Proof. We closely follow the proof of [Reference Frettlöh and Richard24, Theorem 3.11].

Suppose the hull of $\Lambda = (P, \alpha )$ is minimal. Let $R \geq 1$ and fix $\Pi = (Q,\beta ) \in \mathcal {H}_{\Lambda }$ . Assume that $0 <\varepsilon < R$ . We further choose $R^{\prime } \geq R$ such that $B_{R^{\prime }} \cap Q = \overline {B}_R \cap Q$ . We set

$$ \begin{align*} \mathcal{U}_{R,\varepsilon} ( \Pi) = \{ \Theta \in \mathcal{H}_{\Lambda}:\, d_{B_{R^{\prime}}}(\Theta,\Pi)<\varepsilon\}. \end{align*} $$

Then, $\mathcal {U}_{R,\varepsilon } ( \Pi )$ is a neighborhood of $\Pi $ in the weak- $\ast $ -topology, cf. Proposition B.1. By minimality of $\Lambda $ , every element in $\mathcal {H}_{\Lambda }$ has a dense orbit. This gives

$$ \begin{align*} \mathcal{H}_{\Lambda} = \bigcup_{x \in G} x \, \mathcal{U}_{R,\varepsilon} ( \Pi ). \end{align*} $$

By compactness of the hull, we find finitely many $x_1, x_2, \ldots , x_{\ell }$ such that $\mathcal {H}_{\Lambda } \subseteq \bigcup _{j=1}^{\ell } x_j \, \mathcal {U}_{R,\varepsilon } ( \Pi )$ . We now choose $\varrho _1(R)> 0$ such that $x_j \in B_{\varrho _1(R)}$ for all $1 \leq j \leq \ell $ . This means that for all $h \in G$ , there is some $h^{*} \in B_{\varrho _1(R)}$ such that $h.(P,\alpha ) = h^{*}.(D,\gamma )$ and $(D,\gamma )$ is some element in $\mathcal {U}_{R,\varepsilon }(\Pi )$ . This also gives $D \cap B_{R^{\prime }} = h^{*\, -1} h\, P \cap B_{R^{\prime }}$ . Since the patches $(\delta _{(Q \cap B_{R^{\prime }}, \beta _{| Q \cap B_{R^{\prime }}})}, B_{R^{\prime }})$ and $( \delta _{(D \cap B_{R^{\prime }}, \gamma _{| D \cap B_{R^{\prime }}})}, B_{R^{\prime }})$ are $\varepsilon $ -similar, the left-invariance of the metric tells us that also $( \delta _{(Q \cap B_{R^{\prime }}, \beta _{| Q \cap B_{R^{\prime }}})}, B_{R^{\prime }})$ is $\varepsilon $ -similar to the $R^{\prime }$ -patch of P centered at $h^{-1}h^{*}$ . We will now show that $h^{-1}h^{*} B_R B_{\varepsilon } \subseteq B_{\varrho _1(R) + 2R}(h^{-1})$ . The condition $\varepsilon < R$ gives $B_R B_{\varepsilon } \subseteq B_{2R}$ . Further, we have $h^{*}B_{2R} \subseteq B_{\varrho _1(R)}B_{2R} \subseteq B_{\varrho _1(R) + 2R}$ . Hence, we obtain $h^{-1}h^{*}B_RB_{\varepsilon } \subseteq B_{\varrho _1(R) + 2R}(h^{-1})$ , as claimed. Since $\Pi $ was chosen arbitrarily, this shows that for all possible $B_R$ -patterns $[p]$ of $\Lambda $ and for each $h \in G$ , there is some R-patch of $\Lambda $ contained in $B_{\varrho _1(R) + 2R}(h)$ which is $\varepsilon $ -similar to a representative of $[p]$ . A straightforward compactness argument shows that there is a finite number of patterns $[p_k]$ such that every $B_{R}$ -patch of $\Lambda $ is $\varepsilon $ -similar to a representative of one of the $[p_k]$ . Hence, repeating the above procedure finitely many times, we can set $\varrho (R) := \max _{k} \varrho _k(R) + 2R$ and we find that $\Lambda $ is almost repetitive with almost repetitivity function $\varrho $ .

Conversely, assume by contradiction that $\Lambda = (P,\alpha )$ is almost repetitive but $\mathcal {H}_\Lambda $ is not minimal. Let $\Pi = (Q,\beta ) \in \mathcal {H}_{\Lambda }$ and $\{g.\Pi :\,g\in G\}$ is not dense in $\mathcal {H}_{\Lambda }$ . This implies $\Lambda \notin \mathcal {H}_{\Pi }$ as well as $\mathcal {H}_{\Pi } \subsetneq \mathcal {H}_{\Lambda }$ . We fix a compact neighborhood $\mathcal {V} \subseteq \mathcal {H}_{\Lambda }$ of $\Lambda $ containing the set $\{\Theta \in \mathcal {H}_{\Lambda }:\, d_{B_{1/\varepsilon }} ( \Theta ,\, \Lambda ) < \varepsilon \} $ for some small $\varepsilon> 0$ such that $\mathcal {V} \cap \mathcal {H}_{\Pi } = \emptyset $ . We now take a compact set $K \subseteq G$ with $B_{1/\varepsilon }(e) \subseteq \mathring {K}$ and define ${T_{K, \varepsilon }( \Lambda ) = \{g \in G:\, d_{\mathring {K}}(g . \Lambda , \Lambda ) < \varepsilon \}}$ .

Assuming the validity of the claim for a moment, we see

$$ \begin{align*} G.\Lambda = ( K^{\prime}T_{K, \varepsilon}(\Lambda)).\Lambda \subseteq K^{\prime}.\mathcal{V} \end{align*} $$

for some compact set $K^{\prime } \subseteq G$ . By compactness of $\mathcal {V}$ and the continuity of the G-action, we find that the set $K^{\prime }.\mathcal {V}$ is compact. Hence, we arrive at $\mathcal {H}_{\Lambda } \subseteq K^{\prime }.\mathcal {V}$ . We get that $\Pi = x.\Theta $ for some $x \in K^{\prime }$ and some $\Theta \in \mathcal {V}$ . Thus, $\Theta = x^{-1}\Pi $ contradicting the fact that ${\mathcal {V} \cap \mathcal {H}_{\Pi } = \emptyset }$ .

