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\begin{document}
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\begin{center}{\footnotesize Khayyam J. Math. 1 (2015), no. 1, 36--44}\\\end{center}
\noindent\parbox{2.85cm}{\includegraphics*[keepaspectratio=true,scale=0.24]{KJM.jpg}}
\vspace{0.5cm}

\title[Invariant means on CHART groups]{INVARIANT MEANS ON CHART GROUPS}

\author[W.B. Moors]{Warren B. Moors}

\address{Department of Mathematics, The University of Auckland,  Private Bag  92019, Auckland,
New Zealand.}
\email{moors@math.auckland.ac.nz}


%\dedicatory{This paper is dedicated to Professor ABCD}
\dedicatory{{\rm Communicated by A.R. Mirmostafaee}}
\subjclass[2010]{Primary 37B05; Secondary 22C05, 37B40.}

\keywords{Topological group, invariant mean, Furstenberg's fixed point theorem.}

\date{Received: 24 July 2014;  Accepted: 01 August 2014.}

\begin{abstract}
The purpose of this paper is to give a stream-lined proof of the existence and uniqueness of a right-invariant mean on a CHART group.
A CHART group is a slight generalisation of  a compact topological group.  The existence of an invariant mean on a CHART group can be used to prove
Furstenberg's fixed point theorem.
\end{abstract} \maketitle

\section{Introduction and preliminaries}

Given a nonempty set $X$ and a linear subspace $S$ of $\R^X$ that contains all the constant functions we say that a linear functional $m:S \to \R$  is a {\it mean on $S$} if:
\begin{enumerate}
\item[{\rm (i)}] $m(f) \geq 0$ for all $f \in S$ that satisfy $f(x) \geq 0$ for all $x \in X$;
\item[{\rm (ii)}]  $m({\boldsymbol 1}) = 1$, where $\boldsymbol 1$ is the function that is identically equal to $1$.
\end{enumerate}
If all the functions in $S$ are bounded on $X$ then this definition is equivalent to the following: $$1 = m(\boldsymbol{1}) = \|m\|$$
where, $\|m\| := \sup \{m(f): f \in S \mbox{ and } \|f\|_\infty \leq 1\}$.\medskip

If $(X, \cdot)$ is a semigroup then  we can define, for each $g \in X$,  
$L_g:\R^X \to \R^X$ and $R_g:\R^X \to \R^X$ by, 
$$L_g(f)(x) := f(gx) \mbox{ for all $x \in X$ \quad and  \quad } R_g(f)(x) := f(xg) \mbox{ for all $x \in X$.}$$
Note that for all $g, h \in X$, $L_g \circ L_h = L_{hg}$, $R_g \circ R_h = R_{gh}$ and $L_g \circ R_h = R_h \circ L_g$. \\[6.0pt] 
If $S$ is a subspace of $\R^X$ that contains all the constant functions and $L_g(S) \subseteq S$ [$R_g(S) \subseteq S$] for all $g \in X$ then we call a mean $m$ on $S$
{\it left-invariant} [{\it right-invariant}] if, 
$$m(L_g(f)) = m(f) \quad  [m(R_g(f)) = m(f)]   \mbox{\quad for all $g \in X$ and all $f \in S$.}$$

We now need to consider some notions from topology.  Suppose that $X$ and $Y$ are compact Hausdorff spaces and $\pi:X \to Y$ is a continuous surjection.
Then $\pi^{\#}:C(Y) \to C(X)$ defined by, $\pi^{\#}(f) := f \circ \pi$ is an isometric algebra isomorphism into $C(X)$.  Moreover, we know (from topology/functional analysis) that
$f \in \pi^{\#}(C(Y))$ if, and only if, $f \in C(X)$ and $f$ is constant on the fibers of $\pi$ (i.e., $f$ is constant on $\pi^{-1}(y)$ for each $y \in Y$). 

\medskip

The final notion that we need for this section is that of a right topological group (left topological group).  We shall call a triple $(G, \cdot, \tau)$ a {\it right topological group}
({\it left topological group}) if $(G, \cdot)$ is a group, $(G, \tau)$ is a topological space and, for each $g \in G$, the mapping $x \mapsto x\cdot g$ ($x \mapsto g \cdot x$)
is continuous on $G$. If $(G, \cdot, \tau)$ is both a right topological group and a left topological group then we call it a {\it semitopological group}.

