2}+d_0e^{-k\alpha}\delta)^c. } \end{align} $$

Using equation (2.6), for any $0<r<1, a^k Y(r)^c \subset Y(d_0e^{k\alpha }r)^c.$ Since $e^{k\alpha }2^{-i+2}\leq \delta ^{1/2}$ , combining with equation (2.7),

$$ \begin{align*} a^k\partial_{d_0e^{-k\alpha}\delta}P \subset a^k Y(2^{-i+2}+d_0e^{-k\alpha}\delta)^c \subset Y((d_0+d_0^2)\delta^{{1}/{2}})^c. \end{align*} $$

This proves the claim. It follows that

(2.8) $$ \begin{align} { \mu(E_\delta') \leq \mu(Y\setminus Y(C_1 \delta^{{1}/{2}})), } \end{align} $$

where $C_1=d_0 +d_0^{2}$ is a constant depending only on a and G.

Next we estimate $\mu (E_\delta ")$ . It follows from the a-invariance of $\mu $ that

(2.9) $$ \begin{align}\kern-4pt { \mu(E_\delta") \leq\sum_{k=0}^{\infty}\sum_{i=1}^{1+\lceil({\alpha}/{\log 2})k-({\log\delta})/({2\log2})\rceil} \mu(\partial_{d_0e^{-k\alpha}\delta}\mathcal{P}_i)= \sum_{i=1}^{\infty}\sum_{k=k_i}^{\infty}\mu(\partial_{d_0e^{-k\alpha}\delta}\mathcal{P}_i), } \end{align} $$

where $k_i\in \mathbb {N}$ denotes the smallest number of k such that $1+\lceil ({\alpha }/{\log 2})k-({\log \delta })/({2\log 2})\rceil \geq i$ . Note that $k_i\geq ({\log 2}/{\alpha })(i-2)+({\log \delta })/{2\alpha }$ .

However, by Lemma 2.5, we have

(2.10) $$ \begin{align} {\mu(\partial_{d_0e^{-k\alpha}\delta}\mathcal{P}_i)\leq (r_0^{-1}2^{i+4}d_0e^{-k\alpha}\delta)^{{1}/{2}} \mu(Y(2^{-i-1})\setminus Y(2^{-i+3}))} \end{align} $$

for any $k\geq k_i$ , since $d_0e^{-k\alpha }\delta \leq d_02^{-i+2}\delta ^{1/2}<cr_02^{-i-2}$ . By equations (2.9) and (2.10), we have

(2.11) $$ \begin{align} \mu(E_\delta")&\leq \sum_{i=1}^{\infty}\sum_{k=k_i}^{\infty}\mu(\partial_{d_0e^{-k\alpha}\delta}\mathcal{P}_i) \leq \sum_{i=1}^{\infty}\sum_{k=k_i}^{\infty}(r_0^{-1}2^{i+4}d_0e^{-k\alpha}\delta)^{{1}/{2}} \mu(Y(2^{-i-1}){\kern-1pt}\setminus{\kern-1pt} Y(2^{-i+3})) \nonumber\\&=\sum_{i=1}^{\infty}(r_0^{-1}2^{i+4}e^{-k_i\alpha}\delta)^{{1}/{2}}(1-e^{-\alpha/2})^{-1}\mu(Y(2^{-i-1})\setminus Y(2^{-i+3}))\nonumber\\&\leq r_0^{-{1}/{2}}2^{3}\delta^{{1}/{4}}(1-e^{-\alpha/2})^{-1}\sum_{i=1}^{\infty}\mu(Y(2^{-i-1})\setminus Y(2^{-i+3}))\leq C_2 \delta^{{1}/{4}}, \end{align} $$

where $C_2 = 2^{5}r_0^{-1/2}(1-e^{-\alpha /2})^{-1}$ is a constant depending only on $r_0$ , a, and G. Combining equations (2.8) and (2.11), we finally have

$$ \begin{align*}\mu(E_\delta)<\mu(Y\setminus Y(C_1\delta^{{1}/{2}}))+C_2\delta^{{1}/{4}}\end{align*} $$

and the constants $C_1,C_2>0$ depend only on $r_0$ , a, and G.

It remains to check that $B^{G^+}_\delta \cdot y\subset [y]_{\mathcal {P}_0^\infty }$ for any $y\in Y\setminus E_{\delta }$ . Let $h\in B^{G^+}_{\delta }$ and suppose $[hy]_{\mathcal {P}_0^{\infty }}\neq [y]_{\mathcal {P}_0^{\infty }}$ . There is some $k\geq 0$ such that $a^{-k}hy$ and $a^{-k}y$ belong to different elements of the partition $\mathcal {P}$ . Since $a^{-k}ha^{k}\in a^{-k}B_{\delta }^{G^+}a^k \subset B_{d_0e^{-k\alpha }\delta }^G$ by equation (2.6), we have

$$ \begin{align*}d_Y (a^{-k}hy, a^{-k}y) \leq d_G (a^{-k}ha^{k},\mathrm{id}) \leq d_0e^{-k\alpha}\delta.\end{align*} $$

It follows that both $a^{-k}hy$ and $a^{-k}y$ belong to $\partial _{d_0e^{-k\alpha }\delta }\mathcal {P}$ , and hence $y\in E_\delta $ . It concludes that $B^{G^+}_{\delta }\cdot y\subset [y]_{\mathcal {P}_0^{\infty }}$ for any $y\in Y\setminus E_\delta $ .

Proof. For a given a-invariant probability measure $\mu $ on Y, let $\mathcal {P}$ be the countable partition of Y constructed in Lemma 2.7. We will construct a countably generated $\sigma $ -algebra $\mathcal {P}^L$ by taking L-plaques in each $P\in \mathcal {P}$ as atoms of $\mathcal {P}^L$ . Then, $\mathcal {A}^L\overset {\operatorname {def}}{=}(\mathcal {P}^L)_0^\infty $ will be the desired $\sigma $ -algebra.