It remains to show the claim. For this, we proceed along the lines of [Reference Frettlöh and Richard24, proof of Lemma 3.6, implication (iii) ⇒ (i)]. To this end, find some $r \geq 1$ such that $\mathring {K} \subseteq B_r$ and set $R:= \varrho (\varepsilon , r)$ , where $\varrho $ is an almost repetitivity function of $\Lambda $ . We fix a countable cover of G of the form $G = \bigcup _{i=1}^{\infty } B_R(g_i)$ . By almost repetitivity, for all $i \in {\mathbb N}$ , we find $x_i \in G$ such that $x_i^{-1}B_r \subseteq B_{R}(g_i)$ and $d_{B_r}( x_i\Lambda , \Lambda ) = d_{x_i^{-1}B_r} ( \Lambda , x_i^{-1}\Lambda ) < \varepsilon $ . Now to show the claim, it clearly suffices to show that $T_{r,\varepsilon }:= \{x_i:\, i \in {\mathbb N}\} \subseteq T_{K, \varepsilon }(\Lambda )$ is right-relatively dense in G. For each $b \in B_r$ , we have $x_i^{-1}b \subseteq B_R(g_i)$ for all $i \in {\mathbb N}$ . This shows that ${G = \bigcup _{i=1}^{\infty } B_{2R}(x_i^{-1}b) = \bigcup _{i=1}^{\infty } x_i^{-1} B_{2R}(b)}$ . For some fixed $b \in B_r$ , we define

$$ \begin{align*} K^{\prime} := \overline{\{ g \in G:\, g^{-1}b \in B_{2R}(b)\}}. \end{align*} $$

This set is clearly compact. We will show $K^{\prime } T_{r,\varepsilon } = G$ . Take $h \in G$ . This gives $h^{-1}b \in x_i^{-1}B_{2R}(b)$ for some $i \in {\mathbb N}$ . This yields

$$ \begin{align*} h &\in \{ g \in G:\, g^{-1}b \in x_i^{-1}B_{2R}(b)\} = \{ g \in G:\, x_i g^{-1}b \in B_{2R}(b)\} \\ &= \{ y \in G:\, y^{-1}b \in B_{2R}(b)\} \cdot x_i \subseteq K^{\prime} T_{r, \varepsilon}. \end{align*} $$

Since h was chosen arbitrarily, the proof of the claim is finished.

Proof. Without loss of generality, we assume that $e\in U$ . Since $\Lambda = (P,\alpha )$ is of finite local complexity, the set

$$ \begin{align*} P(r):= \{ (x^{-1}.\Lambda)|_{B_{r}} :\, x\in P \} \end{align*} $$

is finite for each $r>0$ . In particular, if $r>0$ is such that the topological boundary $\overline {B_r}\setminus B_r$ is not intersecting $x^{-1}P$ for any $x\in P$ , then there is a $\delta>0$ such that ${x^{-1}P\cap \partial _{B_{\delta _1}}(B_{r}) = \emptyset} $ holds for all $x\in P$ , where

$$ \begin{align*} \partial_{B_{\delta_1}}(B_{r}) := \{g \in G:\, B_{\delta_1}g \cap B_{r} \neq \emptyset \, \wedge\, B_{\delta_1}g \cap (G \setminus B_{r}) \neq \emptyset\} \end{align*} $$

is the $B_{\delta _1}$ boundary of the topological boundary of $B_r$ .

Let $R>0$ be fixed. Since $\Lambda $ is K-relatively dense and of finite local complexity, we can choose an $R_1>0$ and a $\delta _1>0$ such that:

1. (a) for each $x\in G$ , there is a $y_x\in P$ satisfying $x\in B_{R_1}(y_x)$ ;

2. (b) for all $y\in P$ , we have $y^{-1}P\cap \partial _{B_{\delta _1}}(B_{R+R_1}) = \emptyset $ .

Consider an R-patch $p:=\Lambda _{B_R(x)}$ for $x\in G$ . Let $y_x\in P$ be such that $x\in B_{R_1}(y_x)$ . Now consider the patch

$$ \begin{align*} p':=(y_x^{-1}\cdot\Lambda)_{B_{R+R_1}} = y_x^{-1}\cdot (\Lambda|_{B_{R+R_1}(y_x)}). \end{align*} $$

By construction, p occurs in $p'$ . Let $0<\delta <\min \{\delta _1,\sigma ^{-1}\}$ be such that $B_\delta \subseteq U$ and

$$ \begin{align*} q_1,q_2\in P(R+R_1) \text{ with } d_{B_{R+R_1}}(\delta_{(q_1,B_{R+R_1})},\delta_{(q_2,B_{R+R_1})})\leq\delta \Longrightarrow q_1=q_2. \end{align*} $$

The latter is possible as $P(R+R_1)$ is finite.

Let $h\in G$ . Since $\Lambda $ is almost repetitive, $p'\ \delta /2$ -occurs in $B_{\varrho (\delta /2,R+R_1)}(h)$ , where $\varrho $ is an almost repetitivity function for $\Lambda $ . Thus, there is a $g\in G$ such that $gB_{R+R_1}\subseteq B_{\varrho (\delta /2,R+R_1)}(h)$ and $d_{B_{R+R_1}}( \delta _{p'}, \delta _{g^{-1}\cdot q_h}) <\delta /2$ , where $q_h:=\Lambda _{B_{R+R_1}(g)}$ . Thus,

$$ \begin{align*} &d_{B_{R+R_1}}( p', g^{-1} q_h)\\ &\quad = \inf\left\{ \varepsilon>0 :\!\! \begin{array}{c} |\delta_{y_g^{-1}\cdot \Lambda}|_{B_{R+R_1}}(B_{\varepsilon}(y)) - g^{-1}_{\ast}\delta_{\Lambda}|_{B_{R+R_1}(g)}(B_\varepsilon(y))| <\varepsilon\\ \text{ for all } y\in (y_x^{-1}P\cap B_{R+R_1})\cup (g^{-1}P\cap B_{R+R_1}) \end{array} \!\!\right\} <\delta/2 \end{align*} $$

holds. By construction, $\delta _{p'}$ has support on $\{e\}$ , namely, $\delta _{p'}(e)>\sigma ^{-1}$ . Hence, there is a unique $y_g\in P$ and $g'\in B_{\delta /2}$ such that $g=y_g g'$ using that $\delta <\sigma ^{-1}$ , P is $B_\delta $ -uniformly discrete, and $d_{B_{R+R_1}} ( \delta _{p'}, \delta _{g^{-1}.q_h})<\delta /2$ . Due to condition (b), we conclude $\delta _{y_g^{-1}\cdot \Lambda }|_{B_{R+R_1}} = \delta _{y_g^{-1}\cdot \Lambda }|_{B_{R+R_1}(g')}$ and so

$$ \begin{align*} g^{-1}\cdot q_h = g^{\prime-1} ( \delta_{y_g^{-1}\cdot\Lambda}|_{B_{}{R+R_1}(g')}, B_{R+R_1}(g')) = g^{\prime-1} ( \delta_{y_g^{-1}\cdot\Lambda}|_{B_{}{R+R_1}}, B_{R+R_1}(g')). \end{align*} $$

By construction, $(y_g^{-1}\cdot \Lambda )_{B_{R+R_1}}$ occurs in $B_{\varrho (\delta /2,R+R_1)+\delta /2}(h)$ since $g'\in B_{\delta /2}$ . Invoking that $\delta <\sigma ^{-1}$ , P is $B_\delta $ -uniformly discrete, $g'\in B_{\delta /2}$ , and $d_{B_{R+R_1}}(\delta _{p'}, \delta _{g^{-1}\cdot q_h} )<\delta /2$ , we derive

$$ \begin{align*} d_{B_{R+R_1}}\big(\delta_{p'}, \delta_{(y_g^{-1}\cdot\Lambda)_{B_{R+R_1}}}\!\big)\leq\delta. \end{align*} $$

Thus, $(y_g^{-1}\cdot \Lambda )_{B_{R+R_1}}=p'$ follows by the choice of $\delta $ and since $(y_g^{-1}\cdot \Lambda )|_{B_{R+R_1}}, p'\in P(R+R_1)$ . Since p occurs in $p'$ and $(y_g^{-1}\cdot \Lambda )_{B_{R+R_1}}$ occurs in $B_{\varrho (\delta /2,R+R_1)+\delta /2}(h)$ , we conclude that p occurs in $B_{\varrho (\delta /2,R+R_1)+\delta /2}(h)$ .