\medskip

If $(G, \cdot, \tau)$ and $(H, \cdot, \tau')$ are compact Hausdorff right topological groups and $\pi:G \to H$ is a continuous homomorphism then it easy to check that
$$R_g(\pi^{\#}(f)) = \pi^{\#}(R_{\pi(g)}(f)) \mbox{\quad for all $f \in C(H)$ and $g \in G$.}$$
If $\pi:X \to Y$ is surjective then $(\pi^{\#})^{-1}: \pi^{\#}(C(H)) \to C(H)$ exists.  Therefore,
$$(\pi^{\#})^{-1}(R_g(h))  = R_{\pi(g)}((\pi^{\#})^{-1}(h)) \mbox{\quad for all $h \in \pi^{\#}(C(H))$ and $g \in G$.}$$
From these equations we can easily establish our first result.
\begin{proposition} \label{Prop1} Let $(G, \cdot, \tau)$ and $(H, \cdot, \tau')$ be compact Hausdorff right topological groups and let $\pi:G \to H$ be a continuous
epimorphism (i.e., a surjective homomorphism).  If $m$ is a right-invariant mean on $C(H)$ then $m^*:\pi^{\#}(C(H)) \to \R$ defined by,
$m^*(f) := m((\pi^{\#})^{-1}(f))$ for all $f \in \pi^{\#}(C(H))$ is a right-invariant mean on $\pi^{\#}(C(H))$.   If $C(H)$ has a unique right-invariant mean
then $\pi^{\#}(C(H))$ has a unique right-invariant mean.
\end{proposition}

We can now state and prove our main theorem for this section.

\begin{Theo}  \label{Thm1} Let $(G, \cdot, \tau)$ and $(H, \cdot, \tau')$ be compact Hausdorff right topological groups and let $\pi:G \to H$ be a continuous
epimorphism.  If the mapping 
$$m: G \times \mbox{ker}(\pi) \to G \mbox{ defined by, } m(x,y) := x \cdot y \mbox{ for all $(x,y) \in G \times \mbox{ker}(\pi)$}$$
 is continuous and $C(H)$ has a right-invariant mean
then $C(G)$ has a right-invariant mean.  Furthermore, if $C(H)$ has a unique right-invariant mean then so does $C(G)$.
\end{Theo}

