Proof. Consider a Borel measurable subset $A\subseteq X$ and let $\varepsilon>0$ . By regularity of $\mu $ and Urysohn’s lemma, there exists $f\in C(X)$ such that $\Vert \chi _A-f\Vert _1\leq {\varepsilon }/{3}$ . Thanks to a standard argument (see, e.g., [Reference Eisner, Farkas, Haase and NagelEFHN15, Theorem 4.17]), we find a neighborhood U of the neutral element in G such that $\Vert f-f\circ \pi ^g\Vert _\infty \leq {\varepsilon }/{3}$ for every $g \in U$ . In particular, if $g \in U$ , then

$$ \begin{align*} \mu(A\triangle (g^{-1}).A) &=\|\chi_A-\chi_{(g^{-1}).A}\|_1=\|\chi_A-\chi_A\circ \pi^g\|_1\\ &\leq \|\chi_A-f\|_1 + \|f-f\circ \pi^g\|_\infty + \|f \circ \pi^g-\chi_A \circ \pi^g\|_1 \leq \varepsilon. \\[-35pt] \end{align*} $$

Proof. ( $\leq $ ) Let $\varphi \colon{\kern-1.3pt} F {\kern-1.3pt}\to{\kern-3pt} gF$ be a bijection and consider the set $D{\kern-1pt} \mathrel {\mathop :}={\kern-1pt} \{ x {\kern-1.3pt}\in{\kern-1.3pt} F {\kern-1.3pt}\mid{\kern-1.3pt} (x,\varphi (x)) {\kern-1pt}\in{\kern-1pt} U_{\Rsh }{\kern-1pt} \}$ . Of course, $\gamma \colon F \to F, x \mapsto \varphi ^{-1}(gx)$ is a bijection. Note that $\varphi (x) = g\gamma ^{-1}(x)$ for each $x \in F$ . Let $C \mathrel {\mathop :}= \{ x \in F \mid (\gamma (x), gx) \in U_{\Rsh } \}$ . For every $x \in F$ , we have

$$ \begin{align*} \gamma^{-1}(x) \in C &\Longleftrightarrow\ (\gamma (\gamma^{-1}(x)),g\gamma^{-1}(x)) \in U_{\Rsh} \Longleftrightarrow\ (x, g\gamma^{-1}(x)) \in U_{\Rsh} \\ & \Longleftrightarrow\ (x,\varphi (x)) \in U_{\Rsh} \Longleftrightarrow\ x \in D , \end{align*} $$

which means that $\gamma ^{-1}(D) = C$ , which in turn implies that $\vert C \vert = \vert D \vert $ .

( $\geq $ ) Let $\gamma \in \mathrm {Sym}(F)$ and define $C \mathrel {\mathop :}= \{ x \in F \mid (\gamma (x),gx) \in U_{\Rsh } \}$ . We consider the bijection $\varphi \colon F \to gF, x \mapsto g\gamma ^{-1}(x)$ and let $D \mathrel {\mathop :}= \{ x \in F \mid (x,\varphi (x)) \in U_{\Rsh } \}$ . For every $x \in F$ ,

$$ \begin{align*} \gamma(x) \in D &\Longleftrightarrow\ (\gamma (x),\varphi(\gamma (x))) \in U_{\Rsh} \Longleftrightarrow\ (\gamma(x),gx) \in U_{\Rsh} \Longleftrightarrow\ x \in C , \end{align*} $$

which means that $\gamma (C) = D$ . Hence, $\vert C \vert = \vert D \vert $ .

Proof. It is straightforward to show item (i) [Reference HauserHau21, Proposition 2.3]. To see item (ii), we set $B_i:=\overline {\{g\in G; Kg\subseteq A_i\}}$ and follow well-known arguments as in [Reference HauserHau22, Proposition 2.2] to show $\lim _{i\in I}\theta (B_i)/\theta (A_i)=1$ and $KB_i\subseteq A_i$ . Furthermore, for $g\in B_i^c$ , we have $Kg\cap A_i^c\neq \emptyset $ and hence $g\in K^{-1}A_i^c$ .

Proof. We first show that G admits a compact, regular neighborhood of $e_G$ which satisfies $M^{-1}M\cap \omega =\{e_G\}$ . Choose a compact neighborhood $\hat {M}$ of $e_G$ which satisfies $\hat {M}^{-1}\hat {M}\cap \omega =\{e_G\}$ . From [Reference KelleyKel55, Ch. 6], we know that there exist finitely many continuous pseudometrics $d_1,\ldots ,d_J$ on $\hat {M}$ and $\varepsilon>0$ such that

$$ \begin{align*} \eta \mathrel{\mathop:}= \bigcap\nolimits_{j=1}^J \{(x,y)\in \hat{M}^2 \vert d_j(x,y)<\varepsilon \} \in \mathbb{U}_{\hat{M}} \end{align*} $$

and such that $\eta [e_G]$ is contained in the interior of $\hat {M}$ . By a standard argument involving Froda’s theorem, for each $j\in \{1,\ldots , J\}$ , there exists $r_j\in (0,\varepsilon )$ such that the closed $d_j$ -ball $\overline {B}_{r_j}^{d_j}(e_G)$ is regular. Then, $M:=\bigcap _{j=1}^J\overline {B}_{r_j}^{d_j}(e_G)$ is a compact and regular neighborhood of $e_G$ which satisfies $M^{-1}M\cap \omega =\{e_G\}$ .

Let $K\subseteq G$ be a compact subset such that $K\omega =G$ . Then, $\{gU \mid g\in K\}$ is an open cover of K and thus there exists a finite sequence $(g_n)_{n=1}^N$ such that $\bigcup _{n=1}^N g_nU \supseteq K$ . From [Reference HauserHau21, Reference MorrisMor15], we know that $C\mathrel {\mathop :}=\bigcup _{n=1}^N g_nM \setminus (\bigcup _{i=1}^{n-1} g_iM\omega )$ is a precompact fundamental domain. Setting $F\mathrel {\mathop :}=\omega \cap (\bigcup _{n=1}^N g_nM)^{-1}(\bigcup _{n=1}^N g_nM)$ , we obtain a finite set with $C\mathrel {\mathop :}=\bigcup _{n=1}^N g_nM \setminus (\bigcup _{i=1}^{n-1} g_iMF)$ and the regularity of M implies C to be regular. Since M has non-empty interior, so does C.

Proof. Denote by $\mathcal {Z}$ the set of all pairs $(F',\varphi )$ consisting of a finite subset $F' \subseteq G$ and an injective map $\varphi \colon F' \to F$ such that $x\varphi (x)^{-1} \in U$ for all $x \in F'$ and $F' \cap gF' = \emptyset $ for every $g \in E\setminus \{ e_G \}$ . Let $(F',\varphi ) \in \mathcal {Z}$ such that $|F'| = \sup \{ |F"| \mid (F",\psi ) \in \mathcal {Z} \}$ . We claim that $|F'| = |F|$ . In contrast, assume that $|F'| < |F|$ . Then, there exists $y \in F\setminus \varphi (F')$ . Since G is a perfect Hausdorff space, every open non-empty subset of G is infinite. Consequently, there exists some

$$ \begin{align*} x \in Uy \setminus \bigg( F' \cup \bigcup\nolimits_{g \in E} gF' \cup g^{-1}F' \bigg). \end{align*} $$

We define $F" \mathrel {\mathop :}= F' \cup \{ x \}$ and $\psi \colon F" \to F$ such that $\psi |_{F'} = \varphi $ and $\psi (x) = y$ . We observe that $(F",\psi )$ is a member of $\mathcal {Z}$ . Since $|F'| < |F"|$ , this clearly contradicts our hypothesis. Hence, $|F'| = |F|$ and therefore $\varphi $ is a bijection. This finishes the proof.