Hence, the map $\varrho ':(0,\infty )\to (0,\infty )$ defined by $\varrho '(R):=\varrho (\delta /2,R+R_1)+\delta /2$ defines a repetitivity function. Note here that in the previous considerations, $\delta $ and $R_1$ depend on R but not on h.

Proof. Building on the uniform Følner condition proven in [Reference Ornstein and Weiss41], one observes that G admits a strong Følner sequence $(S_n)$ , cf. [Reference Pogorzelski44, Lemma 2.8]. Moreover, the sequence $(S_n)$ can be turned into a strong Følner exhaustion sequence by the usual construction (cf. [Reference Greenleaf26]). By $\sigma $ -compactness of G, we can now choose an exhaustion of G by compact subsets $(K_n)$ containing the identity. Set $T_1 := S_1$ and choose $m_1$ such that $T_1 \subseteq \mathring {K}_{m_1}$ . Then choose $\ell \in {\mathbb N}$ large enough such that $m_G(\partial _{K_{m_1}}(S_{\ell })) \leq \tfrac 12 m_G(S_{\ell })$ . Thus, there is an $h \in G$ such that $K_{m_1}h \subseteq S_{\ell }$ . If we now set $T_2:= S_{\ell }h^{-1}$ , then $T_1 \subseteq \mathring {K}_{m_1} \subseteq T_2$ . One now proceeds by induction to construct the desired strong Følner exhaustion sequence $(T_m)$ .

Proof. We note that by the properties of a strong exhaustion sequence, we find for all $r> 0$ some $\ell (r) \in {\mathbb N}$ such that $\overline {B}_r \subseteq \mathring {T}_{\ell (r)}$ and for each $s \in {\mathbb N}$ , we find some $L(s) \in {\mathbb N}$ such that $T_s \subseteq B_{L(s)}$ . Moreover, by the left-invariance of the metric $d_G$ , we have ${B_l = B_l^{-1}}$ and $\overline {B}_l = \overline {B}_l^{-1}$ for all $l> 0$ . Thus, the condition $\mathcal {R}_{\Lambda }^{\mathcal {T}}(\delta , l) < \infty $ for all $\delta> 0$ and all $l \in {\mathbb N}$ implies for a given $R \geq 1$ , every $B_R$ -pattern $\delta $ -occurs in each ball of radius $L ( \mathcal {R}_{\Lambda }^{\mathcal {T}}(\delta , \ell (R)) )$ . Conversely, for $m \in {\mathbb N}$ , we have that every $\mathring {T}_m^{-1}$ -pattern must $\delta $ -occur in every ball of radius $\varrho (\delta , L(m))$ (with $\varrho $ denoting an almost repetitivity function), whence also in each $h\mathring {T}_M^{-1}$ with $M = \ell ( \varrho (\delta , L(m)))$ .

Proof of Theorem 3.1

Let $w: \mathcal {RK}(G) \times \mathcal {H}_{\Lambda } \to {\mathbb R}$ be an admissible weight function and define $w^+$ and $w^-$ by equation (3).

By Proposition 3.7, the dynamical system $G \curvearrowright X$ is minimal. However, the function w satisfies axiom (W5) and therefore also axiom (W5 $^\ast $ ) by Remark 3.12, and hence the strong form of Lemma 3.17 applies. We conclude that the limit $\nu ^{+} = \lim _{n \to \infty } w^{+}(T_m,\Pi )$ exists uniformly over $\Pi $ and is independent of the choices of $\Pi \in \mathcal {H}_{\Lambda }$ and the strong Følner exhaustion sequence.

Moreover, we define $\nu ^{-}(\Pi ) := \liminf _{n \to \infty } w^{-}(T_n, \Pi )$ for $\Pi \in \mathcal {H}_{\Lambda }$ . It follows from Lemma A.1 in Appendix A that $\nu ^{-} = \nu ^{-}(\Pi )$ is independent of the choice of $\Pi \in \mathcal {H}_{\Lambda }$ and that the limit inferior is realized by one subsequence $(n_l)$ of the integers simultaneously for all $\Pi $ . In light of that, it suffices to prove that

$$ \begin{align*} \nu^{+} = \lim_{n \to \infty} w^{+}(T_m,\Lambda) \leq \liminf_{n \to \infty} w^{-}(T_m, \Lambda) = \nu^{-}. \end{align*} $$

To this end, we fix a subsequence $(m_l)$ of the integers such that $\lim _{l \to \infty } w^{-}(T_{m_l}, \Lambda ) = \nu ^{-}$ . For the sake of simpler notation, we set $T_l^{\prime } := T_{m_l}$ for all $l \in {\mathbb N}$ . We assume by contradiction that $\nu ^{-} < \nu ^{+}$ . Let $0 < \varsigma < ({\nu ^{+} - \nu ^{-}})/{4}$ and choose $L_{\varsigma } \in {\mathbb N}$ such that $|\nu ^{-} - w^{-}(T^{\prime }_l,\, \Lambda )| < \varsigma $ for all $l \geq L_{\varsigma }$ . Then, for $l \geq L_{\varsigma }$ , one finds $g_l \in G$ such that ${w(T_l^{\prime } g_l, \Lambda )}/{m_G(T_l^{\prime })} \leq w^{-}(T_l^{\prime },\Lambda ) + \varsigma $ . Combined with the previous estimate and with the choice of $\varsigma $ , we conclude from $4 \varsigma < \nu ^{+} - \nu ^{-}$ that

(4) $$ \begin{align} \frac{w(T_l^{\prime}g_l,\, \Lambda )}{m_G(T_l^{\prime})} \leq \nu^{-} + 2\varsigma \leq \nu^{+} - 2 \varsigma \end{align} $$

for all $l \geq L_{\varsigma }$ . Writing $\Lambda = (P,\alpha )$ , we define the $S^{-1}_l$ -patches $q_l:= \Lambda _{S_l^{-1}}$ with inverse support $S_l := T_l^{\prime } g_l = T_{m_l} g_l$ . Using axiom (W5), we find some $0< \delta < 1$ and an $m_0 \in {\mathbb N}$ such that for all $l \geq m_0$ , the condition $d_{S^{-1}_{l}}(\Pi , \Phi ) \leq \delta $ implies $|w(S_{l}, \Pi ) - w(S_{l},\Phi )| < \varsigma m_G(S_{l})$ for $\Pi , \Phi \in \mathcal {H}_{\Lambda }$ . We fix this $\delta $ and take an arbitrary $h \in G$ . Since $\Lambda $ is almost tempered repetitive with respect to $\mathcal {T} = (T_m)$ , the pattern arising from the patches $q_l$ must $\delta $ -occur in $h^{-1}T^{-1}_{\mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m_l)}$ , where $\mathcal {R}^{\mathcal {T}}_{\Lambda }$ is the repetitivity index for $\Lambda $ with respect to $\mathcal {T}$ . This means that there must be some element $\widetilde {g}_{l,h} \in G$ such that $S_l \widetilde {g}_{l,h} \subseteq T_{\mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m_l)}h$ and $d_{S_l^{-1}}( \Lambda ,\, \widetilde {g}_{l,h}\cdot \Lambda ) \leq \delta $ . This yields

$$ \begin{align*} | w(S_l, \Lambda) - w(S_l,\, \widetilde{g}_{l,h} \Lambda) | \leq \varsigma m_G(S_l) \end{align*} $$

for $l \geq m_0$ . Furthermore, increasing $m_0$ if necessary and invoking axiom (W4), we find that

