Proof. Let $\{E_m\}_m$ be an exhausting sequence for G. Since, in particular, $E_m \subseteq E_{m+1}$ for every $m \geq 1$ , we have that $0 \leq \delta _{E_{m+1}}(\phi ) \leq \delta _{E_m}(\phi )$ for every $m \geq 1$ . Then, for every $M \geq 1$ ,

$$ \begin{align*} V_F(\phi) &\geq \sum_{m=1}^{M} |E_{m+1}^{-1}F \setminus E_{m}^{-1}F| \cdot\delta_{E_m} (\phi)\\ &\geq \delta_{E_M} (\phi) \cdot \sum_{m=1}^{M} |E_{m+1}^{-1}F \setminus E_{m}^{-1}F| \\ &= \delta_{E_M}(\phi) |E^{-1}_{M+1}F \setminus E_1^{-1}F|, \end{align*} $$

where the last line follows from Remark 2.1. Therefore,

$$ \begin{align*}0 \leq \lim_{M \to \infty} \delta_{E_M}(\phi) \leq \lim_{M \to \infty} \frac{V_F(\phi)}{|E^{-1}_{M+1}F \setminus E_1^{-1}F|} = 0,\end{align*} $$

and the result follows.

Proof. Let $\epsilon> 0 $ . Since $\phi \colon X \to \mathbb {R}$ has summable variation, there exists $m_0 \geq 1$ , such that

$$ \begin{align*}\sum_{m> m_0} |E_{m+1}^{-1} \setminus E_{m}^{-1}| \cdot \delta_{E_m}(\phi) < \epsilon. \end{align*} $$

Then, for every $F \in \mathcal {F}(G)$ ,

$$ \begin{align*} V_F(\phi) &\leq \sum_{m=1}^{m_0} |E_{m+1}^{-1}F \setminus E_{m}^{-1}F| \cdot \delta_{E_m}(\phi) + \sum_{m> m_0} |E_{m+1}^{-1} \setminus E_{m}^{-1}||F|\cdot \delta_{E_m}(\phi) \\ & \leq \sum_{m=1}^{m_0} |E_{m+1}^{-1}F \setminus E_{m}^{-1}F| \cdot \delta_{E_m}(\phi) + |F|\cdot\epsilon. \end{align*} $$

Due to the amenability of G, for any given $m_0 \geq 1$ , we have that, for all $m \leq m_0$ ,

$$ \begin{align*}|F| \leq |E_{m+1}^{-1}F| \leq |E_{m_0+1}^{-1}F| \leq (1+\epsilon)|F| \end{align*} $$

for every $(E_{m_0+1},\epsilon )$ -invariant set F. Therefore, for every $\epsilon> 0$ , there exists $m_0 \geq 1$ and $K \in \mathcal {F}(G)$ , such that for every $(K,\epsilon )$ -invariant set F,

$$ \begin{align*}V_F(\phi) \leq \sum_{m=1}^{m_0} ((1+\epsilon)|F| - |F|) \cdot \delta_{E_m}(\phi) + \epsilon\cdot|F| = \epsilon\cdot|F|\sum_{m=1}^{m_0}\delta_{E_m}(\phi) + \epsilon\cdot|F|, \end{align*} $$

so

$$ \begin{align*}\frac{V_F(\phi)}{|F|} \leq \epsilon\cdot C, \end{align*} $$

where $C = 1 + V(\phi )$ . Since $\epsilon $ was arbitrary, we conclude.

Proof. Let $\epsilon> 0$ . Since $\liminf _{m \to \infty } \delta _{E_m}(\phi ) = 0$ , there exists $m_{0} \geq 1$ , such that $\delta _{E_{m_0}}(\phi ) \leq \epsilon $ . Denote $E_{m_0}$ by K. Due to amenability, we can find $K' \supseteq K$ and $0 < \epsilon ' \leq \epsilon $ , such that if F is $(K',\epsilon ')$ -invariant, we have that

$$ \begin{align*} \left|\mathrm{Int}_K(F) \triangle F\right| < \epsilon\cdot|F|. \end{align*} $$

Considering this, if $x, y \in X$ are such that $x_F = y_F$ , we have that

$$ \begin{align*} |\phi_F(x) - \phi_F(y)| & \leq \sum_{g \in F} |\phi(g \cdot x) - \phi(g \cdot y) |\\ & = \sum_{g \in \mathrm{Int}_{K}(F) \cap F} |\phi(g \cdot x) - \phi(g \cdot y) | + \sum_{g \in F \setminus \mathrm{Int}_{K}(F)} |\phi(g \cdot x) - \phi(g \cdot y) |\\ & \leq \sum_{g \in \mathrm{Int}_{K}(F) \cap F} \delta_{K}(\phi) + \sum_{g \in F \setminus \mathrm{Int}_{K}(F)} \delta(\phi)\\ & \leq |\mathrm{Int}_{K}(F)|\cdot\epsilon + |F \setminus \mathrm{Int}_{K}(F)| \cdot\delta(\phi)\\ & \leq |F| \cdot (1+\epsilon) \cdot \epsilon + |F|\cdot \epsilon \cdot \delta(\phi) \\ & = |F| \cdot \epsilon \cdot (1+\epsilon+\delta(\phi)), \end{align*} $$

and the result follows.

Proof. Given a k-cover $\mathcal {K}$ of F, notice that, since $\left .\phi \right \vert {}_{X \cap A^G}$ is nonnegative,

$$ \begin{align*}\phi_F(x) = \sum_{g \in F} \phi(g \cdot x) \leq \frac{1}{k}\sum_{K \in \mathcal{K}} \sum_{g \in K} \phi(g \cdot x) = \frac{1}{k}\sum_{K \in \mathcal{K}} \phi_K(x) \end{align*} $$

for any $x \in X \cap A^G$ . We proceed by induction on the size of $F \setminus E$ . First, suppose that $|F \setminus E| = 0$ . Then, $F \setminus E = \emptyset $ and

$$ \begin{align*} Z^{u_E}_F & = \mathop{\textrm{exp}}\nolimits \left(\sup \phi_F([u_E])\right) \\ & \leq \mathop{\textrm{exp}}\nolimits\left( \sup \frac{1}{k}\sum_{K \in \mathcal{K}} \phi_K([u_E])\right) \\ & \leq \mathop{\textrm{exp}}\nolimits\left( \sum_{K \in \mathcal{K}}\frac{1}{k} \sup \phi_K([u_E]) \right) \\ & = \prod_{K \in \mathcal{K}} \left(\mathop{\textrm{exp}}\nolimits \sup \phi_K ([u_E])\right)^{1/k} \\ & \leq \prod_{K \in \mathcal{K}} \left(\sum_{w_{K \setminus E}}\mathop{\textrm{exp}}\nolimits \sup \phi_K ([w_{K \setminus E}u_E])\right)^{1/k} \\ & = \prod_{K \in \mathcal{K}} \left(Z^{u_E}_K\right)^{1/k}. \end{align*} $$

Now, suppose that $Z^{u_E}_F \leq \prod _{K \in \mathcal {K}} (Z^{u_E}_K)^{1/k}$ for every $E\subseteq G, u_E \in X_E \cap A^E$ , $F \in \mathcal {F}(G)$ with $|F \setminus E| \leq n$ , and every k-cover $\mathcal {K}$ of F. We will show that the same holds for $E, F$ with $|F \setminus E| = n+1$ . Fix $g \in F \setminus E$ , and notice that $|F \setminus (E \cup \{g\})| = n$ . Then,

$$ \begin{align*} Z^{u_E}_F &= \sum_{a \in A} Z^{a^g u_E}_F\\ &\leq \sum_{a \in A}\prod_{K \in \mathcal{K}} \left(Z^{a^gu_E}_K \right)^{1/k} \\ &= \sum_{a \in A}\prod_{K \in \mathcal{K}: g \notin K} \left(Z^{a^gu_E}_K \right)^{1/k}\cdot \prod_{K \in \mathcal{K}: g \in K} \left(Z^{a^gu_E}_K \right)^{1/k} \\ & \leq \prod_{K \in \mathcal{K}: g \notin K} \left(Z^{u_E}_K \right)^{1/k} \sum_{a \in A} \prod_{K \in \mathcal{K}: g \in K} \left(Z^{a^gu_E}_K \right)^{1/k} \\ & \leq \prod_{K \in \mathcal{K}: g \notin K} \left(Z^{u_E}_K \right)^{1/k} \cdot \prod_{K \in \mathcal{K}: g \in K} \left(\sum_{a \in A} Z^{a^gu_E}_K \right)^{1/k} \\ & = \prod_{K \in \mathcal{K}: g \notin K} \left(Z^{u_E}_K \right)^{1/k}\cdot \prod_{K \in \mathcal{K}: g \in K} \left(Z^{u_E}_K \right)^{1/k} \\ & = \prod_{K \in \mathcal{K}} \left(Z^{u_E}_K \right)^{1/k}. \end{align*} $$

Notice that the first inequality follows from the induction hypothesis and the third inequality follows from the generalized Hölder inequality. Indeed, consider $p \leq 1$ , such that $\sum _{K \in \mathcal {K}: g \in K} \frac {1}{k} = \frac {1}{p}$ and the functions $f_K\colon A \to \mathbb {R}$ given by $f_K(a) = \left (Z^{a^gu_E}_K \right )^{1/k}$ . By the generalized Hölder inequality,

$$ \begin{align*}\left\|\prod_{K \in \mathcal{K}: g \in K} f_K\right\|_p \leq \prod_{K \in \mathcal{K}: g \in K}\|f_K\|_k, \end{align*} $$

where

$$ \begin{align*} \prod_{K \in \mathcal{K}: g \in K}\|f_K\|_k & = \prod_{K \in \mathcal{K}: g \in K} \left(\sum_{a \in A} ((Z^{a^gu_E}_K)^{1/k})^k \right)^{1/k} \\ & = \prod_{K \in \mathcal{K}: g \in K} \left(\sum_{a \in A} Z^{a^gu_E}_K \right)^{1/k} \\ & = \prod_{K \in \mathcal{K}: g \in K} \left(Z^{u_E}_K \right)^{1/k} \end{align*} $$

and, since $\|\cdot \|_p$ is monotonically decreasing in p for any fixed $|A|$ -dimensional vector,

$$ \begin{align*}\left\|\prod_{K \in \mathcal{K}: g \in K} f_K\right\|_p \geq \left\|\prod_{K \in \mathcal{K}: g \in K} f_K\right\|_1 = \sum_{a \in A}\prod_{K \in \mathcal{K}: g \in K} \left(Z^{a^gu_E}_K \right)^{1/k}. \end{align*} $$

Therefore, $Z^{u_E}_F \leq \prod _{K \in \mathcal {K}} \left (Z^{u_E}_K \right )^{1/k}$ . In particular, if $E = \emptyset $ , $Z_F \leq \prod _{K \in \mathcal {K}} \left (Z_K \right )^{1/k}$ .

