Proof. Assume that $f\in C_{c}(E^{0})$ is positive and $\mu $ is vertex-invariant. We then claim that

$$ \begin{align*} \int_{E^{0}} f\,{d} (T^{n}\mu) \leq \int_{E^{0}} f\,{d} (T^{n-1}\mu) \end{align*} $$

for all $n\in \mathbb {N}$ . By assumption, it is true for $n=1$ . Assume that it holds true for all $k\leq n$ for some $n\in \mathbb {N}$ . Then $T^{n} \mu $ and $T^{n-1} \mu $ are regular and $T^{n} \mu \leq T^{n-1} \mu $ , so

$$ \begin{align*} \int_{E^{0}} f\,{d} (T^{n+1}\mu) &= \int_{E^{0}} \sum_{e\in s^{-1}(v)} f\circ r(e)\,{d} (T^{n}\mu)(v) \leq \int_{E^{0}} \sum_{e\in s^{-1}(v)} f\circ r(e)\,{d} (T^{n-1}\mu) (v)\\ &= \int_{E^{0}} f\,{d} (T^{n}\mu), \end{align*} $$

proving the claim. It follows that $T^{n} \mu $ is a regular measure for all $n\in \mathbb {N}_{0}$ when $\mu $ is a vertex-invariant measure. As a consequence, we can define a regular measure $\mu _{1}$ with $\mu _{1} \leq \mu $ by setting

$$ \begin{align*} \int_{E^{0}} f\,{d}\mu_{1} : = \lim_{n\to \infty} \int_{E^{0}} f \,{d}(T^{n}\mu) \quad \text{for } f\in C_{c}(E^{0}). \end{align*} $$

Since $\mu _{1} \leq \mu $ and

$$ \begin{align*} \int_{E^{0}} \sum_{e\in s^{-1}(v)} f \circ r(e)\,{d} \mu <\infty \end{align*} $$

for $f\in C_{c}(E^{0})$ , one can use dominated convergence to prove

$$ \begin{align*} \int_{E^{0}} \sum_{e\in s^{-1}(v)} f \circ r(e) \,{d}\mu_{1} = \lim_{n\to \infty} \int_{E^{0}} \sum_{e\in s^{-1}(v)} f \circ r(e) \,{d}(T^{n}\mu) \end{align*} $$

for $f\in C_{c}(E^{0})$ . This implies that $T\mu _{1} = \mu _{1}$ . Set $\mu _{2}=\mu -\mu _{1}$ , it is then straightforward to check that $\mu _{1}$ and $\mu _{2}$ are vertex-invariant. For any function $f\in C_{c}(E^{0})$ and $\varepsilon>0$ , there then exists an $N\in \mathbb {N}$ such that

$$ \begin{align*} \bigg|{\int_{E^{0}}}\, f \,{d}(T^{n}\mu_{2}) \bigg |= \bigg |{\int_{E^{0}}}\, f \,{d}(T^{n}\mu) - \int_{E^{0}} f \,{d}\mu_{1} \bigg | \leq \varepsilon \quad \text{for all } n\geq N. \end{align*} $$

This proves that the decomposition exists. We leave it to the reader to check that the decomposition is unique.

Proof of Proposition 3.2

Assume first that $\mu $ is a vertex-invariant measure satisfying that

(3.1) $$ \begin{align} \lim_{n\to \infty} \int_{E^{0}} f\,{d} T^{n}\mu =0\quad \text{for all } f\in C_{c}(E^{0}). \end{align} $$

We define a regular positive measure $\eta $ on $E^{0}$ by

$$ \begin{align*} \int_{E^{0}} f\,{d} \eta := \int_{E^{0}} f \,{d} \mu - \int_{E^{1}} f \circ r \,{d} s^{*} \mu \quad \text{for } f\in C_{c}(E^{0}). \end{align*} $$

It is straightforward to check that $E_{\mathrm {sng}}^{n}$ with the subspace topology from $E^{n}$ is homeomorphic with the Borel subset $E_{\mathrm {sng}}^{n}$ of $\partial E$ . Let $s_{n}: E^{n} \to E^{0}$ be the source map, since it is a local homeomorphism, we can therefore consider $s_{n}^{*} \eta $ as a Borel measure on $\partial E$ supported on $E^{n}_{\mathrm {sng}}$ . We now claim that

$$ \begin{align*} \nu:=\sum_{n=0}^{\infty} s_{n}^{*} \eta \end{align*} $$

is a regular invariant measure on $\partial E$ such that $r_{*}\nu = \mu $ . Let $f\in C_{c}(E^{0})$ be positive. The map $f\circ r$ is not necessarily compactly supported, but by approximating it by compactly supported functions from below and using monotone convergence, we get for any $i\in \mathbb {N}_{0}$ that

$$ \begin{align*} \int_{E^{0}} \sum_{\alpha \in s_{i}^{-1}(v) } f\circ r(\alpha)\,{d} T\mu(v)&= \int_{E^{0}} \sum_{e\in s^{-1}(v)}\sum_{\alpha \in s_{i}^{-1}(r(e)) } f\circ r(\alpha) \,{d} \mu(v)\\ &=\int_{E^{0}} \sum_{\beta \in s_{i+1}^{-1}(v) } f\circ r(\beta)\,{d} \mu(v) , \end{align*} $$

from which it follows that

$$ \begin{align*} \int_{\partial E} f\circ r\,{d} s_{i}^{*} \eta &=\int_{E^{0}} \sum_{\alpha \in s_{i}^{-1}(v) } f\circ r(\alpha) \,{d} \eta(v) \\ &= \int_{E^{0}} \sum_{\alpha \in s_{i}^{-1}(v) } f\circ r(\alpha) \,{d} \mu(v) -\int_{E^{0}} \sum_{\alpha \in s_{i}^{-1}(v) } f\circ r(\alpha) \,{d} T\mu(v) \\ &= \int_{E^{0}} f \, {d} T^{i}\mu - \int_{E^{0}} f \,{d} T^{i+1}\mu. \end{align*} $$

By assumption, this implies that

$$ \begin{align*} \int_{\partial E} f\circ r \,{d} \nu = \int_{E^{0}} f \,{d} \mu -\lim_{n\to \infty} \int_{E^{0}} f \,{d} T^{n} \mu = \int_{E^{0}} f\,{d} \mu. \end{align*} $$

If $S\subseteq E^{0}$ , it is easy to check that $1_{S} \circ r = 1_{Z(S)}$ , and since $\partial E$ can be covered by open sets $Z(U)$ with $U\subseteq E^{0}$ open and $\mu (U)<\infty $ , this equality implies that $\nu $ is a regular measure. Since $s_{i}^{*} \eta $ is concentrated on $E^{i}_{\mathrm {sng}}$ , we first see that if $A\subseteq E^{\infty }$ is Borel, then $\nu (A)=0=\nu (\sigma (A))$ . By second countability, it therefore suffices to check that $\nu (A)=\nu (\sigma (A))$ for a Borel set $A \subseteq E^{i}_{\mathrm {sng}}$ with $s_{i}$ and $\sigma $ injective on A, yet for such a set, we have that

$$ \begin{align*} \nu(A) = s_{i}^{*}\eta(A) = \eta(s_{i}(A)) =\eta(s_{i-1}(\sigma(A)))= s_{i-1}^{*}\eta( \sigma(A)) =\nu(\sigma(A)) , \end{align*} $$

which proves the proposition for measures $\mu $ satisfying equation (3.1).