For each $P\in \mathcal {P}$ , by Lemma 2.5, there exist $j\geq 1$ and $z\in P$ such that $P\in Y(2^{-j})\setminus Y(2^{-j+2})$ and $B_{(1/5)r_02^{-j-1}}^G \cdot z \subseteq P \subseteq B_{r_02^{-j-1}}^G \cdot z$ . We can find $B_P \subset G$ with $\operatorname {\mathrm {diam}}(B_P)\leq r_02^{-j}$ such that $P = \pi _Y (B_P)$ , where $\pi _Y : G \to Y$ is the natural quotient map. Let $\mathcal {B}_{G/L}$ be the Borel $\sigma $ -algebra of the quotient $G/L$ . Note that since L is closed, $\mathcal {B}_{G/L}$ is countably generated. Define the $\sigma $ -algebra

$$ \begin{align*}\mathcal{P}^L=\sigma(\{\pi_Y(B_P \cap S): P\in\mathcal{P},\ S\in \mathcal{B}_{G/L}\}).\end{align*} $$

Then, $\mathcal {P}^L$ is a refinement of $\mathcal {P}$ such that atoms of $\mathcal {P}^L$ are open L-plaques, that is, for any $y{\kern-1pt}\in{\kern-1pt} P {\kern-1pt}\in{\kern-1pt} \mathcal {P}$ , $[y]_{\mathcal {P}^L}{\kern-1pt}={\kern-1pt}[y]_{\mathcal {P}} \cap B_{r_02^{-j}}^L\cdot y{\kern-1pt}={\kern-1pt}V_y\cdot y$ , where $V_y{\kern-1pt}\subset{\kern-1pt} B_{r_02^{-j}}^L$ is an open bounded set.

It is clear that $\mathcal {P}^L$ is countably generated, and hence $\mathcal {A}^L=(\mathcal {P}^L)_0^\infty $ is also countably generated. By construction, we have $a\mathcal {A}^L=(\mathcal {P}^L)_1^\infty \subset \mathcal {A}^L$ , which proves the assertion (1).

For any $y\in Y(2^{-k})\setminus Y(2^{-k+2})$ with $k\geq 1$ , take $P\in \mathcal {P}$ such that $y\in P$ . By Lemma 2.5, there exist $j\geq 1$ and $z\in P$ such that $P\in Y(2^{-j})\setminus Y(2^{-j+2})$ and $P \subseteq B_{r_02^{-j-1}}^G \cdot z$ . Observe that $2^{-j+2}> 2^{-k}$ and $2^{-j} < 2^{-k+2}$ , that is, $j-2 < k < j+2 $ . Hence, we have

$$ \begin{align*}[y]_{\mathcal{A}^L}\subset [y]_{\mathcal{P}^L}=V_y\cdot y \subset B_{r_02^{-j}}^L\cdot y \subset B_{r_02^{-k+1}}^L \cdot y,\end{align*} $$

which proves the assertion (2).

For a given $0<\delta <\min (({cr_0}/{16d_0})^{2},1)$ and $y\in Y(\delta )\setminus E_\delta $ , assume that $z=hy$ with $h\in B_{\delta }^L$ . By Lemma 2.7, $B_\delta ^{G^+}\cdot y\subset [y]_{\mathcal {P}_0^\infty }$ . Hence, it follows that for any $k\geq 0$ , $a^{-k}y$ and $a^{-k}z$ belong to the same atom $P_k \subset \mathcal {P}$ . Then, we have

$$ \begin{align*} a^{-k}y,\; a^{-k}z=a^{-k}ha^{k}\cdot(a^{-k}y)\; \in P_k. \end{align*} $$

Note that for any $y\in Y(\delta )$ , the map $B_{\delta }^{G^+}\ni g \mapsto g y$ is injective, and hence the map $a^{-k}B_{\delta }^{G^+}a^k \ni g \mapsto ga^{-k}y$ is injective. Since $a^{-k}ha^{k} \in a^{-k}B_{\delta }^L a^k$ , $a^{-k}y$ and $a^{-k}z$ belong to the same atom of $\mathcal {P}^L$ . This proves the assertion (3).

Proof. Let $\mathcal {P}^L$ be as in the proof of Proposition 2.8. Since $(\mathcal {A}^L)_{-\infty }^\infty =(\mathcal {P}^L)_{-\infty }^{\infty }$ and $Y=\bigcup _{k\geq 1}Y(2^{-k})\setminus Y(2^{-k+2})$ , it is enough to show that for each $k\geq 1$ and for $\mu $ -a.e. $y\in Y(2^{-k})\setminus Y(2^{-k+2})$ , we have $[y]_{(\mathcal {P}^L)_{-\infty }^{\infty }}=\{y\}$ .

For fixed $k\geq 1$ , it follows from Poincaré recurrence (e.g. see [Reference Einsiedler and WardEW11, Theorem 2.11]) that for $\mu $ -a.e. $y\in Y(2^{-k})\setminus Y(2^{-k+2})$ , there exists an increasing sequence $(k_i)_{i\geq 1} \subset \mathbb {N}$ such that

$$ \begin{align*} a^{k_i}y\in Y(2^{-k})\setminus Y(2^{-k+2}) \quad \text{and}\quad k_i \to \infty\ \text{as}\ i\to \infty. \end{align*} $$

By Proposition 2.8(2), it follows that for each $i\geq 1$ ,

$$ \begin{align*}[a^{k_i}y]_{\mathcal{A}^L}=[a^{k_i}y]_{(\mathcal{P}^L)_0^{\infty}}\subset B_{r_02^{-k+1}}^L \cdot a^{k_i}y.\end{align*} $$

Since $[a^{k_i}y]_{(\mathcal {P}^L)_0^{\infty }}=a^{k_i}[y]_{a^{-k_i}(\mathcal {P}^L)_0^{\infty }}=a^{k_i}[y]_{(\mathcal {P}^L)_{-k_i}^{\infty }}$ , using equation (2.6), we have

$$ \begin{align*}[y]_{(\mathcal{P}^L)_{-k_i}^{\infty}} \subset a^{-k_i}B_{r_02^{-k+1}}^L \cdot a^{k_i}y =a^{-k_i}B_{r_02^{-k+1}}^L a^{k_i}\cdot y \subset B^L_{d_0e^{-\alpha k_i}r_02^{-k+1}}\cdot y.\end{align*} $$

Taking $i\to \infty $ , we conclude that $[y]_{(\mathcal {P}^L)_{-\infty }^{\infty }}=\{y\}$ .

