Proof. We first check that $\unicode[STIX]{x1D6F7}$ is well defined. To this end, let $f\in \text{UC}_{b}(X)$ . A standard computation shows that $\unicode[STIX]{x1D6F7}(f):\mathbb{R}X\rightarrow \mathbb{R}$ is linear. We need to show that $\unicode[STIX]{x1D6F7}(f)$ is continuous. Without loss of generality, we may assume that $\Vert f\Vert _{\infty }\leqslant 1$ . As $f$ is uniformly continuous, we obtain a uniformly continuous pseudo-metric $d$ on $X$ by setting

$$\begin{eqnarray}\displaystyle d(x,y):=|f(x)-f(y)|\quad (x,y\in X). & & \displaystyle \nonumber\end{eqnarray}$$

Let $\unicode[STIX]{x1D700}>0$ . Since $f:(X,d)\rightarrow [-1,1]$ is $1$ -Lipschitz continuous, we have $|\unicode[STIX]{x1D6F7}(f)(a)|=|a(f)|\leqslant \unicode[STIX]{x1D700}$ for every $a\in B_{p_{d}}[0,\unicode[STIX]{x1D700}]$ . Hence, $\unicode[STIX]{x1D6F7}(f)(B_{p_{d}}[0,\unicode[STIX]{x1D700}])\subseteq [-\unicode[STIX]{x1D700},\unicode[STIX]{x1D700}]$ . So, $\unicode[STIX]{x1D6F7}$ is indeed well defined.

Moreover, $\unicode[STIX]{x1D6F7}$ is obviously a linear map. Clearly, $\unicode[STIX]{x1D6F7}$ is also injective: if $f\in \text{UC}_{b}(X)$ and $f\neq 0$ , then there exists some $x\in X$ with $f(x)\neq 0$ and we conclude that $\unicode[STIX]{x1D6F7}(f)(\unicode[STIX]{x1D6FF}_{x})=f(x)\neq 0$ and hence $\unicode[STIX]{x1D6F7}(f)\neq 0$ . It remains to show that $\unicode[STIX]{x1D6F7}$ is surjective. For this purpose, consider a continuous linear functional $F:\mathbb{R}X\rightarrow \mathbb{R}$ . Define $f:X\rightarrow \mathbb{R},\,x\mapsto F(\unicode[STIX]{x1D6FF}_{x})$ . Now $f$ is bounded for the following reason: as $F$ is continuous, there is a uniformly continuous pseudo-metric $d$ on $X$ such that $F(B_{p_{d}}[0,1])$ is bounded and, since $\{\unicode[STIX]{x1D6FF}_{x}\mid x\in X\}\subseteq B_{p_{d}}[0,1]$ , it follows that

$$\begin{eqnarray}\displaystyle \sup \{|f(x)|\mid x\in X\}=\sup \{|F(\unicode[STIX]{x1D6FF}_{x})|\mid x\in X\}<\infty . & & \displaystyle \nonumber\end{eqnarray}$$

To see that $f$ is uniformly continuous, let $\unicode[STIX]{x1D700}>0$ . By continuity of $F$ , there exists a uniformly continuous pseudo-metric $d$ on $X$ along with some $\unicode[STIX]{x1D6FF}>0$ such that $F(B_{p_{d}}[0,\unicode[STIX]{x1D6FF}])\subseteq [-\unicode[STIX]{x1D700},\unicode[STIX]{x1D700}]$ . If $x,y\in X$ with $d(x,y)\leqslant \unicode[STIX]{x1D6FF}$ , then $p_{d}(\unicode[STIX]{x1D6FF}_{x}-\unicode[STIX]{x1D6FF}_{y})\leqslant \unicode[STIX]{x1D6FF}$ and thus

$$\begin{eqnarray}\displaystyle |f(x)-f(y)|=|F(\unicode[STIX]{x1D6FF}_{x})-F(\unicode[STIX]{x1D6FF}_{y})|=|F(\unicode[STIX]{x1D6FF}_{x}-\unicode[STIX]{x1D6FF}_{y})|\leqslant \unicode[STIX]{x1D700}. & & \displaystyle \nonumber\end{eqnarray}$$

This shows that $f:X\rightarrow \mathbb{R}$ is uniformly continuous. Finally, we observe that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}(f)(a)=\mathop{\sum }_{x\in X}a(x)f(x)=\mathop{\sum }_{x\in X}a(x)F(\unicode[STIX]{x1D6FF}_{x})=F\biggl(\mathop{\sum }_{x\in X}a(x)\unicode[STIX]{x1D6FF}_{x}\biggr)=F(a) & & \displaystyle \nonumber\end{eqnarray}$$

for all $a\in \mathbb{R}X$ . So, $\unicode[STIX]{x1D6F7}$ is surjective. This completes the proof.◻

Proof. Let $a\in \mathbb{R}X$ with $\Vert a\Vert _{1}=1$ and $a\geqslant 0$ . Put $S:=\operatorname{spt}(a)$ . Furthermore, let $\unicode[STIX]{x1D700}>0$ and let $d$ be a uniformly continuous pseudo-metric on $X$ . Since $\mathbb{Q}$ is dense in $\mathbb{R}$ , there exists $b:X\rightarrow \mathbb{Q}$ such that $\operatorname{spt}(b)=S$ , $b\geqslant 0$ , $\Vert b\Vert _{1}=1$ , and $\Vert a-b\Vert _{1}\leqslant \unicode[STIX]{x1D700}/2$ . Now let $c:X\rightarrow \mathbb{N}$ and $n\geqslant 1$ be such that $b(x)=c(x)/n$ for all $x\in X$ . Since $X$ is a perfect Hausdorff space, every open non-empty subset of $X$ is infinite. For any $x\in X$ , the continuity of $d$ implies that $B_{d}(x,\unicode[STIX]{x1D700}/2)$ is a neighborhood of $x$ in $X$ and thus infinite. Hence, there is a map $\unicode[STIX]{x1D711}:S\rightarrow \mathscr{F}(X)\setminus \{\emptyset \}$ with:

1. (1) $\unicode[STIX]{x1D711}(x)\subseteq B_{d}(x,\unicode[STIX]{x1D700}/2)$ for every $x\in S$ ;

2. (2) $|\unicode[STIX]{x1D711}(x)|=c(x)$ for every $x\in S$ ;

3. (3) $\unicode[STIX]{x1D711}(x)\cap \unicode[STIX]{x1D711}(y)=\emptyset$ for any two distinct $x,y\in S$ .

By (2) and (3), we have $|F|=n$ for $F:=\bigcup \{\unicode[STIX]{x1D711}(x)\mid x\in S\}$ . Moreover, if $f:(X,d)\rightarrow [-1,1]$ is $1$ -Lipschitz continuous, then (1) implies that

$$\begin{eqnarray}\displaystyle |(b-\unicode[STIX]{x1D6FF}_{F})(f)| & = & \displaystyle \displaystyle \frac{1}{n}\biggl|\mathop{\sum }_{x\in S}c(x)f(x)-\mathop{\sum }_{x\in S}\mathop{\sum }_{y\in \unicode[STIX]{x1D711}(x)}f(y)\biggr|\leqslant \frac{1}{n}\mathop{\sum }_{x\in S}\biggl|c(x)f(x)-\mathop{\sum }_{y\in \unicode[STIX]{x1D711}(x)}f(y)\biggr|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \displaystyle \frac{1}{n}\mathop{\sum }_{x\in S}\mathop{\sum }_{y\in \unicode[STIX]{x1D711}(x)}|f(x)-f(y)|\leqslant \frac{1}{n}\mathop{\sum }_{x\in S}\mathop{\sum }_{y\in \unicode[STIX]{x1D711}(x)}d(x,y)\leqslant \frac{\unicode[STIX]{x1D700}}{2n}\mathop{\sum }_{x\in S}c(x)=\frac{\unicode[STIX]{x1D700}}{2}\nonumber\end{eqnarray}$$

and therefore

$$\begin{eqnarray}\displaystyle |(a-\unicode[STIX]{x1D6FF}_{F})(f)| & {\leqslant}|(a-b)(f)|+|(b-\unicode[STIX]{x1D6FF}_{F})(f)|\leqslant \Vert a-b\Vert _{1}+|(b-\unicode[STIX]{x1D6FF}_{F})(f)|\leqslant \unicode[STIX]{x1D700}. & \displaystyle \nonumber\end{eqnarray}$$

Hence, $\unicode[STIX]{x1D6FF}_{F}\in B_{p_{d}}[a,\unicode[STIX]{x1D700}]$ . This finishes the proof.◻

Proof. Let $a\in \mathbb{R}G$ with $\Vert a\Vert _{1}=1$ and $a\geqslant 0$ . For all $x,y\in G$ ,

$$\begin{eqnarray}\displaystyle |R_{a}f(x)-R_{a}f(y)|\leqslant \mathop{\sum }_{g\in G}a(g)|f(xg)-f(yg)|\leqslant \mathop{\sum }_{g\in G}a(g)d(xg,yg)=d(x,y).\Box & & \displaystyle \nonumber\end{eqnarray}$$