Assume now that $\mu $ satisfies that $T\mu = \mu $ . Following [Reference RenaultSch18, §3], we define $\partial E_{n} := E_{\mathrm {sng}}^{0} \cup \cdots \cup E_{\mathrm {sng}}^{n-1}\cup E^{n}$ and set

$$ \begin{align*} Z_{n}(S)=\{ \alpha \in \partial E_{n} \mid r(\alpha)\in S \text{ or } \alpha_{1} \cdots \alpha_{k} \in S \text{ for some}\ 1\leq k \leq n \}, \end{align*} $$

where $S \subseteq E^{*}$ . Then, $\partial E_{n}$ equipped with the topology generated by the basis $Z_{n}(U)\setminus Z_{n}(K)$ with $U\subseteq E^{*}$ open and $K \subseteq E^{*}$ compact satisfies that $\rho _{n}:\partial E_{n} \to \partial E_{n-1}$ , defined by

$$ \begin{align*} \rho_{n}(\alpha_{1} \cdots \alpha_{k})= \begin{cases} \alpha_{1} \cdots \alpha_{n-1} & \text{if } k=n , \\ \alpha_{1} \cdots \alpha_{k} & \text{if } k<n, \end{cases} \end{align*} $$

is a proper, continuous and surjective map, and that we get a homeomorphism

$$ \begin{align*} \partial E = \varprojlim_{n} (\partial E_{n} , \rho_{n}) , \end{align*} $$

cf. [Reference RenaultSch18, Proposition 3.3]. Since $\sigma ^{n}:\partial E_{n} \cap E^{n} \to E^{0}$ is a local homeomorphism by [Reference RenaultSch18, Proposition 3.5], we can define a Borel measure $(\sigma ^{n})^{*}(\mu )$ on $\partial E_{n}$ which is concentrated on $E^{n}$ .

Claim: $r_{*}(\sigma ^{n})^{*} \mu = \mu $ for all $n\in \mathbb {N}$ . Assume that $f\in C_{c}(E^{0})$ is positive and $n\in \mathbb {N}$ . Then

$$ \begin{align*} \int_{E^{0}} f\,{d} r_{*}(\sigma^{n+1})^{*} \mu &= \int_{\partial E_{n+1}} f\circ r \,{d} (\sigma^{n+1})^{*} \mu = \int_{E^{0}} \sum_{\alpha\in (\sigma^{n+1})^{-1} (v) } f\circ r(\alpha) \,{d} \mu (v) \\ &= \int_{E^{0}} \sum_{e \in s^{-1}(v)} \sum_{\alpha\in (\sigma^{n})^{-1} (r(e)) } f\circ r(\alpha e) \,{d} \mu (v)\\ &= \int_{E^{0}} \sum_{\alpha\in (\sigma^{n})^{-1} (v) } f\circ r(\alpha) \,{d} T\mu (v) \\ &= \int_{E^{0}} f \,{d} r_{*}(\sigma^{n})^{*} \mu, \end{align*} $$

since $T\mu = \mu $ . This proves the claim.

Using the claim, we can now see that for $U\subseteq E^{0}$ open with compact closure, then

$$ \begin{align*} (\sigma^{n})^{*}(\mu)(Z_{n}(U)) = r_{*}(\sigma^{n})^{*}(\mu)(U)=\mu(U)<\infty , \end{align*} $$

and hence $(\sigma ^{n})^{*}(\mu )$ is a regular measure on $\partial E_{n}$ . For $f\in C_{c}(\partial E_{n-1})$ , we get

$$ \begin{align*} \int_{\partial E_{n}} f\circ \rho_{n} \,{d} (\sigma^{n})^{*} \mu &= \int_{E^{0}} \sum_{\alpha \in (\sigma^{n})^{-1}(v)} f\circ \rho_{n}(\alpha) \,{d} \mu(v)\\ &= \int_{E^{0}} \sum_{e\in s^{-1}( v)} \sum_{\beta \in (\sigma^{n-1})^{-1}(r(e))} f(\beta) \,{d} \mu(v)\\ &=\int_{E^{0}} \sum_{\beta \in (\sigma^{n-1})^{-1}(v)} f(\beta) \,{d} T\mu(v) = \int_{\partial E_{n-1}} f \,{d} (\sigma^{n-1})^{*}\mu , \end{align*} $$

which proves that $(\rho _{n})_{*} (\sigma ^{n})^{*} \mu = (\sigma ^{n-1})^{*} \mu $ . By [Reference NeshveyevOO78, Theorem 1], there therefore exists a unique regular measure $\nu $ on $\partial E$ such that $\nu \circ p_{n}^{-1} = (\sigma ^{n})^{*} \mu $ , where $p_{n}$ is the projection from $\partial E$ to $\partial E_{n}$ .

To see that $\nu $ is invariant, let $O \subseteq E^{*}$ be an open set such that $\sigma $ is injective on $Z(O) \subseteq \partial E \setminus E^{0}$ . Pick a $k\in \mathbb {N}$ and open subsets $Z(U_{i}) \setminus Z(K_{i}) \subseteq Z(O)$ for $i=1, \dots , k$ as in Lemma 2.1. For sufficiently big n, then $p_{n}^{-1}(Z_{n}(U_{i}) \setminus Z_{n}(K_{i})) = Z(U_{i}) \setminus Z(K_{i})$ for all $1\leq i \leq k$ , so we have that

$$ \begin{align*} \nu\bigg(\bigcap_{i=1}^{k} Z(U_{i}) \setminus Z(K_{i}) \bigg) = (\sigma^{n})^{*} \mu \bigg(\bigcap_{i=1}^{k} Z_{n}(U_{i}) \setminus Z_{n}(K_{i})\bigg) \end{align*} $$

and using that $\sigma $ is injective on $Z(O)$ and item (3) in Lemma 2.1, we get that

(3.2) $$ \begin{align} \nonumber \nu\bigg[\sigma\bigg(\bigcap_{i=1}^{k} Z(U_{i}) \setminus Z(K_{i})\bigg)\bigg] &=\nu\bigg[\bigcap_{i=1}^{k} Z(\sigma(U_{i})) \setminus Z(\sigma(K_{i}))\bigg] \\ &=(\sigma^{n-1})^{*} \mu \bigg(\bigcap_{i=1}^{k} Z_{n-1}(\sigma(U_{i})) \setminus Z_{n-1}(\sigma(K_{i}))\bigg)\nonumber\\ &= (\sigma^{n-1})^{*} \mu \bigg( \sigma\bigg(\bigcap_{i=1}^{k} Z_{n}(U_{i}) \setminus Z_{n}(K_{i})\bigg) \bigg). \end{align} $$