Proof. Let $F \in \mathcal {F}(G)$ , $U \in \mathcal {U}(G)$ , and $x \in G$ . If $E \in U[Fx]$ , then there exists a bijection $\varphi \colon{\kern-1pt} E {\kern-1pt}\to{\kern-1pt} Fx$ such that $y\varphi (y)^{-1} {\kern-1pt}\in{\kern-1pt} U_{\Rsh }$ for all $y {\kern-1pt}\in{\kern-1pt} E$ , and so $\psi \colon Ex^{-1} {\kern-1pt}{\kern-1pt}\to{\kern-1pt}{\kern-1pt} F, y {\kern-1pt}\mapsto{\kern-1pt} \varphi (yx)x^{-1}$ is a well-defined bijection satisfying

$$ \begin{align*} y\psi(y)^{-1} = y( \varphi(yx)x^{-1} )^{-1} = yx\varphi(yx)^{-1} \in U \end{align*} $$

for all $y \in Ex^{-1}$ , whence $Ex^{-1} \in U[F]$ and thus

$$ \begin{align*} E = Ex^{-1}x \in \{ F'x \mid F' \in U[F] \}. \end{align*} $$

Conversely, if $F' \in U[F]$ , then there exists a bijective map $\varphi \colon F' \to F$ with $y\varphi (y)^{-1} \in U$ for all $y \in F'$ , so that $\psi \colon F'x \to Fx, y \mapsto \varphi (yx^{-1})x$ is a well-defined bijection satisfying

$$ \begin{align*} y\psi(y)^{-1} = y( \varphi(yx^{-1})x )^{-1} = yx^{-1}\varphi(yx^{-1})^{-1} \in U \end{align*} $$

for all $y \in F'x$ , which shows that $F'x \in U[Fx]$ . This completes the proof.

Proof. Let $F \in W[F_{0}]$ and $g \in E$ . Fix any bijection $\varphi \colon F \to F_{0}$ such that $(x,\varphi (x)) \in W_{\Rsh }$ for all $x \in F$ . By Lemma 2.4, there is a permutation $\gamma _{0}$ on $F_{0}$ with $\vert D_{0} \vert = {\mathrm {match}}_{\Rsh } (F_{0},gF_{0},W)$ for

$$ \begin{align*} D_{0} \mathrel{\mathop:}= \{ x \in F_{0} \mid (\gamma_{0} (x),gx) \in W_{\Rsh} \}. \end{align*} $$

Consider $\gamma \mathrel {\mathop :}= \varphi ^{-1} \circ \gamma _{0} \circ \varphi \in {\mathrm {Sym}} (F)$ and let $D \mathrel {\mathop :}= \{ x \in F \mid (\gamma (x),gx) \in U_{\Rsh } \}$ . We claim that $\varphi ^{-1}(D_{0}) \subseteq D$ . To prove this, let $x \in \varphi ^{-1}(D_{0})$ . Then, $(\gamma _{0} (\varphi (x)),g\varphi (x)) \in W_{\Rsh }$ . Moreover, by our choice of $\varphi $ , both

$$ \begin{align*} (\gamma(x),\gamma_{0} (\varphi(x))) = (\gamma(x),\varphi(\gamma(x))) \in W_{\Rsh} \end{align*} $$

and $(x,\varphi (x)) \in W_{\Rsh }$ , and thus $gx (g\varphi (x))^{-1} = gx\varphi (x)^{-1}g^{-1} \in gWg^{-1} \subseteq V$ , that is, $(g\varphi (x), gx) \in V_{\Rsh }$ . It follows that $(\gamma (x),gx) \in U_{\Rsh }$ , that is, $x \in D$ . This proves our claim. Hence,

$$ \begin{align*} \operatorname{\mathrm{match}}_{\Rsh}(F,gF,U) \stackrel{\mathrm{Lemma~}2.4}{\geq} \vert D \vert \geq \lvert \varphi^{-1}(D_{0}) \rvert = \vert D_{0} \vert = \operatorname{\mathrm{match}}_{\Rsh}(F_{0},gF_{0},W).\\[-36pt] \end{align*} $$

Proof. To prove the desired convergence, it suffices to show that

(1) $$ \begin{align} \text{ for all } K \in \mathcal{F}_{+}(G), \quad \limsup\nolimits_{i \in I} \frac{f (F_{i})}{\vert F_{i} \vert} \leq \frac{f (K)}{\vert K \vert}. \end{align} $$

For this purpose, let $K \in \mathcal {F}_+(G)$ and $\varepsilon> 0$ . By right invariance of f, we may and will assume that K contains the neutral element of G. Since K is finite and f is continuous, there exists $U \in \mathcal {U}(G)$ such that $U^{-1} = U$ , $KK^{-1} \cap UU = \{ e_G \}$ , and

(2) $$ \begin{align} \text{ for all } K' \subseteq K, \quad \operatorname{\mathrm{diam}} f (U[K']) \leq \frac{\varepsilon}{3}. \end{align} $$

For each $K' \subseteq K$ and every $x \in G$ , right invariance of f readily implies that

(3) $$ \begin{align} \operatorname{\mathrm{diam}} f(U[K'x]) \stackrel{\mathrm{Lemma~}3.2}{=} \operatorname{\mathrm{diam}} \{ f(K"x) \mid K" \in U[K'] \} = \operatorname{\mathrm{diam}} f(U[K']) \stackrel{\mathrm{equation~}(2)}{\leq} \frac{\varepsilon}{3}.\nonumber\\ \end{align} $$

Consider $C \mathrel {\mathop :}= \max \{ 1, \sup \{ f (K') \mid K' \subseteq K \} \}$ and note that, by right invariance of f,

(4) $$ \begin{align} \sup \{ f(K'x) \mid K' \subseteq K, x \in G \} \leq C. \end{align} $$

Let $\tau \mathrel {\mathop :}= 1- {\varepsilon }/{3C\vert K \vert }$ . Furthermore, let $V \in \mathcal {U}(G)$ such that $V^{-1} = V$ and $V^{3} \subseteq U$ . Define

$$ \begin{align*} W \mathrel{\mathop:}= \bigcap\nolimits_{g \in K \cup K^{-1}} g^{-1}Vg. \end{align*} $$

Since $(F_{i})_{i \in I}$ constitutes a thin Følner net in G, there exists $i_{0} \in I$ such that

(5) $$ \begin{align} \text{ for all } i \in I , i_{0} \leq i \ \text{ for all } g \in K, \quad \operatorname{\mathrm{match}}_{\Rsh} (F_{i},gF_{i},W) \geq \tau\vert F_{i} \vert. \end{align} $$