Proof (of Theorem 3.1).

As a consequence of Lemma 3.3, we have that if $\left .\phi \right \vert {}_{X \cap A^G}$ is nonnegative, then $\varphi $ satisfies Shearer’s inequality. Thus, by the Ornstein-Weiss lemma, $p(A,\phi )$ exists, and it satisfies the infimum rule, that is,

$$ \begin{align*}p(A,\phi) = \inf_{E \in \mathcal{F}(G)} \frac{1}{|E|}\log Z_E(A,\phi). \end{align*} $$

Finally, in order to deal with the general case, it suffices to apply the previous result to $\phi + \|\phi \|$ and then observe that $p(A,\phi + C) = p(A,\phi ) + C$ for any constant C.

Proof. Consider a tiling $\mathcal {T}$ made of $(K, \delta )$ -invariant shapes and $\epsilon> 0$ . Suppose that F is $(S_{\mathcal {T}},\epsilon )$ -invariant. Consider the sets $C_F(S) = C(S) \cap \mathrm {Int}_S(F)$ and $\overline {C}_F(S) = C(S) \cap S^{-1}F$ for $S \in \mathcal {S}(\mathcal {T})$ . Notice that, since $\mathcal {T}$ induces a partition, $|SC_F(S)| = |S||C_F(S)|$ , $|S\overline {C}_F(S)| = |S||\overline {C}_F(S)|$ , and

$$ \begin{align*}\bigsqcup_{S \in \mathcal{S}(\mathcal{T})} SC_F(S) \subseteq F \subseteq \bigsqcup_{S \in \mathcal{S}(\mathcal{T})} S\overline{C}_F(S). \end{align*} $$

Therefore,

$$ \begin{align*} F \setminus \bigsqcup_{S \in \mathcal{S}(\mathcal{T})} SC_F(S) & \subseteq \bigsqcup_{S \in \mathcal{S}(\mathcal{T})} S\overline{C}_F(S) \setminus \bigsqcup_{S \in \mathcal{S}(\mathcal{T})} SC_F(S) \\ & = \bigsqcup_{S \in \mathcal{S}(\mathcal{T})} S(\overline{C}_F(S) \setminus C_F(S)) \subseteq \partial_{S_{\mathcal{T}}}(F). \end{align*} $$

Indeed, to check the last inclusion, notice that if $g \in \bigsqcup _{S \in \mathcal {S}(\mathcal {T})} S(\overline {C}_F(S) \setminus C_F(S))$ , then $g = sc$ , where $s \in S$ and $c \in \overline {C}_F(S) \setminus C_F(S)$ for some $S \in \mathcal {S}(\mathcal {T})$ . Therefore, since $c \in \overline {C}_F(S)$ ,

$$ \begin{align*}S_{\mathcal{T}} g \cap F \supseteq SS^{-1}sc \cap F \supseteq Sc \cap F \neq \emptyset. \end{align*} $$

Similarly, since $c \notin C_F(S)$ ,

$$ \begin{align*}S_{\mathcal{T}} g \cap F^c \supseteq SS^{-1}sc \cap F^c \supseteq Sc \cap F^c \neq \emptyset, \end{align*} $$

so that $g \in \partial _{S_{\mathcal {T}}}(F)$ . Then,

$$ \begin{align*}\left|F \setminus \bigsqcup_{S \in \mathcal{S}(\mathcal{T})} SC_F(S)\right| \leq \left|\partial_{S_{\mathcal{T}}}(F)\right| \leq \left|S_{\mathcal{T}} F \triangle F\right| \leq \epsilon\cdot|F|, \end{align*} $$

where we have used that $|\partial _{K}(F)| \leq |(K \cup K^{-1} \cup \{1_G\})F \triangle F|$ for any $K \in \mathcal {F}(G)$ and that $S_{\mathcal {T}}^{-1} = S_{\mathcal {T}}$ and $1_G \in S_{\mathcal {T}}$ .

Proof. Let $\epsilon> 0$ and $F \in \mathcal {F}(G)$ be such that $|F| \geq -\frac {1}{\epsilon }\log (1-\epsilon )$ . For every such F, there exists a finite set of words $W_F \Subset X_F$ , such that

$$ \begin{align*} \sum_{w \in W_F} \mathop{\textrm{exp}}\nolimits \left(\sup \phi_{F}\right) \geq Z_{F}(\phi) \sqrt{1-\epsilon}. \end{align*} $$

On the other hand, since $\phi $ is uniformly continuous, there must be an index $m \geq 1$ for which

$$ \begin{align*} \delta_{E_m}(\phi) \leq \frac{1}{3|F|} \log \left(\frac{1}{\sqrt{1-\epsilon}}\right). \end{align*} $$

For each $w \in W_F$ , pick a word $w' \in \mathbb {N}^{E_mF}$ , such that $w^{\prime }_F=w$ and

$$ \begin{align*} \sup \phi_{F} [w'] \geq \sup \phi_{F} [w] - \frac{1}{3} \log \left(\frac{1}{\sqrt{1-\epsilon}}\right). \end{align*} $$

In addition, for each such $w'$ , pick a configuration $x_w \in [w']$ , such that

$$ \begin{align*} \phi_{F}(x_w) \geq \sup \phi_{F}[w'] - \frac{1}{3} \log \left(\frac{1}{\sqrt{1-\epsilon}}\right). \end{align*} $$

Define $A_F$ to be $\bigcup _{w \in W_F} w'(E_mF)$ , where $w'(E_mF) = \bigcup _{g \in E_mF}\{w'(g)\}$ . It is direct that $A_F$ is a finite subset of $\mathbb {N}$ . Pick $y \in [w'] \cap A_F^G,$ and notice that $(g \cdot x_w)_{E_m} = (g \cdot y)_{E_m}$ for all $g \in F$ . Then, for every $w \in W_F$ ,

$$ \begin{align*} \sup \phi_{F}[[w'] \cap A_F^G] & \geq \phi_{F}(y)\\ & \geq \phi_{F}(x_w) - \sum_{g \in F}\left|\phi(g \cdot x_w) - \phi(g \cdot y)\right|\\ & \geq \phi_{F}(x_w) - \left|F\right| \delta_{E_m}(\phi)\\ & \geq \phi_{F}(x_w) - \frac{1}{3} \log \left(\frac{1}{\sqrt{1-\epsilon}}\right)\\ &\geq \sup \phi_{F}[w'] - \frac{2}{3} \log \left(\frac{1}{\sqrt{1-\epsilon}}\right)\\ & \geq \sup \phi_{F}[w] - \log \left(\frac{1}{\sqrt{1-\epsilon}}\right). \end{align*} $$

Hence,

$$ \begin{align*} Z_{F}(A_F, \phi) & = \sum_{w \in A_{F}^{F}} \mathop{\textrm{exp}}\nolimits \left(\sup \phi_{F}[[w] \cap A_F^G]\right) \\ & \geq \sum_{w \in W_F} \mathop{\textrm{exp}}\nolimits \left(\sup \phi_{F}[[w] \cap A_F^G]\right) \\ & \geq \sum_{w \in W_F} \mathop{\textrm{exp}}\nolimits \left(\sup \phi_{F}[[w'] \cap A_F^G]\right) \\ & \geq \sum_{w \in W_F} \mathop{\textrm{exp}}\nolimits \left(\sup \phi_{F}[w]-\log \left(\frac{1}{\sqrt{1-\epsilon}}\right)\right) \\ & = \sqrt{1-\epsilon}\sum_{w \in W_F} \mathop{\textrm{exp}}\nolimits \left(\sup \phi_{F}[w]\right) \\ & \geq (1-\epsilon) Z_{F}(\phi).\\[-39pt] \end{align*} $$

Proof. Fix $1/2> \epsilon > 0$ and an exhausting sequence $\{E_m\}_m$ for G. Since $\phi $ is uniformly continuous, we have that $\lim _{F \to G} \frac {\Delta _{F}(\phi )}{|F|} = 0$ , by Lemma 2.4. Therefore, there exist ${K' \in \mathcal {F}(G)}$ and $\delta '> 0$ , such that $\Delta _{F}(\phi ) < \epsilon |F|$ for every finite $(K',\delta ')$ -invariant set F.

By Theorem 3.5, there exists a tiling $\mathcal {T}'$ , such that its shapes are $(K',\delta ')$ -invariant. Without loss of generality, by possibly readjusting $K'$ and $\delta '$ , assume that ${|S'| \geq -\frac {1}{\epsilon }\log (1-\epsilon )}$ for every $S' \in \mathcal {S}(\mathcal {T}')$ . Therefore, by Lemma 3.7, for every $S' \in \mathcal {S}(\mathcal {T}')$ , there exists $A_{S'} \Subset \mathbb {N}$ , such that $Z_{S'}(A_{S'}, \phi ) \geq (1-\epsilon ) Z_{S'}(\phi )$ . Define A to be $\bigcup _{S' \in \mathcal {S}(\mathcal {T}')} A_{S'}$ . Then, A is a finite subset of $\mathbb {N}$ . Moreover, since $A_{S'} \subseteq A$ , for each $S' \in \mathcal {S}(\mathcal {T}')$ , we have that

(3.2) $$ \begin{align} Z_{S'}(A, \phi) \geq (1-\epsilon) Z_{S'}(\phi), \end{align} $$

for every $S' \in \mathcal {S}(\mathcal {T}')$ .