(5) $$ \begin{align} | w(S_l, \Lambda) - w(S_l\widetilde{g}_{l,h},\, \Lambda) | \leq \frac{11}{10} \varsigma m_G(S_l) \end{align} $$

for $l \geq m_0$ . We point out that $m_0$ only depends on parameters given by the weight function w and on $\varsigma $ but not on objects constructed in the proof. Given h as above, define ${A_{l,h}:= T_{\mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m_l)}h \setminus S_l \widetilde {g}_{l,h}} $ . We will distinguish the following two cases.

1. (A) There is some $0 < \kappa _{\Lambda }(\delta ) < 1$ such that ${m_G(T_{m_l})}/{m_G(T_{\mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m_l)})} \leq \kappa _{\Lambda }(\delta )$ for all $l \in {\mathbb N}$ .

2. (B) There is some subsequence $(T_{m_{l_k}})_k$ such that $\lim _{k \to \infty } ({m_G(T_{m_{l_k}})}/ {m_G(T_{\mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m_{l_k})})}) = 1$ .

We will cover both of these cases separately and, in both cases, we will obtain $\nu ^{+} \leq \nu ^{+} - c(\varsigma )$ for some $c(\varsigma )> 0$ , which is clearly a contradiction. We start with case (A).

Case (A). Invoking the general relations

$$ \begin{align*} \partial_L(C \setminus D) \subseteq \partial_L(C) \cup \partial_L(D), \quad \partial_L(C)g = \partial_L(Cg) \end{align*} $$

for sets $C,D,L \in \mathcal {RK}(G)$ and $g \in G$ , we find

(6) $$ \begin{align} \frac{m_G(\partial_K(A_{l,h}))}{m_G(A_{l,h})}&\leq \frac{m_G(\partial_K(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)}h)) + m_G(\partial_K(S_l {\tilde{g}}_{l,h}))}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})\, \bigg(1-\frac{m_G(T_{m_l})}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})}\bigg)} \nonumber \\ &\leq \frac{1}{1-\kappa_{\Lambda}(\delta)}\bigg(\frac{m_G(\partial_K(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)}))}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} + \frac{m_G(\partial_K(T^{\prime}_l))}{m_G(T^{\prime}_l)}\bigg) \end{align} $$

for all relatively compact sets $K \subseteq G$ . This shows that the left-hand side of the above inequality tends to $0$ uniformly in h if l tends to infinity. In particular, the sequences $(A_{l,h})_l$ are strong Følner sequences for all $h \in G$ . Now, let $0 < \varepsilon < 1/10$ and choose some ${n_{\varepsilon } \geq \max \{m_0, L_{\varsigma }\}}$ such that

$$ \begin{align*} | \nu^{+} - w^{+}(T_n, \Lambda) | < \varepsilon \end{align*} $$

for all $n \geq n_{\varepsilon }$ . We also make sure that $n_{\varepsilon }$ is chosen large enough such that

(7) $$ \begin{align} \frac{m_G(\partial_L(T_n))}{m_G(T_n)} < \varepsilon \quad \text{for all } n \geq n_{\varepsilon}, \end{align} $$

where $L \in \{ J,B,I\}$ and $J,B,I$ are the compact subsets of G determined by w as of axioms (W2), (W3), and (W4). We fix $n \geq n_{\varepsilon }$ and we apply Lemma 3.16 to find prototile sets

$$ \begin{align*} \{e \} \subseteq T_n \subseteq S_1^{\varepsilon} \subseteq \cdots \subseteq S_{N(\varepsilon)}^{\varepsilon}, \quad S_i^{\varepsilon} \in \{ T_k: k \geq \max\{ i,n \} \}, \end{align*} $$

as well as $M \in {\mathbb N}$ with $M \geq n$ such that for all $l \geq M$ , we have $m_l \geq n$ and for all $h \in G$ , we find a finite set $C_i^{l,h} \subseteq A_{l,h}$ for each $1 \leq i \leq N(\varepsilon )$ such that the properties (T1)–(T4) from Definition 3.13 are satisfied and at the same time, we obtain

$$ \begin{align*} \frac{w(A_{l,h},\, \Lambda )}{m_G(A_{l,h})} \leq \sum_{i=1}^{N(\varepsilon)} \sum_{c \in C_i^{l,h}} \frac{ w( S_i^{\varepsilon}c, \Lambda )}{m_G(A_{l,h})} \, +\, t_1 \cdot \varepsilon, \end{align*} $$

where the constant $t_1 \geq 0$ only depends on the parameters $\eta , \theta , \vartheta $ given by the definition of the weight function w. We emphasize at this point that the fact that the parameters $n, M$ (and hence also l) can be chosen independently of h is justified by the observation that the validity of Lemma 3.16 depends on the invariance properties of $A_{l,h}$ with respect to the prototile sets, combined with the fact that $\lim _l m_G(\partial _{K}(A_{l,h}))/m_G(A_{l,h}) = 0$ uniformly over h for all $K \in \mathcal {RK}(G)$ .

Exploiting the sub-additivity property (W3) of w and recalling $A_{l,h}:= T_{\mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m_l)}h \setminus S_l \widetilde {g}_{l,h}$ , we find with the estimates in equations (6) and (7), as well as with $l \geq M$ , that

$$ \begin{align*} \frac{w( T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)}h, \Lambda )}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})}\!\leq\! \frac{m_G(S_l)}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} \cdot \frac{w( S_l \widetilde{g}_{l,h}, \Lambda )}{m_G(S_l)}\! +\! \frac{m_G(A_{l,h})}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} \cdot \frac{w( A_{l,h}, \Lambda )}{m_G(A_{l,h})}\! +\! t_2 \cdot \varepsilon \end{align*} $$

for some constant $t_2 \geq 0$ which only depends on the sub-additivity parameter $ \theta $ of w and on $\delta $ . Recall that $l \geq M \geq m_0$ . Thus, we can combine the latter inequality with the inequality (5) and $m_G(S_l)=m_G(T^{\prime }_l)$ to obtain

$$ \begin{align*} \frac{w( T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)}h, \Lambda )}{m_G( T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} &\leq \frac{m_G(T^{\prime}_l)}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} \cdot \bigg( \frac{w(S_l, \Lambda)}{m_G(T_l^{\prime})} + \frac{11}{10} \varsigma \bigg)\\ &\quad + \frac{m_G(A_{l,h})}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} \cdot \frac{w( A_{l,h}, \Lambda )}{m_G(A_{l,h})} + t_2 \cdot \varepsilon. \end{align*} $$