Proof. Let $\mathcal {P}^L$ be as in the proof of Proposition 2.8. Since $\mathcal {P}^L$ is countably generated, we can take an increasing sequence of finite partitions $(\mathcal {P}_k^L)_{k\geq 1}$ of Y such that $\mathcal {P}_k^L \nearrow \mathcal {P}^L$ . By Lemma 2.9, we have $\mathcal {B}_Y=(\mathcal {P}^L)_{-\infty }^{\infty }=\bigvee _{k=1}^{\infty }(\mathcal {P}_k^L)_{-\infty }^{\infty }$ modulo $\mu $ , where $\mathcal {B}_Y$ is the Borel $\sigma $ -algebra of Y. It is clear that $(\mathcal {P}_k^L)_{-\infty }^{\infty }\subseteq (\mathcal {P}_{k+1}^L)_{-\infty }^{\infty }$ for all $k\in \mathbb {N}$ . Hence, it follow from Kolmogorov–Sinaı̆ theorem [Reference Einsiedler, Lindenstrauss and WardELW, Proposition 2.20] that

$$ \begin{align*} h_{\mu}(a^{-1}|\mathcal{A}_\infty^L)=\lim_{k\to\infty}h_{\mu}(a^{-1},\mathcal{P}_k^L|\mathcal{A}_\infty^L). \end{align*} $$

Using the future formula [Reference Einsiedler, Lindenstrauss and WardELW, Proposition 2.19(8)], we have

$$ \begin{align*} \lim_{k\to\infty}h_{\mu}(a^{-1},\mathcal{P}_k^L|\mathcal{A}_\infty^L)=\lim_{k\to\infty}H_{\mu}(\mathcal{P}_k^L|(\mathcal{P}_k^L)_1^\infty \vee \mathcal{A}_\infty^L). \end{align*} $$

It follows from monotonicity and continuity of entropy [Reference Einsiedler, Lindenstrauss and WardELW, Propositions 2.10, 2.12, and 2.13] that for any fixed $k\geq 1$ ,

$$ \begin{align*} \lim_{\ell\to\infty}H_{\mu}(\mathcal{P}_k^L|(\mathcal{P}_\ell^L)_1^\infty \vee \mathcal{A}_\infty^L)\leq H_{\mu}(\mathcal{P}_k^L|(\mathcal{P}_k^L)_1^\infty \vee \mathcal{A}_\infty^L) \leq \lim_{\ell\to\infty}H_{\mu}(\mathcal{P}_\ell^L|(\mathcal{P}_k^L)_1^\infty \vee \mathcal{A}_\infty^L), \end{align*} $$

and hence, we have

$$ \begin{align*} H_{\mu}(\mathcal{P}_k^L|(\mathcal{P}^L)_1^\infty \vee \mathcal{A}_\infty^L)\leq H_{\mu}(\mathcal{P}_k^L|(\mathcal{P}_k^L)_1^\infty \vee \mathcal{A}_\infty^L) \leq H_{\mu}(\mathcal{P}^L|(\mathcal{P}_k^L)_1^\infty \vee \mathcal{A}_\infty^L). \end{align*} $$

Taking $k\to \infty $ , it follows that

$$ \begin{align*} \lim_{k\to\infty}H_{\mu}(\mathcal{P}_k^L|(\mathcal{P}_k^L)_1^\infty \vee \mathcal{A}_\infty^L) =H_{\mu}(\mathcal{P}^L|(\mathcal{P}^L)_1^\infty \vee \mathcal{A}_\infty^L)= H_{\mu}(\mathcal{A}^L|a\mathcal{A}^L), \end{align*} $$

which concludes equation (2.13).

Note that $\mathcal {B}_Y=(\mathcal {P}^L)_{-\infty }^{\infty }=\bigvee _{k=1}^{\infty }(\mathcal {P}_k^L)_{-\infty }^{\infty }$ modulo almost every ergodic component of $\mu $ . Thus, following the same argument as above, we can conclude equation (2.13) for almost every ergodic component of $\mu $ .

Proof. By for instance [Reference Einsiedler, Lindenstrauss, Einsiedler, Ellwood, Eskin, Kleinbock, Lindenstrauss, Margulis, Marmi and YoccozEL10, Theorem 5.9], for $\mu $ -a.e. $y\in Y$ , $\mu _y^{a^{j}\mathcal {A}}$ is a probability measure on $[y]_{a^{j}\mathcal {A}}=a^{j}V_{a^{-j}y}a^{-j}\cdot y$ , and $H_\mu (\mathcal {A}|a^{j}\mathcal {A})$ can be written as

$$ \begin{align*} {H_\mu(\mathcal{A}|a^{j}\mathcal{A})=-\int_Y\log \mu_y^{a^{j}\mathcal{A}}([y]_{\mathcal{A}})\,d\mu(y).} \end{align*} $$

Note that $m_L(a^jBa^{-j})=e^{J}m_L(B)$ for any measurable $B\subset L$ . Let

$$ \begin{align*}p(y)\overset{\operatorname{def}}{=} \mu_y^{a^{j}\mathcal{A}}([y]_{\mathcal{A}}) \quad\text{and}\quad p^{\mathrm{Haar}}(y)\overset{\operatorname{def}}{=} \tau_y^{a^{j}\mathcal{A}}([y]_{\mathcal{A}}).\end{align*} $$

Then, we have

$$ \begin{align*}p^{\mathrm{Haar}}(y)=\frac{m_L(V_y)}{m_L(a^jV_{a^{-j}y} a^{-j})}=\frac{m_L(V_y)}{m_L(V_{a^{-j}y})}e^{-J},\end{align*} $$

and hence, applying the ergodic theorem, we have $-\int _Y\log p^{\mathrm {Haar}}(y)\,d\mu (y)=J$ .

Now we estimate an upper bound of $H_\mu (\mathcal {A}|a^{j}\mathcal {A})-J$ following the computation in [Reference Einsiedler, Lindenstrauss, Einsiedler, Ellwood, Eskin, Kleinbock, Lindenstrauss, Margulis, Marmi and YoccozEL10, 7.55]. Following [Reference Einsiedler, Lindenstrauss, Einsiedler, Ellwood, Eskin, Kleinbock, Lindenstrauss, Margulis, Marmi and YoccozEL10, 7.55], we can partition $[y]_{a^{j}\mathcal {A}}$ into a countable union of $\mathcal {A}$ -atoms as follows:

$$ \begin{align*}[y]_{a^{j}\mathcal{A}}=\bigcup_{i=1}^{\infty}[x_i]_{\mathcal{A}}\cup N_y,\end{align*} $$

where $N_y$ is a null set with respect to $\mu _y^{a^{j}\mathcal {A}}$ . Note that $\mu _y^{a^{j}\mathcal {A}}$ is supported on $\operatorname {Supp}\mu $ for $\mu $ -a.e y. Since $B\subset L$ is symmetric, if $x_i\in K\setminus B\operatorname {Supp}\mu $ , then $[x_i]_{\mathcal {A}}\subset B\cdot x_i\subset K\setminus \operatorname {Supp}\mu $ , and hence we have $\mu ^{a^{j}\mathcal {A}}_y([x_i]_{\mathcal {A}})=0$ . If $x_i \in (Y\setminus K) \cup B\operatorname {Supp}\mu $ and , then there exists $x_i' \in [x_i]_{\mathcal {A}}$ such that $x_i' \in K\setminus B\operatorname {Supp}\mu $ , and hence $\mu ^{a^{j}\mathcal {A}}_y([x_i]_{\mathcal {A}})=\mu ^{a^{j}\mathcal {A}}_y([x_i']_{\mathcal {A}})=0$ . Thus, we denote by Z the set of $x_i$ terms in $(Y\setminus K) \cup B\operatorname {Supp}\mu $ such that $[x_i]_{\mathcal {A}} \subset (Y\setminus K)\cup B\operatorname {Supp}\mu $ . It follows that