\begin{proof} Let $L := \mbox{ker}(\pi)$. Then from the hypotheses and \cite[Theorem 2]{E:1957} $(L, \cdot, \tau_L)$ (here $\tau_L$ is the relative $\tau$-topology on $L$) is a compact topological group.
Thus $(L, \cdot, \tau_L)$ admits a unique Borel probability measure $\lambda$ (called the {\it Haar measure} on $L$) such that 
$$\int_L L_g(f)(t) {\rm \ d}\lambda(t) =  \int_L R_g(f)(t) {\rm \ d}\lambda(t) = \int_L f(t) {\rm \ d} \lambda(t) \mbox{ for all $g \in L$ and $f \in C(L)$}.$$
Let $P: C(G) \to \pi^{\#}(C(H))$ be defined by, 
$$P(f)(g) := \int_Lf(g\cdot t) {\rm \ d}\lambda(t) \mbox{\ i.e., $P(f)(g)$ is the ``average'' of $f$ over the coset $gL$.}$$
Firstly, since $m$ is continuous on $G \times L$ (and $L$ is compact) $P(f) \in C(G)$ for each $f \in C(G)$.  Secondly, since $\lambda$ is invariant on $L$ it is routine to check that
$P(f)$ is constant on the fibers of $\pi$. Hence, $P(f) \in \pi^{\#}(C(H))$.  We now show that for each $g \in G$ and $f \in C(G)$, 
$$\int_L L_g(f)(t) {\rm \ d}\lambda(t) =   \int_L f(g\cdot t) {\rm \ d}\lambda(t) = \int_L f(t\cdot g) {\rm \ d}\lambda(t) =   \int_L R_g(f)(t) {\rm \ d}\lambda(t). \mbox{\quad $(*)$}$$
To this end, fixed $g \in G$ and define $G:C(L) \to C(L)$ by, $G(f)(t) := f(g^{-1} \cdot t \cdot g)$. Since $m$ is continuous, 
$t \mapsto (g^{-1}\cdot t)\cdot g$  is continuous and so $G$ is well-defined, i.e., $G(f) \in C(L)$ for each $f \in C(L)$.
We claim that 
$$f \mapsto \int_L G(f)(t) {\rm \ d}\lambda(t)$$
is a right-invariant mean on $C(L)$. Clearly, this mapping is a mean so it remains to show that it is right-invariant. To see this, let $l \in L$.  Then $g \cdot l \cdot g^{-1} \in L$ and
\begin{eqnarray*}
\int_L G(R_l(f))(t) {\rm \ d}\lambda(t) &=& \int_L R_l(f)(g^{-1} \cdot t \cdot g) {\rm \ d}\lambda(t)   \\
&=& \int_L f(g^{-1} \cdot t \cdot g \cdot l){\rm \ d}\lambda(t) \\
&=& \int_L f(g^{-1} \cdot [t \cdot (g \cdot l \cdot g^{-1} )]\cdot g) {\rm \ d}\lambda(t) \\
&=& \int_L G(f)(t \cdot (g \cdot l \cdot g^{-1})) {\rm \ d}\lambda(t) \\
&=& \int_L R_{g\cdot l\cdot g^{-1}}(G(f))(t)  {\rm \ d}\lambda(t) \\
&=& \int_L G(f)(t)  {\rm \ d}\lambda(t) \mbox{ \quad since $\lambda$ is right-invariant.}
\end{eqnarray*}
Now, since there is only one right-invariant mean on $C(L)$ we must have that 
$$\int_L G(f)(t) {\rm \ d}\lambda(t) =   \int_L f(g^{-1}\cdot t \cdot g) {\rm \ d}\lambda(t) =   \int_L f(t) {\rm \ d}\lambda(t)  \mbox{ \quad for all $f \in C(L)$.}$$
It now follows that equation $(*)$ holds. Next, we show that $R_g(P(f)) = P(R_g(f))$ for all $g \in G$ and $f \in C(G)$. To this end, let $g \in G$ and $f \in C(G)$. Then for any $x \in G$,
\begin{eqnarray*}
R_g(P(f))(x) &=& P(f)(x\cdot g) = \int_L f(x \cdot g \cdot t) {\rm \ d}\lambda(t)  = \int_L f(x \cdot t \cdot g) {\rm \ d}\lambda(t) \mbox{ \quad  by $(*)$} \\
&=& \int_L R_g(f)(x \cdot t) {\rm \ d}\lambda(t) = P(R_g(f))(x).
\end{eqnarray*}
Let $\mu$ be the unique right-invariant mean on $\pi^{\#}(C(H))$, given to us by Proposition~\ref{Prop1}. Let $\mu^*:C(G) \to \R$ be defined by, $\mu^*(f) := \mu(P(f))$. It is now
easy to verify that $\mu^*$ is a right-invariant mean on $C(G)$.  

\medskip

So it remains to prove uniqueness.  Suppose that $\mu^*$ and $\nu^*$ are right-invariant means on $C(G)$. Since, by Proposition~\ref{Prop1}, we know that
$\mu^*|_{\pi^{\#}(C(H))} = \nu^*|_{\pi^{\#}(C(H))}$ it will be sufficient to show that $\mu^*(f) = \mu^*(P(f))$ and  $\nu^*(f) = \nu^*(P(f))$ for each $f \in C(G)$.
We shall apply Riesz's representation theorem along with Fubini's theorem.   Let $\mu$ be the probability measure on $G$ that represents $\mu^*$ and let
$f \in C(G)$.  Then
\begin{eqnarray*}
\mu^*(f) &=& \int_G f(s) {\rm \ d}\mu(s) = \int_L \int_G f(s \cdot t) {\rm \ d}\mu(s) {\rm \ d}\lambda(t)  \\
&=& \int_G \int_L f(s \cdot t) {\rm \ d}\lambda(t) {\rm \ d}\mu(s) \\
&=& \int_G P(f)(s) {\rm \ d}\mu(s) = \mu^*(P(f)).
\end{eqnarray*}
A similar argument show that $\nu^*(f) = \nu^*(P(f))$. This completes the proof.
\end{proof}

This paper is the culmination of work done many people, starting with the work of H.~Furstenberg in  \cite{F:1963} on the
existence of invariant measures on distal flows.  This work was later simplified and phrased in terms of CHART groups by I.~Namioka in \cite{N:1972}.
The results of Namioka were further generalised by R. Ellis, \cite{E:1978}.  In 1992, P. Milnes and J. Pym, \cite{MP:1992}  showed that every CHART group (that satisfies
some countability condition) admits a unique right-invariant mean (unique right-invariant measure) called the Haar mean (Haar measure).  Later,
in \cite{MP:1992a}, Milne and Pym managed to remove the countability condition from the proof contained in \cite{MP:1992} by appealing to a result from
\cite{E:1978}. Finally, in \cite{Moors}, a direct proof of the existence and uniqueness of a right-invariant mean on a CHART group was given, however, this proof still relied
upon the results from \cite{MP:1992}. 