Proof. ( $\Longrightarrow$ ) Our proof proceeds by contradiction. To this end, let us consider the set $T:=\{a\in \mathbb{R}G\mid \Vert a\Vert _{1}=1,\,a\geqslant 0\}$ . Suppose that there exist $\unicode[STIX]{x1D700}>0$ , a finite subset $E\subseteq G$ , and a right-invariant continuous pseudo-metric $d$ on $G$ such that $\sup _{g\in E}p_{d}(a-ga)>\unicode[STIX]{x1D700}$ for all $a\in T$ . Then $0$ is not contained in the closure of the convex subset $\{(a-ga)_{g\in E}\mid a\in T\}$ in the locally convex space $(\mathbb{R}G)^{E}$ . Applying the separation theorem for locally convex spaces, we conclude that there exists a continuous linear functional $F\in ((\mathbb{R}G)^{E})^{\prime }$ such that $F((a-ga)_{g\in E})\geqslant 1$ for all $a\in T$ . Since $((\mathbb{R}G)^{E})^{\prime }\cong ((\mathbb{R}G)^{\prime })^{E}$ , Proposition 2.1 asserts the existence of $f_{g}\in \text{RUC}_{b}(G)$ $(g\in E)$ such that $\sum _{g\in E}(a-ga)(f_{g})\geqslant 1$ for all $a\in T$ . That is,

$$\begin{eqnarray}\displaystyle a\biggl(\mathop{\sum }_{g\in E}f_{g}-f_{g}\circ \unicode[STIX]{x1D706}_{g}\biggr)=\mathop{\sum }_{g\in E}a(f_{g})-a(f_{g}\circ \unicode[STIX]{x1D706}_{g})=\mathop{\sum }_{g\in E}(a-ga)(f_{g})\geqslant 1 & & \displaystyle \nonumber\end{eqnarray}$$

for all $a\in T$ . Since $\{\unicode[STIX]{x1D6FF}_{x}\mid x\in G\}\subseteq T$ , it follows that $\sum _{g\in E}f_{g}-f_{g}\circ \unicode[STIX]{x1D706}_{g}\geqslant 1$ . Now, if $\unicode[STIX]{x1D707}$ was a left-invariant mean on $\text{RUC}_{b}(G)$ , then $\unicode[STIX]{x1D707}(\sum _{g\in E}f_{g}-f_{g}\circ \unicode[STIX]{x1D706}_{g})\geqslant 1$ , but also

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D707}\biggl(\mathop{\sum }_{g\in E}f_{g}-f_{g}\circ \unicode[STIX]{x1D706}_{g}\biggr)=\mathop{\sum }_{g\in E}\unicode[STIX]{x1D707}(f_{g})-\unicode[STIX]{x1D707}(f_{g}\circ \unicode[STIX]{x1D706}_{g})=0, & & \displaystyle \nonumber\end{eqnarray}$$

which would clearly constitute a contradiction. Hence, $G$ is not amenable.

( $\Longleftarrow$ ) Consider the set $T:=\{a\in \mathbb{R}G\mid \Vert a\Vert _{1}=1,\,a\geqslant 0\}$ . Our assumption (together with the ultrafilter lemma) implies that there exists a set $I$ along with an ultrafilter $\mathscr{U}$ on $I$ and a family $(a_{i})_{i\in I}\in T^{I}$ such that $\lim _{i\rightarrow \mathscr{U}}p_{d}(a_{i}-ga_{i})=0$ for every $g\in G$ and every right-invariant continuous pseudo-metric $d$ on $G$ . Let us define $\unicode[STIX]{x1D707}:\text{RUC}_{b}(G)\rightarrow \mathbb{R}$ by

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D707}(f):=\lim _{i\rightarrow \mathscr{U}}a_{i}(f)\quad (f\in \text{RUC}_{b}(G)). & & \displaystyle \nonumber\end{eqnarray}$$

It is easy to see that $\unicode[STIX]{x1D707}$ is indeed a well-defined mean on $\text{RUC}_{b}(G)$ . In order to prove that $\unicode[STIX]{x1D707}$ is also left invariant, let $f\in \text{RUC}_{b}(G)$ . Without loss of generality, we assume that $\Vert f\Vert _{\infty }\leqslant 1$ . Consider the right-invariant continuous pseudo-metric $d$ on $G$ given by

$$\begin{eqnarray}\displaystyle d(x,y):=\sup _{g\in G}|f(xg)-f(yg)|\quad (x,y\in G). & & \displaystyle \nonumber\end{eqnarray}$$

As $f:(G,d)\rightarrow [-1,1]$ is $1$ -Lipschitz continuous, we conclude that

$$\begin{eqnarray}\displaystyle |\unicode[STIX]{x1D707}(f)-\unicode[STIX]{x1D707}(f\circ \unicode[STIX]{x1D706}_{g})|=\lim _{i\rightarrow \mathscr{U}}|(a_{i}-ga_{i})(f)|\leqslant \lim _{i\rightarrow \mathscr{U}}p_{d}(a_{i}-ga_{i})=0 & & \displaystyle \nonumber\end{eqnarray}$$

for every $g\in G$ . This completes the proof.◻

Proof. (1) $\Longrightarrow$ (4). This is an immediate consequence of Theorem 3.2.

(4) $\Longrightarrow$ (3). This is obvious.

(2) $\Longrightarrow$ (1). We are going to utilize Theorem 3.2. To this end, let $\unicode[STIX]{x1D700}>0$ , let $E$ be a finite subset of $G$ , and let $d$ be a right-invariant continuous pseudo-metric on  $G$ . By assumption, there exists $a\in \mathbb{R}G$ with $\Vert a\Vert _{1}=1$ and $a\geqslant 0$ such that

$$\begin{eqnarray}\displaystyle \forall g\in E\,\forall f:(G,d)\rightarrow [0,1]~1\text{-Lipschitz}:\quad |a(f)-(ga)(f)|\leqslant \frac{\unicode[STIX]{x1D700}}{2}. & & \displaystyle \nonumber\end{eqnarray}$$

We argue that $p_{d}(a-ga)\leqslant \unicode[STIX]{x1D700}$ for all $g\in E$ . Let $f:(G,d)\rightarrow [-1,1]$ be $1$ -Lipschitz continuous. Consider the two $1$ -Lipschitz functions $f^{+},f^{-}:(G,d)\rightarrow [0,1]$ given by $f^{+}(x):=f(x)\vee 0$ and $f^{-}(x):=-(f(x)\wedge 0)$ for $x\in G$ . It follows that

$$\begin{eqnarray}\displaystyle |a(f)-(ga)(f)|\leqslant |a(f^{+})-(ga)(f^{+})|+|a(f^{-})-(ga)(f^{-})|\leqslant \unicode[STIX]{x1D700} & & \displaystyle \nonumber\end{eqnarray}$$

for every $g\in E$ , which completes the argument.

(3) $\Longrightarrow$ (2). Suppose that (3) holds for some $\unicode[STIX]{x1D700}\in (0,1)$ . Let us first prove the following.

Proof of Claim 1.

Let $E$ be a finite subset of $G$ and let $d$ be a right-invariant continuous pseudo-metric on  $G$ . Consider the right-invariant continuous pseudo-metric $d^{\prime }$ on $G$ defined by $d^{\prime }(x,y):=\sup _{g\in E}d(gx,gy)$ for $x,y\in G$ . Due to (3), there exists $a\in \mathbb{R}G$ with $\Vert a\Vert _{1}=1$ and $a\geqslant 0$ such that

$$\begin{eqnarray}\displaystyle \forall g\in E^{-1}E\,\forall f:(G,d^{\prime })\rightarrow [0,1]~1\text{-Lipschitz}:\quad |a(f)-(ga)(f)|\leqslant \unicode[STIX]{x1D700}. & & \displaystyle \nonumber\end{eqnarray}$$

Now, if $g,h\in E$ and $f:(G,d)\rightarrow [0,1]$ is $1$ -Lipschitz continuous, then $f\circ \unicode[STIX]{x1D706}_{g}:(G,d^{\prime })\rightarrow [0,1]$ is $1$ -Lipschitz continuous and hence

$$\begin{eqnarray}\displaystyle |(ga)(f)-(ha)(f)|=|a(f\circ \unicode[STIX]{x1D706}_{g})-(g^{-1}ha)(f\circ \unicode[STIX]{x1D706}_{g})|\leqslant \unicode[STIX]{x1D700}.\Box & & \displaystyle \nonumber\end{eqnarray}$$

Proof of Claim 2.