If $\sigma $ is injective on a set $A \subseteq \partial E_{n} \setminus E^{0}$ , then by second countability, we can write

$$ \begin{align*} A=A_{0}\sqcup \bigsqcup_{k\in \mathbb{N}} A_{k} \end{align*} $$

such that $\sigma ^{n}$ is injective on $A_{k}$ for $k\geq 1$ and $A_{0} \cap E^{n}=\emptyset $ , and hence

$$ \begin{align*} (\sigma^{n})^{*}\mu(A) = \sum_{k\in \mathbb{N} } \mu(\sigma^{n}(A_{k})) = \sum_{k\in \mathbb{N} } (\sigma^{n-1})^{*}\mu(\sigma(A_{k})) =(\sigma^{n-1})^{*}\mu(\sigma(A)). \end{align*} $$

Using this observation and equation (3.2), we see that

$$ \begin{align*} \nu\bigg[\sigma\bigg(\bigcap_{i=1}^{k} Z(U_{i}) \setminus Z(K_{i})\bigg)\bigg] = (\sigma^{n})^{*} \mu \bigg( \bigcap_{i=1}^{k} Z_{n}(U_{i}) \setminus Z_{n}(K_{i})\bigg) =\nu\bigg[\bigcap_{i=1}^{k} Z(U_{i}) \setminus Z(K_{i})\bigg] , \end{align*} $$

and since intersections of such sets from Lemma 2.1 are a $\pi $ -system, we have that $\nu $ is $\sigma $ -invariant, cf. [Reference CohnCoh13, Corollary 1.6.4].

By linearity and Theorem 3.3, we have now proven Proposition 3.2.

Proof. Let us first check that the map is well defined, that is, that $r_{*} \nu $ satisfies Definition 3.1. Let therefore $\nu $ be a regular measure on $\partial E$ . If $K\subseteq E^{0}$ is compact, then $Z(K)$ is compact in $\partial E$ and hence $\nu (r^{-1}(K))=\nu (Z(K))<\infty $ . So $r_{*}\nu $ is a regular measure. Let $f\in C_{c}(E^{0})$ be positive, then we get

(3.3) $$ \begin{align} \int_{E^{1}} f \circ r \,{d} s^{*}(r_{*}\nu) &= \int_{E^{0}} \sum_{e\in s^{-1}(v) } f \circ r (e)\,{d} r_{*}\nu (v) = \int_{\partial E} \sum_{e\in s^{-1}(r(\alpha)) } f \circ r (e)\,{d} \nu (\alpha) \nonumber\\ &= \int_{\partial E} \sum_{\beta\in \sigma^{-1}(\alpha) } f \circ r (\beta)\,{d} \nu (\alpha). \end{align} $$

When $\nu $ is invariant on $\partial E$ , we know from equation (2.5) that $\sigma ^{*}\nu (A)=\nu (\sigma (A))=\nu (A)$ whenever $\sigma $ is injective on A, and hence it is true for all $A\subseteq \partial E\setminus E^{0}$ . Since $f\circ r$ might not be zero on $E^{0}$ , we get that

$$ \begin{align*} \int_{\partial E} \sum_{\beta\in \sigma^{-1}(\alpha) } f \circ r (\beta)\,{d} \nu (\alpha)= \int_{\partial E\setminus E^{0}} f \circ r \,{d} \sigma^{*}\nu \leq \int_{\partial E } f \circ r \,{d} \nu = \int_{E^{0} } f \,{d} (r_{*}\nu) , \end{align*} $$

with equality if $f\circ r$ is zero on $E^{0}_{\mathrm {sng}}$ . If $\text {supp}(f) \subseteq E_{\mathrm {reg}}^{0}$ , then $f\circ r$ is zero on $E^{0}_{\mathrm {sng}}$ , which proves that the map $\nu \mapsto r_{*} \nu $ is well defined. That it is surjective follows from Proposition 3.2. To check injectivity, assume that $r_{*}\nu _{1} = r_{*}\nu _{2}$ for two invariant measures $\nu _{1}$ and $\nu _{2}$ . If $U \subseteq E^{0}$ is Borel, we have by definition of $Z(U)$ that

$$ \begin{align*} \nu_{1}(Z(U)) = \nu_{1} (r^{-1}(U)) = \nu_{2} (r^{-1}(U)) = \nu_{2} (Z(U)). \end{align*} $$

For a given $n\in \mathbb {N}$ and a Borel set $V \subseteq E^{n}$ satisfying that $\sigma ^{n} : V \to \sigma ^{n}(V) \subseteq E^{0}$ is injective, we have by invariance of the two measures that

$$ \begin{align*} \nu_{1}(Z(V)) = \nu_{1}( \sigma^{n}(Z(V))) =\nu_{1}( Z(\sigma^{n}(V))) =\nu_{2}( Z(\sigma^{n}(V))) = \nu_{2}(Z(V)). \end{align*} $$

If $V_{n} \subseteq E^{n}$ and $V_{m} \subseteq E^{m}$ are two such sets with $m\geq n$ , then the set

$$ \begin{align*} V'=\{\alpha \in E^{m} \mid \alpha =v\alpha ' \text{ with } v\in V_{n} \} \end{align*} $$

is open, and since

$$ \begin{align*} Z(V_{n})\cap Z(V_{m}) = Z(V_{m}\cap V' ) , \end{align*} $$

such sets are a $\pi $ -system that generates the Borel $\sigma $ -algebra. It follows that $\nu _{1} = \nu _{2}$ , which finishes the proof.

Proof. Remark 3.4 implies the if implication in both statements. For the only if implication in item (1), assume that $\nu $ is concentrated on $E^{\infty }$ and decompose $r_{*}\nu =\mu _{1}+\mu _{2}$ as in Theorem 3.3. Let $\nu _{i}$ be the invariant measure with $r_{*}\nu _{i}=\mu _{i}$ for $i=1,2$ . It follows that $\nu =\nu _{1}+\nu _{2}$ and that $\nu _{2}$ is concentrated on $E^{*}_{\mathrm {sng}}$ , and hence $\nu _{2}=0$ since $\nu (E^{*}_{\mathrm {sng}})=0$ . This implies that $\mu _{2}=0$ and proves the only if implication. The only if implication in item (2) follows similarly.