Now, by Theorem 3.1, $p(A,\phi ) =\lim _{F \to G} \frac {1}{|F|}\log Z_F(A,\phi )$ exists, so we can pick $K \in \mathcal {F}(G)$ and $\delta> 0$ , such that $K \supseteq K'$ , $\delta < \delta '$ , and

(3.3) $$ \begin{align} \log Z_F(A,\phi) \leq |F| (p(A,\phi) + \epsilon) \end{align} $$

for every $(K,\delta )$ -invariant set $F \in \mathcal {F}(G)$ .

Next, by Theorem 3.5, we can obtain a tiling $\mathcal {T}$ of $(K,\delta )$ -invariant sets, such that every tile in $\mathcal {T}$ is a union of tiles in $\mathcal {T}'$ , that is, $S = \bigsqcup _{S' \in \mathcal {S}(\mathcal {T}')}\bigsqcup _{c' \in C_{S}(S')} S'c'$ . Furthermore, by Lemma 3.6, for every $(S_{\mathcal {T}},\epsilon )$ -invariant set $F \in \mathcal {F}(G)$ , there exist centre sets $C_F(S) \subseteq C(S) \in \mathcal {C}(\mathcal {T})$ for $S \in \mathcal {S}(\mathcal {T})$ , such that

$$ \begin{align*}F \supseteq T_F \quad \text{and} \quad |F \setminus T_F| \leq \epsilon|F|, \end{align*} $$

where $T_F = \bigsqcup _{S \in \mathcal {S}(\mathcal {T})} SC_F(S)$ .

Furthermore, for every $S \in \mathcal {S}(\mathcal {T})$ , we have that

$$ \begin{align*} Z_S(A,\phi) &= \sum_{w_{S} \in A^{S}} \mathop{\textrm{exp}}\nolimits \left(\sup\phi_{S} ([w_{S}] \cap A^G)\right)\\ &\geq \sum_{w_{S} \in A^{S}} \mathop{\textrm{exp}}\nolimits\left(\inf\phi_{S} \left([w_{S}] \cap A^G\right)\right)\\ &\geq \prod_{S' \in \mathcal{S}(\mathcal{T}')} \prod_{c' \in C_S(S')} \sum_{w_{S'c'} \in A^{S'c'}} \mathop{\textrm{exp}}\nolimits \left(\inf\phi_{S'c'} \left([w_{S'c'}] \cap A^G\right)\right) \\ & \geq \prod_{S' \in \mathcal{S}(\mathcal{T}')} \prod_{c' \in C_{S}(S')} \sum_{w_{S'c'} \in A^{S'c'}} \mathop{\textrm{exp}}\nolimits\left(\sup\phi_{S'c'} \left([w_{S'c'}] \cap A^G\right) - \Delta_{S'c'}(\phi)\right) \\ & = \prod_{S' \in \mathcal{S}(\mathcal{T}')} \mathop{\textrm{exp}}\nolimits\left(-\Delta_{S'}(\phi)|C_{S}(S')|\right) \prod_{c' \in C_{S}(S')} Z_{S'c'}(A,\phi) \\ & = \prod_{S' \in \mathcal{S}(\mathcal{T}')} \mathop{\textrm{exp}}\nolimits\left(-\Delta_{S'}(\phi)|C_{S}(S')|\right) Z_{S'}(A,\phi)^{|C_{S}(S')|}, \end{align*} $$

where we used that, for every $g \in G$ , $Z_F(A, \phi ) = Z_{Fg}(A, \phi )$ and that $\Delta _{F}(\phi ) = \Delta _{Fg}(\phi )$ . Thus,

(3.4) $$ \begin{align} \prod_{S' \in \mathcal{S}(\mathcal{T}')} Z_{S'}(A,\phi)^{|C_{S}(S')|} \leq Z_{S}(A,\phi) \cdot \prod_{S' \in \mathcal{S}(\mathcal{T}')} \mathop{\textrm{exp}}\nolimits\left(|C_{S}(S')|\Delta_{S'}(\phi)\right). \end{align} $$

Now, given a $(S_{\mathcal {T}},\epsilon )$ -invariant set $F \in \mathcal {F}(G)$ , we have that

$$ \begin{align*} Z_{T_F}(\phi) & \leq \prod_{S \in \mathcal{S}(\mathcal{T})} \prod_{c \in C_F(S)} Z_{Sc}(\phi) \\ &= \prod_{S \in \mathcal{S}(\mathcal{T})} Z_{S}(\phi)^{|C_F(S)|} \\ &\leq \prod_{S \in \mathcal{S}(\mathcal{T})} \left(\prod_{S' \in \mathcal{S}(\mathcal{T}')} \prod_{c' \in C_{S}(S')} Z_{S'c'}(\phi)\right)^{|C_F(S)|} \\ &= \prod_{S \in \mathcal{S}(\mathcal{T})} \prod_{S' \in \mathcal{S}(\mathcal{T}')} (Z_{S'}(\phi)^{|C_{S}(S')|})^{|C_F(S)|}. \end{align*} $$

Therefore, from equation (3.2), we obtain that

$$ \begin{align*} \prod_{S \in \mathcal{S}(\mathcal{T})} &\prod_{S' \in \mathcal{S}(\mathcal{T}')} (Z_{S'}(\phi)^{|C_{S}(S')|})^{|C_F(S)|}\\ & \leq \prod_{S \in \mathcal{S}(\mathcal{T})} \prod_{S' \in \mathcal{S}(\mathcal{T}')} \left(\frac{1}{1-\epsilon} Z_{S'}(A,\phi)\right)^{|C_{S}(S')||C_F(S)|} \\ & \leq \left(\frac{1}{1-\epsilon}\right)^{|T_F|} \prod_{S \in \mathcal{S}(\mathcal{T})} \left(Z_{S}(A,\phi) \mathop{\textrm{exp}}\nolimits\left(\sum_{S' \in \mathcal{S}(\mathcal{T}')} |C_{S}(S')|\Delta_{S'}(\phi)\right)\right)^{|C_F(S)|} \\ & \leq \left(\frac{1}{1-\epsilon}\right)^{|F|} \prod_{S \in \mathcal{S}(\mathcal{T})} \mathop{\textrm{exp}}\nolimits\left(|S|(p(A,\phi) + \epsilon) + \sum_{S' \in \mathcal{S}(\mathcal{T}')} |C_{S}(S')|\Delta_{S'}(\phi)\right)^{|C_F(S)|}, \end{align*} $$

where the second inequality follows from equation (3.4) and the third from equation (3.3). Hence, if $0 < \epsilon < \frac {1}{2}$ , we have that $\log \left (\frac {1}{1-\epsilon }\right ) \leq 2\epsilon $ , so

$$ \begin{align*} \frac{1}{|F|} &\log Z_{T_F}(\phi)\\ & \leq \log\left(\frac{1}{1-\epsilon}\right) + \frac{1}{|F|}\sum_{S \in \mathcal{S}(\mathcal{T})}|C_F(S)| \left(|S|(p(A,\phi) + \epsilon) + \sum_{S' \in \mathcal{S}(\mathcal{T}')} |C_{S}(S')|\Delta_{S'}(\phi)\right) \\ & = \log\left(\frac{1}{1-\epsilon}\right) + \frac{|T_F|}{|F|}(p(A,\phi) + \epsilon) + \sum_{S \in \mathcal{S}(\mathcal{T})} \sum_{S' \in \mathcal{S}(\mathcal{T}')} \frac{|C_F(S)||C_{S}(S')||S'|}{|F|} \frac{\Delta_{S'}(\phi)}{|S'|} \\ & \leq 2\epsilon + (p(A,\phi) + \epsilon) + \sum_{S \in \mathcal{S}(\mathcal{T})} \sum_{S' \in \mathcal{S}(\mathcal{T}')} \frac{|C_F(S)||C_{S}(S')||S'|}{|F|} \epsilon \\ & = p(A,\phi) + 3\epsilon + \frac{|T_F|}{|F|} \epsilon \\ & \leq p(A,\phi) + 4\epsilon. \end{align*} $$

In addition,

$$ \begin{align*} Z_F(\phi) \leq Z_{T_F}(\phi)Z_{F \setminus T_F}(\phi) \leq Z_{T_F}(\phi)Z_{1_G}(\phi)^{|F \setminus T_F|} \leq Z_{T_F}(\phi)Z_{1_G}(\phi)^{\epsilon|F|}, \end{align*} $$

so, considering that $p(A,\phi ) = \inf _{E \in \mathcal {F}(G)} \frac {1}{|E|} \log Z_E(A,\phi )$ by Theorem 3.1, we have that

$$ \begin{align*} \frac{1}{|F|} \log Z_F(\phi) \leq \inf_{E \in \mathcal{F}(G)} \frac{1}{|E|} \log Z_E(A,\phi) + 4\epsilon + \epsilon\cdot\log Z_{1_G}(\phi). \end{align*} $$

We conclude that, for every $0 < \epsilon < \frac {1}{2}$ , there exist $A \in \mathcal {F}(\mathbb {N})$ , $K \in \mathcal {F}(G)$ , and $\delta> 0$ , such that for every $(K,\delta )$ -invariant set $F \in \mathcal {F}(G)$ ,

$$ \begin{align*} \frac{1}{|F|} \log Z_F(\phi) \leq \inf_{E \in \mathcal{F}(G)} \frac{1}{|E|} \log Z_E(A,\phi) + \epsilon \cdot C, \end{align*} $$

where $C = 4 + \log Z_{1_G}(\phi )$ . Since $\epsilon $ was arbitrary, we conclude the result.