Our proof will proceed by downward induction starting at $i=n-2$ . Since $d_{n-1}=d$ , our choice of $a_{n-1}$ readily implies that

$$\begin{eqnarray}\displaystyle 0\leqslant (R_{a_{n-1}}f)(g)-\min _{h\in E_{n-1}}(R_{a_{n-1}}f)(h)=\max _{h\in E_{n-1}}(ga_{n-1})(f)-(ha_{n-1})(f)\leqslant \unicode[STIX]{x1D700} & & \displaystyle \nonumber\end{eqnarray}$$

for all $g\in E_{n-1}$ , which means that $f_{n-2}(E_{n-1})\subseteq [0,1]$ and therefore $f_{n-2}|_{E_{n-1}}=f_{n-2}^{\ast }|_{E_{n-1}}$ . For the inductive step, let us assume that $f_{i+1}|_{E_{i+2}}=f_{i+1}^{\ast }|_{E_{i+2}}$ for some $i\in \{0,\ldots ,n-3\}$ . Since $E_{i+1}(\operatorname{spt}a_{i+1})\subseteq E_{i+2}$ , this implies that $(ga_{i+1})(f_{i+1})=(ga_{i+1})(f_{i+1}^{\ast })$ for all $g\in E_{i+1}$ . By our choice of $a_{i+1}$ , it follows that

$$\begin{eqnarray}\displaystyle 0 & {\leqslant} & \displaystyle R_{a_{i+1}\cdots a_{n-1}}f(g)-\min _{h\in E_{i+1}}R_{a_{i+1}\cdots a_{n-1}}f(h)\nonumber\\ \displaystyle & = & \displaystyle \max _{h\in E_{i+1}}(ga_{i+1})(R_{a_{i+2}\cdots a_{n-1}}f)-(ha_{i+1})(R_{a_{i+2}\cdots a_{n-1}}f)\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D700}^{n-i-2}\Bigl(\max _{h\in E_{i+1}}(ga_{i+1})(f_{i+1})-(ha_{i+1})(f_{i+1})\Bigr)\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D700}^{n-i-2}\Bigl(\max _{h\in E_{i+1}}(ga_{i+1})(f_{i+1}^{\ast })-(ha_{i+1})(f_{i+1}^{\ast })\Bigr)\leqslant \unicode[STIX]{x1D700}^{n-i-1}\nonumber\end{eqnarray}$$

whenever $g\in E_{i+1}$ . Hence, $f_{i}(E_{i+1})\subseteq [0,1]$ and therefore $f_{i}|_{E_{i+1}}=f_{i}^{\ast }|_{E_{i+1}}$ . This completes our induction and thus proves Claim 2.◻

Proof. (2) $\Longrightarrow$ (3). This is obvious.

(1) $\Longrightarrow$ (2). Since the discrete case is covered by Theorem 4.1 already, we will insist on $G$ being non-discrete. Let $\unicode[STIX]{x1D703}\in (0,1)$ , $E\in \mathscr{F}(G)$ , and $U\in \mathscr{U}_{e}(G)$ . Without loss of generality, we may assume that $e\in E$ . Due to Urysohn’s lemma for uniform spaces, there exists a right-uniformly continuous function $f:G\rightarrow [0,1]$ such that $\operatorname{spt}(f)\subseteq U$ and $f(e)=1$ . Consider the continuous right-invariant pseudo-metrics $d$ and $d^{\prime }$ on $G$ given by

$$\begin{eqnarray}\displaystyle d(x,y):=\sup _{g\in G}|f(xg)-f(yg)|,\quad d^{\prime }(x,y):=\sup _{g\in E}d(gx,gy)\quad (x,y\in G). & & \displaystyle \nonumber\end{eqnarray}$$

By Theorem 3.2, there is $a\in \mathbb{R}G$ with $\Vert a\Vert _{1}=1$ , $a\geqslant 0$ , and $p_{d}(a-ga)\leqslant (1-\unicode[STIX]{x1D703})/3$ for all  $g\in E$ . Since $G$ is not discrete, it constitutes a perfect Hausdorff space. Due to Lemma 2.2, there thus exists a finite non-empty subset $F\subseteq G$ such that $p_{d^{\prime }}(a-\unicode[STIX]{x1D6FF}_{F})\leqslant (1-\unicode[STIX]{x1D703})/3$ . It follows that

$$\begin{eqnarray}\displaystyle p_{d}(\unicode[STIX]{x1D6FF}_{F}-\unicode[STIX]{x1D6FF}_{gF})\leqslant p_{d}(\unicode[STIX]{x1D6FF}_{F}-a)+p_{d}(a-ga)+p_{d}(ga-g\unicode[STIX]{x1D6FF}_{F})\leqslant p_{d}(a-ga)+2p_{d^{\prime }}(a-\unicode[STIX]{x1D6FF}_{F})\leqslant 1-\unicode[STIX]{x1D703} & & \displaystyle \nonumber\end{eqnarray}$$

for all $g\in E$ . We show that $\unicode[STIX]{x1D707}(F,gF,U)\geqslant \unicode[STIX]{x1D703}|F|$ for every $g\in E$ . So, let $g\in E$ and consider the bipartite graph $\mathscr{B}:=\mathscr{B}(F,gF,U)$ . Let $S\subseteq F$ . Since $d$ is right invariant, $f\circ \unicode[STIX]{x1D70C}_{h}:(G,d)\rightarrow [0,1]$ is $1$ -Lipschitz continuous for every $h\in G$ . Hence, the function $f_{S}:(G,d)\rightarrow [0,1]$ given by

$$\begin{eqnarray}\displaystyle f_{S}(x):=\mathop{\vee }_{y\in S}f(xy^{-1})\quad (x\in G) & & \displaystyle \nonumber\end{eqnarray}$$

is $1$ -Lipschitz continuous, too. Therefore, $|\unicode[STIX]{x1D6FF}_{F}(f_{S})-\unicode[STIX]{x1D6FF}_{gF}(f_{S})|\leqslant 1-\unicode[STIX]{x1D703}$ . Also note that $f_{S}|_{S}=1$ . As $\operatorname{spt}(f)\subseteq U$ , it now follows that

$$\begin{eqnarray}\displaystyle |S| & {\leqslant} & \displaystyle |F|\unicode[STIX]{x1D6FF}_{F}(f_{S})\leqslant (1-\unicode[STIX]{x1D703})|F|+|F|\unicode[STIX]{x1D6FF}_{gF}(f_{S})=(1-\unicode[STIX]{x1D703})|F|+\mathop{\sum }_{x\in gF}f_{S}(x)\nonumber\\ \displaystyle & = & \displaystyle (1-\unicode[STIX]{x1D703})|F|+\mathop{\sum }_{x\in N_{\mathscr{B}}(S)}f_{S}(x)\leqslant (1-\unicode[STIX]{x1D703})|F|+|N_{\mathscr{B}}(S)|,\nonumber\end{eqnarray}$$

that is, $|S|-|N_{\mathscr{B}}(S)|\leqslant (1-\unicode[STIX]{x1D703})|F|$ . Applying Theorem 4.2, we arrive at

$$\begin{eqnarray}\displaystyle \frac{\unicode[STIX]{x1D707}(F,gF,U)}{|F|} & =\frac{|F|-\sup _{S\subseteq F}(|S|-|N_{\mathscr{B}}(S)|)}{|F|}\geqslant \frac{|F|-(1-\unicode[STIX]{x1D703})|F|}{|F|}=\unicode[STIX]{x1D703}. & \displaystyle \nonumber\end{eqnarray}$$

(3) $\Longrightarrow$ (1). We are going to prove that if (2) holds for some $\unicode[STIX]{x1D703}\in (0,1)$ , then the third condition of Theorem 3.3 is satisfied for $\unicode[STIX]{x1D700}:=1-\unicode[STIX]{x1D703}/2$ . Consider a finite subset $E\subseteq G$ and a right-invariant continuous pseudo-metric $d$ on $G$ . Then $U:=\{x\in G\mid d(x,e)\leqslant {\textstyle \frac{1}{2}}\}$ is an identity neighborhood in  $G$ . According to (2), there exists a finite non-empty subset $F\subseteq G$ such that $\unicode[STIX]{x1D707}(F,gF,U)\geqslant \unicode[STIX]{x1D703}|F|$ for all $g\in E$ . We show that $p_{d}(\unicode[STIX]{x1D6FF}_{F}-g\unicode[STIX]{x1D6FF}_{F})\leqslant \unicode[STIX]{x1D700}$ for every $g\in E$ . To this end, let $g\in E$ and consider an injective map $\unicode[STIX]{x1D711}:D\rightarrow gF$ with $D\subseteq F$ , $|D|=\unicode[STIX]{x1D707}(F,gF,U)$ , and $\unicode[STIX]{x1D711}(x)\in Ux$ for all $x\in D$ . Let $\bar{\unicode[STIX]{x1D711}}:F\rightarrow gF$ be any bijection with $\bar{\unicode[STIX]{x1D711}}|_{D}=\unicode[STIX]{x1D711}$ . Now, if $f:(G,d)\rightarrow [0,1]$ is $1$ -Lipschitz continuous, then

$$\begin{eqnarray}\displaystyle |\unicode[STIX]{x1D6FF}_{F}(f)-(g\unicode[STIX]{x1D6FF}_{F})(f)| & = & \displaystyle \frac{1}{|F|}\biggl|\mathop{\sum }_{x\in F}f(x)-\mathop{\sum }_{x\in F}f(gx)\biggr|\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{|F|}\biggl|\mathop{\sum }_{x\in D}(f(x)-f(\bar{\unicode[STIX]{x1D711}}(x)))+\mathop{\sum }_{x\in F\setminus D}(f(x)-f(\bar{\unicode[STIX]{x1D711}}(x)))\biggr|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{1}{|F|}\mathop{\sum }_{x\in D}|f(x)-f(\unicode[STIX]{x1D711}(x))|+\frac{1}{|F|}\mathop{\sum }_{x\in F\setminus D}|f(x)-f(\bar{\unicode[STIX]{x1D711}}(x))|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{|D|}{2|F|}+\frac{|F|-|D|}{|F|}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 1-\frac{\unicode[STIX]{x1D703}}{2}=\unicode[STIX]{x1D700}.\Box \nonumber\end{eqnarray}$$

Proof. ( $\Longleftarrow$ ) This is due to Theorem 4.5.