\medskip

In the present paper we give a stream-lined proof (that does not require knowledge from topological dynamics) of the existence and uniqueness of a right-invariant mean
on a CHART group. 

\section{Groups}

Let $(G,\cdot, \tau)$ be a right topological group and let $H$ be a subgroup of $G$. We shall denote by $(H,\tau_H)$ the set $H$ equipped with the relative $\tau$-topology. 
It is easy to see that $(H, \cdot, \tau_H)$ is also a right topological group. 

\medskip

Now let $G/H$ be the set $\{xH:x\in G\}$ of all left cosets of $H$ in $G$ and give $G/H$ the quotient topology $q(\tau)$ 
induced from $(G,\tau)$ by the map $\pi : G \rightarrow G/H$ defined by $\pi(x) := xH$. 
\medskip

Note that $\pi$ is an open mapping because, if $U$ is an open subset of $G$ then 
$$\pi^{-1}(\pi (U)) = UH=\mbox{$\bigcup$}\{Ux:x\in H\}$$
and this last set is open since right multiplication is a homeomorphism on $G$. 

\medskip

If $H$ is a normal subgroup of a right(left)[semi] topological group $(G, \cdot, \tau)$ then one can check that $(G/H, \cdot, q(\tau))$ is also a right(left)[semi] topological group.

\medskip

In order to continue our investigations further we need to introduce a new topology.
 
\subsection{The $\sigma$-topology}
Let $(G, \cdot, \tau)$ be a right topological group and let $\varphi : G\times G \rightarrow G$ be the map defined by 
$$\varphi (x,y) := x^{-1}\cdot y.$$ 
Then the quotient topology on $G$ induced from $(G\times G, \tau\times\tau)$ by the map $\varphi$ is called the $\sigma(G,\tau)$-topology or $\sigma$-topology.

\medskip

The proof of the next  result can be found in \cite[Theorem~1.1,Theorem~1.3]{N:1972} or~\cite[Lemma~4.3]{N:2011}.

\begin{Lem} \label{th:1.3} Let $(G, \cdot, \tau)$ be a right topological group. Then,
\begin{enumerate}
\item[{\rm (i)}]  $(G,\sigma)$ is a semitopological group.
\item[{\rm (ii)}]  $\sigma \subseteq \tau$.
\item[{\rm (iii)}] $(G/H,q(\tau))$ is Hausdorff provided the subgroup $H$ is closed with respect to the $\sigma$-topology on $G$.
\end{enumerate}
\end{Lem}

\subsection{Admissibility and CHART groups}
Let $(G,\cdot, \tau)$ be a right topological group and let $\Lambda(G,\tau)$ be the set of all $x\in G$ such that the map $y\mapsto x\cdot y$ is $\tau$ continuous. 
If $\Lambda(G, \tau)$ is $\tau$-dense in $G$ then $(G,\tau)$ is said to be {\it admissible}.  

\medskip

The proof for the following proposition may be found in \cite[Theorem 1.2, Corollary 1.1]{N:1972} or \cite[Proposition 4.4, Proposition 4.5]{N:2011}.
\begin{proposition}\label{th:1.5b} Let $(G, \cdot, \tau)$ be an admissible right topological group.
\begin{enumerate} 
\item[{\rm (i)}]  If $\mathcal{U}$ is the family of all $\tau$-open neighbuorhoods of $e$ in $G$ then \\
$\{U^{-1}U:U\in\mathcal{U}\}$ is a base of open neighbuorhoods of $e$ in $(G,\sigma)$.
\item[{\rm (ii)}] If $N(G, \tau) :=\bigcap\{U^{-1}U:U\in\mathcal{U}\}$ then $N(G, \tau) = \overline{\{e\}}^\sigma$.
\end{enumerate}
\end{proposition}


A compact Hausdorff admissible right topological group $(G,\cdot, \tau)$ is called a {\it CHART group}.