Proof. By Proposition 3.8, for every $\frac {1}{2}> \epsilon > 0$ , there exist $A \in \mathcal {F}(\mathbb {N})$ , $K \in \mathcal {F}(G)$ , and $\delta> 0$ , such that for every $(K,\delta )$ -invariant set $F \in \mathcal {F}(G)$ ,

$$ \begin{align*} \frac{1}{|F|} \log Z_F(\phi) \leq \inf_{E \in \mathcal{F}(G)} \frac{1}{|E|} \log Z_E(A,\phi) + \epsilon. \end{align*} $$

Therefore, for every such F,

$$ \begin{align*} \inf_{E \in \mathcal{F}(G)} \frac{1}{|E|} \log Z_E(\phi) & \leq \frac{1}{|F|} \log Z_F(\phi) \\ & \leq \inf_{E \in \mathcal{F}(G)} \frac{1}{|E|} \log Z_E(A,\phi) + \epsilon \\ & \leq \inf_{E \in \mathcal{F}(G)} \frac{1}{|E|} \log Z_E(\phi) + \epsilon. \end{align*} $$

Thus, $\lim _{F \to G} \frac {1}{|F|} \log Z_F(\phi ) = \inf _{E \in \mathcal {F}(G)} \frac {1}{|E|} \log Z_E(\phi )$ , $p(\phi )$ exists, and there exists $A \in \mathcal {F}(\mathbb {N})$ , such that

$$ \begin{align*} p(\phi) \leq p(A,\phi) + \epsilon \leq p(\phi) + \epsilon, \end{align*} $$

so $p(\phi ) = \sup _{A \in \mathcal {F}(\mathbb {N})} p(A,\phi )$ .

Proof. Let $K, E, F \in \mathcal {F}(G)$ and $\tau _K \in \mathcal {E}_K$ be as in the statement of the lemma. Then, it is easy to verify that, for any $x \in X$ , $(\phi ^{\tau _K}_{E} - \phi ^{\tau _K}_{F})(x) = \sum _{g \in E \setminus F} \left [\phi \left (\tau _{Kg^{-1}}^{-1}(g \cdot x)\right ) - \phi (g \cdot x)\right ]$ . Thus,

$$ \begin{align*} \Vert \phi_{E}^{\tau_{K}} - \phi_{F}^{\tau_K}\Vert_{\infty} &= \sup_{x \in X} \left|\phi_{E}^{\tau_K}(x) - \phi_{F}^{\tau_K}(x)\right|\\ &= \sup_{x \in X} \left|\sum_{g \in E \setminus F} \left[\phi\left(\tau_{Kg^{-1}}^{-1}(g \cdot x)\right) - \phi(g \cdot x)\right]\right|\\ &\leq \sup_{x \in X} \sum_{g \in E \setminus F} \left| \phi\left(\tau_{Kg^{-1}}^{-1}(g \cdot x)\right) - \phi(g \cdot x)\right|\\ &\leq \sum_{g \in E \setminus F} \sup_{x \in X} \left|\phi\left(\tau_{Kg^{-1}}^{-1}(g \cdot x)\right) - \phi(g \cdot x)\right|\\ &= \sum_{g \in E \setminus F} \left\Vert \phi \circ \tau_{Kg^{-1}}^{-1} - \phi\right\Vert{}_{\infty}\\ &\leq \sum_{g \in G \setminus F} \left\Vert \phi \circ \tau_{Kg^{-1}}^{-1} - \phi\right\Vert{}_{\infty}.\\[-49pt] \end{align*} $$

Proof. First, suppose that K is a singleton $\{h\}$ for some $h \in G$ , and let $\epsilon> 0$ . Since $\phi $ has summable variation according to $\{E_m\}_m$ , there exists $m_0 \in \mathbb {N}$ , such that $\sum _{m \geq m_0} |E_{m+1}^{-1} \setminus E_m^{-1}| \delta _{E_m}(\phi ) < \epsilon $ . Now, consider another (possibly different) exhausting sequence. Then, there exists $m_1 \geq m_0$ , such that , for all $m \geq m_1$ . On the other hand, since $\{E_m\}_m$ is an exhausting sequence, for every $m \geq m_1$ , there exists $k_m \in \mathbb {N}$ , such that for all $k \geq k_m$ , . Therefore, by Lemma 4.1, for every $m \geq m_1$ and every $k \geq k_m$ ,

Moreover, since , we obtain that , so that

Therefore, for every $\epsilon> 0$ , there exists $m_1 \geq m_0$ , such that for every $m \geq m_1$ , there exists $k_m$ , such that for every $k \geq k_m$ ,

Notice that, in the particular case that is the same as $\{E_m\}_m$ , one just needs to take $k_m = m$ and the same inequality would follow. This proves that is a Cauchy sequence for any $\tau _h \in \mathcal {E}_{\{h\}}$ , which implies that the uniform limit $\phi _{*}^{\tau _h} = \lim _{m \to \infty } \phi ^{\tau _h}_{E_m}$ exists. On the other hand, if is another exhausting sequence, this proves that , that is, the limit is independent of the exhausting sequence, provided $\phi $ has summable variation according to some exhausting sequence.

Now, let’s consider a general $K \in \mathcal {F}(G)$ and write $K = \{h_1,\dots ,h_{|K|}\}$ . Then, for each $m \in \mathbb {N}$ ,

$$ \begin{align*} \phi_{E_m} \circ \tau^{-1}_K - \phi_{E_m} & = \sum_{i=0}^{|K|-1} \left(\phi_{E_m} \circ \tau^{-1}_{\{h_1,\dots,h_{i+1}\}} - \phi_{E_m} \circ \tau^{-1}_{\{h_1,\dots,h_i\}}\right) \\ & = \sum_{i=0}^{|K|-1} \phi^{\tau_{h_{i+1}}}_{E_m} \circ \tau^{-1}_{\{h_1,\dots,h_i\}}, \end{align*} $$

where we regard $\tau _{\emptyset }$ as the identity, so the first equality follows from the fact that the considered sum is telescopic. Therefore, by considering the uniform convergence for singletons,

$$ \begin{align*} \lim_{m \to \infty} \phi^{\tau_K}_{E_m} &= \lim_{m \to \infty} \sum_{i=0}^{|K|-1} \phi^{\tau_{h_{i+1}}}_{E_m} \circ \tau^{-1}_{\{h_1,\dots,h_i\}} \\ &= \sum_{i=0}^{|K|-1} \lim_{m \to \infty} \phi^{\tau_{h_{i+1}}}_{E_m} \circ \tau^{-1}_{\{h_1,\dots,h_i\}} \\ & = \sum_{i=0}^{|K|-1} \phi^{\tau_{h_{i+1}}}_{*} \circ \tau^{-1}_{\{h_1,\dots,h_i\}}, \end{align*} $$

which concludes the result.

Proof. Notice that, given $g \in G$ and $x \in X$ , we have that $\tau _K^{-1}(g \cdot x) = g \cdot \tau ^{-1}_{Kg}(x)$ , so that

$$ \begin{align*}\phi_{*}^{\tau_K}(g \cdot x) = \lim_{m \to \infty} \phi_{E_m}^{\tau_K}(g \cdot x) = \lim_{m \to \infty} \phi_{E_mg}^{\tau_{Kg}}(x) = \phi_{*}^{\tau_{Kg}}(x), \end{align*} $$

since $\{E_mg\}_m$ is also an exhausting sequence.

Proof. Let $F \in \mathcal {F}(G)$ . From Lemma 4.1, we know that

$$ \begin{align*}\left\Vert\phi^{\tau_F}_{E_m} - \phi^{\tau_F}_F\right\Vert{}_{\infty} \leq \sum_{g \in G \setminus F} \left\Vert\phi \circ \tau_{Fg^{-1}}^{-1} - \phi\right\Vert{}_{\infty}, \end{align*} $$

for every $m \in \mathbb {N}$ , such that $F \subseteq E_m$ . Therefore, by Theorem 4.2,

$$ \begin{align*}\left\Vert\phi_{*}^{\tau_F} - \phi^{\tau_F}_F\right\Vert{}_{\infty} = \lim_{m \to \infty} \left\Vert\phi^{\tau_F}_{E_m} - \phi^{\tau_F}_F\right\Vert{}_{\infty} \leq \sum_{g \in G \setminus F} \left\Vert\phi \circ \tau_{Fg^{-1}}^{-1} - \phi\right\Vert{}_{\infty}. \end{align*} $$

Now, given $m \in \mathbb {N}$ , notice that $g \in (E_m^{-1}F)^c \iff Fg^{-1} \cap E_m = \emptyset $ , so that $\left \Vert \phi \circ \tau _{Fg^{-1}}^{-1} - \phi \right \Vert {}_{\infty } \leq \delta _{E_m}(\phi )$ . Considering this, we have that

$$ \begin{align*} \sum_{g \in G \setminus F} \left\Vert\phi \circ \tau_{Fg^{-1}}^{-1} - \phi\right\Vert{}_{\infty} &=\sum_{m=1}^{\infty} \sum_{g \in E_{m+1}^{-1}F \setminus E_{m}^{-1}F} \left\Vert\phi \circ \tau_{Fg^{-1}}^{-1} - \phi\right\Vert{}_{\infty} \\ & \leq \sum_{m=1}^{\infty} |E^{-1}_{m+1}F \setminus E^{-1}_{m}F| \cdot \delta_{E_m}(\phi) \\ & = V_F(\phi).\\[-32pt] \end{align*} $$

Proof. Let $K \in \mathcal {F}(G)$ and $x,y \in X$ be such that $x_{G \setminus K} = y_{G \setminus K}$ . Notice that for any $g \in G$ , $(g \cdot x)_{G \setminus Kg^{-1}} = (g \cdot y)_{G \setminus Kg^{-1}}$ . In addition, given $m \in \mathbb {N}$ , we have that $g \in \left (E_{m}^{-1} K\right )^{c} \iff K g^{-1} \cap E_{m} = \emptyset $ . In particular, if $g \in \left (E_{m}^{-1} K\right )^{c}$ , we have that $|\phi (g \cdot x) - \phi (g \cdot y)| \leq \delta _{E_{m}}(\phi )$ . Considering this, we obtain that

$$ \begin{align*} \sum_{g \in G \setminus K}|\phi(g \cdot x) - \phi(g \cdot y)| & = \sum_{m=1}^{\infty} \sum_{g \in E_{m+1}^{-1}K \setminus E_{m}^{-1}K} |\phi(g \cdot x) - \phi(g \cdot y)| \\ & \leq \sum_{m=1}^{\infty} \sum_{g \in E_{m+1}^{-1}K \setminus E_{m}^{-1} K}\delta_{E_m}(\phi) \\ & = \sum_{m=1}^{\infty} |E_{m+1}^{-1}K \setminus E_{m}^{-1} K|\cdot\delta_{E_m}(\phi) \\ & = V_K(\phi). \end{align*} $$