Invoking the sub-additivity induced by the above $\varepsilon $ -quasi tilings and using the basic inequality $m_G(A_{l,h}) \leq m_G(T_{\mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m_l)})$ , the latter inequality transforms to

$$ \begin{align*} \frac{w(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)}h, \Lambda )}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} &\leq \frac{m_G(T^{\prime}_l)}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} \cdot \bigg(\frac{w(S_l, \Lambda)}{m_G(T_l^{\prime})} + \frac{11}{10} \varsigma \bigg) \\ & \quad + \frac{m_G(A_{l,h})}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} \cdot \sum_{i=1}^{N(\varepsilon)} \sum_{c \in C_i^{l,h}} \frac{m_G(S_i^{\varepsilon}c)}{m_G(A_{l,h})} \frac{ w( S_i^{\varepsilon}c, \Lambda )}{m_G(S_i^{\varepsilon}c)} + (t_1 + t_2) \cdot \varepsilon. \end{align*} $$

Note that the tiling properties (T1)–(T4) from Definition 3.13 yield

$$ \begin{align*} (1- 2 \varepsilon) \leq \sum_{i=1}^{N(\varepsilon)} \sum_{c \in C_i^{l,h}} \frac{m_G(S_i^{\varepsilon}c)}{m_G(A_{l,h})} \leq \frac{1}{1-\varepsilon}. \end{align*} $$

Furthermore, inequality (4) leads to

$$ \begin{align*} \frac{m_G(T^{\prime}_l)}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} \cdot \bigg( \frac{w(S_l, \Lambda)}{m_G(T_l^{\prime})} + \frac{11}{10} \varsigma \bigg) \leq \frac{m_G(T^{\prime}_l)}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} \cdot \bigg( \nu^{+} - \frac{9}{10} \varsigma \bigg) \end{align*} $$

for $l \geq M \geq L_{\varsigma }$ . Recall that we have chosen $n \geq n_{\varepsilon }$ and $S_i^{\varepsilon } \in \{ T_k: k \geq \max \{ i,n \} \}$ . Thus,

$$ \begin{align*} \frac{w(S_i^{\varepsilon}c,\, \Lambda )}{m_G(S_i^{\varepsilon}c)} \leq w^+(S_i^{\varepsilon},\Lambda) \leq \nu^{+} + \varepsilon \end{align*} $$

holds for all $1 \leq i \leq N(\varepsilon )$ , all $l \geq M$ , each $h \in G$ , and each $c \in C_i^{l,h}$ .

Since $\Lambda $ is almost tempered repetitive, the repetitivity portion $\zeta :[0,1) \to [0,1]$ of $\Lambda $ with respect to $\mathcal {T} = (T_m)$ is positive everywhere and, in particular, we have

$$ \begin{align*} \inf_{m \in {\mathbb N}} \frac{m_G(T_m)}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m)})} = \zeta(\delta)> 0. \end{align*} $$

We set $a_{\varepsilon } := (1-2\varepsilon )$ if $\nu ^+ \leq 0$ , and $a_{\varepsilon } := (1-\varepsilon )^{-1}$ if $\nu ^{+}> 0$ and $t_3 := t_1 + t_2$ . Since $a_{\varepsilon } \nu ^{+} - \nu ^{+}\geq 0$ , we finally derive from the above estimates

$$ \begin{align*} \frac{w( T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)}h, \Lambda )}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} &\leq \frac{m_G(T^{\prime}_l)}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} ( \nu^{+} - \frac{9}{10} \varsigma )\\ &\quad + \frac{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)}) - m_G(T_l^{\prime})}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} \bigg( a_{\varepsilon} \nu^{+} + \frac{\varepsilon}{1-\varepsilon} \bigg) + t_3 \varepsilon \\ &\leq a_{\varepsilon}\nu^{+} - \frac{m_G(T^{\prime}_l)}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})}\bigg(a_{\varepsilon} \nu^{+} - \bigg( \nu^{+} - \frac{9}{10}\varsigma \bigg) \bigg) + (t_3 + 2)\varepsilon \\ &\leq a_{\varepsilon}\nu^{+} - \zeta(\delta) \cdot \bigg( a_{\varepsilon} \nu^{+} - \nu^{+} + \frac{9}{10}\varsigma \bigg) + (t_3 + 2)\varepsilon \end{align*} $$

for all $l \geq M$ and all $h \in G$ , where the tempered repetitivity was used in the last step. Thus, taking the supremum over all $h \in G$ on the left-hand side of the latter inequality and sending $l \to \infty $ , we find that

$$ \begin{align*} \nu^{+} \leq a_{\varepsilon}\nu^{+} - \zeta(\delta) \cdot \bigg( a_{\varepsilon} \nu^{+} - \nu^{+} + \frac{9}{10}\varsigma \bigg) + (t_3 + 2)\varepsilon. \end{align*} $$

By sending $\varepsilon \to 0$ , we obtain $\nu ^{+} \leq \nu ^{+} - 9/10 \zeta (\delta ) \varsigma $ and since both $\varsigma $ and $\zeta (\delta )$ are positive numbers, we have arrived at a contradiction. Hence, we conclude $\nu ^{+} = \nu ^{-}$ in case (A).

Case (B) is much easier to handle. We assume with no loss of generality that

$$ \begin{align*} \lim_{l \to \infty} \frac{m_G(T_{m_l})}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})}~=~1. \end{align*} $$

Then, there is some $M_{\varsigma } \in {\mathbb N}$ such that $\sup _{h \in G} m_G(A_{l,h}) (m_G(T_{\mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m_l)}))^{-1} < {\varsigma }/{5\eta }$ for all $l \geq M_{\varsigma }$ , where $\eta> 0$ is the boundedness constant of w as of axiom (W2). Increasing $M_{\varsigma }$ if necessary, we can assume that $\sup _{h \in G} m_G( \partial _L(A_{l,h})) (m_G(T_{\mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m_l)}))^{-1} < ({1}/{10})\varsigma (\max \{1,\eta ,\theta \})^{-1}$ and ${m_G(\partial _L(S_l))}/{m_G(S_l)} < {1}/{10} \varsigma (\max \{1,\theta \})^{-1}$ for all ${L \in \{ J,B,I \}}$ and each $l \geq M_{\varsigma }$ , where $J,B,I \subseteq G$ are compact and $\theta \geq 0$ depend on the weight function $w$ . We now fix $l \geq M_{\varsigma }$ . By the choice of $\widetilde {g}_{l,h}$ , inequalities (4) and (5) together with $m_G(S_l)=m_G(T^{\prime }_l)$ imply

$$ \begin{align*} \frac{w( S_l \widetilde{g}_{l,h} ,\, \Lambda )}{m_G(S_l)} \leq \frac{w(S_l,\, \Lambda )}{m_G(S_l)} + \frac{11}{10} \varsigma \leq \nu^{+} - \frac{9}{10}\varsigma \end{align*} $$