Proof. Let for each $i \in \mathbb {N}$ the set $U_{i} \subseteq E^{0}$ satisfy that $\overline {U_{i}}$ is compact, $U_{i}$ is open and $E^{0} = \bigcup _{i\in \mathbb {N}} U_{i}$ . Define $V_{1}=U_{1}$ and for each $n>1$ , define

$$ \begin{align*} V_{n} = U_{n} \setminus ( U_{1} \cup \cdots \cup U_{n-1} ). \end{align*} $$

The sets $\{V_{n}\}_{n\in \mathbb {N}}$ are disjoint with union $E^{0}$ , so we get

$$ \begin{align*} E^{*}_{\mathrm{sng}} = \bigsqcup_{n \in \mathbb{N}} E^{*}_{\mathrm{sng}}V_{n}. \end{align*} $$

Since $\nu $ is extremal and hence ergodic, it follows from Lemma 2.2 that there exists a unique $m \in \mathbb {N}$ such that $\nu ((E^{*}_{\mathrm {sng}} V_{m} )^{C}) =0$ . The relative topology on $\overline {U_{m}}$ makes it a compact second countable Hausdorff space, and hence a metrisable space. Using standard arguments, we can therefore find a sequence $\{W_{i}\}_{i=1}^{\infty }$ of Borel subsets of $\overline {U_{m}}$ with $W_{i+1} \subseteq W_{i}$ such that the diameter of $W_{i}$ is at most $1/i$ , and such that $\nu ( (E^{*}_{\mathrm {sng}} W_{i} )^{C}) =0$ for all i. We now have that

$$ \begin{align*} 0= \nu \bigg( \bigcup_{i\in \mathbb{N} }(E^{*}_{\mathrm{sng}} W_{i})^{C} \bigg) =\nu \bigg(\bigg(\bigcap_{i\in \mathbb{N} } E^{*}_{\mathrm{sng}} W_{i} \bigg )^{C} \bigg) = \nu \bigg(\bigg(E^{*}_{\mathrm{sng}} \bigg(\bigcap_{i\in \mathbb{N} } W_{i} \bigg) \bigg)^{C}\bigg). \end{align*} $$

Since $\nu \neq 0$ , it follows that $\bigcap _{i\in \mathbb {N} } W_{i}= \{v \}$ for some $v\in E^{0}$ . Hence, we have that $0=\nu ((E^{*}_{\mathrm {sng}} v)^{C})$ , so $v\in E^{0}_{\mathrm {sng}}$ . Since the topology is second countable and s is a local homeomorphism, the set $E^{*}_{\mathrm {sng}} v=E^{*}v$ contains at most countably many elements. Now take a path $\alpha \in E^{*} v$ , we then have by equation (2.5) that $\nu (\{v\})=\nu (\sigma ^{|\alpha |}(\{\alpha \}) )=\nu (\{\alpha \})$ . This implies that

$$ \begin{align*} \nu= \sum_{\alpha \in E^{*}v} \nu(\{\alpha\}) \delta_{\alpha} = \nu(\{v\}) \sum_{\alpha \in E^{*}v} \delta_{\alpha} , \end{align*} $$

where $\nu (\{v\})>0$ by assumption. Since $\nu $ is regular and $Z(K)$ is a compact set for any compact $K\subseteq E^{0}$ , we can for any $w\in E^{0}$ find an open neighbourhood $V_{w}$ of w in $E^{0}$ with $\nu (Z(V_{w})) <\infty $ . Since

$$ \begin{align*} \nu(Z(V_{w})) = \nu(\{v\}) \sum_{\alpha \in E^{*}v} \delta_{\alpha} (Z(V_{w})) = \nu(\{v\}) \cdot | V_{w} E^{*}v | , \end{align*} $$

this finishes the proof.

Proof. Assume that $v\in E^{0}_{\mathrm {sng}}$ satisfies that for any $w\in E^{0}$ , there exists an open neighbourhood $V_{w} \subseteq E^{0}$ of w satisfying $|V_{w}E^{*}v|< \infty $ , and define $\nu :=\sum _{\alpha \in E^{*}v} \delta _{\alpha }$ . If $V\subseteq E^{0}$ is an open set, then

$$ \begin{align*} \nu(Z(V)) = \sum_{\alpha \in E^{*}v} \delta_{\alpha}(Z(V)) =|VE^{*}v|. \end{align*} $$

Since the set $Z(V_{w})$ with $w\in E^{0}$ is an open cover of $\partial E$ , it follows that $\nu $ is a regular measure. Now assume that $A \subseteq \partial E \setminus E^{0}$ with $\sigma $ injective on A. Since $\sigma (\alpha ) \in E^{*}v$ if and only if $\alpha \in E^{*}v$ for all $\alpha \in A$ , we get that

$$ \begin{align*} \nu(\sigma(A)) &= |\sigma(A) \cap E^{*}v| =|\sigma (A \cap E^{*} v)| =|A \cap E^{*}v| = \nu(A) , \end{align*} $$

which by equation (2.5) proves that $\nu $ is an invariant boundary measure. The measure is clearly extremal, because any invariant set A either contains $E^{*}v$ or is disjoint from it, depending on whether it contains v. It follows that the map which maps such a vertex v to the extremal ray of boundary measures $\{ r \cdot \nu \mid r>0\}$ is well defined, and by Proposition 4.1, it is also surjective. To see that it is injective, notice that $\nu (\{w\})>0$ for a $w\in E^{0}$ if and only if $w=v$ , which proves that the map is a bijection.

Proof. Let us first argue that there will also exist an extremal invariant boundary measure. Set $\eta = \int _{\partial E} P(\cdot )\,{d} \nu $ which is a tracial weight on $C^{*}(E)$ . Take an $f\in C_{c}(E^{0})_{+}$ with

$$ \begin{align*} \int_{E^{0}} f\,{d} T(r_{*}\nu) < \int_{E^{0}}f\,{d} (r_{*}\nu), \end{align*} $$

which we can rewrite using equation (3.3) to get

(4.1) $$ \begin{align} \int_{\partial E} \sum_{\beta \in \sigma^{-1}(\alpha)} f\circ r(\beta)\,{d} \nu(\alpha) < \int_{\partial E} f\circ r \,{d} \nu. \end{align} $$

Since $r^{-1}(\text {supp}(f))=Z(\text {supp}(f))$ , then $f\circ r \in C_{c}(\partial E)_{+}$ . We now want to use Choquet theory to find an extremal measure satisfying equation (4.1), but since $f\circ r$ does not a priori lie in $C_{c}(\partial E \setminus E^{0})$ , we take increasing sequence $\{ g_{n} \}_{n\in \mathbb {N}} \subseteq C_{c}(\partial E \setminus E^{0} )$ of positive functions with $\lim _{n\to \infty } g_{n} = f\circ r|_{\partial E \setminus E^{0}}$ . Using that $\mathcal {G}_{E}$ is a minimal groupoid, [Reference ChristensenChr23, Proposition 6.10] implies that the tracial weights $\psi $ satisfying $\psi (f\circ r)=\eta (f\circ r)$ is a simplex. Since

$$ \begin{align*} \lim_{n\to \infty} \int_{\partial E} \sum_{\beta \in \sigma^{-1}(\alpha)} g_{n}(\beta)\,{d} \nu(\alpha) = \int_{\partial E} \sum_{\beta \in \sigma^{-1}(\alpha)} f\circ r(\beta) \,{d} \nu(\alpha) < \int_{\partial E} f\circ r \,{d} \nu \end{align*} $$

and $\alpha \mapsto \sum _{\beta \in \sigma ^{-1}(\alpha )} g_{n}(\beta )$ lies in $C_{c}(\partial E)$ , then Choquet theory implies that there is an extremal tracial weight $\psi $ with $\psi (f\circ r)=\eta (f\circ r)$ and

$$ \begin{align*} \int_{\partial E} \sum_{\beta \in \sigma^{-1}(\alpha)} f\circ r(\beta)\,{d} \nu_{\psi}(\alpha) =\lim_{n\to \infty} \int_{\partial E} \sum_{\beta \in \sigma^{-1}(\alpha)} g_{n}(\beta) \,{d} \nu_{\psi}(\alpha) < \int_{\partial E} f\circ r \,{d} \nu_{\psi} , \end{align*} $$

where $\nu _{\psi }$ is the measures associated to $\psi $ . This implies that

$$ \begin{align*} \int_{E^{0}} f\,{d} T(r_{*}\nu_{\psi}) =\int_{\partial E} \sum_{\beta \in \sigma^{-1}(\alpha)} f\circ r(\beta)\,{d} \nu_{\psi}(\alpha) < \psi(f\circ r)=\int_{E^{0}}f\,{d} (r_{*}\nu_{\psi}), \end{align*} $$

from which it follows that $\nu _{\psi }$ is an extremal boundary measure.