( $\Longrightarrow$ ) Assume that $G$ is amenable. Let $\unicode[STIX]{x1D703}\in (0,1)$ , $U\in \mathscr{U}_{e}(G)$ , and consider a finite subset $E\subseteq G$ . Then $V:=\bigcap _{g\in E}g^{-1}Ug$ is an identity neighborhood in $G$ , too. Hence, Theorem 4.5 asserts the existence of a finite non-empty subset $F\subseteq G$ such that $\unicode[STIX]{x1D707}(F,gF,V)\geqslant \unicode[STIX]{x1D703}|F|$ for all $g\in E^{-1}E$ . We prove that $\unicode[STIX]{x1D707}(gF,hF,U)\geqslant \unicode[STIX]{x1D703}|F|$ for all $g,h\in E$ . To this end, let $g,h\in E$ . Consider an injective map $\unicode[STIX]{x1D711}:D\rightarrow g^{-1}hF$ such that $D\subseteq F$ , $|D|=\unicode[STIX]{x1D707}(F,g^{-1}hF,V)$ , and $\unicode[STIX]{x1D711}(x)\in Vx$ for all $x\in D$ . Define $D^{\prime }:=gD$ . Clearly,

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D713}:D^{\prime }\rightarrow hF,\quad x\mapsto g\unicode[STIX]{x1D711}(g^{-1}x) & & \displaystyle \nonumber\end{eqnarray}$$

is injective and $\unicode[STIX]{x1D713}(x)=g\unicode[STIX]{x1D711}(g^{-1}x)\in gVg^{-1}x\subseteq Ux$ for every $x\in D^{\prime }$ . Hence,

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D707}(gF,hF,U)\geqslant |D^{\prime }|=|D|=\unicode[STIX]{x1D707}(F,g^{-1}hF,V)\geqslant \unicode[STIX]{x1D703}|F|.\Box & & \displaystyle \nonumber\end{eqnarray}$$

Proof. The implication from (2) to (1) is an immediate consequence of Rickert’s fixed point theorem [Reference RickertRic67, Theorem 4.2] and the fact that continuous actions of topological groups on compact spaces are bounded. To prove the converse, consider any bounded action of $G$ by affine homeomorphisms on a bounded convex subset $C$ of a locally convex space $X$ . Fix some $x\in C$ . By Theorem 4.5, there is a net $(F_{\unicode[STIX]{x1D704}})_{\unicode[STIX]{x1D704}\in I}$ of finite non-empty subsets of $G$ such that

$$\begin{eqnarray}\displaystyle \forall g\in G\,\forall U\in \mathscr{U}_{e}(G):\quad \frac{\unicode[STIX]{x1D707}(F_{\unicode[STIX]{x1D704}},gF_{\unicode[STIX]{x1D704}},U)}{|F_{\unicode[STIX]{x1D704}}|}\rightarrow 1. & & \displaystyle \nonumber\end{eqnarray}$$

For each $\unicode[STIX]{x1D704}\in I$ , let us consider

$$\begin{eqnarray}\displaystyle x_{\unicode[STIX]{x1D704}}:=\frac{1}{|F_{\unicode[STIX]{x1D704}}|}\mathop{\sum }_{h\in F_{\unicode[STIX]{x1D704}}}hx\in C. & & \displaystyle \nonumber\end{eqnarray}$$

We claim that $x_{\unicode[STIX]{x1D704}}-gx_{\unicode[STIX]{x1D704}}\rightarrow 0$ for every $g\in G$ . To see this, let $g\in G$ and let $N$ be a zero neighborhood in $X$ . Since $X$ is locally convex, there is a balanced, convex zero neighborhood $V$ in $X$ such that $V+V\subseteq N$ . As $C$ is bounded, there exists $\unicode[STIX]{x1D6FC}\in \mathbb{K}$ such that $C-C\subseteq \unicode[STIX]{x1D6FC}V$ . Since $G$ acts on $C$ in a bounded manner, we can find some $U\in \mathscr{U}_{e}(G)$ so that $hx\in V+x$ for all $x\in C$ and $h\in U$ . By our hypothesis about $(F_{\unicode[STIX]{x1D704}})_{\unicode[STIX]{x1D704}\in I}$ , there exists $\unicode[STIX]{x1D704}_{0}\in I$ such that

$$\begin{eqnarray}\displaystyle \forall \unicode[STIX]{x1D704}\in I,\,\unicode[STIX]{x1D704}\geqslant \unicode[STIX]{x1D704}_{0}:\quad \frac{\unicode[STIX]{x1D707}(F_{\unicode[STIX]{x1D704}},gF_{\unicode[STIX]{x1D704}},U)}{|F_{\unicode[STIX]{x1D704}}|}\geqslant 1-|\unicode[STIX]{x1D6FC}|^{-1}. & & \displaystyle \nonumber\end{eqnarray}$$

We show that $x_{\unicode[STIX]{x1D704}}-gx_{\unicode[STIX]{x1D704}}\in V$ for all $\unicode[STIX]{x1D704}\in I$ with $\unicode[STIX]{x1D704}\geqslant \unicode[STIX]{x1D704}_{0}$ . So, let $\unicode[STIX]{x1D704}\in I$ , $\unicode[STIX]{x1D704}\geqslant \unicode[STIX]{x1D704}_{0}$ . Due to the above, there exists a subset $D_{\unicode[STIX]{x1D704}}\subseteq F_{\unicode[STIX]{x1D704}}$ along with a bijection $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D704}}:F_{\unicode[STIX]{x1D704}}\rightarrow gF_{\unicode[STIX]{x1D704}}$ such that $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D704}}(h)\in Uh$ for each $h\in D_{\unicode[STIX]{x1D704}}$ and moreover

$$\begin{eqnarray}\displaystyle \frac{|D_{\unicode[STIX]{x1D704}}|}{|F_{\unicode[STIX]{x1D704}}|}\geqslant 1-|\unicode[STIX]{x1D6FC}|^{-1},\quad \text{i.e.,}\quad \frac{|F_{\unicode[STIX]{x1D704}}\setminus D_{\unicode[STIX]{x1D704}}|}{|F_{\unicode[STIX]{x1D704}}|}|\unicode[STIX]{x1D6FC}|\leqslant 1. & & \displaystyle \nonumber\end{eqnarray}$$

It follows that

$$\begin{eqnarray}\displaystyle x_{\unicode[STIX]{x1D704}}-gx_{\unicode[STIX]{x1D704}} & = & \displaystyle \frac{1}{|F_{\unicode[STIX]{x1D704}}|}\biggl(\mathop{\sum }_{h\in F_{\unicode[STIX]{x1D704}}}hx-\mathop{\sum }_{h\in F_{\unicode[STIX]{x1D704}}}ghx\biggr)\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{|F_{\unicode[STIX]{x1D704}}|}\biggl(\mathop{\sum }_{h\in D_{\unicode[STIX]{x1D704}}}hx-\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D704}}(h)x+\mathop{\sum }_{h\in F_{\unicode[STIX]{x1D704}}\setminus D_{\unicode[STIX]{x1D704}}}hx-\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D704}}(h)x\biggr)\nonumber\\ \displaystyle & = & \displaystyle \frac{|D_{\unicode[STIX]{x1D704}}|}{|F_{\unicode[STIX]{x1D704}}|}\biggl(\frac{1}{|D_{\unicode[STIX]{x1D704}}|}\mathop{\sum }_{h\in D_{\unicode[STIX]{x1D704}}}hx-\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D704}}(h)x\biggr)+\frac{|F_{\unicode[STIX]{x1D704}}\setminus D_{\unicode[STIX]{x1D704}}|}{|F_{\unicode[STIX]{x1D704}}|}\biggl(\frac{1}{|F_{\unicode[STIX]{x1D704}}\setminus D_{\unicode[STIX]{x1D704}}|}\mathop{\sum }_{h\in F_{\unicode[STIX]{x1D704}}\setminus D_{\unicode[STIX]{x1D704}}}hx-\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D704}}(h)x\biggr)\nonumber\\ \displaystyle & \in & \displaystyle \frac{|D_{\unicode[STIX]{x1D704}}|}{|F_{\unicode[STIX]{x1D704}}|}V+\frac{|F_{\unicode[STIX]{x1D704}}\setminus D_{\unicode[STIX]{x1D704}}|}{|F_{\unicode[STIX]{x1D704}}|}(C-C)\,\subseteq \,V+\frac{|F_{\unicode[STIX]{x1D704}}\setminus D_{\unicode[STIX]{x1D704}}|}{|F_{\unicode[STIX]{x1D704}}|}\unicode[STIX]{x1D6FC}V\,\subseteq \,V+V\,\subseteq \,N,\nonumber\end{eqnarray}$$