Let $i \in I$ with $i_{0} \leq i$ . Since f is continuous, there exists some $W_{0} \in \mathcal {U}(G)$ such that $W_{0} \subseteq W$ and ${\mathrm {diam}} f (W_{0}[F_{i}]) \leq {\varepsilon }/{3}$ . By Lemma 3.1 and since we assume G to be non-discrete, there exists $F \in W_{0}[F_{i}]$ with $F \cap gF = \emptyset $ for every $g \in K\setminus \{ e_G \}$ . In particular, we deduce from ${\mathrm {diam}} f (W_{0}[F_{i}]) \leq {\varepsilon }/{3}$

(6) $$ \begin{align} \vert f(F_{i}) - f(F) \vert \leq \frac{\varepsilon}{3}. \end{align} $$

Furthermore, thanks to Lemma 3.3, equation (5), and our choice of $W_{0} \subseteq W$ , it follows that ${\mathrm {match}}_{\Rsh } (F,gF,U) \geq {\mathrm {match}}_{\Rsh }(F_{i},gF_{i},W)$ whenever $g \in K$ . Hence, for each $g \in K\setminus \{ e_G \}$ , there exists an injection $\psi _{g} \colon D_{g} \to gF$ such that $D_{g} \subseteq F$ , $\vert D_{g} \vert \geq \tau \vert F \vert $ , and

(7) $$ \begin{align} \text{ for all } x \in D_{g}, \quad (x,\psi_{g}(x)) \in U_{\Rsh}. \end{align} $$

For each $g \in K\setminus \{ e_G \}$ , consider the bijection $\varphi _{g} \colon G \to G$ defined by

$$ \begin{align*} \varphi_{g}(x) \mathrel{\mathop:}= \begin{cases} \psi_{g}^{-1}(x) & \text{if } x \in \psi_{g}(D_{g}) , \\ \psi_{g}(x) & \text{if } x \in D_{g} , \\ x & \text{otherwise} \end{cases} \quad (x \in G). \end{align*} $$

Put $D_{e_G} \mathrel {\mathop :}= F$ , $\psi _{e_G} \mathrel {\mathop :}= {\mathrm {id}}_{F}$ , and $\varphi _{e_G} \mathrel {\mathop :}= {\mathrm {id}}_{G}$ . Let

$$ \begin{align*} \gamma \colon K \times G \longrightarrow G, \quad (g,x) \longmapsto \varphi_{g}(gx). \end{align*} $$

We now claim that

(8) $$ \begin{align} \text{ for all } g_{0},g_{1} \in K \text{ for all } x \in G, \quad\gamma (g_{0},x) = \gamma(g_{1},x) \ \Longrightarrow \ g_{0} = g_{1}. \end{align} $$

Indeed, if $x \in G$ and $g_{0},g_{1} \in K$ with $\gamma (g_{0},x) = \gamma (g_{1},x)$ , then from $U^{-1} = U$ and

$$ \begin{align*} (g_{0}x,\gamma (g_{0},x)) &= (g_{0}x,\varphi_{g_{0}}(g_{0}x)) \stackrel{\mathrm{equation~}(7)}{\in} U_{\Rsh} ,\\(g_{1}x,\gamma (g_{1},x)) &= (g_{1}x,\varphi_{g_{1}}(g_{1}x)) \stackrel{\mathrm{equation~}(7)}{\in} U_{\Rsh}, \end{align*} $$

we infer that $(g_{0}x,g_{1}x) \in (UU)_{\Rsh }$ and therefore $g_{0}g_{1}^{-1} = (g_{0}x)(g_{1}x)^{-1} \in UU$ , whence $g_{0} = g_{1}$ as $KK^{-1} \cap UU = \{ e_G \}$ . Next, we prove that

(9) $$ \begin{align} \text{ for all } y \in G, \quad \vert \{ x \in G \mid \text{there exists } g \in K \colon \gamma (g,x) = y \} \vert = \vert K \vert. \end{align} $$

To see this, let $y \in G$ . For each $g \in K$ , the map $G \to G, x \mapsto \gamma (g,x)$ is a bijection, whence

$$ \begin{align*} \vert \{ x \in G \mid \gamma (g,x) = y \} \vert = 1. \end{align*} $$

Since

it follows that

$$ \begin{align*} \vert \{ x \in G \mid \text{there exists } g \in K \colon \gamma (g,x) = y \} \vert = \sum\nolimits_{g \in K} \vert \{ x \in G \mid \gamma (g,x) = y \} \vert = \vert K \vert. \end{align*} $$

This proves equation (9). Now, we consider the finite set

$$ \begin{align*} H \mathrel{\mathop:}= \{ x \in G \mid \gamma (K \times \{x \}) \cap F \ne \emptyset \} = \bigcup\nolimits_{g \in K} g^{-1}\varphi_{g}^{-1}(F) \end{align*} $$

and note that

(10) $$ \begin{align} \vert H \vert \leq \sum\nolimits_{g \in K} \lvert g^{-1}\varphi_{g}^{-1}(F) \rvert = \vert K \vert \vert F \vert. \end{align} $$

We observe that, by equation (9),

$$ \begin{align*} \text{ for all } y \in F, \quad \vert \{ x \in H \mid y \in \gamma(K \times \{ x \}) \cap F \} \vert = \vert K \vert , \end{align*} $$

which entails that $(\gamma (K\times \{x\})\cap F)_{x\in H}$ is a $|K|$ -covering of F. Since f satisfies Shearer’s inequality, we observe

(11) $$ \begin{align} f (F) \leq \frac{1}{\vert K \vert} \sum\nolimits_{x \in H} f (\gamma (K \times \{x \}) \cap F). \end{align} $$

Moreover, for each $x \in G$ , we have $\vert \gamma (K \times \{ x \}) \vert = \vert K \vert $ due to equation (8), and then

(12) $$ \begin{align} \gamma (K \times \{x \}) \in U[Kx] \end{align} $$

by equation (7) and $U=U^{-1}$ , wherefore

(13) $$ \begin{align} \vert f(\gamma (K \times \{x \})) - f(Kx) \vert \stackrel{\mathrm{equation~}(3)}{\leq} \frac{\varepsilon}{3}. \end{align} $$

Also, if $x \in G$ , then equation (12) implies that $\gamma (K \times \{ x\}) \cap F \in U[K'x]$ for some $K' \subseteq K$ , so that

$$ \begin{align*} \vert f(\gamma (K \times \{x \}) \cap F) - f(K'x) \vert \stackrel{\mathrm{equation~}(3)}{\leq} \frac{\varepsilon}{3} \end{align*} $$

and thus

(14) $$ \begin{align} f(\gamma (K \times \{x \}) \cap F) \stackrel{\mathrm{equation~}(4)}{\leq} C + \frac{\varepsilon}{3}. \end{align} $$

Furthermore, since $e_G \in K$ and $\varphi _{e_G} = {\mathrm {id}}_{G}$ ,

$$ \begin{align*} H_{0} \mathrel{\mathop:}= \{ x \in G \mid \gamma (K \times \{x \}) \subseteq F \} = \bigcap\nolimits_{g \in K} g^{-1}\varphi_{g}^{-1}(F) = \bigcap\nolimits_{g \in K} \{ x {\kern-1pt}\in{\kern-1pt} F \mid \varphi_{g}(gx) {\kern-1pt}\in{\kern-1pt} F \}. \end{align*} $$