for all $l \geq M_{\varsigma }$ and each $h \in G$ . Then, $m_G(\partial _L(C)) = m_G(\partial _L(Cg))$ , the sub-additivity property (W3) with constant $\theta \geq 0$ (given by $w$ ), and the imposed invariance conditions imply

$$ \begin{align*} \frac{w( T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)}h ,\, \Lambda)}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} &\leq \frac{w(S_l \widetilde{g}_{l,h},\, \Lambda )}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} + \frac{w(A_{l,h},\, \Lambda )}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})}\\ &\quad + \theta \cdot \frac{m_G(\partial_{B}(S_l))}{m_G(S_l)} + \theta \cdot \frac{m_G(\partial_B(A_{l,h}))}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} \\ &\leq \frac{w(S_l \widetilde{g}_{l,h},\, \Lambda )}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} + \frac{w(A_{l,h},\, \Lambda )}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} + \frac{1}{5}\varsigma \\ &\leq \frac{w(A_{l,h},\, \Lambda )}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} + \nu^{+} - \frac{9}{10}\varsigma + \frac{1}{5}\varsigma \end{align*} $$

for all $l \geq M_{\varsigma }$ and all $h \in G$ . Invoking Remark 3.10 with constant $\eta> 0$ applied to $w(A_{l,h},\, \Lambda )$ , then the invariance condition on $A_{l,h}$ and the implications explained at the beginning of the proof of Case (B) yield

$$ \begin{align*} \frac{w( T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)}h ,\, \Lambda)}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} &\leq \eta \cdot \frac{m_G(A_{l,h})}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} + \eta \cdot \frac{m_G(\partial_J(A_{l,h}))}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m_l)})} + \nu^{+} - \frac{9}{10}\varsigma + \frac{1}{5}\varsigma \\ &\leq \frac{1}{5} \varsigma + \frac{1}{10} \varsigma + \nu^{+} - \frac{9}{10} \varsigma + \frac{1}{5} \varsigma = \nu^{+} - \frac{2}{5} \varsigma \end{align*} $$

for all $l \geq M_{\varsigma }$ and all $h \in G$ . Thus, by taking first the supremum over all $h \in G$ and then sending $l \to \infty $ , we arrive at

$$ \begin{align*} \nu^{+} \leq \nu^{+} - \frac{2}{5}\varsigma, \end{align*} $$

which is a contradiction since $\varsigma> 0$ .

Proof. We have to check axioms (W1)–(W4): axiom (W1) holds by the convention of the empty integral being equal to $0$ . Furthermore, axiom (W2) holds with $J = \emptyset $ and ${\eta = \| f\|_{\infty }}$ . The additivity of the integral yields axiom (W3) (which holds as an equality) with $B = \emptyset $ and $\theta = 0$ . As for axiom (W4), a straightforward integral substitution allows us to choose $\vartheta = 0$ and $I = \emptyset $ .

Proof. We fix $f\in \mathcal {C}(\mathcal {H}_\Lambda )$ , a strong Følner exhaustion sequence $(T_m)$ , and some $\varepsilon> 0$ throughout. We then have to show the existence of some $\delta>0$ and an $m_0\in {\mathbb N}$ such that for all $m\geq m_0$ , we have the implication

(9) $$ \begin{align} d_{T_m^{-1}}(\Pi,\Theta)\leq \delta \Longrightarrow |w_f(T_m, \Pi) -w_f(T_m, \Theta)|\leq \varepsilon \cdot m_G(T_m)\,. \end{align} $$

Since f is uniformly continuous on the compact space $\mathcal {H}_{\Lambda }$ , there is a $0<\delta <c_\ast $ such that the condition $d_{\ast }(\Pi ,\Theta ) \leq \delta $ implies $|f(\Pi ) - f(\Theta )| \leq {\varepsilon }/{3}$ . Since $(T_m)$ is a strong Følner exhaustion sequence and $B:=B_{1/\delta }\in \mathcal {RK}(G)$ , we find some $m_0 \in {\mathbb N}$ such that ${m_G(\partial _B(T_m))}/{m_G(T_m)}\leq {\varepsilon }/{3\|f\|_\infty }$ and $B \subseteq T_m$ for all $m\geq m_0$ . It follows that for all $\Theta \in \mathcal {H}_{\Lambda }$ ,

$$ \begin{align*} &\bigg|\!\int_{T_m} f(g.\Theta)\, dg-\int_{T_m\setminus\partial_B(T_m)} f(g.\Theta)\, dg\bigg| \leq \int_{\partial_B(T_m)} |f(g.\Theta)|\, dg\\ &\quad \leq \|f\|_\infty m_G(\partial_B(T_m)) \leq \frac{\varepsilon}{3} m_G(T_m). \end{align*} $$

Note that since $e \in B$ and $B=B^{-1}$ , we deduce $Bg \subseteq T_m$ for $g \in T_m \setminus \partial _B(T_m)$ . Let $\Theta ,\Pi \in \mathcal {H}_\Lambda $ such that $d_{T^{-1}_m}(\Pi ,\Theta )\leq \delta $ . Thus, we conclude $d_{g^{-1}B^{-1}}(\Pi ,\Theta )\leq \delta $ , and hence $d_{B}(g.\Pi , g.\Theta )=d_{B^{-1}}(g.\Pi , g.\Theta )\leq \delta $ for all $g\in T_m\setminus \partial _B(T_m)$ . Since $\delta <c_\ast $ , $d_{\ast }(g.\Pi ,g.\Theta )<\delta $ follows by the definition of $d_\ast $ for all $g\in T_m\setminus \partial _B(T_m)$ . Thus, $g\in T_m\setminus \partial _B(T_m)$ yields $|f(g.\Pi )-f(g.\Theta )|\leq {\varepsilon }/{3}$ and we arrive at

$$ \begin{align*} \bigg|\!\int_{T_m\setminus\partial_B(T_m)} f(g.\Pi)\, dg-\int_{T_m\setminus\partial_B(T_m)} f(g.\Theta)\, dg\bigg| &\leq \int_{T_m\setminus\partial_B(T_m)} |f(g.\Pi)-f(g.\Theta)| \, dg \\ &\leq \frac{\varepsilon}{3}\, m_G(T_m\setminus\partial_B(T_m)) \leq \frac{\varepsilon}{3}\, m_G(T_m){.} \end{align*} $$

Now, equation (9) follows from the triangle inequality.