So we can assume without loss of generality that $\nu $ is an extremal invariant boundary measure. It follows that there exists a $v \in E_{\mathrm {sng}}^{0}$ such that $\nu $ is concentrated on the countable set $E^{*}v$ . If we can prove that $\partial E = E^{*}v$ , this will prove the corollary, because then $Z(V_{w})=V_{w}E^{*}v$ and hence $\partial E$ is discrete, and the groupoid corresponding to the graph will be isomorphic to the equivalence relation $E^{*}v \times E^{*}v$ , which exactly gives rise to the C $^{*}$ -algebra $\mathcal {K}(l^{2}(E^{*}v))$ . Since $E^{*}v$ is invariant and $C^{*}(E)$ is simple, it suffices to prove that $E^{*}v$ is a closed set to see that $\partial E = E^{*} v$ , because a non-trivial closed invariant set in the unit space of a groupoid gives rise to a non-trivial ideal.

Assume now that $\alpha \in \partial E$ with $\alpha \notin E^{*}v$ . By existence of $\nu $ , there is an open set $U \subseteq E^{0}$ such that $r(\alpha ) \in U$ yet $|U E^{*} v| < \infty $ . Write $U E^{*} v = \{ \beta ^{1}, \dots , \beta ^{n} \}$ . We consider two cases. First, assume that $r(\alpha _{j})\neq v$ for all $1\leq j \leq |\alpha |$ . By definition, each $\beta ^{i}$ has $s(\beta ^{i}) = v$ , so it follows that $\alpha \notin Z(\{\beta ^{1}, \dots , \beta ^{n} \})$ . Since $\{\beta ^{1}, \dots , \beta ^{n} \}$ is compact, the set $Z(U)\setminus Z(\{\beta ^{1}, \dots , \beta ^{n} \})$ is an open set containing $\alpha $ , yet a $\gamma \in E^{*}v \cap Z(U)$ has to be an element of $U E^{*} v$ , proving that $Z(U)\setminus Z(\{\beta ^{1}, \dots , \beta ^{n} \})$ is disjoint from $E^{*} v$ .

Assume for the other case that $r(\alpha _{i})=v$ for some $i\in \mathbb {N}$ . If this was true for more than one i, we would have $|U E^{*} v|=\infty $ . So this i is unique. It follows that there exists a unique $\beta ^{k}\in U E^{*} v$ with $\alpha =\beta ^{k}\alpha '$ . Since $\alpha \notin E^{*}v$ , we have $|\alpha |>|\beta ^{k}|$ . Now let $W\subseteq E^{|\beta ^{k}|+1}$ be an open set containing the path $\alpha _{1}\dots \alpha _{|\beta ^{k}|+1}$ such that $r(W)\subseteq U$ , which can be arranged since $r:E^{*} \to E^{0}$ is continuous. The set

$$ \begin{align*} Z(W)\setminus Z(\{ \beta^{1}, \dots , \beta^{k-1}, \beta^{k+1}, \dots, \beta^{n} \}) \end{align*} $$

is an open set in $\partial E$ containing $\alpha $ . Assume for a contradiction that $\gamma $ lies in this open set and $E^{*}v$ . By choice of W, then $r(\gamma )\in r(W)\subseteq U$ , proving that $\gamma =\beta ^{i}$ for some $1\leq i \leq n$ . Since $\gamma \notin Z(\{ \beta ^{1}, \dots , \beta ^{k-1}, \beta ^{k+1}, \dots , \beta ^{n} \})$ , then $\gamma =\beta ^{k}$ . Since $W\subseteq E^{|\beta ^{k}|+1}$ , then $\gamma \notin Z(W)$ , and hence the intersection is empty. In conclusion, $E^{*}v$ is a closed set.

Proof. Assume for a contradiction that $\gamma \in r(\alpha )E^{*}$ is a loop such that $\gamma \notin \{ \alpha ^{n} \}_{n\in \mathbb {N}}$ . Write $\gamma = \gamma '\alpha ^{n}$ with $n\geq 0$ as big as possible. It follows that $|\gamma '|>0$ and that $\gamma '$ is a loop. We can also decompose $\gamma '= \gamma " \delta $ , where $\delta \in r(\alpha ) E^{*}$ is a loop and $r(\delta _{i})\neq r(\alpha )$ for $2\leq i\leq |\delta |$ . Since n was chosen as big as possible, we have that $\delta \in E_{\alpha }^{*} $ .

We now claim that $\delta ^{m} \in E_{\alpha }^{*}$ for all $m\geq 2$ . If $|\delta | \geq |\alpha |$ , this follows directly from the fact that $\delta \in E_{\alpha }^{*} $ . If $|\delta |<|\alpha |$ and $\delta ^{m}=\beta \alpha $ , then we can assume (by possibly picking a smaller m) that $|\beta | < |\delta |$ . By assumption (a), then $|\beta |\neq 0$ , which means that $\delta =\beta \gamma $ with $\gamma $ and $\beta $ loops, which can not be the case since $r(\delta _{i})\neq r(\alpha )$ for $2\leq i\leq |\delta |$ . So $\{ \delta ^{m}\}_{m\in \mathbb {N}} \subseteq V_{r(\alpha )}E_{\alpha }^{*}$ , which contradicts assumption (b).

If $\alpha $ is not simple, then there exists an $i<j$ such that $r(\alpha _{i})=r(\alpha _{j})$ , but then $\gamma :=\alpha _{1}\cdots \alpha _{i-1}\alpha _{j}\cdots \alpha _{|\alpha |} \in r(\alpha )E^{*}$ is a loop such that $\gamma \notin \{ \alpha ^{n}\}_{n\in \mathbb {N}}$ .