which completes the argument. ◻

Proof. Denote by $\mathscr{Z}$ the set of all pairs $(D,\unicode[STIX]{x1D711})$ consisting of a subset $D\subseteq F$ and an injective map $\unicode[STIX]{x1D711}:D\rightarrow G$ such that $\unicode[STIX]{x1D711}(x)\in Ux$ for all $x\in D$ and $\unicode[STIX]{x1D711}(D)\cap g\unicode[STIX]{x1D711}(D)=\emptyset$ for every $g\in E\setminus \{e\}$ . Let $(D,\unicode[STIX]{x1D711})\in \mathscr{Z}$ be such that $|D|=\sup \{|D^{\prime }|\mid (D^{\prime },\unicode[STIX]{x1D713})\in \mathscr{Z}\}$ . We claim that $|D|=|F|$ . For contraposition, assume that $|D|<|F|$ . Then there exists $x\in F\setminus D$ . Since $G$ is a perfect Hausdorff space, every open non-empty subset of $G$ is infinite. Consequently, there exists some

$$\begin{eqnarray}\displaystyle y\in Ux\bigg\backslash\biggl(\unicode[STIX]{x1D711}(D)\cup \mathop{\bigcup }_{g\in E}g\unicode[STIX]{x1D711}(D)\cup g^{-1}\unicode[STIX]{x1D711}(D)\biggr). & & \displaystyle \nonumber\end{eqnarray}$$

We define $D^{\prime }:=D\cup \{x\}$ and $\unicode[STIX]{x1D713}:D^{\prime }\rightarrow F$ such that $\unicode[STIX]{x1D713}|_{D}=\unicode[STIX]{x1D711}$ and $\unicode[STIX]{x1D713}(x)=y$ . We observe that $(D^{\prime },\unicode[STIX]{x1D713})$ is a member of $\mathscr{Z}$ . Since $|D|<|D^{\prime }|$ , this clearly contradicts our hypothesis. Hence, $|D|=|F|$ . This finishes the proof.◻

Proof. ( $\Longleftarrow$ ) This is an immediate consequence of Theorem 4.5.

( $\Longrightarrow$ ) Let $\unicode[STIX]{x1D703}\in (0,1)$ , $E\subseteq \mathscr{F}(G)$ , and $U\in \mathscr{U}_{e}(G)$ . Without loss of generality, suppose that $e\in E=E^{-1}$ . Choose $V\in \mathscr{U}_{e}(G)$ with $V^{-1}=V$ and $V^{3}\subseteq U$ and let $W:=\bigcap _{g\in E}g^{-1}Vg$ . By Theorem 4.5, there exists a finite non-empty subset $F_{0}\subseteq G$ such that

$$\begin{eqnarray}\displaystyle \forall g\in E:\quad \unicode[STIX]{x1D707}(F_{0},gF_{0},V)\geqslant \biggl(1-\frac{1-\unicode[STIX]{x1D703}}{|E|}\biggr)|F_{0}|. & & \displaystyle \nonumber\end{eqnarray}$$

By Lemma 5.5, there is an injection $\unicode[STIX]{x1D6FC}:F_{0}\rightarrow G$ with $\unicode[STIX]{x1D6FC}(x)\in Wx$ for all $x\in F_{0}$ . For each $g\in E$ , choose a subset $D_{g}\subseteq F_{0}$ and an injection $\unicode[STIX]{x1D713}_{g}:D_{g}\rightarrow gF_{0}$ such that $|D_{g}|=\unicode[STIX]{x1D707}(F_{0},gF_{0},V)$ and $\unicode[STIX]{x1D713}_{g}(x)\in Vx$ for all $x\in D_{g}$ . Define $F:=\unicode[STIX]{x1D6FC}(F_{0})$ and $D:=\bigcap _{g\in E}\unicode[STIX]{x1D6FC}(D_{g})$ . Clearly, $F\cap gF=\emptyset$ for every $g\in E\setminus \{e\}$ . Furthermore, note that

$$\begin{eqnarray}\displaystyle |D| & = & \displaystyle |F|-|F\setminus D|=|F_{0}|-\biggl|\mathop{\bigcup }_{g\in E}\unicode[STIX]{x1D6FC}(F_{0}\setminus D_{g})\biggr|\geqslant |F_{0}|-\mathop{\sum }_{g\in E}|F_{0}\setminus D_{g}|\nonumber\\ \displaystyle & {\geqslant} & \displaystyle |F_{0}|-|E|\biggl(|F_{0}|-\biggl(1-\frac{1-\unicode[STIX]{x1D703}}{|E|}\biggr)|F_{0}|\biggr)=\unicode[STIX]{x1D703}|F_{0}|=\unicode[STIX]{x1D703}|F|.\nonumber\end{eqnarray}$$

Finally, for each $g\in E$ , we consider the injection $\unicode[STIX]{x1D711}_{g}:D\rightarrow gF$ given by

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D711}_{g}(x):=g\unicode[STIX]{x1D6FC}(g^{-1}\unicode[STIX]{x1D713}_{g}(\unicode[STIX]{x1D6FC}^{-1}(x)))\quad (x\in D) & & \displaystyle \nonumber\end{eqnarray}$$

and we observe that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D711}_{g}(x)\in gWg^{-1}\unicode[STIX]{x1D713}_{g}(\unicode[STIX]{x1D6FC}^{-1}(x))\subseteq V\unicode[STIX]{x1D713}_{g}(\unicode[STIX]{x1D6FC}^{-1}(x))\subseteq V^{2}\unicode[STIX]{x1D6FC}^{-1}(x)\subseteq V^{3}x\subseteq Ux & & \displaystyle \nonumber\end{eqnarray}$$

for all $x\in D$ . This completes the argument.◻

Proof of Theorem 5.3.

Note that the discrete case is trivial, which is why we will assume that $G$ is not discrete.

(1) $\Longrightarrow$ (2). Due to being a non-discrete Hausdorff space, $G$ is infinite. Thus, $|I|=|G|$ for $I:=\{(E,n)\in \mathscr{F}(G)\times (\mathbb{N}\setminus \{0\})\mid e\in E\}$ . Denote by $\unicode[STIX]{x1D705}$ the cardinality of $I$ , i.e., the least ordinal equinumerous to $I$ . Fix a bijection $\unicode[STIX]{x1D705}\rightarrow I,\,\unicode[STIX]{x1D714}\mapsto (E_{\unicode[STIX]{x1D714}},n_{\unicode[STIX]{x1D714}})$ . Let $U\in \mathscr{U}_{e}(G)$ . Without loss of generality, we assume that $U^{-1}=U$ . By transfinite recursion, we will construct a family $(F_{\unicode[STIX]{x1D714}},D_{\unicode[STIX]{x1D714}},(\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D714}.g})_{g\in E_{\unicode[STIX]{x1D714}}\setminus \{e\}})_{\unicode[STIX]{x1D714}<\unicode[STIX]{x1D705}}$ such that the following hold:

1. (i) $D_{\unicode[STIX]{x1D714}}\subseteq F_{\unicode[STIX]{x1D714}}\in \mathscr{F}(G)\setminus \{\emptyset \}$ and $|D_{\unicode[STIX]{x1D714}}|\geqslant (1-1/n_{\unicode[STIX]{x1D714}})|F_{\unicode[STIX]{x1D714}}|$ for all $\unicode[STIX]{x1D714}<\unicode[STIX]{x1D705}$ ;

2. (ii) $F_{\unicode[STIX]{x1D714}}\cap gF_{\unicode[STIX]{x1D714}}=\emptyset$ for all $\unicode[STIX]{x1D714}<\unicode[STIX]{x1D705}$ and $g\in E_{\unicode[STIX]{x1D714}}\setminus \{e\}$ ;

3. (iii) $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D714},g}:D_{\unicode[STIX]{x1D714}}\rightarrow gF_{\unicode[STIX]{x1D714}}$ is an injection for all $\unicode[STIX]{x1D714}<\unicode[STIX]{x1D705}$ and $g\in E_{\unicode[STIX]{x1D714}}\setminus \{e\}$ ;

4. (iv) $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D714},g}(x)\in Ux$ for all $\unicode[STIX]{x1D714}<\unicode[STIX]{x1D705}$ , $g\in E_{\unicode[STIX]{x1D714}}\setminus \{e\}$ , and $x\in D_{\unicode[STIX]{x1D714}}$ ;

5. (v) $E_{\unicode[STIX]{x1D714}}F_{\unicode[STIX]{x1D714}}\cap E_{\unicode[STIX]{x1D714}^{\prime }}F_{\unicode[STIX]{x1D714}^{\prime }}=\emptyset$ for any two $\unicode[STIX]{x1D714},\unicode[STIX]{x1D714}^{\prime }<\unicode[STIX]{x1D705}$ with $\unicode[STIX]{x1D714}\neq \unicode[STIX]{x1D714}^{\prime }$ .