\medskip

The proof for the following result may be found \cite[Proposition 2.1]{N:1972} or \cite[Proposition 4.6]{N:2011}.
\begin{proposition}\label{th:1.7a} \label{th:1.7b}
Let $(G,\cdot, \tau)$ be a CHART group. Then the following hold:
\begin{enumerate}
\item[{\rm (i)}]  If $L$ is a $\sigma$-closed normal subgroup of $G$, then so is $N(L, \sigma_L)$.
\item[{\rm (ii)}]  If $m:(G/N(L,\sigma_L),q(\tau))\times(L/N(L,\sigma_L),q(\tau))\rightarrow(G/N(L, \sigma_L),q(\tau))$ is defined by 
$$m(xN(L, \sigma_L),yN(L,\sigma_L)) := x\cdot yN(L,\sigma_L) \mbox{\quad for all $(x,y) \in G \times L$}$$
then $m$ is well-defined and continuous.
\end{enumerate}
\end{proposition}

\begin{Rem}\label{Rem} By considering the mapping $\pi:G/N(L, \sigma_L) \to G/L$, Theorem~\ref{Thm1} and Proposition~\ref{th:1.7a} we see that if $(G/L, q(\tau))$ admits a unique right-invariant
mean then so does $(G/N(L, \sigma_L), q(\tau))$. Hence if $N(L, \sigma_L)$ is a proper subset of $L$ then we have made some progress towards showing that $G \cong G/\{e\}$
admits a unique right-invariant mean. 
\end{Rem}



\section{$N(L, \sigma_L) \not= L$}
In this section we will show that if $L$ is a nontrivial $\sigma$-closed normal subgroup of a CHART group $(G, \cdot, \tau)$ then $N(L, \sigma_L)$
is a proper subset of $L$.

\begin{Lem} \label{Lem1} Let $(H, \cdot)$ be a group and $X$ be a nonempty set.   Then for any $f :H \to X$, $S := \{s \in H: f(hs) = f(h) \mbox{ for all $h \in H$}\}$ is a subgroup of $H$.
\end{Lem}
\begin{proof} Clearly, $e \in S$.  Now suppose that, $s_1, s_2 \in S$. Let $h$ be any element of $H$ then 
$$f(h(s_1s_2)) = f((hs_1)s_2) = f(hs_1) = f(h)$$
Therefore, $s_1s_2 \in S$.  Next, let $s$ be any element of $S$ and $h$ be any element of $H$ then
$$f(h) = f(h(s^{-1}s))   =  f((hs^{-1})s)   =  f(hs^{-1}) .$$ 
Therefore, $s^{-1} \in S$.
\end{proof}


\begin{Lem}\label{th:3.1}
Let $(G,\cdot, \tau)$ be a compact right topological group and let $\sigma$ be a topology on $G$ weaker than $\tau$ such that $(G,\cdot, \sigma)$ is also a right topological group. 
If $U$ is a dense open subset of $(G,\sigma)$ then $U$ is also a dense subset of $(G, \tau)$.
\end{Lem}
\begin{proof}
Let $C :=G \backslash U$. Then $C$ is a $\sigma$-closed (hence $\tau$-closed) nowhere-dense subset of $G$. If $U$ is not $\tau$-dense in $G$ then 
$C$ contains a nonempty $\tau$-open subset. By the compactness of $(G,\tau)$ there exists a finite subset $F$ of $G$ such that $G=\bigcup \{Cg: g\in F\}$. 
Now each $Cg$ is nowhere dense in $(G,\sigma)$ since each right multiplication is a homeomorphism. This forms a contradiction since a nonempty topological 
space can never be the union of a finite number of nowhere dense subsets.
\end{proof}

\begin{Lem}\label{th:3.2}
Let $(G,\cdot, \tau)$ be a CHART group and let $\Lambda=\Lambda(G, \tau)$. If $A$ and $B$ are nonempty open subsets of $(G,\tau)$, then $A^{-1}B=(A\cap\Lambda)^{-1}B$.
\end{Lem}
\begin{proof}
Let $x\in A^{-1}B$. Then for some $a\in A, ax\in B$. Since $B$ is open and $A\cap\Lambda$ is dense in $A$ there is a $c\in A\cap\Lambda$ such that $cx\in B$. 
Hence $x \in c^{-1}B \subseteq (A \cap \Lambda)^{-1}B$. Thus, $A^{-1}B \subseteq (A \cap \Lambda)^{-1}B$.  
The reverse inclusion is obvious.
\end{proof}