Proof. Assume that E is not a free graph and let $\alpha \in vE^{*}$ be the simple loop from $(\text {F}1)$ . Since $\alpha $ is simple, it clearly satisfies condition (a) in Definition 5.1. By $(\text {F}3)$ , we can find an open set $V_{v} \subseteq E^{0}$ such that $r(\beta )\in V_{v}$ and $\beta \in E^{*}v$ implies that $r(\beta )=v$ . Now assume for a contradiction that $\gamma \in V_{v} E_{\alpha }^{*}$ with $|\gamma | \neq 0$ . Then $r(\gamma )\in V_{v}$ and $\gamma \in E^{*}v$ , and hence $\gamma $ is a loop in $vE^{*}$ . Then $s(\gamma _{1})\in \{r(\beta ) \mid \beta \in E^{*}v\}$ and $r(\gamma _{1})=r(\alpha _{1})$ , so by $(\text {F}2)$ , then $\gamma _{1}=\alpha _{1}$ . If $|\gamma |\geq 2$ , then $s(\gamma _{2})\in \{r(\beta ) \mid \beta \in E^{*}v\}$ and $r(\gamma _{2})=s(\gamma _{1})=s(\alpha _{1})=r(\alpha _{2})$ , so $\gamma _{2}=\alpha _{2}$ . We can continue like this to write $\gamma =\alpha ^{n}\alpha _{1}\cdots \alpha _{k}$ for some $0\leq k <|\alpha |$ . However, $s(\gamma )=r(\alpha )$ , so since $\alpha $ is simple, we get that $k=0$ . Since $|\gamma |\neq 0$ and $\gamma \in E_{\alpha }^{*}$ , we reach a contradiction. So $V_{v} E_{\alpha }^{*} = \{v\}$ , and hence $\alpha $ also satisfies condition (b) in Definition 5.1 which makes it an isolated loop.

Assume now that there exists an isolated loop $\alpha $ in $E^{*}$ . Set $v=r(\alpha )$ . By Lemma 5.2, $\alpha $ is simple. Let now $e\in E^{1}$ satisfy that $s(e)=r(\beta )$ with $\beta \in E^{*}v$ and $r(e)=r(\alpha _{k})$ for some k. It follows that we have a loop $\alpha _{1}\cdots \alpha _{k-1}e\beta \in r(\alpha )E^{*}$ . By Lemma 5.2, then $\alpha _{1}\cdots \alpha _{k-1}e\beta =\alpha ^{n}$ . So $e=\alpha _{k}$ , which proves $(\text {F}2)$ . Lastly, let us prove that v is isolated in $\{r(\beta ) \mid \beta \in E^{*}v\}$ . By assumption, there exists an open neighbourhood $V_{v}$ of v in $E^{0}$ with $|V_{v} E^{*}_{\alpha } |< \infty $ . It follows that

$$ \begin{align*} N=\{r(\beta) \mid \beta \in V_{v} E^{*} v \text{ with } r(\beta)\neq v \} \end{align*} $$

is a finite set in $E^{0}$ , and hence the set $V_{v} \setminus N$ is an open set in $E^{0}$ containing v yet which does not contain any other elements of $\{r(\beta ) \mid \beta \in E^{*}v\}$ . In conclusion, the vertex v ensures that E is not a free graph which proves the lemma.

Proof. Assume that $|\alpha |=|\beta |=n$ and $\beta =\alpha _{k+1}\cdots \alpha _{n}\alpha _{1} \cdots \alpha _{k}$ . If $\gamma \in VE_{\alpha }^{*}$ , then $\gamma \alpha _{1}\cdots \alpha _{k} \in VE^{*}r(\beta )$ . If $\gamma \alpha _{1}\cdots \alpha _{k}=\delta \beta ^{2}$ for some $\delta $ , then we have that

$$ \begin{align*} \gamma \alpha_{1}\cdots \alpha_{k}= \delta \alpha_{k+1}\cdots \alpha_{n}\alpha_{1} \cdots \alpha_{k} \alpha_{k+1}\cdots \alpha_{n}\alpha_{1} \cdots \alpha_{k} \Rightarrow \gamma = (\delta \alpha_{k+1}\cdots \alpha_{n} )\alpha , \end{align*} $$

which contradicts that $\gamma \in E_{\alpha }^{*}$ . Hence, we have an injective map

$$ \begin{align*} VE_{\alpha}^{*} \ni \gamma \mapsto \gamma \alpha_{1}\cdots \alpha_{k} \in VE_{\beta}^{*} \sqcup \{ \delta \beta \mid \delta \in VE_{\beta}^{*} \} , \end{align*} $$

and hence $|VE_{\alpha }^{*}|<\infty $ if $|VE_{\beta }^{*}|<\infty $ . By symmetry, it follows that $\alpha $ satisfies condition (b $'$ ) if and only if $\beta $ does. If $\alpha =\gamma ^{l}$ with $l\geq 2$ and $\sigma ^{k}(\alpha ^{\infty } )=\beta ^{\infty }$ , then

$$ \begin{align*} \sigma^{|\gamma|}(\beta^{\infty}) = \sigma^{|\gamma|}(\sigma^{k}(\alpha^{\infty} ))=\sigma^{k}(\alpha^{\infty} )=\beta^{\infty} , \end{align*} $$

which implies that $\beta $ also does not satisfy condition (a) in Definition 5.1. This proves the lemma.

Proof. It is straightforward to check that $\mathcal {C}S$ is invariant under $\sigma $ . Take an $n\in \mathbb {N}$ . The map $r\circ \sigma ^{i}: E^{\infty } \to E^{0}$ is continuous for any $i\in \mathbb {N}$ , so the set

$$ \begin{align*} M_{n}:=\{x\in E^{\infty} \mid \sigma^{n}(x)=x \} \cap \bigcup_{i=0}^{n-1} \{ x\in E^{\infty} \mid r\circ \sigma^{i}(x) \in S \} \end{align*} $$

is Borel in $E^{\infty }$ and hence in $\partial E$ by Lemma 2.2. Since

$$ \begin{align*} \mathcal{C} S = \bigcup_{n=1}^{\infty} \bigcup_{m=0}^{\infty} \sigma^{-m}(M_{n}) \;, \end{align*} $$

then $\mathcal {C}S$ is a Borel subset of $\partial E$ , proving the lemma.

Proof. Let $\{V_{i} \}_{i\in \mathbb {N}}$ be a countable basis of $E^{0}$ of open sets with compact closures. Then

$$ \begin{align*} \mathcal{C} = \bigcup_{i=1}^{\infty} \mathcal{C} V_{i}. \end{align*} $$

Since $\nu $ is extremal and $\mathcal {C} V_{i} $ is an invariant Borel set by Lemma 5.6, there exists some j with $\nu ( (\mathcal {C} \overline {V_{j}})^{C} )=0$ . Since $\overline {V_{j}}$ is a second countable compact Hausdorff space in the relative topology, we can view the topology on it as generated by a metric. We can therefore find a sequence $\{W_{i} \}_{i=1}^{\infty }$ of Borel sets in $\overline {V_{j}}$ such that $\nu ((\mathcal {C} W_{i})^{C})=0$ for all i, such that $W_{i+1} \subseteq W_{i}$ for all i and such that the diameter of $W_{i}$ is less than $1/i$ . It follows that

$$ \begin{align*} 0= \nu\bigg( \bigcup_{i\in \mathbb{N}} (\mathcal{C} W_{i})^{C} \bigg) = \nu \bigg ( \bigg( \bigcap_{i\in \mathbb{N}} \mathcal{C} W_{i} \bigg)^{C} \bigg) = \nu \bigg ( \bigg( \mathcal{C} \bigcap_{i\in \mathbb{N}} W_{i} \bigg)^{C} \bigg) , \end{align*} $$