Now, let $m_0 \in \mathbb {N}$ be the smallest index, such that $K \subseteq E_{m_0}$ . Then, for every $m \geq m_0$ , every $x \in X$ , and every $v,w \in X_K$ , we have that

$$ \begin{align*} \phi_{E_m}^{\tau_{w,v}}(wx_{K^c}) &= \phi_{E_m}(vx_{K^c}) - \phi_{E_m}(wx_{K^c})\\ &\leq \phi_K(vx_{K^c}) - \phi_{K}(wx_{K^c})+ \sum_{g \in G \setminus K}|\phi(g \cdot (vx_{K^c})) - \phi(g \cdot (wx_{K^c}))|\\ &\leq \phi_K(vx_{K^c}) - \phi_{K}(wx_{K^c}) + V_K(\phi)\\ &\leq \sup\phi_{K}[v] - \sup\phi_{K}[w] + \Delta_K(\phi)+ V_K(\phi), \end{align*} $$

and, similarly,

$$ \begin{align*} \phi_{E_m}^{\tau_{w,v}}(wx_{K^c})) &\geq \sup\phi_{K}[v]-\sup\phi_{K}[w]-\Delta_K(\phi)-V_K(\phi), \end{align*} $$

so we conclude that

$$ \begin{align*}|\phi_{E_m}^{\tau_{w,v}}(wx_{K^c}) - (\sup\phi_{K}[v] - \sup\phi_{K}[w])| \leq \Delta_K(\phi)+ V_K(\phi).\\[-37pt] \end{align*} $$

Proof. Given $K \in \mathcal {F}(G)$ , $w \in X_K$ , and $x \in X$ , consider the sequence of functions $f_m\colon X_K \to \mathbb {R}$ given by $f_m(v) := \mathop {\textrm {exp}}\nolimits (\phi ^{\tau _{w, v}}_{E_m}(w x_{K^c})).$ By Theorem 4.2, we have that $\{f_m\}_m$ converges pointwise (in v) to $\mathop {\textrm {exp}}\nolimits (\phi _{*}^{\tau _{w, v}}(w x_{K^c}))$ , uniformly on X. In addition, by Lemma 4.7, there exist $m_0 \in \mathbb {N}$ and a constant $C = \mathop {\textrm {exp}}\nolimits (\Delta _K(\phi )+V_K(\phi ))> 0$ , such that for every $m \geq m_0$ and for every $v \in X_K$ ,

$$ \begin{align*}C^{-1} \cdot h(v) \leq f_m(v) \leq C \cdot h(v), \end{align*} $$

where $h(v) := \mathop {\textrm {exp}}\nolimits (-\sup \phi _{K}[w]) \cdot \mathop {\textrm {exp}}\nolimits (\sup \phi _{K}[v])$ . Notice that

$$ \begin{align*}\sum_{v \in X_K} h(v) = \mathop{\textrm{exp}}\nolimits(-\sup\phi_{K}[w]) \cdot Z_K(\phi), \end{align*} $$

so h (and, therefore, $C \cdot h$ ) is integrable with respect to the counting measure in $X_K$ . Therefore, by the Dominated Convergence Theorem, if follows that

$$ \begin{align*} \sum_{v \in X_K} \mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{w,v}}(wx_{K^c})\right) & = \sum_{v \in X_K} \lim_m \mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}^{\tau_{w,v}}(wx_{K^c})\right) \\ & = \lim_m \sum_{v \in X_K} \mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}^{\tau_{w,v}}(wx_{K^c})\right) \\ & \geq \lim_m \sum_{v \in X_K} C^{-1} \cdot h(v) \\ & = C^{-1} \cdot \mathop{\textrm{exp}}\nolimits(-\sup\phi_{K}[w]) \cdot Z_K(\phi)> 0.\\[-37pt] \end{align*} $$

Proof Proof (of Proposition 4.6).

First, note that for any given $K \in \mathcal {F}(G)$ ,

$$ \begin{align*} \sum_{v \in X_{K}} \mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}(v x_{K^c})\right)> 0, \end{align*} $$

for all $m \in \mathbb {N}$ and, due to Lemma 4.8, the left-hand side is bounded away from zero uniformly in m. Furthermore, for each $w \in X_K$ ,

$$ \begin{align*} \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}(w x_{K^c})\right)}{\sum_{v \in X_{K}} \mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}(v x_{K^c})\right)} & = \frac{1}{\sum_{v \in X_{K}} \mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}(v x_{K^c}) - \phi_{E_m}(w x_{K^c}))\right)} \\ & = \frac{1}{\sum_{v \in X_{K}} \mathop{\textrm{exp}}\nolimits(\phi^{\tau_{w, v}}_{E_m}(w x_{K^c}))}. \end{align*} $$

Therefore, uniformly on X,

$$ \begin{align*} \lim_{m \to \infty} \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}(w x_{K^c})\right)}{\sum_{v \in X_{K}} \mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}(v x_{K^c})\right)} & = \frac{1}{\lim_{m \to \infty}\sum_{v \in X_{K}} \mathop{\textrm{exp}}\nolimits(\phi^{\tau_{w, v}}_{E_m}(w x_{K^c}))} \\ & = \frac{1}{\sum_{v \in X_{K}} \mathop{\textrm{exp}}\nolimits(\phi_{*}^{\tau_{w, v}}(w x_{K^c}))}, \end{align*} $$

again, due to Lemma 4.8. Now, let $B \in \mathcal {B}$ and $x \in X$ . Then, uniformly on X,

where the exchange of the limit and the sum follows from Lemma 4.8.

Proof. Let $K \in \mathcal {F}(G)$ , $w \in X_K$ , and $x \in X$ . Then, for any $v \in X_K$ ,

$$ \begin{align*} \lim_{m \to \infty} \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}(w x_{K^c})\right)}{\sum_{{w'}\in X_{K}}\mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}({w'} x_{K^c})\right)} &=\lim_{m \to \infty} \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}\circ\tau_{w,v}^{-1} - \phi_{E_m}\right)(vx_{K^c})}{\sum_{{w'}\in X_{K}} \mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}\circ\tau_{w',v}^{-1} - \phi_{E_m}\right)(vx_{K^c})}\\[4pt] &=\frac{\lim_{m \to \infty}\mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}\circ\tau_{w,v}^{-1} - \phi_{E_m}\right)(vx_{K^c})}{\sum_{{w'}\in X_{K}}\lim_{m \to \infty} \mathop{\textrm{exp}}\nolimits\left(\phi_{E_m}\circ\tau_{w',v}^{-1} - \phi_{E_m}\right)(vx_{K^c})}\\&= \frac{\mathop{\textrm{exp}}\nolimits\left(\lim_{m \to \infty} \left(\phi_{E_m}\circ\tau_{w,v}^{-1} - \phi_{E_m}\right)(vx_{K^c})\right)}{\sum_{{w'}\in X_{K}}\mathop{\textrm{exp}}\nolimits \left(\lim_{m \to \infty}\left(\phi_{E_m}\circ\tau_{w',v}^{-1} - \phi_{E_m}\right)(vx_{K^c})\right)}\\ &= \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{w, v}}(vx_{K^c})\right)}{\sum_{w^{\prime} \in X_{K}}\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{w^{\prime}, v}}(vx_{K^c})\right)}, \end{align*} $$

where the last equality follows from Theorem 4.2. Also, if $m_0 \in \mathbb {N}$ is such that $K \subseteq E_{m_0}$ , the exchange of limit and sum in the denominator from the first to the second line follows from Lemma 4.8.

Proof. Let $K \in \mathcal {F}(G)$ . Given $v \in X_K$ , let $y^v \in X$ be arbitrary and such that $y^v_K = v$ . Then,

$$ \begin{align*} \gamma_{Kg^{-1}}(g \cdot [w],g \cdot x) & = \gamma_{Kg^{-1}}([(g \cdot y^w)_{Kg^{-1}}],g \cdot x) \\ &= \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{Kg^{-1}}}((g \cdot y^w)_{Kg^{-1}}(g \cdot x)_{(Kg^{-1})^c})\right)}{\sum_{w' \in X_{K}}\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{Kg^{-1}}}((g \cdot y^{w'})_{Kg^{-1}}(g \cdot x)_{(Kg^{-1})^c})\right)} \\ &= \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{Kg^{-1}}}(g \cdot (y^w_K x_{K^c}))\right)}{\sum_{w' \in X_{K}}\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{Kg^{-1}}}(g \cdot (y^{w'}_K x_{K^c}))\right)} \\ &= \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_K}(y^w_K x_{K^c})\right)}{\sum_{w' \in X_{K}}\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_K}(y^{w'}_K x_{K^c})\right)} \\ & = \gamma_{K}([w],x), \end{align*} $$

where we have used the property of $\phi _{*}^{\tau }$ from Corollary 4.3.