Now, if $g \in K$ and $x \in g^{-1}\psi _{g}(D_{g})$ , then $\varphi _{g}(gx) = \psi ^{-1}_{g}(gx) \in F$ . This entails that

$$ \begin{align*} \bigcap\nolimits_{g \in K} g^{-1}\psi_{g}(D_{g}) \subseteq \bigcap\nolimits_{g \in K} \{ x \in F \mid \varphi_{g}(gx) \in F \} = H_{0} \end{align*} $$

and, in turn,

$$ \begin{align*} \vert H_{0} \vert &\geq \bigg\lvert \bigcap\nolimits_{g \in K} g^{-1}\psi_{g}(D_{g}) \bigg\rvert = \vert F \vert - \bigg\lvert \bigcup\nolimits_{g \in K} F \setminus g^{-1}\psi_{g}(D_{g}) \bigg\rvert \\ & \geq \vert F \vert - \sum\nolimits_{g \in K} \lvert F \setminus g^{-1}\psi_{g}(D_{g}) \rvert = \vert F \vert - \sum\nolimits_{g \in K} \vert F \setminus D_{g} \vert \\ & \geq \vert F \vert - \vert K \vert (1-\tau)\vert F \vert = \bigg(1-\frac{\varepsilon}{3C}\bigg) \vert F \vert. \end{align*} $$

Hence, considering the set

$$ \begin{align*} H_{1} \mathrel{\mathop:}= H\setminus H_{0} = \bigcup\nolimits_{g \in K} (g^{-1}\varphi_{g}^{-1}(F))\setminus \bigcap\nolimits_{{g}' \in K} (g^{\prime-1}\varphi_{g}^{\prime-1}(F)) , \end{align*} $$

we conclude that

(15) $$ \begin{align} \vert H_{1} \vert \leq \sum\nolimits_{g \in K} \lvert (g^{-1}\varphi_{g}^{-1}(F))\setminus H_{0} \rvert \leq \vert K \vert ( \vert F\vert - \vert H_{0} \vert ) \leq \frac{\varepsilon \vert F \vert \vert K \vert}{3C}. \end{align} $$

Consequently, it follows that

$$ \begin{align*} f (F_{i}) &\stackrel{\mathrm{equation~}(6)}{\leq} f (F) + \frac{\varepsilon}{3} \stackrel{\mathrm{equation~}(11)}{\leq} \frac{1}{\vert K \vert} \sum\nolimits_{x \in H} f (\gamma (K \times \{ x \}) \cap F) + \frac{\varepsilon}{3} \\ &= \frac{1}{\vert K \vert} \sum\nolimits_{x \in H_{0}} f (\gamma (K \times \{ x \}) \cap F) + \frac{1}{\vert K \vert} \sum\nolimits_{x \in H_{1}} f(\gamma (K \times \{ x \}) \cap F) + \frac{\varepsilon}{3} \\ &= \frac{1}{\vert K \vert} \sum\nolimits_{x \in H_{0}} f (\gamma (K \times \{ x \})) + \frac{1}{\vert K \vert} \sum\nolimits_{x \in H_{1}} f(\gamma (K \times \{ x \}) \cap F) + \frac{\varepsilon}{3} \\ &\stackrel{\mathrm{equations~}(13)+(14)}{\leq} \frac{1}{\vert K \vert} \sum\nolimits_{x \in H_{0}} f (Kx) + \frac{\varepsilon \vert H_{0} \vert}{3\vert K\vert} + \frac{(C+{\varepsilon}/{3}) \vert H_{1} \vert}{\vert K \vert} + \frac{\varepsilon}{3} \\ &= \frac{1}{\vert K \vert} \sum\nolimits_{x \in H_{0}} f (K) + \frac{\varepsilon \vert H \vert}{3\vert K\vert} + \frac{C \vert H_{1} \vert}{\vert K \vert} + \frac{\varepsilon}{3} \\ &\stackrel{\mathrm{equations~}(10)+(15)}{\leq} \frac{f (K)\vert H_{0} \vert}{\vert K \vert} + \frac{\varepsilon \vert F \vert}{3} + \frac{\varepsilon \vert F \vert}{3} + \frac{\varepsilon}{3} \\ &\leq \frac{f (K)\vert H_{0} \vert}{\vert K \vert} + \varepsilon \vert F \vert \leq \frac{f (K)\vert F \vert}{\vert K \vert} + \varepsilon \vert F \vert , \end{align*} $$

and therefore,

$$ \begin{align*} \frac{f (F_{i})}{\vert F_{i} \vert} = \frac{f (F_{i})}{\vert F \vert} \leq \frac{f (K)}{\vert K \vert} + \varepsilon. \end{align*} $$

This proves equation (1) and hence the theorem.

Proof. As $(A_i)_{i\in F}$ is $\varepsilon $ -disjoint, there exist disjoint compact subsets $B_i\subseteq A_i\ (i \in F)$ such that $\theta (B_i)> (1-\varepsilon )\theta (A_i)$ for each $i \in F$ . By our assumptions,

$$ \begin{align*}\theta\bigg(A\setminus \bigg(\bigcup\nolimits_{i\in F} A_i\bigg)\bigg) \geq \theta(A)-\theta\bigg(\bigcup\nolimits_{i\in F}A_i\bigg)> (1-\varepsilon)\theta(A). \end{align*} $$

Thus, there is $\rho>0$ such that $(1-\rho )\theta (A\setminus \bigcup _{i\in F} A_i)>(1-\varepsilon )\theta (A)$ . As $\theta $ is regular, there exists a compact subset $B\subseteq A\setminus \bigcup _{i\in F}A_i$ such that $\theta (B)\geq (1-\rho )\theta (A\setminus \bigcup _{i\in F}A_i)\geq (1-\varepsilon )\theta (A)$ . For each $i \in I$ , the set B is disjoint from $A_i$ , and thus disjoint from $B_{i}$ . Hence, $(A_i)_{i \in I} \sqcup (A)$ is $\varepsilon $ -disjoint.

Proof. Let us abbreviate $M\mathrel {\mathop :}=\max _{i} \alpha (A_i,K)=\max _i \theta (\partial _K A_i)/\theta (A_i)$ . A straightforward argument shows that $\partial _K (\bigcup _{i\in F}A_i)\subseteq \bigcup _{i\in F}\partial _K A_i$ , and hence

$$ \begin{align*} \theta \bigg(\partial_K \bigg(\bigcup\nolimits_{i\in F}A_i\bigg)\bigg) &\leq \theta\bigg( \bigcup\nolimits_{i\in F}\partial_K A_i\bigg) \\ & \leq \sum\nolimits_{i\in F}\theta(\partial_K A_i) = \sum\nolimits_{i\in F}\theta(A_i)\frac{\theta(\partial_K A_i)}{\theta(A_i)} \leq M \sum\nolimits_{i\in F}\theta(A_i). \end{align*} $$

Since the considered family is $\varepsilon $ -disjoint, we obtain that $(1-\varepsilon ) \sum _{i\in F}\theta (A_i)\leq \theta (\bigcup _{i\in F} A_i)$ and thus conclude that

$$ \begin{align*}\alpha\bigg(\bigcup\nolimits_{i\in F} A_i,K\bigg) = \frac{\theta(\partial_K \bigcup\nolimits_{i\in F} A_i )}{\theta(\bigcup\nolimits_{i\in F} A_i )} \leq \frac{M \sum_{i\in F}\theta(A_i)}{(1-\varepsilon)\sum_{i\in F}\theta(A_i)} = \frac{M}{1-\varepsilon}.\\[-44pt] \end{align*} $$