\begin{Lem}\label{th:3.3}
Let $(G,\cdot, \tau)$ be a compact Hausdorff right topological group. If S is a nonempty subsemigroup of $\Lambda(G,\tau)$ then $\overline{S}$  is a subgroup of $G$.
\end{Lem}
\begin{proof} In this proof we shall repeatedly use the following fact, \cite[Lemma 1]{E:1958}  ``Every nonempty compact right topological semigroup admits an idempotent element (i.e., an element
$u$ such that $u\cdot u = u$).  Firstly, it is easy to see that $\overline{S}$ is a subsemigroup of $G$.  Hence, $(\overline{S}, \cdot)$ is a nonempty compact right topological semigroup
and so has an idempotent element $u$. However, since $G$ is a group it has only one idempotent element, namely $e$.  Therefore, $e = u \in \overline{S}$.  Next, let $s$ be any element of
$\overline{S}$. Then $\overline{S} \cdot s$ is a  nonempty compact right topological semigroup of $\overline{S}$.  Therefore, there exists an element
$s' \in \overline{S}$ such that $(s' \cdot s)\cdot (s' \cdot s) = (s' \cdot s)$. Again, since $G$ is a group, $s' \cdot s =e$.  By multiplying both sides of this equation by $s^{-1}$
we see that $s^{-1} = s' \in \overline{S}$.
\end{proof}

The following lemma is a simplified form of the structure theorem found in \cite{Moors}.
\begin{Lem} \label{th:3.4}
Let $(G,\cdot, \tau)$ be a CHART group and let $\sigma$ denote its $\sigma$-topology. Suppose $L$ is a nontrivial $\sigma$-closed subgroup of $G$. 
Then $N(L, \sigma_L)$ is a proper subset of $L$.
\end{Lem}
\begin{proof}
Let $\mathcal{U}$ denote the family of all open neighbuorhoods of $e$ in $(G,\tau)$. Then it follows from Proposition \ref{th:1.5b} that $\mathcal{V}=\{U^{-1}U:U\in \mathcal{U} \}$ is a base for the system of open neighbourhoods of $e$ in $(G,\sigma)$. Then $\{V \cap L:V\in \mathcal{V}\}$ is a basis for the system of neighbourhoods of $e$ in $(L,\sigma_L)$. From the definition of $N(L, \sigma_L)$  
(see Proposition \ref{th:1.5b} part (ii)) it follows that 
$$N(L,\sigma_L)=\mbox{$\bigcap$} \{(V\cap L)^{-1}(V\cap L):V\in \mathcal{V}\}.$$
The proof is by contradiction. So assume that $N(L, \sigma_L) = L$. Then 
$$L=\mbox{$\bigcap$} \{(V\cap L)^{-1}(V\cap L):V\in \mathcal{V}\}.$$ 
Hence, for each $V\in \mathcal{V}$, $(V\cap L)^{-1}(V\cap L)=L$, or equivalently, for each $V\in \mathcal{V}$, $(V\cap L)$ is dense in $(L,\sigma_L)$. That is, for each $U\in \mathcal{U}$,
$(U^{-1}U\cap L)$ is open and dense in $(L,\sigma_L)$ and hence, by Lemma \ref{th:3.1},  dense in $(L,\tau_L)$.

\medskip

Since $L\neq \{e\}$, there exists a point $a\in L$ such that $a \neq e$. Note that since $(G,\tau)$ is compact and Hausdorff there is a continuous function $f$ 
on $(G,\tau)$ such that $f(e)=0$ and $f \equiv 1$ on a $\tau$-neighbuorhood of $a$.

\medskip


For the rest of the proof, the topology always refers to $\tau$ and we shall denote $\Lambda(G,\tau)$ by $\Lambda$.
By induction on $n$, we construct a sequence $\{U_n: n\in\N\}$ in $\mathcal U$, a sequence
$\{V_n:n\in\N\}$ of nonempty open subsets of $G$, each of which intersects $L$ and
sequences $\{u_n:n\in\N\}$ and $\{v_n:n\in\N\}$ in $G$ which satisfy the following conditions:
\begin{enumerate}
\item[(i)] $v_n\in U_{n-1}^{-1}U_{n-1}\cap (V_{n-1}\cap \Lambda) = (U_{n-1}\cap\Lambda)^{-1}
U_{n-1}\cap(V_{n-1}\cap \Lambda)$;  by Lemma~\ref{th:3.2}.
\item[(ii)] $u_n\in U_{n-1}\cap\Lambda$;
\item[(iii)] $V_n\subset\overline{V_n}\subset V_{n-1}\subset f^{-1}(1)$ \ and \ $V_n \cap L \not= \emptyset$;
\item[(iv)] $u_n V_n\subset U_{n-1}$;
\item[(v)] if $H_n$ denotes the semigroup generated by $\{u_1, v_1,u_2, v_2,\ldots,u_n, v_n\}$; which we enumerate as: 
$H_n := \{h^n_j:j\in\N\}$ and   
$$U_n:= \{t \in G: |f(h^i_j\,t)-f(h^i_j)| < 1/n\quad \text{for}\quad 1\le i,j\le n\} $$
then $H_n \subset \Lambda$ and  $e\in U_n\subset \overline{U_n}\subset U_{n-1}$.
\end{enumerate}