Proof. Let $h \in \mathcal {L}$ , and let $\epsilon> 0$ . Given any $K \in \mathcal {F}(G)$ , first notice that

$$ \begin{align*} |\gamma_{K}h(x)| \leq \sum_{w \in X_{K}}\gamma_{K}(w,x)|h(wx_{K^c})| \leq \|h\|_{\infty}\sum_{w \in X_{K}}\gamma_{K}(w,x) = \|h\|_{\infty}, \end{align*} $$

so $\|\gamma _{K}h\|_{\infty } \leq \|h\|_{\infty }$ . In addition, if $x, y \in X$ are such that $x_{E_n} = y_{E_n}$ for n to be determined, we have that

$$ \begin{align*} |\gamma_{K}&h(x) - \gamma_{K}h(y)|\\ &\leq \sum_{w \in X_{K}}|\gamma_{K}(w,x)h(wx_{K^c}) - \gamma_{K}(w,y)h(wy_{K^c})| \\ &= \sum_{w \in X_{K}}\gamma_{K}(w,x) \left|h(wx_{K^c}) - \frac{\gamma_{K}(w,y)}{\gamma_{K}(w,x)}h(wy_{K^c})\right| \\ &\leq \sum_{w \in X_{K}}\gamma_{K}(w,x) \left|h(wx_{K^c}) - e^{\pm 2\epsilon}h(wy_{K^c})\right| \\ &\leq \sum_{w \in X_{K}}\gamma_{K}(w,x) \left|h(wx_{K^c}) - h(wy_{K^c})\right| + \sum_{w \in X_{K}}\gamma_{K}(w,x)(1-e^{\pm 2\epsilon}) \left|h(wy_{K^c})\right| \\ &\leq \sum_{w \in X_{K}}\gamma_{K}(w,x) \left|h(wx_{K^c}) - h(wy_{K^c})\right| + \sum_{w \in X_{K}}\gamma_{K}(w,x)(1-e^{\pm 2\epsilon})\left\|h\right\|_{\infty} \\ &\leq \sup_{x',y': x^{\prime}_{E_n} = y^{\prime}_{E_n}}|h(x') - h(y')| + (1-e^{\pm 2\epsilon}) \left\|h\right\|_{\infty}. \end{align*} $$

To justify the second inequality, first observe that, for every $w,v \in X_K$ , $\phi _{*}^{\tau _{v,w}}$ is uniformly continuous, since it is a uniform limit of uniformly continuous potentials, namely, $\phi _{E_m}$ . Then, there exists $n_0 \in \mathbb {N}$ , such that for every $n \geq n_0$ , every $w,v \in X_K$ , and every $x, y \in X$ with $x_{E_n} = y_{E_n}$ ,

$$ \begin{align*}|\phi_{*}^{\tau_{v, w}}(wx_{K^c}) - \phi_{*}^{\tau_{v, w}}(wy_{K^c})| < \epsilon, \end{align*} $$

so

$$ \begin{align*}\gamma_{K}(w,y) = \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{w, w}}(wy_{K^c})\right)}{\sum_{v \in X_{K}}\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{v, w}}(wy_{K^c})\right)} \leq \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{w, w}}(wx_{K^c}) + \epsilon\right)}{\sum_{v \in X_{K}}\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{v, w}}(wx_{K^c})-\epsilon\right)} = e^{2\epsilon}\gamma_{K}(w,x). \end{align*} $$

Now, since h is local, we have that $\lim _{n \to \infty } \sup _{\substack {x,y \in X\\ x_{E_n} = y_{E_n}}} |h(x) - h(y)| = 0$ , so that there exists $n_1 \in \mathbb {N}$ , such that for all $n \geq n_1$ , $\sup _{\substack {x,y \in X\\ x_{E_n} = y_{E_n}}} |h(x) - h(y)| < \epsilon $ . Taking $n=\max \{n_0,n_1\}$ , we obtain that

$$ \begin{align*} |\gamma_{K}h(x) - \gamma_{K}h(y)| \leq \epsilon + (1-e^{\pm 2\epsilon}) \left\|h\right\|_{\infty}, \end{align*} $$

and since $\epsilon $ was arbitrary, we conclude.

Proof. Indeed, given $\epsilon> 0$ , there exist $K \in \mathcal {F}(G)$ and $\delta> 0$ so that

$$ \begin{align*}\mathop{\textrm{exp}}\nolimits\left(-\epsilon\cdot|F|\right) \mathop{\textrm{exp}}\nolimits\left(\phi_F(x)\right) \leq \mu([x_F])\mathop{\textrm{exp}}\nolimits\left(p |F|\right) \leq \mathop{\textrm{exp}}\nolimits\left(\epsilon\cdot|F|\right) \mathop{\textrm{exp}}\nolimits\left(\phi_F(x)\right) \end{align*} $$

for every $(K,\delta )$ -invariant set $F \in \mathcal {F}(G)$ and every $x \in X$ . Since x is arbitrary, we have that

$$ \begin{align*}\mathop{\textrm{exp}}\nolimits\left(-\epsilon\cdot|F|\right) \mathop{\textrm{exp}}\nolimits\left(\sup \phi_F[x_F]\right) \leq \mu([x_F])\mathop{\textrm{exp}}\nolimits\left(p |F|\right) \leq \mathop{\textrm{exp}}\nolimits\left(\epsilon\cdot|F|\right) \mathop{\textrm{exp}}\nolimits\left(\sup \phi_F[x_F]\right), \end{align*} $$

and, since $\mu $ is a probability measure, adding over all $x_F \in X_F$ , we get

$$ \begin{align*}\mathop{\textrm{exp}}\nolimits\left(-\epsilon\cdot|F|\right) Z_F(\phi) \leq \mathop{\textrm{exp}}\nolimits\left(p |F|\right) \leq \mathop{\textrm{exp}}\nolimits\left(\epsilon\cdot|F|\right) Z_F(\phi). \end{align*} $$

Then, if we take logarithms and divide by $|F|$ , we obtain that

$$ \begin{align*}-\epsilon + \frac{\log Z_F(\phi)}{|F|} \leq p \leq \frac{\log Z_F(\phi)}{|F|} + \epsilon, \end{align*} $$

so, taking the limit as F becomes more and more invariant, we obtain that

$$ \begin{align*}-\epsilon + p(\phi) \leq p \leq p(\phi) + \epsilon, \end{align*} $$

and since $\epsilon $ was arbitrary, we conclude that $p = p(\phi )$ .

Proof. Let $\{A_n\}_n$ be an exhausting sequence of finite alphabets and $F \in \mathcal {F}(G)$ . Consider $X^{F,n} = \{x \in X: x_F \in A_n^F\} \in \mathcal {B}_F$ . Since $\phi $ is exp-summable, then it is bounded from above. Without loss of generality, suppose that it is bounded from above by $0$ . Thus, so is $\phi _F$ . Define

$$ \begin{align*}\phi^{F,n}(x) = \begin{cases} \phi_F(x), & x \in X^{F,n};\\ 0, & \text{otherwise}. \end{cases} \end{align*} $$

Notice that, for every $x \in X$ , $\phi _F(x) = \lim _{n\to \infty }\phi _{F,n}(x)$ and, for every $n \in \mathbb {N}$ , $\phi _F(x) \leq \phi ^{F, n+1}(x) \leq \phi ^{F, n}(x)$ . Therefore, by the Monotone Convergence Theorem, we can conclude that

$$ \begin{align*} \int \phi_F d\nu = \lim_{n\to\infty} \int \phi^{F,n}(x) d\nu. \end{align*} $$

For each $n\in \mathbb {N}$ , let $H_{F, n}(\nu ) = -\sum _{w \in A_n^F} \nu ([w])\log \nu ([w])$ . Then, $\lim _{n\to \infty } H_{F,n}(\nu ) = H_F(\nu )$ . Also, for each $n\in \mathbb {N}$ and $F \in \mathcal {F}(G)$ , notice that $\phi ^{F,n } \leq \sum _{w \in A_n^{F}} 1_{[w]} \sup \phi _{F}([w])$ . Therefore, for every $n\in \mathbb {N}$ and $F \in \mathcal {F}(G)$ ,

$$ \begin{align*} H_{F,n}(\nu) + \int \phi^{F,n} d\nu &= -\sum_{w \in A^{F}_n} \nu([w])\log \nu([w]) + \int \phi^{F,n} d\nu\\ &\leq -\sum_{w \in A_n^{F}} \nu([w])\log \nu([w]) + \sum_{w \in A_n^{F}} \nu([w])\sup \phi_{F}([w])\\ &= \sum_{w \in A_n^{F}} \nu([w])\log\left(\frac{\mathop{\textrm{exp}}\nolimits\left(\sup\phi_F([w])\right)}{\nu([w])}\right)\\ &\leq \log\left(\sum_{w \in A_n^{F}} \mathop{\textrm{exp}}\nolimits \sup \phi_F([w])\right)\\ &= \log Z_{F}(A_n,\phi), \end{align*} $$

where we assume that all the sums involved are over cylinder sets with positive measure. The second inequality follows from Jensen’s inequality. In addition, notice that, in the case that $\nu $ is G-invariant, it follows that

$$ \begin{align*} H_{F}(\nu) &= \lim_{n \to \infty} H_{F,n}(\nu) \\ &\leq \lim_{n \to \infty} \left(\log Z_{F}(A_n,\phi) - \int \phi^{F,n} d\nu\right) \\ &= \log Z_F(\phi) - \int \phi_F d\nu \\ & \leq |F|\left(\log Z_{1_G}(\phi) - \int \phi d\nu\right), \end{align*} $$

where we have used that $\log Z_{F}(\phi ) \leq |F|\log Z_{1_G}(\phi )$ and $\int \phi _F d\nu = |F|\int \phi d\nu $ . Therefore, $H_F(\nu ) < \infty $ and, in particular, $H(\nu ) = H_{\{1_G\}}(\nu ) < \infty $ .

Proof. If $\mu $ is a DLR measure for $\phi $ , then for every $K \in \mathcal {F}(G)$ , $B \in \mathcal {B}$ , and $x \in X$ ,

$$ \begin{align*} \mu\left(B\,|\,\mathcal{B}_{K^c}\right)(x) = \gamma_{K}(B,x) \quad \mu(x)\text{-a.s.} \end{align*} $$

Thus, in particular, if $w \in X_K$ , it holds that

$$ \begin{align*} \mu([{w}] \,\vert \, \mathcal{B}_{K^c})(x) = \gamma_K([w],x) \qquad \mu(x)\text{-a.s.}, \end{align*} $$

and the result follows from Proposition 4.9.

On the other hand, if we assume that for every $K \in \mathcal {F}(G)$ , $w \in X_{K}$ , and $x \in X$ , equation (5.5) holds $\mu (x)$ -almost surely for every $v \in X_{K}$ , then, from equation (5.4) and Proposition 4.9, $\mu (x)$ -a.s., it holds that

Proof. Indeed, by Proposition 4.9, for every $x \in X$ ,

$$ \begin{align*} \gamma_K([w],x) &= \frac{\mathop{\textrm{exp}}\nolimits \left(\phi_{*}^{\tau}(v x_{K^c})\right)}{\sum_{w' \in X_{K}}\mathop{\textrm{exp}}\nolimits \left(\phi_{*}^{\tau_{w^{\prime}, v}}(v x_{K^c})\right)}\\ &= \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{v, v}}(v x_{K^c})\right)}{\sum_{w' \in X_{K}}\mathop{\textrm{exp}}\nolimits \left(\phi_{*}^{\tau_{w^{\prime}, v}}(v x_{K^c})\right)} \cdot\frac{\mathop{\textrm{exp}}\nolimits \left(\phi_{*}^{\tau}(v x_{K^c})\right)}{\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{v, v}}(v x_{K^c})\right)}\\ &= \gamma_K([v],x)\mathop{\textrm{exp}}\nolimits \left(\phi_{*}^{\tau}(v x_{K^c}) - \phi_{*}^{\tau_{v, v}}(v x_{K^c})\right). \end{align*} $$

Now, notice that $\phi _{*}^{\tau }(v x_{K^c}) - \phi _{*}^{\tau _{v, v}}(v x_{K^c}) = \phi _{*}^{\tau }(v x_{K^c})$ , and the result follows.