Proof. A straightforward argument shows that $\partial _K(R\setminus B)\subseteq \partial _K R\cup \partial _K B$ . Thus, if $\varepsilon>0$ and $\theta (R\setminus B)\geq \varepsilon \theta (R)$ , then we conclude that

$$ \begin{align*}\frac{\theta(\partial_K (R\setminus B))}{\theta(R\setminus B)} \leq \frac{1}{\varepsilon}\frac{\theta(\partial_K R)+\theta(\partial_K B)}{\theta(R)} \leq \frac{1}{\varepsilon}\bigg(\frac{\theta(\partial_K R)}{\theta(R)}+\frac{\theta(\partial_K B)}{\theta(B)}\bigg).\\[-41pt] \end{align*} $$

Proof. As $(Ag)_{g \in C}$ is an $\varepsilon $ -disjoint family, we obtain

$$ \begin{align*}\theta(R) \geq \theta(AC) = \theta\bigg(\bigcup\nolimits_{g\in C}Ag\bigg) \geq (1-\varepsilon) \sum\nolimits_{g\in C} \theta(Ag) = (1-\varepsilon)|C|\theta(A).\\[-44pt] \end{align*} $$

Proof. Since $\theta (A)>0$ , there exists $a\in G$ such that $Aa$ contains the identity $e_G$ . Then, $\theta (\partial _{(Aa)^{-1}}R)=\theta (a^{-1}\partial _{A^{-1}} R)=\theta (\partial _{A^{-1}} R)$ , and upon translating C, we may and will assume without lost of generality that A contains $e_G$ . For $g\in R\setminus \partial _{A^{-1}}R$ , we have $g\in R\subseteq A^{-1}\overline {R}$ and thus $g\notin A^{-1}\overline {R^c}$ . In particular, $Ag\cap \overline {R^c}$ is empty, and we deduce that $Ag\subseteq R$ . If $g\notin C$ , then necessarily $\theta (Ag\cap AC)\geq \varepsilon \theta (Ag)$ , as otherwise, by Lemma 4.1, $(Ag')_{g'\in C\cup \{g\}}$ would be an $\varepsilon $ -disjoint family, a contradiction to the maximal cardinality of C. For $g\in C$ , we furthermore obtain $Ag\subseteq AC$ and thus, $\theta (Ag\cap AC)=\theta (Ag)\geq \varepsilon \theta (A)$ from $\varepsilon \in (0,1)$ . This shows that, for every $g\in R\setminus \partial _{A^{-1}}R$ ,

$$ \begin{align*} \theta(Ag\cap AC) \geq \varepsilon\theta(A). \end{align*} $$

Let now $\chi _M$ denote the characteristic function of a subset $M\subseteq G$ . For any $g'\in G$ , we compute that

$$ \begin{align*}\theta(A) = \theta(A^{-1}) = \int_G \chi_{A}(g^{-1})\,d\theta(g) = \int_G \chi_{A}(g'g^{-1})\,d\theta(g).\end{align*} $$

Thus, Tonelli’s theorem implies

$$ \begin{align*} &\theta(A)\theta(AC)\\ &\quad=\int_G \theta(A)\chi_{AC}(g') \,d\theta(g') =\int_G \int_G \chi_{A}(g'g^{-1})\,d\theta(g) \chi_{AC}(g') \,d\theta(g')\\ &\quad=\int_G \int_G \chi_{A}(g'g^{-1}) \chi_{AC}(g') \,d\theta(g') \,d\theta(g) =\int_G \int_G \chi_{Ag\cap AC}(g') \,d\theta(g')\,d\theta(g)\\ &\quad=\int_G\theta(Ag\cap AC)\,d\theta(g) \geq \int_{R\setminus \partial_{A^{-1}}R} \varepsilon\theta(A)\,d\theta(g)=\varepsilon\theta(A)\theta(R\setminus \partial_{A^{-1}}R). \end{align*} $$

Consequently, we arrive at

$$ \begin{align*}\theta(AC) \geq \varepsilon \theta(R\setminus \partial_{A^{-1}}R) \geq \varepsilon(\theta(R)-\theta(\partial_{A^{-1}}R)) = \varepsilon\bigg(1-\frac{\theta(\partial_{A^{-1}} R)}{\theta(R)}\bigg) \theta(R).\\[-43pt] \end{align*} $$

Proof. As $\varepsilon \in (0,1/2)$ , there exists $N\in \mathbb {N}$ such that $(1-{\varepsilon }/2)^{N}\leq \varepsilon /2$ . As $(A_i)_{i\in J}$ is a van Hove net, we can choose inductively $i_n\in J$ for $n={{N}},\ldots ,1$ such that $A_{i_n}$ has positive Haar measure, such that the $i_n$ are pairwise distinct and such that

$$ \begin{align*} \alpha\bigg(A_{i_n},K\cup \bigg(\bigcup\nolimits_{m=n+1}^{{{N}}}A_{i_m}^{-1}\bigg)\bigg) \leq \varepsilon^{2(N-n)+4}. \end{align*} $$

We abbreviate $K_n\mathrel {\mathop :}= K\cup (\bigcup _{m=n+1}^{{{N}}}A_{i_m}^{-1})$ and observe that $A_{i_n}$ is $(\varepsilon ^{2(N-n)+4}, K_n)$ -invariant for $n=0,\ldots ,{{N}}$ . Furthermore, we have $K_0\supseteq K_1\supseteq \cdots \supseteq K_N=K$ and $A_{i_n}^{-1}\subseteq K_{n-1}$ for $n=1,\ldots ,N$ .

We set $F\mathrel {\mathop :}= \{i_1,\ldots ,i_N\}$ , $\delta \mathrel {\mathop :}= \varepsilon ^{2{N}+1}$ , and $D\mathrel {\mathop :}= K_0$ , and consider a compact and $(\delta ,D)$ -invariant subset A of G. Set $R_0\mathrel {\mathop :}= A$ . Using Lemma 4.4, we now choose inductively for $n=1,\ldots ,M$ finite $(\varepsilon ,A_{i_n})$ -fillings $C_n$ of $R_{n-1}$ of maximal cardinality, where we abbreviate $R_n\mathrel {\mathop :}= R_{n-1}\setminus A_{i_n}C_n$ and $M\leq N$ is the smallest integer where our choices lead to $\theta (R_M)=\theta (R_{M-1}\setminus A_{i_M}C_M)\leq \varepsilon \theta (R_{M-1})$ and N if we never encounter this situation. Thus, in particular, for $n=1,\ldots ,M-1$ , there holds $\theta (R_n)>\varepsilon \theta (R_{n-1})$ . For $n\in \{M+1,\ldots ,N\}$ , we set $C_n\mathrel {\mathop :}= \emptyset $ . Note that $R_n$ is precompact and measurable for $n=0,\ldots ,M$ and that there holds

$$ \begin{align*}R = R_M \subseteq R_{M-1} \subseteq R_0 = A.\end{align*} $$

We will now show that defining $C_{i_n}\mathrel {\mathop :}= C_n$ , we get that $(C_i)_{i\in F}=(C_n)_{n=1}^N$ is a family of $\varepsilon $ -quasi-tiling centers that fulfills the required properties.