$$ \begin{align*} H_{\mu}(\mathcal{A}|a^j \mathcal{A})-J &= -\int_Y (\log p(z) - \log p^{\mathrm{Haar}}(z) )\,d\mu(z)\\&= \int_Y \int_Y (\log p^{\mathrm{Haar}}(z) - \log p(z) )\,d\mu_y^{a^j \mathcal{A}}(z)\,d\mu(y)\\&= \int_Y \sum_{x_i \in Z} \int_{z\in [x_i]_{\mathcal{A}}} (\log p^{\mathrm{Haar}}(z) - \log p(z) )\,d\mu_y^{a^j \mathcal{A}}(z)\,d\mu(y)\\&= \int_Y \sum_{x_i \in Z} \log\bigg(\frac{\tau^{a^{j}\mathcal{A}}_y([x_i]_{\mathcal{A}})}{\mu^{a^{j}\mathcal{A}}_y([x_i]_{\mathcal{A}})}\bigg)\mu^{a^{j}\mathcal{A}}_y([x_i]_{\mathcal{A}})\,d\mu(y)\\&\leq \int_Y \log\bigg(\sum_{x_i\in Z}\tau_y^{a^{j}\mathcal{A}}([x_i]_{\mathcal{A}})\bigg) \,d\mu(y)\\&\leq \int_Y \log \tau_y^{a^{j}\mathcal{A}}((Y\setminus K) \cup B \operatorname{Supp}\mu) \,d\mu(y). \end{align*} $$

The second last inequality follows from the convexity of the logarithm. This proves the proposition.

Proof. Writing the ergodic decomposition $\mu =\int \mu _{z}^{\mathcal {E}}\,d\mu (z)$ , we have

$$ \begin{align*}h_\mu(a^j|\mathcal{A}^L_\infty)=\int h_{\mu_z^{\mathcal{E}}}(a^j|\mathcal{A}^L_\infty)\,d\mu(z),\end{align*} $$

where $\mathcal {A}^L_\infty $ is the $\sigma $ -algebra as in equation (2.12). By Proposition 2.10, we also have

$$ \begin{align*}H_\mu(\mathcal{A}^L|a^j\mathcal{A}^L)=\int H_{\mu_z^{\mathcal{E}}}(\mathcal{A}^L|a^j\mathcal{A}^L)\,d\mu(z).\end{align*} $$

It follows from the a-invariance of $\mu $ and $\mu _z^{\mathcal {E}}$ that

$$ \begin{align*}H_\mu(\mathcal{A}|a^j\mathcal{A})=\int H_{\mu_z^{\mathcal{E}}}(\mathcal{A}|a^j\mathcal{A})\,d\mu(z).\end{align*} $$

Applying Theorem 2.12 for each $\mu _{z}^{\mathcal {E}}$ , we obtain

$$ \begin{align*} H_\mu(\mathcal{A}|a^j\mathcal{A}) &= \int H_{\mu_z^{\mathcal{E}}}(\mathcal{A}|a^j\mathcal{A})\,d\mu(z)\\ &\leq J+\int_Y\int_Y \log\tau_y^{a^{j}\mathcal{A}}(B^2\operatorname{Supp}\mu_z^{\mathcal{E}})\,d\mu_z^{\mathcal{E}}(y)\,d\mu(z)\\ &\leq J+\int_Y \log\tau_y^{a^{j}\mathcal{A}}(B^2\operatorname{Supp}\mu)\,d\mu(y).\\[-3.4pc] \end{align*} $$

Proof. Suppose that there exist arbitrarily large t terms satisfying $a_t y_{A,b}\notin \mathcal {L}_\epsilon $ . Denote $e^{\mathbf {r} t}\overset {\operatorname {def}}{=} \textrm {diag}(e^{r_1t},\ldots ,e^{r_mt})\in M_{m,m}(\mathbb {R})$ and $e^{\mathbf {s}t}\overset {\operatorname {def}}{=}\textrm {diag}(e^{s_1t},\ldots ,e^{s_nt})\in M_{n,n}(\mathbb {R}).$ Then, the vectors in the grid $\Lambda _{a_t y_{A,b}}$ can be represented as

$$ \begin{align*}a_t \left(\left(\begin{matrix} I_m & A \\ 0 & I_n \\ \end{matrix}\right) \left(\begin{matrix} p \\ q \\ \end{matrix}\right) +\left(\begin{matrix} -b \\ 0 \\ \end{matrix}\right)\right) =\left(\begin{matrix} e^{\mathbf{r} t}(Aq+p-b)\\ e^{-\mathbf{s} t}q\\ \end{matrix}\right)\end{align*} $$

for $(p,q)\in \mathbb {Z}^m\times \mathbb {Z}^n$ . Therefore, $a_t x_{A,b}\notin \mathcal {L}_\epsilon $ implies that for some $q\in \mathbb {Z}^n$ ,

(3.3) $$ \begin{align} {e^{t}\langle Aq-b\rangle_{\mathbf{r}}<\epsilon^{{m}/{d}} \quad \text{and}\quad e^{-t}\|q\|_{\mathbf{s}}<\epsilon^{{n}/{d}},} \end{align} $$

and thus $\|q\|_{\mathbf {s}}\langle Aq-b\rangle _{\mathbf {r}}<\epsilon $ . Since $\langle Aq-b\rangle _{\mathbf {r}}\neq 0$ for all q, we use the condition $\langle Aq-b\rangle _{\mathbf {r}}<e^{-t}\epsilon ^{{m}/{d}}$ for arbitrarily large t to conclude that $\|q\|_{\mathbf {s}}\langle Aq-b\rangle _{\mathbf {r}}<\epsilon $ holds for infinitely many q terms. This is a contradiction to the assumption that $(A,b)\in \mathbf {Bad}(\epsilon )$ .