Proof of Theorem 1.3

Suppose that $\Lambda $ is almost tempered repetitive with respect to the strong Følner exhaustion sequence $(T_m)$ . Let $f \in C(\mathcal {H}_{\Lambda })$ . Since $w_f$ , as considered above, is an admissible weight function, we deduce from Theorem 3.1 that there is a number $I(f) \in {\mathbb R}$ such that

$$ \begin{align*} I(f) = \lim_{n \to \infty} \frac{w_f(T_m, \Pi)}{m_G(T_m)} = \lim_{m \to \infty} \frac{1}{m_G(T_m)} \int_{T_m} f(g.\Pi)\, dg \end{align*} $$

for all $\Pi \in \mathcal {H}_{\Lambda }$ . We claim that for an arbitrary G-invariant Borel probability measure $\nu $ on $\mathcal {H}_{\Lambda }$ , we get $I(f) = \int _{\mathcal {H}_{\Lambda }} f\, d\nu $ . Indeed, combining the latter limit relation with the dominated convergence theorem, Fubini’s theorem, and the G-invariance of $\nu $ , we arrive at

$$ \begin{align*} I(f) &= \int_{\mathcal{H}_{\Lambda}} \lim_{m \to \infty} \frac{1}{m_G(T_m)} \int_{T_m} f(g.\Pi)\, dg \, d\nu(\Pi)\\ &= \lim_{m \to \infty} \frac{1}{m_G(T_m)} \int_{T_m} \int_{\mathcal{H}_{\Lambda}} f(g.\Pi)\, d\nu(\Pi)\, dg \\ &= \lim_{m \to \infty} \frac{1}{m_G(T_m)} \int_{T_m} \int_{\mathcal{H}_{\Lambda}} f(\Pi)\, d\nu(\Pi) \, dg = \int_{\mathcal{H}_{\Lambda}} f(\Pi)\, d\nu(\Pi). \end{align*} $$

This shows that for any two G-invariant Borel probability measures $\nu _1$ and $\nu _2$ , we obtain

$$ \begin{align*} \int_{\mathcal{H}_{\Lambda}} f(\Pi)\, d\nu_1(\Pi) = I(f) = \int_{\mathcal{H}_{\Lambda}} f(\Pi)\, d\nu_2(\Pi) \end{align*} $$

for all $f \in C(\mathcal {H}_{\Lambda })$ . This, in turn, is equivalent to $\nu _1 = \nu _2$ . Therefore, the hull of $\Lambda $ is uniquely ergodic. It is also minimal by Proposition 3.7.

Proof. To deal with both cases simultaneously, we set $r_0 :=0$ in Case (a). Note first that in either case, the sets $T_n$ are compact by continuity, respectively properness of d (whereas closed balls need not be topologically closed, let alone compact, in Case (b)). Moreover, we have $T_n \subseteq \mathring {T}_{n+1}$ by equations (12) and (11), respectively, and hence $(T_n)$ is a strong exhaustion sequence in either case.

Moreover, by the previous remarks, we can find for every compact subset $K \subseteq G$ some $r_K> 0$ with $K \subseteq B_{r_K}$ . This yields

$$ \begin{align*} \partial_KT_n = K^{-1} T_n \cap K^{-1}( G \setminus T_n ) \subseteq B_{r_K}T_n \cap B_{r_K}( G \setminus T_n ). \end{align*} $$

We claim that if $x \in B_{r_n-r_K}$ , then $x \not \in \partial _K T_n$ . Indeed, otherwise, we would find $y \in B_{r_K}$ and $z \in G \setminus T_n$ such that $x = yz$ and hence $z = y^{-1}x \in B_{r_n} \subseteq T_n$ . However, we deduce that in Case (a), we have $\partial _KT_n \subseteq B_{r_K} \overline {B}_{r_n} \subseteq \overline {B}_{r_K + r_n} \subseteq B_{r_n + r_K + 1}$ . In Case (b), we can use equation (11) to similarly deduce

$$ \begin{align*} \partial_KT_n \subseteq B_{r_K} \overline{B_{r_n}} \subseteq \overline{B_{r_K + r_n}} \subseteq B_{r_K + r_n + r_0}. \end{align*} $$

Thus, in either case, we have

$$ \begin{align*} \partial_KT_n \subseteq B_{r_n + r_K + r_0 + 1} \setminus B_{r_n -r_K}. \end{align*} $$

Using equation (13), we obtain

$$ \begin{align*} \limsup_{n \to \infty} \frac{m_G(\partial_KT_n)}{m_G(T_n)} &\leq \limsup_{n \to \infty} \frac{m_G(B_{r_n + r_K + r_0 + 1} \setminus B_{r_n -r_K})}{m_G({B}_{r_n})} \\ &\leq \lim_{n \to \infty} \frac{m_G(B_{r_n+(r_K+r_0+1)})}{m_G(B_{r_n})}- \lim_{n \to \infty} \frac{m_G(B_{r_n-r_K})}{m_G(B_{(r_n-r_K) + r_K})} = 1 - 1 = 0. \end{align*} $$

This finishes the proof.

Proof. Assume that G has exact polynomial growth with respect to d with parameters say $C> 0$ and $q \geq 0$ , and let $(t_m)$ be any strictly increasing sequence such that $t_m \nearrow \infty $ as $m \to \infty $ . We recall from equation (13) that for all $c> 0$ , there exists a constant $\gamma> 1$ such that for all $m \in {\mathbb N}$ ,

$$ \begin{align*} m_G(B_{ct_m + r_0}) = \frac{ m_G(B_{ct_m + r_0})}{m_G(B_{ct_m})} \cdot m_G(B_{ct_m}) \leq \gamma m_G(B_{ct_m}). \end{align*} $$

Then, for every $c^{\prime }> 0$ , we find a constant $\beta \geq 1$ such that for all $m \in {\mathbb N}$ and $c \in \{1, c'\}$ ,

$$ \begin{align*} \beta^{-1} C (ct_{m})^{q} \leq m_G(B_{ct_m}) \leq m_G(\overline{B_{c t_{m}}}) \leq m_G(B_{ct_m + r_0})\leq \gamma m_G(B_{ct_m}) \leq \beta C (ct_{m})^q.\end{align*} $$

The same argument also holds for $\overline {B}_{ct_m}$ instead of $\overline {B_{c t_{m}}}$ (one can set $r_0 = 1$ in this situation).

(i) $\Longrightarrow $ (ii). Assume that $\Lambda $ is almost linearly repetitive with almost repetitivity function $\varrho (\delta ,R) = c_{\Lambda }(\delta ) \cdot R$ (where $c_{\Lambda } = c_{\Lambda }(0)$ in the FLC case) and let $\mathcal T = (T_m)$ , where ${T_m:= \overline {B_{r_m}}}$ . We set $a:= \inf _m (r_{m+1} - r_m)> 0$ , as well as $A := \sup _m (r_{m+1} - r_m) < \infty $ . Then, by the given almost linear repetitivity, we get $\mathcal {R}_{\Lambda }^{\mathcal {T}}(\delta , m) \leq a^{-1} c_{\Lambda }(\delta ) r_{m}$ for all $m \in {\mathbb N}$ . Further, we have that $T_{\mathcal {R}_{\Lambda }^{\mathcal {T}}(\delta , m)} \subseteq \overline {B_{A \mathcal {R}_{\Lambda }^{\mathcal {T}}(\delta , m)}} $ for all $m \in {\mathbb N}$ . In line with the remark made at the beginning of the proof (applied with $t_m = r_m$ and with $c^{\prime } = A a^{-1} c_{\Lambda }(\delta )$ ), we find $\beta = \beta (\delta ) \geq 1$ such that

$$ \begin{align*} \inf_{m \in {\mathbb N}} \frac{m_G(T_m)}{m_G(T_{\mathcal{R}_{\Lambda}^{\mathcal{T}}(\delta, m)})} \geq \inf_{m \in {\mathbb N}} \frac{\beta(\delta)^{-1} \cdot C r_m^{q} }{m_G(\overline{B_{A a^{-1} c_{\Lambda}(\delta)\cdot r_m}})} \geq \beta(\delta)^{-2} (Aa^{-1})^{-q} c_{\Lambda}(\delta)^{-q}> 0. \end{align*} $$

Hence, the repetitivity portion of $\Lambda $ is bounded from below by $\zeta (\delta ) \geq \beta (\delta )^{-2} \cdot (Aa^{-1})^{-q} c_{\Lambda }(\delta )^{-q}$ , which is positive. This shows that $\Lambda $ is almost tempered repetitive with respect to $\mathcal {T}$ . In the continuous case, the same argument works with $T_m := \overline {B}_{r_m}$ instead of $T_m:= \overline {B_{r_m}}$ .