We first show inductively for $n=0,\ldots ,M-1$ that $R_n$ is $(\varepsilon ^{2(N-n)+1},K_n)$ -invariant, that is,

(16) $$ \begin{align} \alpha(R_n,K_n) \leq \varepsilon^{2(N-n)+1}. \end{align} $$

This is clearly satisfied for $n=0$ , as $R_0=A$ , $K_0=D$ , and $\varepsilon ^{2(N-0)+1}=\delta $ . To proceed inductively, we assume $R_n$ to be $(\varepsilon ^{2(N-n)+1},K_n)$ -invariant for some $n<M-1$ and as $K_{n+1}\subseteq K_n$ , we obtain

$$ \begin{align*}\alpha(R_n,K_{n+1})\leq \varepsilon^{2(N-n)+1}.\end{align*} $$

Now recall that $A_{i_{n+1}}$ is $(\varepsilon ^{2(N-n)+4},K_{n+1})$ -invariant. As $C_{n+1}$ is an $(\varepsilon ,A_{i_{n+1}})$ -filling, we obtain that $(A_{i_{n+1}}g)_{g \in C_{n+1}}$ is $\varepsilon $ -disjoint and apply Lemma 4.2 to see

$$ \begin{align*} \alpha(A_{i_{n+1}}C_{n+1},K_{n+1}) &\leq \frac{1}{1-\varepsilon}\max_{g\in C_{n+1}}\alpha(A_{i_{n+1}} g,K_{n+1})\\ &= \frac{\alpha(A_{i_{n+1}},K_{n+1})}{1-\varepsilon}\\ &\leq \frac{\varepsilon^{2(N-n)+4}}{1-\varepsilon} \leq \varepsilon^{2(N-n)+1}. \end{align*} $$

For this, we have used that $\varepsilon <1/2$ yields that $\varepsilon /(1-\varepsilon )<1$ . As we assume $n<M-1$ , we obtain $\theta (R_{n}\setminus A_{i_{n+1}}C_{n+1})=\theta (R_{n+1})> \varepsilon \theta (R_{n})$ . Thus, Lemma 4.3 yields that

$$ \begin{align*} \alpha(R_{n+1},K_{n+1}) &= \alpha(R_{n}\setminus A_{i_{n+1}}C_{n+1},K_{n+1})\\ &\leq \frac{\alpha(R_{n},K_{n+1})+\alpha(A_{i_{n+1}}C_{n+1},K_{n+1})}{\varepsilon}\\ &\leq 2\varepsilon^{2(N-n)} \leq \varepsilon^{2(N-n)-1} = \varepsilon^{2(N-(n+1))+1} , \end{align*} $$

and we have completed the induction to show equation (16).

We next show that

(17) $$ \begin{align} \theta(R) \leq \frac{\varepsilon}{2}\theta(A). \end{align} $$

This statement is satisfied whenever $M<N$ , as then we obtain that

$$ \begin{align*}\theta(R)=\theta(R_{M})\leq \varepsilon \theta(R_{M-1})\leq \varepsilon\theta (A).\end{align*} $$

To show equation (17), we thus assume without lost of generality that $M=N$ and that $\theta (R)>0$ . Then, $\theta (R_N)=\theta (R)>0$ and, for $n\in \{1,\ldots ,N-1\}$ , we obtain

$$ \begin{align*} \theta(R_n) \geq \varepsilon\theta(R_{n-1}) \geq \varepsilon^n \theta(R_0) = \varepsilon^n\theta(A)> 0.\end{align*} $$

For $n\leq M=N$ , we have chosen $C_n$ to be an $(\varepsilon ,A_{i_n})$ -filling of $R_{n-1}$ of maximal cardinality. Thus, we obtain from Lemma 4.5, equation (16), and $A_{i_n}^{-1}\subseteq K_{n-1}$ that

$$ \begin{align*}\frac{\theta(A_{i_n} C_n)}{\theta(R_{n-1})} \geq \varepsilon(1-\alpha(R_{n-1},A_{i_n}^{-1})) \geq \varepsilon(1-\alpha(R_{n-1},K_{n-1})) \geq \varepsilon(1-\varepsilon^{2(N-(n-1))+1}) \geq \frac{\varepsilon}{2}.\end{align*} $$

Thus,

$$ \begin{align*}\theta(R_n) = \theta(R_{n-1})-\theta(A_{i_n}C_n) < \bigg(1-\frac{\varepsilon}{2}\bigg)\theta(R_{n-1})\end{align*} $$

and we obtain from our choice of N that

$$ \begin{align*}\theta(R) = \theta(R_{N}) < \bigg(1-\frac{\varepsilon}{2}\bigg)^{N}\theta(R_0) \leq \frac{\varepsilon}{2}\theta(A).\end{align*} $$

This shows the claimed statement in equation (17).

As $C_n$ is an $(\varepsilon ,A_{i_n})$ -filling of $R_{n-1}$ , we obtain from the construction $R_n\mathrel {\mathop :}= R_{n-1}\setminus A_{i_n}C_n$ that $(A_{i_n}g)_{g\in C_n}$ is $\varepsilon $ -disjoint for all $n\leq M$ and that $(A_{i}C_i)_{i\in F} = (A_{i_n}C_n)_{n=1}^{M} \sqcup (\emptyset )_{n=M+1}^{N}$ is a disjoint family. In particular, we observe that $(A_ig)_{g\in C_i}$ is $\varepsilon $ -disjoint for all $i\in F$ . Furthermore, one obtains

$$ \begin{align*}\bigcup_{i\in F} A_iC_i=\bigcup_{n=1}^M A_{i_n} C_n\subseteq \bigcup_{n=0}^M R_n= R_0=A.\end{align*} $$

Thus, equation (17) allows to compute

$$ \begin{align*}\theta\bigg(A\cap \bigcup\nolimits_{i\in F}A_iC_i\bigg)=\theta(A)-\theta(R)\geq (1-\varepsilon)\theta(A).\end{align*} $$

This shows that $(A_i)_{i\in F}$ is an $\varepsilon $ -quasi-tiling of A and that $\bigcup _{i\in F} A_iC_i\subseteq A$ .

By equation (17), it remains to show that $\theta (\partial _K R)\leq (\varepsilon /2) \theta (A).$ Recall that $A_{i_M}$ is $(\varepsilon ^{2(N-M)+4}, K_{M})$ -invariant. As $K\subseteq K_M$ , we obtain

$$ \begin{align*}\alpha(A_{i_M},K)\leq \varepsilon^4.\end{align*} $$

Since $C_{M}$ is an $(\varepsilon ,A_{i_M})$ -filling, we observe that $(A_{i_M}g)_{g\in C_{M}}$ is $\varepsilon $ -disjoint and apply Lemma 4.2 to see

$$ \begin{align*} \alpha(A_{i_M}C_{M},K) &\leq \frac{1}{1-\varepsilon}\max_{g\in C_{M}}\alpha(A_{i_M} g,K)\\ &=\frac{\alpha(A_{i_M},K)}{1-\varepsilon}\\ &\leq \frac{\varepsilon^{4}}{1-\varepsilon}\\ &\leq \varepsilon^3. \end{align*} $$