For the recursive step, let $\unicode[STIX]{x1D714}<\unicode[STIX]{x1D705}$ and suppose that $(F_{\unicode[STIX]{x1D70E}},D_{\unicode[STIX]{x1D70E}},(\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D70E},g})_{g\in E_{\unicode[STIX]{x1D70E}}\setminus \{e\}})_{\unicode[STIX]{x1D70E}<\unicode[STIX]{x1D714}}$ has been constructed. Thanks to Lemma 5.6, there exist finite non-empty subsets $D\subseteq F\subseteq G$ and injections $\unicode[STIX]{x1D711}_{g}:D\rightarrow gF$ ( $g\in E_{\unicode[STIX]{x1D714}}\setminus \{e\}$ ) with:

• ∙ $|D|\geqslant (1-1/n_{\unicode[STIX]{x1D714}})|F|$ ;

• ∙ $F\cap gF=\emptyset$ for every $g\in E_{\unicode[STIX]{x1D714}}\setminus \{e\}$ ;

• ∙ $\unicode[STIX]{x1D711}_{g}(x)\in Ux$ for all $g\in E_{\unicode[STIX]{x1D714}}\setminus \{e\}$ and $x\in D$ .

Let $B:=\bigcup \{E_{\unicode[STIX]{x1D70E}}F_{\unicode[STIX]{x1D70E}}\mid \unicode[STIX]{x1D70E}<\unicode[STIX]{x1D714}\}$ . We notice that either $\unicode[STIX]{x1D714}$ is finite and thus $(E_{\unicode[STIX]{x1D714}}F)^{-1}B$ is finite as well, or $\unicode[STIX]{x1D714}$ is infinite and then $|(E_{\unicode[STIX]{x1D714}}F)^{-1}B|=|\unicode[STIX]{x1D714}|<\unicode[STIX]{x1D705}=|G|$ . Since $G$ is infinite, either way we have $|(E_{\unicode[STIX]{x1D714}}F)^{-1}B|<|G|$ . Therefore, $(E_{\unicode[STIX]{x1D714}}F)^{-1}B\neq G$ . Consequently, we may choose $z\in G\setminus ((E_{\unicode[STIX]{x1D714}}F)^{-1}B)$ . Clearly, $E_{\unicode[STIX]{x1D714}}Fz\cap B=\emptyset$ . Define $F_{\unicode[STIX]{x1D714}}:=Fz$ , $D_{\unicode[STIX]{x1D714}}:=Dz$ , and

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D711}_{\unicode[STIX]{x1D714},g}:D_{\unicode[STIX]{x1D714}}\rightarrow gF_{\unicode[STIX]{x1D714}},\quad x\mapsto \unicode[STIX]{x1D711}_{g}(xz^{-1})z & & \displaystyle \nonumber\end{eqnarray}$$

for $g\in E_{\unicode[STIX]{x1D714}}\setminus \{e\}$ . Let us remark that, if $\unicode[STIX]{x1D70E}\leqslant \unicode[STIX]{x1D714}$ , then either $\unicode[STIX]{x1D70E}<\unicode[STIX]{x1D714}$ and thus $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D70E},g}(x)\in Ux$ for all $g\in E_{\unicode[STIX]{x1D70E}}\setminus \{e\}$ and $x\in D_{\unicode[STIX]{x1D70E}}$ , or $\unicode[STIX]{x1D70E}=\unicode[STIX]{x1D714}$ and then

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D711}_{\unicode[STIX]{x1D70E},g}(x)=\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D714},g}(x)=\unicode[STIX]{x1D711}_{g}(xz^{-1})z\in Uxz^{-1}z=Ux & & \displaystyle \nonumber\end{eqnarray}$$

for all $g\in E_{\unicode[STIX]{x1D70E}}\setminus \{e\}$ and $x\in D_{\unicode[STIX]{x1D70E}}$ . It follows by transfinite induction that any thus constructed family $(F_{\unicode[STIX]{x1D714}},D_{\unicode[STIX]{x1D714}},(\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D714}.g})_{g\in E_{\unicode[STIX]{x1D714}}\setminus \{e\}})_{\unicode[STIX]{x1D714}<\unicode[STIX]{x1D705}}$ will satisfy the conditions (i)–(v) listed above.

Finally, let us define a map $\unicode[STIX]{x1D713}:G\rightarrow \text{Sym}(G)$ by setting $\unicode[STIX]{x1D713}(e):=\operatorname{id}_{G}$ and

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D713}(g)(x):=\left\{\begin{array}{@{}ll@{}}\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D714},g}(x)\quad & \text{if }g\in E_{\unicode[STIX]{x1D714}}\text{ and }x\in D_{\unicode[STIX]{x1D714}}\text{ for some }\unicode[STIX]{x1D714}<\unicode[STIX]{x1D705},\\ \unicode[STIX]{x1D711}_{\unicode[STIX]{x1D714},g}^{-1}(x)\quad & \text{if }g\in E_{\unicode[STIX]{x1D714}}\text{ and }x\in \unicode[STIX]{x1D711}_{\unicode[STIX]{x1D714},g}(D_{\unicode[STIX]{x1D714}})\text{ for some }\unicode[STIX]{x1D714}<\unicode[STIX]{x1D705},\\ x\quad & \text{otherwise},\end{array}\right. & & \displaystyle \nonumber\end{eqnarray}$$

for $g,x\in G$ , $g\neq e$ . We note that $\unicode[STIX]{x1D713}$ is well defined due to clauses (i)–(iii) and (v) above. Moreover, $\{\unicode[STIX]{x1D713}(g)(x)x^{-1}\mid g,x\in G\}\subseteq U$ by clause (iv) and symmetry of $U$ . Hence,

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FC}:G\rightarrow \text{Sym}(G),\quad g\mapsto \unicode[STIX]{x1D713}(g)\circ \unicode[STIX]{x1D706}_{g} & & \displaystyle \nonumber\end{eqnarray}$$

is a member of $\mathscr{N}_{G}(U)$ . We are left to prove that the action of $\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FC})$ on $G$ is amenable. To this end, we will verify the amenability criterion in Remark 5.2 with respect to the generating set $S:=\{\unicode[STIX]{x1D6FC}(g)\mid g\in G\}$ . Let $\unicode[STIX]{x1D703}>1$ and consider a finite subset $E^{\prime }\subseteq S$ . Take a finite subset $E\subseteq G$ with $e\in E$ and $E^{\prime }\subseteq \{\unicode[STIX]{x1D6FC}(g)\mid g\in E\}$ . Choose $n\in \mathbb{N}\setminus \{0\}$ such that $1+|E|/n\leqslant \unicode[STIX]{x1D703}$ . Let $\unicode[STIX]{x1D714}<\unicode[STIX]{x1D705}$ with $(E,n)=(E_{\unicode[STIX]{x1D714}},n_{\unicode[STIX]{x1D714}})$ . Then

$$\begin{eqnarray}\displaystyle E^{\prime }F_{\unicode[STIX]{x1D714}}\subseteq \bigcup \{\unicode[STIX]{x1D6FC}(g)(F_{\unicode[STIX]{x1D714}})\mid g\in E\}\subseteq F_{\unicode[STIX]{x1D714}}\cup \bigcup \{(gF_{\unicode[STIX]{x1D714}})\setminus \unicode[STIX]{x1D711}_{\unicode[STIX]{x1D714},g}(D_{\unicode[STIX]{x1D714}})\mid g\in E\setminus \{e\}\} & & \displaystyle \nonumber\end{eqnarray}$$

and hence

$$\begin{eqnarray}\displaystyle |E^{\prime }F_{\unicode[STIX]{x1D714}}|\leqslant |F_{\unicode[STIX]{x1D714}}|+|E|(|F_{\unicode[STIX]{x1D714}}|-|D_{\unicode[STIX]{x1D714}}|)\leqslant \biggl(1+\frac{|E|}{n}\biggr)|F_{\unicode[STIX]{x1D714}}|\leqslant \unicode[STIX]{x1D703}|F_{\unicode[STIX]{x1D714}}| & & \displaystyle \nonumber\end{eqnarray}$$

by clauses (i) and (iii) above. Thus, the action is amenable by Theorem 5.1.