However, if $(A,b)\notin \mathbf {Bad}(\epsilon )$ , then since $(A,b)$ is irrational, there are infinitely many $q\in \mathbb {Z}^n$ such that $\|q\|_{\mathbf {s}}\langle Aq-b\rangle _{\mathbf {r}}<\epsilon $ . Thus, we can choose arbitrarily large t so that equation (3.3) holds, which contradicts the assumption that the $a_t$ -orbit of the point $y_{A,b}$ is eventually in $\mathcal {L}_\epsilon $ .

Proof. For $s\in \{0,\ldots , T_{D}-1\}$ and $k\in \mathbb {Z}_{\geq 0}$ , let us denote $I_{s,k}(T_{D})= \{s,s+T_{D},\ldots , s+kT_{D}\}$ and

$$ \begin{align*} E_{y,k}^{s}= \{z\in B_{r_D}^{L}\cdot y : \text{ for all } t \in I_{s,k}(T_{D}) \smallsetminus I, d_{Y}(a_{t}y,a_{t}z)\leq r_D \}. \end{align*} $$

Following the proof of [Reference Lim, de Saxcé and ShapiraLSS19, Lemma 2.4] with $E_{y,k}^{s}$ instead of $E_{y,T}$ , we obtain the following claim.

Proof of the claim

We prove the claim by induction on k. Since the number of $d_{\mathbf {c}}$ -balls of radius $C_0 r_D^{{1}/{\max \mathbf {c}}} e^{-s}$ needed to cover $B_{r_D}^L \cdot y$ is bounded by a constant $C_s$ depending on $Q_\infty ^0$ , D, and s, the claim holds for $k=0$ .

Suppose that $E_{y,k-1}^s$ can be covered by $N_{k-1}=C_s e^{(J_L(T_{D}-1)+D)|I\cap I_{s,k-1}(T_{D})|}$ $d_{\mathbf {c}}$ -balls $\{B_j:j=1,\ldots , N_{k-1}\}$ of radius $C_0 r_D^{{1}/{\max \mathbf {c}}}e^{-(s+(k-1)T_{D})}$ . By the inequality in equation (3.4), any $d_{\mathbf {c}}$ -ball of radius $C_0 r_D^{{1}/{\max \mathbf {c}}}e^{-(s+(k-1)T_D)}$ can be covered by

$$ \begin{align*} \prod_{i=1}^{\dim\mathfrak{l}} \bigg\lceil \frac{e^{-(s+(k-1)T_D) c_i }}{e^{-(s+kT_D) c_i }} \bigg\rceil &= \prod_{i=1}^{\dim\mathfrak{l}}\lceil e^{T_D c_i} \rceil \leq \prod_{i=1}^{\dim\mathfrak{l}} e^{c_{i}T_{D}}e^{({D-J_L})/{\dim \mathfrak{l}}}\\ &= e^{J_L T_D}e^{D-J_L}= e^{J_L(T_D-1)+D} \end{align*} $$

$d_{\mathbf {c}}$ -balls of radius $C_0 r_D^{{1}/{\max \mathbf {c}}}e^{-(s+kT_D)}$ . Thus, if $s+kT_D \in I$ , then $E_{y,k}^s$ can be covered by $N_k=e^{J_L(T_D-1)+D}N_{k-1} d_{\mathbf {c}}$ -balls of radius $C_0 r_D^{{1}/{\max \mathbf {c}}}e^{-(s+kT_D)}$ .

Suppose that $s+kT_D \notin I$ . Since $E_{y,k}^s \subset E_{y,k-1}^s$ , the sets $E_{y,k}^s \cap B_j$ with $j=1,\ldots , N_{k-1}$ cover $E_{y,k}^s$ . We now claim that for any $1\leq j \leq N_{k-1}$ and $x_1,x_2\in E_{y,k}^s \cap B_j$ , we have

$$ \begin{align*}d_{L,\mathbf{c}}(x_1,x_2)\leq 2^{{1}/{\min\mathbf{c}}}C_0 r_D^{{1}/{\max\mathbf{c}}}e^{-(s+kT_D)}.\end{align*} $$

Indeed, since $B_j$ is a $d_{L,\mathbf {c}}$ -ball of radius $C_0 r_D^{{1}/{\max \mathbf {c}}}e^{-(s+(k-1)T_{D})}$ and $x_1,x_2 \in B_j \subset B_{r_D}^L \cdot y$ , there are $h\in B_{r_D}^L$ and $h_1,h_2\in B_{C_0 r_D^{{1}/{\max \mathbf {c}}}e^{-(s+(k-1)T_{D})}}^{L,\mathbf {c}}$ such that $x_1=h_1 hy$ and $x_2=h_2 hy$ . It follows from $s+kT_d \notin I$ and $x_1,x_2\in E_{y,k}^s$ that $a^{s+kT_d}y \subset Y \smallsetminus Q_\infty $ and $d_Y(a^{s+kT_D} y, a^{s+kT_D} x_\ell )\leq r_D$ for $\ell =1,2$ , and hence by equation (3.5), we have $a^{s+kT_D} x_1 \in B_{r_D}^G (Y \smallsetminus Q_\infty ) \subset Y(\tfrac 12 r(Q_\infty ))$ and $d_Y(a^{s+kT_D} x_1, a^{s+kT_D} x_2)\leq 2r_D$ . Observe that by equation (3.5),

$$ \begin{align*} a^{s+kT_D}h_1 h_2^{-1} a^{-(s+kT_D)} &\subset a^{s+kT_D}B_{2^{{1}/{\min\mathbf{c}}}C_0 r_D^{{1}/{\max\mathbf{c}}}e^{-(s+(k-1)T_{D})}}^{L,\mathbf{c}}a^{-(s+kT_D)}\\ &= B_{2^{{1}/{\min\mathbf{c}}}C_0 r_D^{{1}/{\max\mathbf{c}}}e^{T_{D}}}^{L,\mathbf{c}} \subset B_{\min(r_0,({1}/{2})r(Q_\infty))}^L. \end{align*} $$

Thus, it follows from equation (3.1) and the above observations that

$$ \begin{align*} 2r_D \geq d_Y (a^{s+kT_D}x_1,a^{s+kT_D}x_2) &= d_L(a^{s+kT_D}h_1h_2^{-1}a^{-(s+kT_D)},\mathrm{id}) \\ &\geq \frac{1}{C_0} d_\infty (a^{s+kT_D}h_1h_2^{-1}a^{-(s+kT_D)},\mathrm{id})\\ &= \frac{1}{C_0} \max_{i=1,\ldots,\dim\mathfrak{l}} e^{c_i(s+kT_D)} |(\log h_1h_2^{-1})_i|, \end{align*} $$

where $(\log h_1h_2^{-1})_i$ is the ith coordinate of $\log h_1h_2^{-1}$ with respect to the standard basis $\{e_i: 1\leq i\leq \dim \mathfrak {l}\}$ of $\mathfrak {l}$ . Since $L=U$ or $L=W$ , that is, a commutative subgroup of G, for each $i=1,\ldots ,\dim \mathfrak {l}$ , we have

$$ \begin{align*} |(\log h_1h_2^{-1})_i|=|(\log h_1 - \log h_2)_i|\leq 2r_D C_0 e^{-c_i(s+kT_D)}. \end{align*} $$