Furthermore, a straightforward argument shows that $\partial _K R_M=\partial _K (R_{M-1}\setminus A_{i_M}C_M)\subseteq \partial _K R_{M-1} \cup \partial _K (A_{i_M}C_M)$ . As $K\subseteq D$ , we obtain that A is $(K,\delta )$ -invariant and as $\delta \in (0,1)$ , we have $\theta (K)\leq \theta (A)$ . Thus, equation (16) and $K\subseteq K_M \subseteq K_{M-1}$ allow to compute

$$ \begin{align*} \frac{\theta(\partial_K R)}{\theta(A)} &\leq \frac{\theta(\partial_K R_{M-1})}{\theta(A)}+ \frac{\theta(\partial_K A_{i_M}C_M)}{\theta(A)}\\ &\leq \alpha(R_{M-1},K)+\alpha(A_{i_M}C_M,K)\\ &\leq \alpha(R_{M-1},K_{M-1})+\varepsilon^3\\ &\leq \varepsilon^{3} +\varepsilon^3 \leq \frac{\varepsilon}{2}.\\[-3.3pc] \end{align*} $$

Proof. Note first that the subadditivity and the monotonicity of f imply that f only takes values in $[0,\infty )$ . Let V be a compact and symmetric neighborhood of $e_G$ that satisfies $V\subseteq K$ and set $c_K\mathrel {\mathop :}= f(VV)/\theta (V)$ . For a non-empty and precompact subset $R\subseteq G$ , we let $F\subseteq \overline {R}$ be a V-separated subset, that is, such that $(Vg)_{g\in F}$ is a disjoint family. Then, $VF=\bigcup _{g\in F}Vg$ is a disjoint union and thus

$$ \begin{align*}\theta(V\overline{R}) \geq \theta(VF) = \sum_{g\in F}\theta(Vg) = |F|\theta(V).\end{align*} $$

In particular, we obtain the cardinality of F to be bounded by $\theta (V\overline {R})/\theta (V)< \infty $ . We thus assume without lost of generality that F is a V-separated subset of $\overline {R}$ of maximal cardinality. Then, for any $g\in \overline {R}$ , there is $g'\in F$ such that $Vg$ and $Vg'$ intersect. Thus, $g\in VVg'\subseteq VVF$ and we observe that $\overline {R}\subseteq VVF$ . We compute that

$$ \begin{align*} f(\overline{R}) \leq f(VVF) &\leq \sum\nolimits_{g\in F}f(VVg) = |F|f(VV) \\ & \leq \frac{\theta(V\overline{R})}{\theta(V)}f(VV) = c_K\theta(V\overline{R}) \leq c_K\theta(K\overline{R}).\\[-3.5pc] here \end{align*} $$

Proof. Let $\varepsilon \in (0,1/2)$ and choose an arbitrary compact neighborhood K of $e_G$ . Then, by Lemma 4.7, there exists a constant c such that $f(\overline {A})\leq c \theta (K\overline {A})$ for every non-empty, precompact, measurable subset $A\subseteq G$ . Consider $\unicode{x3bb} \mathrel {\mathop :}= \liminf _{i\in I}{f(A_i)}/{\theta (A_i)}$ . As $(KA_i)_{i\in I}$ is a van Hove net in G with $\lim _{i\in I}\theta (KA_i)/\theta (A_i)=1$ , it follows that

$$ \begin{align*} \unicode{x3bb} = \liminf_{i\in I}{f(A_i)}/{\theta(KA_i)} \leq c < \infty. \end{align*} $$

In particular, there is a cofinal subset $J\subseteq I$ such that $(f(A_i))/\theta (A_{i}))_{i\in J}$ converges to $\unicode{x3bb} $ . Thus, there exists $j\in J$ such that for all $i\in J$ with $i\geq j$ , we have

(18) $$ \begin{align} \frac{f(A_i)}{\theta(A_i)} \leq \unicode{x3bb}+\varepsilon. \end{align} $$

Since also $\{i\in J \mid i\geq j\}$ is cofinal in I, we assume without lost of generality that equation (18) holds for all $i\in J$ . From Proposition 4.6, we obtain the existence of a finite subset $F\subseteq J$ , $\delta>0$ , as well as a compact, non-empty subset $D\subseteq G$ such that any $(\delta ,D)$ -invariant, compact subset $A\subseteq G$ can be $\varepsilon $ -quasi-tiled and the $\varepsilon $ -quasi-tiling centers can be chosen with the additional properties as in Proposition 4.6.

We now consider a compact and $(\delta ,D)$ -invariant subset $A\subseteq G$ and choose $\varepsilon $ -quasi-tiling centers $C_i$ such that these additional properties are satisfied, that is, such that $\bigcup _{i\in F}A_iC_i\subseteq A$ and such that $R\mathrel {\mathop :}= A\setminus \bigcup _{i\in F}A_iC_i$ satisfies $\theta (R)+\theta (\partial _KR)\leq \varepsilon \theta (A)$ . From $F\subseteq J$ and equation (18), we obtain $f(A_i)/\theta (A_i)\leq \unicode{x3bb} +\varepsilon $ for every $i\in F$ , and we compute

$$ \begin{align*} \frac{f\bigg(\bigcup_{i\in F} A_{i}C_i\bigg)}{\theta(A)} &\leq \sum_{i\in F}\sum_{g\in C_i}\frac{f(A_ig)}{\theta(A)}\\ &=\sum_{i\in F}\sum_{g\in C_i}\frac{f(A_{i})}{\theta(A_{i})}\frac{\theta(A_{i})}{\theta(A)}\\ &\leq (\unicode{x3bb}+\varepsilon)\sum_{i\in F}\sum_{g\in C_i}\frac{\theta(A_{i})}{\theta(A)}. \end{align*} $$

By the properties of $\varepsilon $ -quasi-tiling centers, $((A_{i}g)_{g \in C_{i}})_{i \in F}$ is an $\varepsilon $ -disjoint family. Therefore, $\bigcup _{i\in F}A_iC_i\subseteq A$ implies that

$$ \begin{align*}\sum_{i\in F}\sum_{g\in C_i}\frac{\theta(A_{i})}{\theta(A)} = \sum_{i\in F}\sum_{g\in C_i}\frac{\theta(A_{i}g)}{\theta(A)} \leq \frac{1}{1-\varepsilon}\frac{\theta(\bigcup_{i\in J} A_{i}C_i)}{\theta(A)} \leq \frac{1}{1-\varepsilon}.\end{align*} $$

We have shown that

(19) $$ \begin{align} \frac{f\bigg(\bigcup_{i\in F} A_{i}C_i\bigg)}{\theta(A)} &\leq \frac{\unicode{x3bb}+\varepsilon}{1-\varepsilon}. \end{align} $$

Note that $e_G\in K$ implies $K\overline {R}\subseteq R \cup \partial _KR. $ As we require the $\varepsilon $ -quasi-tiling centers to satisfy $\theta (R)+\theta (\partial _KR)\leq \varepsilon \theta (A)$ , we obtain from the choice of the constant c at the beginning of the proof that

$$ \begin{align*}f(\overline{R}) \leq c \theta(K\overline{R}) \leq c(\theta(R)+\theta(\partial_KR)) \leq \varepsilon c \theta(A).\end{align*} $$