Proof. Suppose, first, that $\mu \in \mathcal {M}(X)$ is a conformal measure for $\phi $ , and let $K \in \mathcal {F}(G)$ . Begin by noticing that if $B \in \mathcal {B}_{K^c}$ and, for some $w \in X_{K}$ and $x \in X$ , $w x_{K^c} \in B$ , then $vx_{K^c} \in B$ , for every $v \in X_{K}$ . As a consequence, we have that, for all $\tau \in \mathcal {E}_{K}$ and all $B \in \mathcal {B}_{K^c}$ , $B = \tau ^{-1}(B)$ .

For $w,v \in X_{K}$ , consider $\tau _{w,v} \in \mathcal {E}_{K}$ . Thus, $\tau _{w,v}^{-1}([v])=[w]$ and, for every $B \in \mathcal {B}_{K^c}$ , $\tau _{w,v}^{-1}([v]\cap B) = \tau _{w,v}^{-1}([v])\cap \tau _{w,v}^{-1}(B) = [w]\cap B$ . Furthermore,

On the other hand,

Therefore, for any $w, v \in X_K$ , $\mu (x)$ -almost surely it holds that

(5.6)

Now, let $A \in \mathcal {F}(\mathbb {N})$ be a finite alphabet and $v \in A^{K}$ . For any $w' \in A^K$ , we have that $\tau _{w',v} \in \mathcal {E}_{K, A}$ . Summing equation (5.6) over all $w' \in A^{K}$ , we obtain that $\mu (x)$ -almost surely it holds that

(5.7)

(5.8)

If $\{A_n\}_n$ is an exhausting sequence of finite alphabets, then $\bigcap _{n \geq 1} \left (A_n^{K}\times X_{K^c}\right )^{c} = \emptyset $ . Moreover, for each $n \in \mathbb {N}$ ,

Therefore, . Since conditional expectation given $\mathcal {B}_{K^c}$ is a continuous linear operator on $L^2(\mu )$ , we have , $\mu (x)$ -almost surely in $L^2(\mu )$ as $n \to \infty $ . Therefore, for any fixed $v \in X_K$ , there exists $n_0 \in \mathbb {N}$ such that $v \in A^K_{n_0}$ and, consequently, $v \in A_n^K$ , for all $n \geq n_0$ . Therefore, $\mu (x)$ -almost surely it holds that

Moreover, equation (5.6) yields that, for any $w \in X_K$ , $\mu (x)$ -almost surely

so that, for any $w \in X_K$ , $\mu (x)$ -almost surely it holds that

$$ \begin{align*} \mu\left([w] \middle\vert \mathcal{B}_{K^c}\right)(x) = \frac{\mathop{\textrm{exp}}\nolimits\left(\phi_{*}^{\tau_{w,v}}(v x_{K^c})\right)}{\sum_{w^{\prime} \in X_{K}}\mathop{\textrm{exp}}\nolimits \phi_{*}^{\tau_{w',v}} (v x_{K^c})}. \end{align*} $$

Therefore, due to Corollary 5.6, $\mu $ is a DLR measure.

Conversely, suppose that $\mu \in \mathcal {M}(X)$ is a DLR measure for $\phi $ , and let $A \in \mathcal {F}(\mathbb {N})$ , $K \in \mathcal {F}(G)$ , and $\tau \in \mathcal {E}_{K, A}$ . For any $v \in X_{K}$ and $w = \left [\tau ^{-1}([v])\right ]_{K}$ , due to Lemma 5.7, we obtain

which concludes the result.

Proof. Let $F \in \mathcal {F}(G)$ and $\tau \in \mathcal {E}_F$ . From Proposition 4.4, we have that for every $x \in X$ ,

(5.9) $$ \begin{align} |\phi_{*}^{\tau}(x) - \phi^{\tau}_F(x)| \leq V_F(\phi), \end{align} $$

which, in particular, yields that, for every $x \in X$ ,

(5.10) $$ \begin{align} 0< \mathop{\textrm{exp}}\nolimits(-V_F(\phi)) \mathop{\textrm{exp}}\nolimits\phi_{*}^{\tau}(x) \leq \mathop{\textrm{exp}}\nolimits\phi^{\tau}_F(x) \leq \mathop{\textrm{exp}}\nolimits(V_F(\phi)) \mathop{\textrm{exp}}\nolimits\phi_{*}^{\tau}(x). \end{align} $$

For a fixed $v \in X_F$ and for every $w' \in X_F$ , the map $\tau _{w',v}$ belongs to $\mathcal {E}_F$ . Thus, inequality (5.10) holds for any such $\tau _{w',v}$ and, summing over all those such maps, we obtain that, for every $x \in X$ ,

$$ \begin{align*} \mathop{\textrm{exp}}\nolimits(-V_F(\phi)) \sum_{w' \in X_F} \mathop{\textrm{exp}}\nolimits\phi_{*}^{\tau_{w',v}}(x) \leq \sum_{w' \in X_F} \mathop{\textrm{exp}}\nolimits\phi^{\tau_{w',v}}_F(x) \leq \mathop{\textrm{exp}}\nolimits(V_F(\phi)) \sum_{w' \in X_F} \mathop{\textrm{exp}}\nolimits\phi_{*}^{\tau_{w',v}}(x). \end{align*} $$

Therefore, for every $F \in \mathcal {F}(G)$ , $v \in X_F$ , and $x \in X$ , we have

(5.11) $$ \begin{align} \mathop{\textrm{exp}}\nolimits(-V_F(\phi)) \leq \frac{\sum_{w' \in X_F}\mathop{\textrm{exp}}\nolimits\phi^{\tau_{w',v}}_F(x)}{\sum_{w' \in X_F}\mathop{\textrm{exp}}\nolimits\phi_{*}^{\tau_{w',v}}(x)} \leq \mathop{\textrm{exp}}\nolimits(V_F(\phi)). \end{align} $$

On the other hand, inequality (5.9) also yields that for every $F \in \mathcal {F}(G)$ , $w, v \in X_F$ , and $x \in X$ ,

(5.12) $$ \begin{align} \mathop{\textrm{exp}}\nolimits(-V_F(\phi)) \leq \frac{\mathop{\textrm{exp}}\nolimits\phi_{*}^{\tau_{w,v}}(x)}{\mathop{\textrm{exp}}\nolimits\phi^{\tau_{w,v}}_F(x)} \leq \mathop{\textrm{exp}}\nolimits(V_F(\phi)). \end{align} $$

Then, from inequalities (5.11) and (5.12), we obtain that, for every $F \in \mathcal {F}(G)$ , $w, v \in X_F$ , and $x \in X$ ,

(5.13) $$ \begin{align} \mathop{\textrm{exp}}\nolimits(-2 V_F(\phi)) \leq \frac{\sum_{w' \in X_F} \mathop{\textrm{exp}}\nolimits\phi^{\tau_{w',v}}_F(x)}{\sum_{w' \in X_F} \mathop{\textrm{exp}}\nolimits \phi_{*}^{\tau_{w',v}}(x)} \cdot\frac{\mathop{\textrm{exp}}\nolimits\phi_{*}^{\tau_{w,v}}(x)}{\mathop{\textrm{exp}}\nolimits\phi^{\tau_{w,v}}_F(x)} \leq \mathop{\textrm{exp}}\nolimits(2 V_F(\phi)). \end{align} $$

So, if $x \in [v]$ , inequality (5.13) can be rewritten as

(5.14) $$ \begin{align} \mathop{\textrm{exp}}\nolimits(-2 V_F(\phi)) \leq \frac{\sum_{w' \in X_F} \mathop{\textrm{exp}}\nolimits\phi^{\tau_{w',v}}_F(v x_{F^c})}{\sum_{w' \in X_{F}} \mathop{\textrm{exp}}\nolimits \phi_{*}^{\tau_{w',v}}(v x_{F^c})} \cdot\frac{\mathop{\textrm{exp}}\nolimits\phi_{*}^{\tau_{w,v}}(v x_{F^c})}{\mathop{\textrm{exp}}\nolimits\phi^{\tau_{w,v}}_F(vx_{F^c})} \leq \mathop{\textrm{exp}}\nolimits(2 V_F(\phi)). \end{align} $$

Since $\mu $ is a DLR measure for $\phi $ , from Corollary 5.6, we obtain that $\mu (x)$ -almost surely it holds that

(5.15) $$ \begin{align} \mathop{\textrm{exp}}\nolimits(-2 V_F(\phi)) \leq \mu\left([w] \,|\, \mathcal{B}_{F^c}\right)(x) \frac{\sum_{w' \in X_F} \mathop{\textrm{exp}}\nolimits\phi^{\tau_{w', v}}_{F}(v x_{F^c})}{\mathop{\textrm{exp}}\nolimits\phi^{\tau_{w,v}}_{F}(v x_{F^c})} \leq \mathop{\textrm{exp}}\nolimits(2 V_{F}(\phi)). \end{align} $$

Furthermore, notice that

$$ \begin{align*} \frac{\sum_{w' \in X_{F}} \mathop{\textrm{exp}}\nolimits\phi^{\tau_{w',v}}_{F}(v x_{F^c})}{\mathop{\textrm{exp}}\nolimits\phi^{\tau_{w,v}}_{F}(vx_{F^c})} &= \frac{\sum_{w' \in X_{F}} \mathop{\textrm{exp}}\nolimits \phi_{F}(w' x_{F^c}) }{\mathop{\textrm{exp}}\nolimits \phi_{F}(w x_{F^c})}, \end{align*} $$

so that inequality (5.15) can be rewritten as

(5.16) $$ \begin{align} \mathop{\textrm{exp}}\nolimits(-2 V_F(\phi)) \leq \mu\left([w] \,|\, \mathcal{B}_{F^c}\right)(x) \frac{\sum_{w' \in X_{F}} \mathop{\textrm{exp}}\nolimits \phi_{F}(w' x_{F^c}) }{\mathop{\textrm{exp}}\nolimits \phi_{F}(w x_{F^c})} \leq \mathop{\textrm{exp}}\nolimits(2 V_{F}(\phi)). \end{align} $$