Note that

$$ \begin{align*} d_{L,\mathbf{c}}(x_1,x_2)=d_{L,\mathbf{c}}(h_1,h_2) =\max_{i=1,\ldots,\dim\mathfrak{l}} |(\log h_1 - \log h_2)_i|^{{1}/{c_i}}. \end{align*} $$

Therefore, we have

$$ \begin{align*} d_{L,\mathbf{c}}(x_1,x_2) \leq \max_{i=1,\ldots,\dim\mathfrak{l}} (2r_D C_0)^{{1}/{c_i}}e^{-(s+kT_D)}\leq 2^{{1}/{\min\mathbf{c}}}C_0 r_D^{{1}/{\max\mathbf{c}}}e^{-(s+kT_D)}. \end{align*} $$

It follows from the claim that $E_{y,k}^s \cap B_j$ is contained in a single $d_{L,\mathbf {c}}$ -ball of radius $C_0 r_D^{{1}/{\max \mathbf {c}}} e^{-(s+kT_D)}$ for each $j=1,\ldots ,N_{k-1}$ . Hence, $E_{y,k}^s$ can be covered by $N_k=N_{k-1} d_{L,\mathbf {c}}$ -balls of radius $C_0 r_D^{{1}/{\max \mathbf {c}}} e^{-(s+kT_D)}$ .

Now, for any non-negative integer T, we can find $s\in \{0,\ldots ,T_{D}-1\}$ and $k\in \mathbb {Z}_{\geq 0}$ such that

$$ \begin{align*} { T_{D}|I\cap I_{s,k}(T_{D})| \leq |I\cap \{1,\ldots,T\}|\quad\text{and}\quad T-T_{D}<s+kT_{D}\leq T } \end{align*} $$

from the pigeon hole principle. By the above observation, $E_{y,T}\subset E_{y,k}^{s}$ can be covered by $C_{s}e^{(J_L(T_{D}-1)+D)|I\cap I_{s,k}(T_{D})|} d_{\mathbf {c}}$ -balls of radius $C_0 r_D^{{1}/{\max \mathbf {c}}} e^{-(s+kT_D)}$ . Since $T-T_{D}<s+kT_{D}\leq T$ and $D>J_L$ , $E_{y,T}$ can be covered by $(\max _{0\leq s\leq T_{D}-1}C_s) e^{D|I\cap \{1,\ldots ,T\}|} d_{\mathbf {c}}$ -balls of radius $C_0 e^{T_D} r_D^{{1}/{\max \mathbf {c}}} e^{-T}$ . Hence, there exists a constant $C>0$ depending on $Q_\infty ^0$ , r, and D, but independent of T such that $E_{y,T}$ can be covered by $C e^{D|I\cap \{1,\ldots ,T\}|} d_{\mathbf {c}}$ -balls of radius $r_D^{{1}/{\max \mathbf {c}}} e^{-T}$ .

Proof. Since $x_{A}$ has $\eta _A$ -escape of mass on average but no more than $\eta _A,$ we may fix an increasing sequence of integers $\{k_i\}_{i\geq 1}$ such that

$$ \begin{align*}\frac{1}{k_i}\sum_{k=0}^{k_i-1}\delta_{a^k x_A}\overset{\operatorname{w}^*}{\longrightarrow}\mu_A\in\mathscr{P}(\overline{X})\end{align*} $$

with $\mu _A(X)= 1-\eta _A$ .

Let us denote by $\mathbb {T}^{m}= [0,1]^{m}/\!\sim $ the torus in $\mathbb {R}^{m}$ , where the equivalence relation is modulo $1.$ Consider the increasing family of sets

$$ \begin{align*}R^{A,T}\overset{\operatorname{def}}{=}\{b\in \mathbb{T}^m|\text{ for all } t\ge T,a_t y_{A,b}\in\mathcal{L}_\epsilon\}\cap\mathbf{Bad}_A(\epsilon).\end{align*} $$

By Proposition 3.2 and Remark 3.3, $\bigcup _{T=1}^\infty R^{A,T}$ has Hausdorff dimension equal to $\dim _H \mathbf {Bad}_A(\epsilon )$ . For any $\gamma>0$ , it follows that there exists $T_\gamma \in \mathbb {N}$ satisfying $\dim _H R^{A,T_\gamma }\geq \dim _H \mathbf {Bad}_A(\epsilon )-\gamma $ .

Let $\phi _A:\mathbb {T}^{m}\to Y$ be the map defined by $\phi _A(b)=y_{A,b}$ . Note that $\phi _A$ is a one-to-one Lipschitz map between $\mathbb {T}^m$ and $\phi _A(\mathbb {T}^m)$ , so we may consider a quasinorm on $\phi _A(\mathbb {T}^m)$ induced from the $\mathbf {r}$ -quasinorm on $\mathbb {R}^m$ and denote it again by $\|\cdot \|_{\mathbf {r}}$ .

For each $k_i\ge T_\gamma $ , let $S_i$ be a maximal $e^{-k_i}$ -separated subset of $R^{A,T_\gamma }$ with respect to the $\mathbf {r}$ -quasinorm. By Lemma 3.1(3),

$$ \begin{align*} {\liminf_{i\to\infty}\frac{\log|S_i|}{k_i}\ge \underline{\dim}_{\mathbf{r}}(R^{A,T_\gamma})\ge 1-r_{1}(m+\gamma- \dim_H \mathbf{Bad}_A(\epsilon)).} \end{align*} $$

Let $\nu _i\overset {\operatorname {def}}{=}({1}/{|S_i|})\sum _{b\in S_i}\delta _{y_{A,b}}$ be the normalized counting measure on the set $D_i\overset {\operatorname {def}}{=}\{y_{A,b}: b\in S_i\}\subset Y$ . Extracting a subsequence if necessary, we may assume that

$$ \begin{align*}\mu_i\overset{\operatorname{def}}{=}\frac{1}{k_i}\sum_{k=0}^{k_i-1}a_*^k\nu_i\overset{\operatorname{w}^*}{\longrightarrow}\mu^\gamma\in\mathscr{P}(\overline{Y}).\end{align*} $$

The measure $\mu ^{\gamma }$ is a-invariant since $a_* \mu _i -\mu _i$ goes to zero measure.