Thus, equation (19) yields that

$$ \begin{align*}\frac{f(A)}{\theta(A)} \leq \frac{f(\bigcup_{i\in F}A_iC_i)}{\theta(A)}+\frac{f(\overline{R})}{\theta(A)} \leq \frac{\unicode{x3bb}+\varepsilon}{1-\varepsilon}+\varepsilon c\end{align*} $$

for every $(\delta ,K)$ -invariant, compact subset $A\subseteq G$ . Hence, considering another van Hove net $(B_\iota )_{\iota \in \tilde {I}}$ , we see that

$$ \begin{align*}\limsup_{\iota \in \tilde{I}}\frac{f(B_\iota)}{\theta(B_\iota)} \leq \frac{\unicode{x3bb}+\varepsilon}{1-\varepsilon}+\varepsilon c.\end{align*} $$

Since $\varepsilon>0$ was arbitrary, this shows that, for any two van Hove nets $(A_i)_{i\in I}$ and $(B_\iota )_{\iota \in \tilde {I}}$ in G,

$$ \begin{align*}\limsup_{\iota \in \tilde{I}}\frac{f(B_\iota)}{\theta(B_\iota)} \leq \unicode{x3bb}=\liminf_{i\in I}\frac{f(A_i)}{\theta(A_i)} < c,\end{align*} $$

which clearly implies the statement of the theorem.

Proof of Proposition 5.2

To construct the topological model $\varphi $ and the algebra isomorphism $\iota $ for $\pi $ , we essentially follow the arguments from [Reference Eisner, Farkas, Haase and NagelEFHN15, §12.3]. We will then add to these arguments and argue that $\varphi $ is indeed a continuous action. (This is trivially satisfied if G is discrete.) For the convenience of the reader, we present the full construction.

Note that $L^\infty (X,\mu )$ is a commutative $C^*$ -algebra with unit $\chi _X$ . Thus, by the Gelfand–Naimark theorem, there exists a compact Hausdorff space K and a unital $C^*$ -algebra isomorphism $\Phi \colon C(K)\to L^\infty (X,\mu )$ . By the Riesz–Markov–Kakutani representation theorem, there exists a unique regular Borel probability measure $\nu $ on K such that $\int _K f\,d\nu =\int _X\Phi (f)\,d\mu $ for all $f\in C(K)$ . It is straightforward to observe that the canonical operator $C(K)\to L^\infty (K,\nu )$ is a homeomorphism and that $\Phi $ preserves the respective $L^1$ -norms. In particular, $\nu $ has full support. With standard arguments, one then obtains that $\Phi $ extends uniquely to a Markov isomorphism $L^1(K,\nu )\to L^1(X,\mu )$ , which we also denote by $\Phi $ . The restriction of $\Phi $ to the set of all characteristic functions in the corresponding $L^1$ -spaces then induces an algebra isomorphism $\iota \colon \Sigma (K)\to \Sigma (X)$ .

For $g\in G$ , we consider the Markov automorphism $\Phi ^{-1}\circ \pi ^g_*\circ \Phi \colon C(K)\to C(K)$ . From [Reference Eisner, Farkas, Haase and NagelEFHN15, Theorem 7.23], we know that there exists a unique homeomorphism $\varphi ^g$ such that $\Phi ^{-1}\circ \pi ^g_*\circ \Phi =\varphi ^g_*$ . It is straightforward to check that this defines an action $\varphi \colon G\times K\to K$ , that $\nu $ is invariant, and that $\iota $ establishes an algebra isomorphism between the measure-preserving action $\varphi $ of G (equipped with the discrete topology) on $(K,\nu )$ and $\pi $ .

It remains to show that $\varphi $ is continuous with respect to the original topology of G. Since $\pi $ is measure preserving, any characteristic function $f\in L^1(X,\mu )$ satisfies

$$ \begin{align*} \|\pi^g_*f-f\|_1 \longrightarrow 0 \quad\text{as } g\to e_G. \end{align*} $$

By a straightforward density argument, this readily entails the continuity of the Koopman representation $\pi _*\colon G\to \operatorname {MAut}^1(X,\mu )$ . Since restriction induces a homeomorphism between $\operatorname {MAut}^1(X,\mu )$ and $\operatorname {MAut}^\infty (X,\mu )$ , the Koopman representation $\pi _*\colon G\to \operatorname {MAut}^\infty (X,\mu )$ is continuous, too. For every $f\in C(K)$ , we know that $\Phi (f)\in L^\infty (X,\nu )$ and hence,

$$ \begin{align*} (\pi^g_*\circ \Phi)(f) = (\pi_*^g)(\Phi(f)) \longrightarrow \Phi(f) \quad\text{as } g\to e_G. \end{align*} $$

Thus, $\varphi _*^g(f)=(\Phi ^{-1}\circ \pi ^g_*\circ \Phi )(f)\rightarrow f$ as $g\to e_G$ . This proves the continuity of the Koopman representation $\varphi _*\colon G\to \mathcal {L}(C(K))$ , which implies the continuity of $\varphi $ .

Proof. Let $\alpha $ be a finite measurable partition of X. It is standard to show that the considered map is right-invariant and satisfies Shearer’s inequality (see, e.g., [Reference Downarowicz, Frej and RomagnoliDFR16]). To show the continuity, consider any $F\in \mathcal {F}_+(G)$ . Since $\pi $ is measure preserving, there exists an open neighborhood V of $e_G$ such that $\mu (A\triangle g.A)<\varepsilon /(|F||\alpha ^F|)$ for each $A\in \alpha $ . Now, let $E\in V[F]$ and choose a bijection $b\colon F\to E$ such that $b(g) \in Vg$ for each $g\in F$ . For any map $A^{(\cdot )}\colon F\to \alpha , g\mapsto A^{(g)}$ , we observe that

$$ \begin{align*} \mu\bigg(\! \bigg(\!\bigcap\nolimits_{g\in F}g^{-1}.A^{(g)}\!\bigg) \triangle \bigg(\!\bigcap\nolimits_{g\in F}(b(g))^{-1}.A^{(g)}\!\bigg)\! \bigg) &\leq \mu \bigg(\!\bigcup\nolimits_{g\in F}g^{-1}.A^{(g)}\triangle (b(g))^{-1}.A^{(g)} \!\bigg)\\&\leq \sum\nolimits_{g\in F} \mu (g^{-1}.A^{(g)}\triangle (b(g))^{-1}.A^{(g)} )\\&= \sum\nolimits_{g\in F} \mu((b(g)g^{-1}).A^{(g)}\triangle A^{(g)} )\\&\leq \frac{|F|\varepsilon}{|F||\alpha^F|} = \frac{\varepsilon}{|\alpha^F|}. \end{align*} $$

This shows that

$$ \begin{align*}\mathcal{F}_+(G) \longrightarrow \mathbb{R}_{\geq 0}, \quad F \longmapsto \sum\nolimits_{A^{(\cdot)}\in \alpha^F}\mu\bigg(\!\bigg(\!\bigcap\nolimits_{g\in F}g^{-1}.A^{(g)}\!\bigg) \triangle \bigg(\!\bigcap\nolimits_{g\in F}(b(g))^{-1}.A^{(g)}\!\bigg)\!\bigg)\end{align*} $$

is continuous. By a standard argument (as for example contained in [Reference WaltersWal82, Lemma 4.15]), we thus obtain the continuity of $\mathcal {F}_+(G) \to \mathbb {R}_{\geq 0} , F\mapsto H_\mu (\alpha _F)$ .