but since $W_{i}$ has diameter less than $1/i$ and $\nu \neq 0$ , this implies that $\bigcap _{i\in \mathbb {N}} W_{i}=\{v\}$ for some $v\in E^{0}$ . We now have that

$$ \begin{align*} 0= \nu((\mathcal{C} \{v\} )^{C} ). \end{align*} $$

This implies in particular that $\mathcal {C} \{v\} \neq \emptyset $ . Since s is a local homeomorphism and E is second countable, there exists an at most countable collection $\{\alpha ^{i}\}_{i\in I}$ of loops satisfying that $s(\alpha ^{i})=v$ . It follows by definition of $\mathcal {C}$ that we can write

$$ \begin{align*} \mathcal{C} \{v\} = \bigcup_{i\in I} \{ \beta (\alpha^{i})^{\infty} \mid \beta \in E^{*}v \} , \end{align*} $$

and since each of these sets is invariant, there is a loop $\alpha $ such that $r(\alpha )=s(\alpha )=v$ and

$$ \begin{align*} \nu ( \{ \beta \alpha^{\infty} \mid \beta \in E^{*}r(\alpha) \}^{C} ) =0. \end{align*} $$

It follows that $\nu $ is concentrated on $\{ \beta \alpha ^{\infty } \mid \beta \in E^{*}r(\alpha ) \}$ . Let $n\in \mathbb {N}$ be the smallest number such that $\sigma ^{n}(\alpha ^{\infty })=\alpha ^{\infty }$ . By possible substituting $\alpha $ , we can assume that $|\alpha |=n$ . We claim that the surjective map

$$ \begin{align*} E_{\alpha}^{*} \ni \beta \; \mapsto \; \beta \alpha^{\infty} \in \{ \beta \alpha^{\infty} \mid \beta \in E^{*}r(\alpha) \} \end{align*} $$

is also injective. If $\beta \alpha ^{\infty }=\delta \alpha ^{\infty }$ with $|\delta |<|\beta |$ , then $\beta = \delta \beta '$ , and hence using $\sigma ^{|\delta |}$ on $\beta \alpha ^{\infty }=\delta \alpha ^{\infty }$ , we get that $\beta '\alpha ^{\infty }=\alpha ^{\infty }$ . Since $\beta '\in E_{\alpha }^{*}$ , we can write $\beta '=\alpha ^{k} \zeta $ with $0<|\zeta |< |\alpha |$ , which implies that $\zeta \alpha ^{\infty }=\alpha ^{\infty }$ . Iterating this equality, we see that $\zeta ^{\infty }=\alpha ^{\infty }$ , which contradicts our choice of $\alpha $ . Hence, the map is surjective, and since $\nu (\{ \beta \alpha ^{\infty }\})=\nu (\{ \alpha ^{\infty }\})$ , we get that

$$ \begin{align*} \nu= \sum_{\beta \in E_{\alpha}^{*}} \delta_{\beta \alpha^{\infty}} \; \nu(\{ \beta \alpha^{\infty}\}) = \nu(\{ \alpha^{\infty}\}) \cdot \sum_{\beta \in E_{\alpha}^{*}} \delta_{\beta \alpha^{\infty}}. \end{align*} $$

We now only have to prove that $\alpha $ is a summable loop. Fix therefore a $w\in E^{0}$ . Since $\nu $ is regular on $\partial E$ , there exists an open neighbourhood $V_{w}$ of w in $E^{0}$ with $\overline {V_{w}}$ compact, and hence $\nu (Z(V_{w}))<\infty $ . Since

$$ \begin{align*} \nu(Z(V_{w})) = r\cdot \sum_{\beta \in E_{\alpha}^{*}} \delta_{\beta \alpha^{\infty}}(Z(V_{w})) = r \cdot | V_{w} E_{\alpha}^{*} | , \end{align*} $$

then $\alpha $ is in fact a summable loop.

Proof. Let $\alpha \in E^{*}$ be a summable loop. The set $E_{\alpha }^{*} $ is countable by second countability of E. We define the measure $\nu $ as in equation (6.1). It follows as in the proof of Theorem 4.2 that $\nu $ is non-zero, regular, extremal, invariant and satisfies that $\nu (\mathcal {C}^{C})=0$ . If two loops $\alpha $ and $\beta $ are equivalent, they clearly give rise to the same measure. Hence, the map which sends an equivalence class of a summable loop $\alpha $ to a measure $\nu $ is well defined, and it is surjective by Proposition 6.1. To see that it is injective, let $\nu _{i}$ be defined by the loop $\alpha ^{i}$ for $i=1,2$ and assume $\nu _{1}=\nu _{2}$ . Then $\nu _{2}((\alpha ^{1})^{\infty })>0$ , and hence $(\alpha ^{1})^{\infty }=\beta (\alpha ^{2})^{\infty }$ for some $\beta \in E_{\alpha _{2}}^{*}$ . This implies that $\alpha _{1}$ and $\alpha _{2}$ are equivalent because $\sigma ^{|\beta |}((\alpha ^{1})^{\infty })=(\alpha ^{2})^{\infty }$ . Hence, our map is a bijection which proves the theorem.

Proof. Let $\omega $ be a tracial weight on $C^{*}(E)$ which is not gauge-invariant. Since $\partial E$ is second countable, we can find a strictly positive function $h\in C_{0}(\partial E)$ such that

$$ \begin{align*} \omega(h) = \int_{\partial E} h\,{d} \nu_{\omega} =1. \end{align*} $$

Letting $\mathcal {W}(C^{*}(E))$ denote the tracial weights on $C^{*}(E)$ , we make the following claim.

Let $\{E_{n}\}_{n\in \mathbb {N}} \subseteq C_{c}(\partial E)_{+}$ be an approximate identity as chosen in [Reference ChristensenChr23, §4]. To see that $\mathcal {N}$ is a closed set notice, by e.g. [Reference ChristensenChr23, Corollary 4.8], that

$$ \begin{align*} \mathcal{N}= \bigcap_{n\in \mathbb{N}} \{\psi \in \mathcal{W}(C^{*}(E)) \mid \psi(E_{n}hE_{n})\leq 1\}. \end{align*} $$

Since h is strictly positive, there exists an $\varepsilon _{n}>0$ for each $n\in \mathbb {N}$ such that $h(x) \geq \varepsilon _{n}$ for $x\in \text {supp}(E_{n})$ . If $\psi \in \mathcal {N}$ , it follows that

$$ \begin{align*} \psi(E_{n}^{2})\cdot \varepsilon_{n} = \psi(\varepsilon_{n} E_{n}^{2}) \leq \psi( h) \leq 1 , \end{align*} $$

and hence $\psi (E_{n}^{2}) \leq \varepsilon _{n}^{-1}$ . It follows that the elements of $\mathcal {N}$ restricted to $\mathcal {A}_{n}:=\overline {E_{n} C^{*}(E)E_{n} }$ are contained in the ball in $\mathcal {A}_{n}^{*}$ of radius $\varepsilon _{n}^{-1}$ . The topological space we embed $\mathcal {W}(C^{*}(E))$ in is $\varprojlim _{n} \mathcal {A}_{n}^{*}$ , cf. [Reference ChristensenChr23, §4], and hence $\mathcal {N}$ is compact in this space. This proves the claim.