Choose any sequence of positive real numbers $(\gamma _j)_{j\geq 1}$ converging to zero and let $\{\mu ^{\gamma _j}\}$ be a family of a-invariant probability measures on $\overline {Y}$ obtained from the above construction for each $\gamma _j$ . Extracting a subsequence again if necessary, we may take a $\text {weak}^*$ -limit measure $\overline {\mu }\in \mathscr {P}(\overline {Y})$ of $\{\mu ^{\gamma _j}\}$ . We prove that $\overline {\mu }$ is the desired measure. The measure $\overline {\mu }$ is clearly a-invariant.

(1) We show that for all $\gamma>0$ , $\mu ^\gamma (Y\setminus \mathcal {L}_\epsilon )=0$ . For any $b\in S_i\subseteq R^{A,T_\gamma }$ , $a_T y_{A,b}\in \mathcal {L}_\epsilon $ holds for $T>T_\gamma $ . Thus, we have

$$ \begin{align*} \mu_i(Y\setminus\mathcal{L}_\epsilon)&=\frac{1}{k_i}\sum_{k=0}^{k_i-1}a^k_{*}\nu_i(Y\setminus\mathcal{L}_\epsilon)= \frac{1}{k_i}\sum_{k=0}^{T_\gamma}a^k_{*}\nu_i(Y\setminus\mathcal{L}_\epsilon)\\ &= \frac{1}{k_{i} |S_{i}|}\sum_{y\in D_{i}, 0\leq k \leq T_{\gamma}} \delta_{a^{k}y}(Y\setminus\mathcal{L}_\epsilon) \leq \frac{T_\gamma}{k_i}. \end{align*} $$

By taking $k_i\to \infty $ , we have $\mu ^{\gamma }(Y\setminus \mathcal {L}_\epsilon )=0$ for arbitrary $\gamma>0$ , and hence

$$ \begin{align*}\overline{\mu}(Y\setminus\mathcal{L}_\epsilon)=\lim_{j\to\infty}\mu^{\gamma_j}(Y\setminus\mathcal{L}_\epsilon)=0.\end{align*} $$

(2) For all $\gamma>0$ , $\pi _*\mu ^\gamma =\mu _A$ since $\pi _*\nu _i=\delta _{x_{A}}$ for all $i\ge 1$ . It follows that $\pi _*\overline {\mu }=\mu _A$ . Hence,

$$ \begin{align*}\overline{\mu}(\overline{Y}\setminus Y)=\lim_{j\to\infty}\mu^{\gamma_j}(\overline{Y}\setminus Y)=\mu_A(\overline{X}\setminus X)=\eta_A,\end{align*} $$

so we have a decomposition $\overline {\mu }=(1-\eta _A)\mu +\eta _A\delta _\infty $ for some a-invariant $\mu \in \mathscr {P}(Y)$ .

(3) We first fix any $D>J_W =1$ and $Q^0_\infty \subset X$ such that $X\smallsetminus Q_\infty ^0$ has compact closure. As in [Reference Lim, de Saxcé and ShapiraLSS19, Proof of Theorem 4.2, Claim 2], we can construct a finite partition $\mathcal {Q}$ of Y satisfying:

• • $\mathcal {Q}$ contains an atom $Q_\infty $ of the form $\pi ^{-1}(Q_\infty ^0)$ ;

• • $\text { for all } Q\in \mathcal {Q}\smallsetminus \{Q_\infty \}$ , $\operatorname {\mathrm {diam}} Q<r_D=r_D(Q_\infty ^0)$ , where $r_D$ is from equation (3.5);

• • $\text { for all } Q\in \mathcal {Q},\text { for all } j\geq 1,\; \mu ^{\gamma _j}(\partial Q)=0$ .

Remark that for all $i\geq 1$ , $D_i \subset \phi _A(\mathbb {T}^m)$ , which is a compact set in Y; therefore, we can choose $Q^0_{\infty }$ so that

(4.1) $$ \begin{align} { Q_\infty \cap D_i = \varnothing. } \end{align} $$

We claim that it suffices to show the following statement. For all $q\ge 1$ ,

(4.2) $$ \begin{align} {\frac{1}{q}H_{\overline{\mu}}(\overline{\mathcal{Q}}^{(q)}|\overline{\mathcal{A}^W_\infty})\ge 1-r_{1}(m-\dim_H \mathbf{Bad}_A(\epsilon))-D\overline{\mu}(\overline{Q_\infty}).} \end{align} $$

Indeed, by taking $q\to \infty $ , we have

$$ \begin{align*}h_{\overline{\mu}}(a|\overline{\mathcal{A}^W})\ge 1-r_{1}(m- \dim_H \mathbf{Bad}_A(\epsilon))-D\overline{\mu}(\overline{Q_\infty}).\end{align*} $$

Taking $D\to 1$ and $Q_\infty ^0\subset X$ such that $\overline {\mu }(\overline {Q_\infty }) \to \overline {\mu }(\overline {Y}\setminus Y)=\eta _A$ and $D\to 1$ , we conclude equation (3).

In the rest of the proof, we show the inequality in equation (4.2). It is clear if $\overline {\mu }(Q_\infty )=1$ , so assume that $\overline {\mu }(Q_\infty )<1$ , and hence for all large enough $j\geq 1$ , $\mu ^{\gamma _j}(Q_\infty )<1$ . Now, we fix such $j\geq 1$ and write temporarily $\gamma =\gamma _j$ .