(2) $\Longrightarrow$ (3). Let $U\in \mathscr{U}_{e}(G)$ . Take any $\unicode[STIX]{x1D6FC}\in \mathscr{N}_{G}(U)$ and consider the associated injection

$$\begin{eqnarray}\displaystyle f_{\unicode[STIX]{x1D6FC}}:G\rightarrow G\times \mathscr{N}_{G}(U),\quad x\mapsto (x,\unicode[STIX]{x1D6FC}). & & \displaystyle \nonumber\end{eqnarray}$$

Evidently, $f_{\unicode[STIX]{x1D6FC}}(G)=G\times \{\unicode[STIX]{x1D6FC}\}$ is $\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FC})$ -invariant and $h:\unicode[STIX]{x1D6E4}_{U}(G)\rightarrow \unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FC}),\unicode[STIX]{x1D6FE}\mapsto f_{\unicode[STIX]{x1D6FC}}^{-1}\circ \unicode[STIX]{x1D6FE}\circ f_{\unicode[STIX]{x1D6FC}}$ is a well-defined homomorphism satisfying $f_{\unicode[STIX]{x1D6FC}}\circ h(\unicode[STIX]{x1D6FE})=\unicode[STIX]{x1D6FE}\circ f_{\unicode[STIX]{x1D6FC}}$ for every $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{U}(G)$ . Therefore, if the action of $\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FC})$ on $G$ is amenable, then so is the action of $\unicode[STIX]{x1D6E4}_{U}(G)$ on $f_{\unicode[STIX]{x1D6FC}}(G)$ and hence on $G\times \mathscr{N}_{G}(U)$ . This proves the desired implication.

(3) $\Longrightarrow$ (1). We are going to utilize Theorem 4.5. To this end, let $\unicode[STIX]{x1D703}\in (0,1)$ , $U\in \mathscr{U}_{e}(G)$ , and $E\in \mathscr{F}(G)$ . By assumption, $\unicode[STIX]{x1D6E4}_{U}(G)$ acts amenably on $G\times \mathscr{N}_{G}(U)$ . According to Theorem 5.1, there thus exists a finite non-empty subset $F^{\prime }\subseteq G\times \mathscr{N}_{G}(U)$ such that $|E^{\prime }F^{\prime }|\leqslant (2-\unicode[STIX]{x1D703})|F^{\prime }|$ for $E^{\prime }:=\{(\bigoplus _{\unicode[STIX]{x1D6FC}\in \mathscr{N}_{G}(U)}\unicode[STIX]{x1D6FC})(g)\mid g\in E\}$ . It follows that there exists some $\unicode[STIX]{x1D6FC}\in \mathscr{N}_{G}(U)$ such that $F:=f_{\unicode[STIX]{x1D6FC}}^{-1}(F^{\prime })$ is non-empty and $|E^{\prime \prime }F|\leqslant (2-\unicode[STIX]{x1D703})|F|$ for $E^{\prime \prime }:=\{\unicode[STIX]{x1D6FC}(g)\mid g\in E\}$ . Let $g\in E$ . Consider $D:=F\cap \unicode[STIX]{x1D6FC}(g)(F)$ and the injection $\unicode[STIX]{x1D711}:D\rightarrow gF,\,x\mapsto g\unicode[STIX]{x1D6FC}(g)^{-1}(x)$ . Observe that

$$\begin{eqnarray}\displaystyle |D|=|F|-|\unicode[STIX]{x1D6FC}(g)(F)\setminus F|\geqslant |F|-(|E^{\prime \prime }F|-|F|)\geqslant \unicode[STIX]{x1D703}|F|. & & \displaystyle \nonumber\end{eqnarray}$$

For all $x\in G$ , we have $gx\in U\unicode[STIX]{x1D6FC}(g)(x)$ and so $g\unicode[STIX]{x1D6FC}(g)^{-1}(x)\in Ux$ . Hence, $\unicode[STIX]{x1D711}(x)\in Ux$ for all $x\in D$ . Thus, $\unicode[STIX]{x1D707}(F,gF,U)\geqslant \unicode[STIX]{x1D703}|F|$ . Due to Theorem 4.5, this means that $G$ is amenable.◻

Proof. ( $\Longrightarrow$ ) Without loss of generality, we assume that $G$ is infinite. Let $U\in \mathscr{U}_{e}(G)$ . Fix $V\in \mathscr{U}_{e}(G)$ such that $V^{-1}=V$ and $V^{3}\subseteq U$ . Since every precompact topological group is SIN, we may also assume that $gV=Vg$ for all $g\in G$ . By $G$ being precompact, there exists a finite subset $F\subseteq G$ such that $FF^{-1}\cap V=\{e\}$ and

$$\begin{eqnarray}\displaystyle |F|=\sup \{|E||\,E\subseteq G,\,EE^{-1}\cap V=\{e\}\}. & & \displaystyle \nonumber\end{eqnarray}$$

Let $g\in G$ . As $gV=Vg$ , it follows that $gF(gF)^{-1}\cap V=g(FF^{-1}\cap V)g^{-1}=\{e\}$ . We prove that the bipartite graph $\mathscr{B}:=\mathscr{B}(F,gF,V)$ admits a perfect matching. To this end, let $S\subseteq F$ and $T:=N_{\mathscr{B}}(S)$ . Consider the set $E:=((gF)\setminus T)\cup S$ . If $x,y\in E$ and $xy^{-1}\in V$ , then either $\{x,y\}\subseteq gF$ and thus $x=y$ by the above, or $\{x,y\}\cap S\neq \emptyset$ and hence $\{x,y\}\subseteq S$ , which also implies that $x=y$ . Therefore, $EE^{-1}\cap V=\{e\}$ . It follows that

$$\begin{eqnarray}\displaystyle |F|\geqslant |E|=|(gF)\setminus T|+|S|=|F|-|T|+|S|, & & \displaystyle \nonumber\end{eqnarray}$$

i.e., $|S|\leqslant |T|$ . Consequently, $\mathscr{B}$ admits a perfect matching $\unicode[STIX]{x1D711}_{g}:F\rightarrow gF$ by Corollary 4.3. Let $W\in \mathscr{U}_{e}(G)$ be such that $W^{-1}=W$ and $W^{2}\subseteq V$ . Since $G=VF$ and $Wx\cap Wy=\emptyset$ for any two distinct $x,y\in F$ , we may choose a map $\unicode[STIX]{x1D70B}:G\rightarrow F$ such that:

• ∙ $Wx\subseteq \unicode[STIX]{x1D70B}^{-1}(x)$ for every $x\in F$ ;

• ∙ $\unicode[STIX]{x1D70B}(x)\in Vx$ for all $x\in G$ .

Since $G$ is infinite and precompact, every non-empty open subset of $G$ has the same cardinality as $G$ . This readily implies that $|\unicode[STIX]{x1D70B}^{-1}(x)|=|G|$ for every $x\in F$ . For each $x\in F$ , choose a bijection $\unicode[STIX]{x1D6FD}_{x}:\unicode[STIX]{x1D70B}^{-1}(x)\rightarrow G$ . Let us define a map $h:\text{Sym}(F)\rightarrow \text{Sym}(G)$ by setting

$$\begin{eqnarray}\displaystyle h(\unicode[STIX]{x1D6FE})(x):=\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70B}(x)}^{-1}(\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D70B}(x)}(x))\quad (\unicode[STIX]{x1D6FE}\in \text{Sym}(F),\,x\in G). & & \displaystyle \nonumber\end{eqnarray}$$

Note that $\unicode[STIX]{x1D70B}(h(\unicode[STIX]{x1D6FE})(x))=\unicode[STIX]{x1D70B}(\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70B}(x)}^{-1}(\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D70B}(x)}(x)))=\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70B}(x)$ for all $\unicode[STIX]{x1D6FE}\in \text{Sym}(F)$ and $x\in G$ . It follows that $h$ is a homomorphism: if $\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D6FE}^{\prime }\in \text{Sym}(F)$ and $x\in G$ , then

$$\begin{eqnarray}\displaystyle h(\unicode[STIX]{x1D6FE}^{\prime }\unicode[STIX]{x1D6FE})(x) & = & \displaystyle \unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D6FE}^{\prime }\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70B}(x)}^{-1}(\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D70B}(x)}(x))=\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D6FE}^{\prime }\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70B}(x)}^{-1}(\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70B}(x)}(\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70B}(x)}^{-1}(\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D70B}(x)}(x))))\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D6FE}^{\prime }\unicode[STIX]{x1D70B}(h(\unicode[STIX]{x1D6FE})(x))}^{-1}(\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D70B}(h(\unicode[STIX]{x1D6FE})(x))}(h(\unicode[STIX]{x1D6FE})(x)))=h(\unicode[STIX]{x1D6FE}^{\prime })(h(\unicode[STIX]{x1D6FE})(x)).\nonumber\end{eqnarray}$$

Now, for each $g\in G$ , let us define $\unicode[STIX]{x1D6FE}(g)\in \text{Sym}(F)$ by

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FE}(g)(x):=\unicode[STIX]{x1D711}_{g}^{-1}(gx)\quad (x\in F) & & \displaystyle \nonumber\end{eqnarray}$$

and put $\unicode[STIX]{x1D6FC}(g):=h(\unicode[STIX]{x1D6FE}(g))\in \text{Sym}(G)$ . Then $\{\unicode[STIX]{x1D6FC}(g)\mid g\in G\}=\{h(\unicode[STIX]{x1D6FE}(g))\mid g\in G\}\subseteq h(\text{Sym}(F))$ . Hence, $\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FC})$ is contained in the finite group $h(\text{Sym}(F))$ and therefore finite itself. It remains to prove that $\unicode[STIX]{x1D6FC}\in \mathscr{N}_{G}(U)$ . Let $g,x\in G$ . By choice of $\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D6FE}(g)\unicode[STIX]{x1D70B}(g^{-1}x)}$ , $\unicode[STIX]{x1D711}_{g}$ , and $\unicode[STIX]{x1D70B}$ , it follows that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FC}(g)(x)=h(\unicode[STIX]{x1D6FE}(g))(x)=\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D6FE}(g)\unicode[STIX]{x1D70B}(x)}^{-1}(\unicode[STIX]{x1D6FD}_{\unicode[STIX]{x1D70B}(x)}(x))\in V\unicode[STIX]{x1D6FE}(g)\unicode[STIX]{x1D70B}(x) & = & \displaystyle V\unicode[STIX]{x1D711}_{g}^{-1}(g\unicode[STIX]{x1D70B}(x))\subseteq V^{2}g\unicode[STIX]{x1D70B}(x)\nonumber\\ \displaystyle & \subseteq & \displaystyle V^{2}gVx=V^{3}gx\subseteq Ugx.\nonumber\end{eqnarray}$$

This proves that $\unicode[STIX]{x1D6FC}$ is contained in $\mathscr{N}_{G}(U)$ , which completes the argument.