\noindent {\bf Construction.} We let $U_0 :=G$ and let $V_0$ be the interior of $f^{-1}(1)$ and $u_0,\,v_0$ are not defined.
Assume that $n\in\N$ and that $U_k,\,V_k$ are defined for $0\le k<n$ and $v_k,\,u_k$ are
defined for $0<k<n$.  By our assumption there exists an $x \in (U_{n-1}\cap\Lambda)^{-1} U_{n-1}\cap(V_{n-1}\cap L).$
So there is a $u_n \in U_{n-1}\cap\Lambda$ such that $u_nx \in U_{n-1}$.  Since $u_n \in \Lambda,\; x \in V_{n-1}$ and
$U_{n-1}$ is open, there is an open neighbourhood $V_n$ of $x $ such that $x \in V_n\subset
\overline{V_n}\subset V_{n-1}$ and $u_n V_n\subset U_{n-1}$. Then $V_n\cap L\ne\emptyset$ since $x \in V_n\cap L$.
Thus (ii)-(iv) are satisfied. Let $v_n$ be any element of $V_n \cap \Lambda$, then by (iv) and (ii),
(i) is satisfied and $H_n\subset \Lambda$ is defined. Finally, since the map
$t\mapsto |f(gt)-f(g)|$ is continuous for $g\in\Lambda$, the set $U_n$ is an open neighbourhood of $e$ and so condition (v) is satisfied. 
This completes the construction.

We let
$$U_{\infty}=\mbox{$\bigcap$}\{\overline{U_n}:n\in\N\}  \mbox{ \quad and \quad }H = \mbox{$\bigcup$} \{H_n:n \in \N\}$$
and let $u_{\infty},\,v_{\infty}$ be cluster points of the sequences $\{u_n:n\in\N\},\;\{v_n:n \in \N\}$ respectively. Clearly
$u_{\infty}\in U_{\infty}$, $v_{\infty}\in V_{0}$ and $\overline{H}$ is a subgroup of $G$, by Lemma \ref{th:3.3}.
Moreover, by the construction, $f(ht)=f(h)$ for each
$h\in H$ and each $t\in\bigcap\{\overline{U_n}:n\in\N\}$.
Therefore, if we let
\begin{eqnarray*}
S &=& \{s \in \overline{H} : f(hs)=f(h)  \mbox{ for each $h \in H$} \} \\
&=&   \{s \in \overline{H} : f(hs)=f(h)
  \mbox{ for each $h \in \overline{H}$}\}
 \end{eqnarray*}
then $\bigcap\{\overline{U_n}:n\in\N\}\cap\overline{H}\subset S$ and $S$ is a subgroup of $G$ by Lemma \ref{Lem1}. Furthermore, by (ii), 
$u_{\infty} \in U_{\infty} \cap \overline{H} \subset S$ and by (iv) $u_n v_{\infty} \in \overline{U_{n-1}} \cap \overline{H}$ for each $n \in \N$.
Hence $$u_{\infty} v_{\infty} \in \bigcap_{n \in \N}\overline{U_{n-1}} \cap \overline{H} \subset S.$$ Therefore, $v_{\infty} = u_{\infty}^{-1}(u_{\infty}v_{\infty}) \in S^{-1}S \subset S$.
Now, $f(s) =0$ for all $s \in S$ since 
$$f(es) =f(e) =0 \mbox{\quad for all $s \in S$.}$$ 
Therefore, $f(v_{\infty}) =0$. On the other hand, since $v_{\infty}\in V_0\subset f^{-1}(1),\; f(v_{\infty})=1$. This contradiction completes the proof.
\end{proof}



\section{Invariant means on CHART groups}

In this section we will show that every CHART group admits a unique right-invariant mean. 
 