(ii) $\Longrightarrow $ (iii). This is obvious since the sequence $(r_n)$ with $r_n = r_0n$ (or $r_n = n$ in the continuous case) for all $n \in {\mathbb N}$ satisfies the growth condition given in assertion (ii).

(iii) $\Longrightarrow $ (i). Since the statements are invariant under rescaling the metric, we may assume that $r_0 = 1$ to simplify notation. Assume that $\Lambda $ is almost tempered repetitive with respect to $\mathcal {T} = (\overline {B_n}) $ . Now let $R \geq 1$ and $m := \lceil R \rceil + 1$ such that $m \in {\mathbb N}$ with $m \geq 2$ and $m-2 < R \leq m - 1$ . By assumption, there is a function $\zeta :[0,1) \to [0,1]$ with $\zeta (\delta )>0$ for all $\delta> 0$ (and in the FLC case, $\zeta (0)> 0$ ) such that

$$ \begin{align*} 0 < \zeta(\delta) \leq \inf_{m \in {\mathbb N}} \frac{m_G(T_m)}{m_G(T_{\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m)})}. \end{align*} $$

Using the considerations made at the beginning of the proof (applied with $t_m =m$ and $c^{\prime } = 1$ ), we obtain $\beta (\delta ) \geq 1$ such that for all $m \in {\mathbb N}$ ,

$$ \begin{align*} {\mathcal{R}^{\mathcal{T}}_{\Lambda}(\delta, m)} \leq \bigg(\frac{\beta(\delta)^2}{\zeta(\delta)} \bigg)^{1/q} \cdot m. \end{align*} $$

Note that $m = (m-2) + 2 \leq R + 2R = 3R$ . By the first inclusion given in equation (11) and $R+1 \leq m$ , we get $B_R \subseteq \mathring {\overline {B_{R+1}}} \subseteq \mathring {\overline {B_m}}$ . It follows from the definition of the repetitivity index that every $B_R$ -pattern $\delta $ -occurs in $h \mathring {T}^{-1}_{\mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m)} = \mathring {\overline {B_{{\mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m)}}(h)}} \subseteq B_{ \mathcal {R}^{\mathcal {T}}_{\Lambda }(\delta , m) + 1}(h) $ for all $h \in G$ , where the latter inclusion follows from the second inclusion in equation (11). We can now choose $\kappa (\delta )> 0$ large enough such that $( \beta (\delta )^2/\zeta (\delta ))^{1/q} 3R + 1 \leq \kappa (\delta )R$ for all $R \geq 1$ , and then $\varrho (\delta , R):= \kappa (\delta ) R$ is an almost repetitivity function for $\Lambda $ . The special case where d is a continuous metric follows along similar lines, using the inclusions given in equation (12).

Proof. The unique ergodicity follows from combining Proposition 4.7 with Theorem 1.3. Furthermore, $\mathcal {H}_\Lambda $ is minimal by Proposition 2.4 since $\Lambda $ is almost (linearly) repetitive.

Proof of Theorem 1.1

In view of Theorem 4.9 and the fact that linear repetitivity implies almost linear repetitivity, we only have to show that every homogeneous metric d is adapted and that G has exact polynomial growth with respect to d. The former follows from [Reference Fischer and Ruzhansky21, Proposition 3.1.37] and the latter follows from Proposition 4.11.

Proof. (ii) ${\implies}$ (i). Since the balls $B_n$ exhaust G and G is compactly generated, it is generated by some ball $B_N$ ; rescaling the metric by a positive constant if necessary, we may thus assume that G is generated by $B_1$ . Now, let $q>0$ be chosen such that d-balls satisfy the condition $\lim _{t \to \infty } ({m_G(B_t)}/{C t^{q}}) = 1$ and pick an integer $k> q$ . Then, we can find $C'> 0$ such that $m_G(B_t) < C' t^{k}$ for all $t\geq 1$ and hence

$$ \begin{align*} m_G(B_1^n) \leq m_G(B_n) < C' n^k, \end{align*} $$

which shows that G has polynomial growth.

(i) ${\implies}$ (ii). Let $\Omega $ be a compact generating set of G. It then follows from Theorem 4.12 that the word metric $d_\Omega $ with respect to $\Omega $ has exact polynomial growth, and this metric is obviously adapted.

Proof. Let $x \in \overline {B}_n$ with $\alpha := \|x\|$ and $m := \lceil \alpha \rceil - 1 \leq n-1$ . Choose a geodesic $\gamma $ in G with $\gamma (0) = e$ and $\gamma (\alpha ) = x$ , and for $k = 1, \ldots , n$ , define $y_k := \gamma (k-1)^{-1}\gamma (k)$ . Then, by the left-invariance of the metric d and the fact that $d(e, y_k) = d(\gamma (k-1),\gamma (k)) = 1$ , we get

$$ \begin{align*}x = y_1 \cdots y_m (\gamma(m)^{-1}x) \in \overline{B}_1^{m+1} \subseteq \overline{B}_1^n,\end{align*} $$

and hence $ \overline {B}_n \subseteq \overline {B}_1^n$ . The converse inclusion is immediate from left-invariance and the triangle inequality.

Proof of Theorem 4.14

We apply equation (15) to $\Omega := \overline {B}_1$ and use that $m_G(\overline {B}_r)= m_G(B_r)$ for all $r> 0$ . For $q=d(G)$ and $c=c(\Omega )$ as in Theorem 4.12, we obtain with Lemma 4.15 that

$$ \begin{align*} \lim_{n \to \infty} \frac{m_G(B_n)}{c \cdot n^q} = \lim_{n \to \infty} \frac{m_G(\overline{B}_n)}{c \cdot n^q} = \lim_{n \to \infty} \frac{m_G(\Omega^n)}{c \cdot n^q} = 1. \end{align*} $$

Now, for an arbitrary sequence $(t_n)$ with $\lim _{n \to \infty } t_n = \infty $ , we have

$$ \begin{align*} 1 = \lim_{n \to \infty} \frac{m_G(B_{\lfloor t_n \rfloor})}{c \cdot ( \lfloor t_n \rfloor )^q} \cdot \lim_{n \to \infty} \frac{c \cdot ( \lfloor t_n \rfloor )^q}{c \cdot ( \lceil t_n \rceil )^q} \leq \liminf_{n \to \infty} \frac{m_G(B_{t_n})}{c \cdot t_n^q}. \end{align*} $$

The converse inequality follows analogously.