Choose $\beta>0$ such that $\mu ^\gamma (Q_\infty )<\beta <1$ . For large enough $i\geq 1$ , we have

$$ \begin{align*} { \mu_i(Q_\infty)=\frac{1}{k_i|S_i|}\sum_{y\in D_i, 0\le k<k_i}\delta_{a^k y}(Q_\infty) =\frac{1}{k_i}\sum_{0\le k<k_i}\delta_{a^k x_{A}}(Q^0_\infty) < \beta. } \end{align*} $$

In other words, there exist at most $\beta k_i$ number of $a^k x_A$ terms in $Q^0_\infty $ , and thus for any $y\in D_{i}$ , we have

$$ \begin{align*} |\{k\in\{0,\ldots,k_{i}-1\}: a^{k}y \in Q_{\infty}\}| < \beta k_{i}. \end{align*} $$

From Lemma 3.4 with $L=W$ and equation (4.1), if Q is any non-empty atom of $\mathcal {Q}^{(k_i)}$ , fixing any $y\in D_i\cap Q$ , the set

$$ \begin{align*} { D_{i}\cap Q = D_{i}\cap [y]_{\mathcal{Q}^{(k_i)}} \subset E_{y,k_{i}-1} } \end{align*} $$

can be covered by $Ce^{D\beta k_{i}}$ many $r_D^{1/r_1}e^{-k_{i}}$ -balls for $d_{\mathbf {r}}$ , where C is a constant depending on $Q_\infty ^0$ and D, but not on $k_i$ . Since $D_{i}$ is $e^{-k_{i}}$ -separated with respect to $d_{\mathbf {r}}$ and $r_D^{1/r_{1}}<\tfrac 12$ , we get

(4.3) $$ \begin{align} {\mathrm{Card}(D_i \cap Q)\leq Ce^{D\beta k_i }.} \end{align} $$

Now let $\mathcal {A}^W=(\mathcal {P}^W)_0^\infty =\bigvee _{i=0}^{\infty }a^i \mathcal {P}^W$ be as in Proposition 2.8 for $\mu $ , $r_0$ , and $L=W$ , and let $\mathcal {A}^W_\infty $ be as in equation (2.12).

Proof of the claim

Using the continuity of entropy, we have

$$ \begin{align*} {H_{\nu_i}(\mathcal{Q}^{(k_i)}|\mathcal{A}^W_\infty)=\lim_{\ell\to\infty}H_{\nu_i}(\mathcal{Q}^{(k_i)}|(\mathcal{P}^W)_\ell^\infty).} \end{align*} $$

Now we show $H_{\nu _i}(\mathcal {Q}^{(k_i)}|(\mathcal {P}^W)_\ell ^\infty ) = H_{\nu _i}(\mathcal {Q}^{(k_i)})$ for all large enough $\ell \geq 1$ . Let $E_\delta $ be the dynamical $\delta $ -boundary of $\mathcal {P}$ as in Lemma 2.7 for $\mu $ and $r_0$ . As mentioned in Remark 2.4, we may assume that there exists $y\in \phi _A(\mathbb {T}^m)$ such that $y\notin \partial \mathcal {P}$ . Since $E_\delta = \bigcup _{k=0}^\infty a^k \partial _{d_0e^{-k\alpha }\delta }\mathcal {P}$ , there exists $\delta>0$ such that $y\in Y \setminus E_{\delta }$ . For any $\ell \geq 1$ , we have $a^{-\ell }y \in Y\setminus a^{-\ell }E_\delta \subset Y\setminus E_\delta $ . Hence, it follows from equation (2.6) and Proposition 2.8 that

$$ \begin{align*} [y]_{(\mathcal{P}^W)_\ell^\infty} = a^\ell[a^{-\ell}y]_{(\mathcal{P}^W)_0^\infty} = a^\ell[a^{-\ell}y]_{\mathcal{A}^W} \supset a^{\ell} B_{\delta}^{W} a^{-\ell} y \supset B_{d_0 e^{\alpha\ell}\delta}^{W} y. \end{align*} $$

Since the support of $\nu _i$ is a set of finite points on a single compact W-orbit $\phi _A(\mathbb {T}^m)$ , $\nu _i$ is supported on a single atom of $(\mathcal {P}^W)_\ell ^\infty $ for all large enough $\ell \geq 1$ . This proves the claim.

Combining equation (4.3) and the above claim, it follows that

(4.4) $$ \begin{align} { H_{\nu_i}(\mathcal{Q}^{(k_i)}|\mathcal{A}^W_\infty)= H_{\nu_i}(\mathcal{Q}^{(k_i)}) \geq \log |S_i|-D\beta k_i-\log C. } \end{align} $$

For any $q\ge 1$ , write the Euclidean division of large enough $k_i-1$ by q as

$$ \begin{align*}k_i-1=qk'+s \ \textrm{with} \ s\in\{0,\ldots,q-1\}.\end{align*} $$

By subadditivity of the entropy with respect to the partition, for each $p\in \{0,\ldots ,q-1\}$ ,

$$ \begin{align*}H_{\nu_i}(\mathcal{Q}^{(k_i)}|\mathcal{A}^W_\infty)\leq H_{a^{p}\nu_i}(\mathcal{Q}^{(q)}|\mathcal{A}^W_\infty)+\cdots+H_{a^{p+qk'}\nu_i}(\mathcal{Q}^{(q)}|\mathcal{A}^W_\infty)+2q\log |\mathcal{Q}|.\end{align*} $$

Summing those inequalities for $p=0,\ldots ,q-1$ , and using the concave property of entropy with respect to the measure, we obtain

(4.5) $$ \begin{align} qH_{\nu_i}(\mathcal{Q}^{(k_i)}|\mathcal{A}^W_\infty)&\leq\sum_{k=0}^{k_i-1}H_{a^k \nu_i}(\mathcal{Q}^{(q)}|\mathcal{A}^W_\infty)_0^M+2q^2\log |\mathcal{Q}|\nonumber\\&\leq k_iH_{\mu_i}(\mathcal{Q}^{(q)}|\mathcal{A}^W_\infty)+2q^2\log |\mathcal{Q}|, \end{align} $$

and it follows from equation (4.4) that

$$ \begin{align*} \frac{1}{q}H_{\mu_i}(\mathcal{Q}^{(q)}|\mathcal{A}^W_\infty)&\ge \frac{1}{k_i}H_{\nu_i}(\mathcal{Q}^{(k_i)}|\mathcal{A}^W_\infty)-\frac{2q\log |\mathcal{Q}|}{k_i}\\ &\ge \frac{1}{k_i}(\log |S_i|-D\beta k_i -\log C - 2q\log|\mathcal{Q}|). \end{align*} $$

Now we can take $i\to \infty $ because the atoms Q of $\overline {\mathcal {Q}}$ and hence of $\overline {\mathcal {Q}}^{(q)}$ satisfy $\mu ^\gamma (\partial Q)=0$ . Also, the constants C and $|\mathcal {Q}|$ are independent to $k_i$ . Thus, we obtain

$$ \begin{align*} \frac{1}{q}H_{\mu^\gamma}(\overline{\mathcal{Q}}^{(q)}|\overline{\mathcal{A}^W_\infty}) \ge 1-r_{1}(m+\gamma- \dim_H \mathbf{Bad}_A(\epsilon))-D\beta. \end{align*} $$

By taking $\beta \to \overline {\mu }(\overline {Q_\infty })$ and $\gamma =\gamma _j \to 0$ , the inequality in equation (4.2) follows.