Let now $f\in C_{c}(\mathcal {G}_{E})$ be supported in $\Phi ^{-1}(\{k\})$ for some $k\in \mathbb {Z} \setminus \{0\}$ satisfying that $\omega (f)\neq 0$ . Since the $\psi \in \mathcal {N}$ with $\psi (f) \neq 0$ is an open set by definition of the topology on $\varprojlim _{n} \mathcal {A}_{n}^{*}$ , we have by the Krein–Milman theorem that there exists an element $\psi ' \in \mathcal {N}$ which is a finite convex combination of extremal elements in $\mathcal {N}$ with $\psi '(f) \neq 0$ . In particular, there exists an extremal element $\psi \in \mathcal {N}$ with $\psi (f)\neq 0$ . It follows that $\psi \neq 0$ , so since $h>0$ on $\partial E$ , then $\psi (h)>0$ . Since $\psi $ is extremal in $\mathcal {N}$ , we must have $\psi (h)=1$ .

To see that $\psi $ also has to be an extremal tracial weight on $C^{*}(E)$ , write $\psi =\psi _{1}+\psi _{2}$ for two non-zero tracial weights $\psi _{1}$ and $\psi _{2}$ on $C^{*}(E)$ . Since $\psi (h)=1$ , then $\psi _{i}(h)<\infty $ for $i=1,2$ . Since $h>0$ , then $\psi _{i}(h)=0$ would imply that the associated measure $\nu _{\psi _{i}}=0$ which would contradict that $\psi _{i}\neq 0$ . Hence, we can write

$$ \begin{align*} \psi=\psi_{1}(h) \frac{\psi_{1}}{\psi_{1}(h)}+\psi_{2}(h) \frac{\psi_{2}}{\psi_{2}(h)} , \end{align*} $$

which implies that $\psi _{1} =\psi \cdot \psi _{1}(h)$ since $\psi $ is extremal in $\mathcal {N}$ . In conclusion, we have an extremal tracial weight $\psi $ with $\psi (f)\neq 0$ . Since $\text {supp}(f) \subseteq \Phi ^{-1}(\{k\})$ , then $\gamma _{z}(f)=z^{k} f$ , so $\psi $ is not gauge-invariant.

By Theorem 2.3, it follows that the subgroup $H\subseteq \mathbb {Z}$ with

$$ \begin{align*} \nu_{\psi}(\{ x \in \partial E\mid H=\mathrm{Per}(x) \}^{C} )=0 \end{align*} $$

is not $\{0\}$ . If $\text {Per}(x)\neq \{0\}$ , then there exists $k>l$ such that $\sigma ^{k}(x)=\sigma ^{l}(x)$ . Then $\sigma ^{k-l}(\sigma ^{l}(x)) = \sigma ^{l}(x)$ and it follows that $x\in \mathcal {C}$ . In conclusion, $\nu _{\psi }(\mathcal {C}^{C})=0$ , and by Proposition 6.1, there exists a summable loop in E.

Assume now that $\omega $ is a tracial state and let $f\in C_{c}(\mathcal {G}_{E})$ be supported in $\Phi ^{-1}(\{k\})$ for some $k\in \mathbb {Z} \setminus \{0\}$ satisfying that $\omega (f)\neq 0$ . The set

$$ \begin{align*} M_{r} = \{ \psi \in C^{*}(E)^{*} \mid \lVert \psi \rVert\leq 1, \psi\text{ is a trace and } |\psi(f)|\geq r \} \end{align*} $$

is a weak $^{*}$ compact subset of the dual for all $r\geq 0$ . Since $M_{r_{2}} \subseteq M_{r_{1}}$ for $r_{2} \geq r_{1}$ , there exists a unique $r_{0}>0$ such that $M_{r_{0}}\neq \emptyset $ and $M_{r} = \emptyset $ for $r>r_{0}$ . Assume that $\psi $ is an extremal element in $M_{r_{0}}$ , and that $\psi =\psi _{1}'+\psi _{2}'$ for two non-zero tracial weights $\psi _{1}'$ and $\psi _{2}'$ . Set $\unicode{x3bb} = \nu _{\psi _{1}'}(\partial E)$ . Since $\nu _{\psi }(\partial E)=1$ , defining $\psi _{1}=\unicode{x3bb} ^{-1} \psi _{1}'$ and $\psi _{2}=(1-\unicode{x3bb} )^{-1} \psi _{2}'$ gives

$$ \begin{align*} \psi= \unicode{x3bb} \psi_{1}+(1-\unicode{x3bb}) \psi_{2}. \end{align*} $$

This implies that

$$ \begin{align*} r_{0}=|\psi(f)|\leq \unicode{x3bb} |\psi_{1}(f) |+(1-\unicode{x3bb})| \psi_{2}(f)| \leq \unicode{x3bb} r_{0}+(1-\unicode{x3bb}) r_{0} = r_{0}, \end{align*} $$

by definition of $r_{0}$ . Hence, we have equality throughout, and $\psi _{i} \in M_{r_{0}}$ for $i=1,2$ . This proves that $\psi $ is a tracial state which is extreme in the set of tracial weights and which satisfies $\psi (f)\neq 0$ . The rest of the proof follows as in the tracial weight case and by invoking Corollary 6.4.

Proof. If there exists a summable loop $\alpha $ in E, then the measure $\nu =\sum _{\beta \in E_{\alpha }^{*}} \delta _{\beta \alpha ^{\infty }}$ is an extremal harmonic measure by Theorem 6.2. The unique subgroup H corresponding to $\nu $ in Theorem 2.3 is $|\alpha | \mathbb {Z}$ . Let the state $\phi $ on $C^{*}(|\alpha | \mathbb {Z})$ satisfy that $\phi (u_{g})=1$ for all $g\in |\alpha | \mathbb {Z}$ . Recall that the map $\Phi $ on $\mathcal {G}_{E}$ given by $(x, k, y)\mapsto k$ gives rise to the gauge-action $\{\gamma _{z}\}_{z\in \mathbb {T}}$ on $C^{*}(\mathcal {G}_{E})$ . If $f\in C_{c}(\mathcal {G}_{E})$ is a positive function which is supported on a bisection inside $\Phi ^{-1}(|\alpha |)$ and satisfies $f(\alpha ^{\infty }, |\alpha |, \alpha ^{\infty })=1$ , then $\psi _{\nu , \phi }(f)>0$ . Hence, $\psi _{\nu , \phi }(\gamma _{z}(f))=z^{|\alpha |}\psi _{\nu , \phi }(f)$ and there exists a tracial weight which is not gauge-invariant. If $|E_{\alpha }^{*} |<\infty $ , then $\nu $ is finite, and the weight $\psi _{\nu , \phi }$ can be normalised to a tracial state.

Proof of Theorem 7.5

We will prove that if there exists a non-gauge-invariant tracial weight, then the graph is not free. If there exists a non-gauge-invariant tracial weight, it follows from Proposition 7.1 that there exists a summable loop $\alpha $ in $E^{*}$ . We have already noted that summable loops are isolated, so by Lemma 5.4, the graph is not free.