( $\Longleftarrow$ ) Let $U\in \mathscr{U}_{e}(G)$ . By our hypothesis, there is $\unicode[STIX]{x1D6FC}\in \mathscr{N}_{G}(U^{-1})$ such that $\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FC})$ is finite. Consider the finite subset $F:=\{\unicode[STIX]{x1D6FC}(g)(e)\mid g\in G\}\subseteq G$ . Clearly, if $g\in G$ , then $\unicode[STIX]{x1D6FC}(g)(e)\in U^{-1}g$ and thus $g\in U\unicode[STIX]{x1D6FC}(g)(e)\subseteq UF$ . That is, $G=UF$ . This shows that $G$ is precompact.◻

Proof. It is not difficult to see that Whyte’s proof of Theorem 5.10 works just as well for group actions: a group action $(G,X)$ is non-amenable if and only if the free group $F_{2}$ is isomorphic to a semi-regular subgroup of $W(G,X)$ (see [Reference SchneiderSch17, Corollary 2.3] for the details). This observation combined with our Theorem 5.3 implies the desired corollary.◻

Proof. Clearly, $W(G,X)$ -invariance implies $G$ -invariance. To prove the converse, suppose that $\unicode[STIX]{x1D707}$ is $G$ -invariant. Let $h\in W(G,X)$ . Then there is a finite partition $\mathscr{P}$ of $G$ along with $(g_{P})_{P\in \mathscr{P}}\in G^{\mathscr{P}}$ such that $h|_{P}=g_{P}|_{P}$ for each $P\in \mathscr{P}$ . For $f\in \ell ^{\infty }(X)$ , it follows that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D707}(f\circ h) & = & \displaystyle \mathop{\sum }_{P\in \mathscr{P}}\unicode[STIX]{x1D707}((f\circ h)\cdot \mathbf{1}_{P})=\mathop{\sum }_{P\in \mathscr{P}}\unicode[STIX]{x1D707}((f\circ g_{P})\cdot \mathbf{1}_{P})=\mathop{\sum }_{P\in \mathscr{P}}\unicode[STIX]{x1D707}(f\cdot (\mathbf{1}_{P}\circ g_{P}^{-1}))\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{P\in \mathscr{P}}\unicode[STIX]{x1D707}(f\cdot (\mathbf{1}_{g_{P}P}))=\mathop{\sum }_{P\in \mathscr{P}}\unicode[STIX]{x1D707}(f\cdot \mathbf{1}_{\unicode[STIX]{x1D6FC}(P)})=\unicode[STIX]{x1D707}(f).\Box \nonumber\end{eqnarray}$$

Proof. By Theorem 5.3, there exists $\unicode[STIX]{x1D6FC}\in \mathscr{N}_{G}(U)$ such that the action of $\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FC})$ on $G$ is amenable. Let $\unicode[STIX]{x1D707}:\ell ^{\infty }(G)\rightarrow \mathbb{R}$ be a $\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FC})$ -invariant mean. By Lemma 5.13, it follows that $\unicode[STIX]{x1D707}$ is $W(\unicode[STIX]{x1D6FC})$ -invariant. By assumption, there is an embedding $\unicode[STIX]{x1D711}:H\rightarrow W(\unicode[STIX]{x1D6FC})$ such that $\unicode[STIX]{x1D711}(H)$ is semi-regular. Hence, there exists a map $\unicode[STIX]{x1D713}:G\rightarrow H$ such that $\unicode[STIX]{x1D713}(\unicode[STIX]{x1D711}(h)(x))=h\unicode[STIX]{x1D713}(x)$ for all $h\in H$ and $x\in G$ . Now define $\unicode[STIX]{x1D708}:\ell ^{\infty }(H)\rightarrow \mathbb{R}$ by

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D708}(f):=\unicode[STIX]{x1D707}(f\circ \unicode[STIX]{x1D713})\quad (f\in \ell ^{\infty }(H)). & & \displaystyle \nonumber\end{eqnarray}$$

It is easy to see that $\unicode[STIX]{x1D708}$ is a mean. Furthermore, the $W(\unicode[STIX]{x1D6FC})$ -invariance of $\unicode[STIX]{x1D707}$ implies that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D708}(f\circ \unicode[STIX]{x1D706}_{h})=\unicode[STIX]{x1D707}(f\circ \unicode[STIX]{x1D706}_{h}\circ \unicode[STIX]{x1D713})=\unicode[STIX]{x1D707}(f\circ \unicode[STIX]{x1D713}\circ \unicode[STIX]{x1D711}(h))=\unicode[STIX]{x1D707}(f\circ \unicode[STIX]{x1D713})=\unicode[STIX]{x1D708}(f) & & \displaystyle \nonumber\end{eqnarray}$$

for all $f\in \ell ^{\infty }(H)$ and $h\in H$ . Thus, $\unicode[STIX]{x1D708}$ is $H$ -invariant, as desired.◻

Proof. Fix an open identity neighborhood $V\in \mathscr{U}_{e}(G)$ such that $V^{2}\subseteq U$ . By density, the amenability of $G$ is equivalent to the amenability of $H$ (carrying the respective subspace topology). Hence, Theorem 5.3 implies the existence of a map $\unicode[STIX]{x1D6FD}:H\rightarrow \text{Sym}(H)$ such that:

• ∙ $\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FD})$ acts amenably on the set $H$ ;

• ∙ $\unicode[STIX]{x1D6FD}(g)(h)\in Vgh$ for all $g,h\in H$ .

Extend $\unicode[STIX]{x1D6FD}$ to a map $\unicode[STIX]{x1D6FD}^{\prime }:H\rightarrow \text{Sym}(G)$ by setting

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FD}^{\prime }(h)(x):=\left\{\begin{array}{@{}ll@{}}\unicode[STIX]{x1D6FD}(h)(x)\quad & \text{if }x\in H,\\ hx\quad & \text{otherwise},\end{array}\right.\quad (h\in H,\,x\in G). & & \displaystyle \nonumber\end{eqnarray}$$

Evidently, $H$ is $\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FD}^{\prime })$ -invariant, and $\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D6FD}^{\prime })$ acts amenably on $H$ and acts by left $H$ -translations on $G\setminus H$ . As $H$ is countable, $\unicode[STIX]{x1D6FD}^{\prime }(h)$ is an automorphism of the Borel space $G$ for each $h\in H$ and it moreover follows that $H\times G\rightarrow G,\,(h,x)\mapsto \unicode[STIX]{x1D6FD}^{\prime }(h)(x)$ is Borel measurable. Due to $H$ being countable and dense in $G$ , a straightforward recursion also provides us with a Borel measurable map $\unicode[STIX]{x1D70B}:G\rightarrow H$ such that $\unicode[STIX]{x1D70B}(x)\in Vx$ for all $x\in G$ . Hence, the resulting maps $\unicode[STIX]{x1D6FC}:=\unicode[STIX]{x1D6FD}^{\prime }\circ \unicode[STIX]{x1D70B}:G\rightarrow \text{Sym}(G)$ and $G\times G\rightarrow G,\,(g,h)\mapsto \unicode[STIX]{x1D6FC}(g)(h)$ are Borel measurable, too. Clearly, we are only left to note that $\unicode[STIX]{x1D6FC}$ is a member of $\mathscr{N}_{G}(U)$ : indeed, if $g,h\in G$ , then

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FC}(g)(h)=\unicode[STIX]{x1D6FD}^{\prime }(\unicode[STIX]{x1D70B}(g))(h)\in V\unicode[STIX]{x1D70B}(g)h\subseteq V^{2}gh\subseteq Ugh.\Box & & \displaystyle \nonumber\end{eqnarray}$$

Proof. This follows from Theorem 5.3, Proposition 5.15, and [Reference Marks and UngerMU16, Theorem 1.1]. ◻

Proof. This follows from Theorem 5.3, Proposition 5.15, and [Reference Marks and UngerMU16, Theorem 1.2]. ◻