\begin{Theo}  Every CHART group $(G, \cdot, \tau)$ possesses  a unique right-invariant mean $m$ on $C(G)$.
\end{Theo}
\begin{proof} Let $\mathcal L$ be the family of all $\sigma$-closed normal subgroups $L$ of $G$ for which $C(G/L)$
has a unique right-invariant mean. Clearly, $\mathcal{L} \not= \emptyset$ as $G \in \mathcal{L}$. Now, $(\mathcal{L}, \subseteq )$
is a partially ordered set. We claim that $(\mathcal{L}, \subseteq )$ possesses a minimal element.  To prove this, it is sufficient
to show that every totally ordered subfamily $\mathcal M$ of $\mathcal L$ has a lower bound (in $\mathcal L$).  To this end, let $\mathcal{M} := \{M_\alpha:\alpha \in A\}$ be
a nonempty totally ordered subfamily of $\mathcal{L}$. Let 
$$M_0 := \mbox{$\bigcap$}\{M_\alpha : \alpha \in A\}.$$  Then $M$ is a $\sigma$-closed
normal subgroup of $G$ and $M_0 \subseteq M_\alpha$ for every $\alpha \in A$. 
Thus, to complete the proof of the claim we must show that $M_0 \in \mathcal{L}$,
i.e., show that $C(G/M_0)$ admits a unique right-invariant mean.  For each $\alpha \in A$, let $\pi_\alpha :G/M_0 \to G/M_\alpha$ be defined by,
$\pi_\alpha(gM_0) := gM_\alpha$. Then $\pi_\alpha$ is a continuous, open and onto map and its dual map $\pi_\alpha^{\#}:C(G/M_\alpha) \to C(G/M_0)$
is an isometric algebra isomorphism of $C(G/M_\alpha)$ into $C(G/M_0)$.  By Proposition \ref{Prop1}, for each $\alpha \in A$, there exists a
unique right-invariant mean $m_\alpha$ on $\pi_\alpha^{\#}(C(G/M_\alpha))$.  From the Hahn-Banach extension theorem it follows that each mean $m_\alpha$ has
an extension to a mean  $m_\alpha^*$ on $C(G/M_0)$. Let $\mathscr {A} := \bigcup \{\pi_\alpha^{\#}(C(G/M_\alpha)) :\alpha \in A\}$. 
Then $\mathscr{A}$ is a sub-algebra of $C(G/M_0)$, that contains all the constant functions and separates the point of $G/M_0$ since
$M_0 := \bigcap\{M_\alpha : \alpha \in A\}$.  Therefore, by the Stone-Weierstrass theorem, $\mathscr{A}$ is dense in $C(G/M_0)$.
Let $m$ be a weak$^*$ cluster-point of the net $(m^*_\alpha:\alpha \in A)$ in $B_{C(G/M_0)^*}$.  Clearly, $m$ is a mean on $C(G/M_0)$.
Furthermore, it is routine to show that (i) $m|_{\mathscr{A}}$ is a right-invariant mean on $\mathscr{A}$ and (ii) $m|_{\mathscr{A}}$ is the only (unique) right-invariant mean on 
$\mathscr{A}$. It now follows from continuity that $m$ is the one and only right-invariant mean on $C(G/M_0)$, i.e., $M_0 \in \mathcal L$. 

\medskip

Let $L_0$ be a minimal element of $\mathcal L$.  Then by Remark \ref{Rem}, $N(L_0, \sigma_{L_0}) \in \mathcal{L}$. However, since $N(L, \sigma_{L_0}) \subseteq L_0$ and
$L_0$ is a minimal element of $\mathcal L$ we must have that $N(L, \sigma_L) = L_0$.  Thus, by Lemma \ref{th:3.4}, it must be the case that $L_0 = \{e\}$.  This completes the proof.
\end{proof} 
 
\medskip

Let us now note that the unique right-invariant mean given above is also partially left invariant in the sense that for each $g \in \Lambda(G, \tau)$,
$m(L_g(f)) = m(f)$ for all $f \in C(G)$.  To see why this is true, consider the mean $m^*$ on $C(G)$ defined by, $m^*(f) := m(L_g(f))$ for each $f \in C(G)$
and some $g \in \Lambda(G, \tau)$. Then for any $h \in G$, 
$$m^*(R_h(f)) = m(L_g(R_h(f))) = m(R_h(L_g(f))) = m(L_g(f)) = m^*(f).$$
Therefore, $m^*$ is a right-invariant mean on $C(G)$. Thus, $m^* =m$ and so 
$$m(L_g(f)) = m^*(f) = m(f) \mbox{ for all $f \in C(G)$ and all $g \in \Lambda(G, \tau)$.} $$


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