For $F \in \mathcal {F}(G)$ and $x \in X$ , define the following auxiliary probability measure over $X_F$ :

$$ \begin{align*}\pi_{F}^x(w):=\frac{\mathop{\textrm{exp}}\nolimits \phi_{F}(wx_{F^c})}{\sum_{w' \in X_F} \mathop{\textrm{exp}}\nolimits \phi_{F}(w'x_{F^c})}, \quad \text{ for } w \in X_F.\end{align*} $$

Thus, inequality (5.16) yields that $\mu (x)$ -almost surely it holds that

$$ \begin{align*} \mathop{\textrm{exp}}\nolimits(-2 V_{F}(\phi)) \pi_{F}^{x}(w) \leq \mu\left([w] \,|\, \mathcal{B}_{F^c}\right)(x) \leq \mathop{\textrm{exp}}\nolimits(2 V_{F}(\phi)) \pi_{F}^{x}(w). \end{align*} $$

Now, given $y \in X$ , notice that the tail configuration $x_{F^c}$ can be replaced by $y_{F^c}$ with a penalty of $2\Delta _{F}(\phi )$ as follows

$$ \begin{align*} \pi_{F}^y(w) \mathop{\textrm{exp}}\nolimits(-2\Delta_{F}(\phi)) \leq \pi_{F}^{x}(w) \leq \pi_{F}^y(w)\mathop{\textrm{exp}}\nolimits(2\Delta_{F}(\phi)), \end{align*} $$

so that

(5.17) $$ \begin{align} \mathop{\textrm{exp}}\nolimits\left(-2(V_{F}(\phi) + \Delta_{F}(\phi))\right) \leq \frac{\mu\left([w] \,|\, \mathcal{B}_{F^c}\right)(x)}{\pi^y_{F}(w)}\leq \mathop{\textrm{exp}}\nolimits\left(2( V_{F}(\phi) + \Delta_{F}(\phi))\right). \end{align} $$

Moreover, it is easy to verify that

$$ \begin{align*} \mathop{\textrm{exp}}\nolimits\left(-\Delta_{F}(\phi)\right) \leq \frac{\pi_{F}^y(w)}{\mathop{\textrm{exp}}\nolimits\left(\phi_{F}(wy_{F^c}) - \log Z_{F}(\phi)\right)} \leq \mathop{\textrm{exp}}\nolimits\left( \Delta_{F}(\phi)\right). \end{align*} $$

Therefore, for every $w \in X_{F}$ , $y \in X$ , it holds $\mu (x)$ -almost surely that

$$ \begin{align*} \mu\left([w] \,|\, \mathcal{B}_{F^c}\right)(x) &\geq \mathop{\textrm{exp}}\nolimits\left(-2\left(V_F(\phi) + \Delta_{F}(\phi)\right)\right)\mathop{\textrm{exp}}\nolimits\left(\phi_F(wy_{F^c}) - \log Z_F(\phi) - \Delta_F(\phi)\right) \\ &= \mathop{\textrm{exp}}\nolimits\left(-2 V_F(\phi) - 3\Delta_F(\phi)\right) \mathop{\textrm{exp}}\nolimits\left(\phi_{F}(wy_{F^c}) - \log Z_F(\phi)\right) \end{align*} $$

and that

$$ \begin{align*} \mu\left([w] \,|\, \mathcal{B}_{F^c}\right)(x) &\leq \mathop{\textrm{exp}}\nolimits\left(2\left( V_F(\phi) + \Delta_F(\phi)\right)\right) \mathop{\textrm{exp}}\nolimits\left(\phi_F (wy_{F^c}) - \log Z_F(\phi)+ \Delta_F(\phi)\right)\\ &= \mathop{\textrm{exp}}\nolimits\left(2V_F(\phi) + 3\Delta_F(\phi)\right)\mathop{\textrm{exp}}\nolimits\left(\phi_{F}(wy_{F^c}) - \log Z_F(\phi)\right). \end{align*} $$

Thus,

$$ \begin{align*} \mathop{\textrm{exp}}\nolimits\left(-2V_F(\phi) - 3\Delta_F(\phi)\right) \leq \frac{\mu\left([w] \,|\, \mathcal{B}_{F^c}\right)(x)}{\mathop{\textrm{exp}}\nolimits\left(\phi_{F}(wy_{F^c}) - \log Z_F(\phi)\right)} \leq \mathop{\textrm{exp}}\nolimits\left(2V_F(\phi) + 3\Delta_F(\phi)\right), \end{align*} $$

concluding the proof.

Proof. Indeed, for every $\epsilon> 0$ , we obtain, from Proposition 2.3, Lemma 2.4, and Theorem 3.9, that there exist $K \in \mathcal {F}(G)$ and $\delta> 0$ , such that, for every $(K, \delta )$ -invariant set ${F \in \mathcal {F}(G)}$ ,

$$ \begin{align*} \Delta_F(\phi) \leq \epsilon\cdot|F|,~ V_F(\phi) \leq \epsilon\cdot|F|, \text{ and } \left|\log Z_F(\phi) - p(\phi)|F|\right| \leq \epsilon\cdot|F|, \end{align*} $$

respectively. Considering a sufficiently large K and sufficiently small $\delta $ so that the three conditions are satisfied at the same time, we obtain from Proposition 5.9 that

$$ \begin{align*} \mathop{\textrm{exp}}\nolimits\left(-\epsilon \cdot|F|\right) \leq \frac{\mu\left([w] \,|\, \mathcal{B}_{F^c}\right)(x)}{\mathop{\textrm{exp}}\nolimits\left(\phi_F(x) - p(\phi) \cdot|F|\right)} \leq \mathop{\textrm{exp}}\nolimits\left(\epsilon\cdot |F|\right). \end{align*} $$

Integrating this inequality with respect to $d\mu (x)$ , it follows that $\mu $ is a Bowen-Gibbs measure for $\phi $ .

Proof. Fix some $n \in \mathbb {N}$ . Then, for any $a \in \mathbb {N}$ and any $y \in A_n^{\{1_G\}^c}$ , Proposition 5.9 yields that

$$ \begin{align*} \mathop{\textrm{exp}}\nolimits\left(-C(\phi^n) \right) \leq \frac{\mu_n([a])}{\mathop{\textrm{exp}}\nolimits\left(\phi^n(ay) - \log Z_{E_1}(\phi^n)\right)} \leq \mathop{\textrm{exp}}\nolimits\left(C(\phi^n) \right), \end{align*} $$

where $\phi ^n = \phi |_{A_n^G}$ and $C(\phi ^n) = 2 V_{E_1}(\phi ^n) + 3\delta (\phi ^n)$ . Furthermore, $0 < Z_{E_1}(\phi ^n) \leq Z_{E_1}(\phi ^{n+1}) < \infty $ and $\{Z_{E_1}(\phi ^n)\}_n$ converges monotonically to $Z_{E_1}(\phi )$ . In particular, there exists $c = -\log Z_{E_1}(\phi ^1)$ , such that $c \geq -\log Z_{E_1}(\phi ^n)$ , for all $n \in \mathbb {N}$ .

If $a \notin A_n$ , then . On the other hand, if $a \in A_n$ , then for every $y \in A_n^{\{1_G\}^c}$ ,

where $C(\phi ) = 2 V_{E_1}(\phi ) + 3\delta (\phi )$

Now, let $\epsilon> 0$ . Since $\phi $ is exp-summable, for each $m \in \mathbb {N}$ , there must exist a finite alphabet $A_{\epsilon , m} \in \mathcal {F}(\mathbb {N})$ , such that

(5.18) $$ \begin{align} \sum_{b \in \mathbb{N} \setminus A_{\epsilon, m}} \mathop{\textrm{exp}}\nolimits \left(\sup_{x \in [b]} \phi(x) \right) < \frac{\epsilon\cdot\mathop{\textrm{exp}}\nolimits\left(-C(\phi)- c\right)}{2^m|E_m\setminus E_{m-1}|}. \end{align} $$

Let

$$ \begin{align*} K_{\epsilon} = A_{\epsilon, 1}^{E_1} \times A_{\epsilon, 2}^{E_2 \setminus E_1} \times A_{\epsilon, 3}^{E_3 \setminus E_2} \times \cdots. \end{align*} $$

By Tychonoff’s theorem (see [Reference Munkres47]), $K_{\epsilon }$ is compact. Moreover, notice that

$$ \begin{align*} K_{\epsilon} = \bigcap_{m=1}^{\infty} \bigcap_{g \in E_m\setminus E_{m-1}} \bigcup_{a \in A_{\epsilon, m}} [a^g], \end{align*} $$

where $[a^g] = \{x \in X : x(g) = a\}$ . Therefore, for each $n \in \mathbb {N}$ ,

Since all the measures considered here are G-invariant, it follows that, for any $y \in A_n^{\{1_G\}^c}$ ,

where the fifth line follows from estimate (5.18). Therefore, for all $n \in \mathbb {N}$ , , so that , which proves the tightness of .

Proof. Let $\{\mu _n\}_{n}$ be such that, for each $n \in \mathbb {N}$ , $\mu _n$ is a G-invariant conformal measure for $\phi ^n: A_n^G \to \mathbb {R}$ (or, equivalently, $\mu _n$ is a G-invariant DLR measure for $\phi ^n$ ). Thus, for each $n \in \mathbb {N}$ , any $K \in \mathcal {F}(G)$ , and any $\tau \in \mathcal {E}_{K, A_n}$ ,

(5.19) $$ \begin{align} \mathop{\textrm{exp}}\nolimits\left((\phi^n)^{\tau_n}_{*}\right) = \frac{d (\mu_n \circ (\tau_n)^{-1})}{d\mu_n}, \end{align} $$

where $\tau _n = \tau |_{A^G_n}$ . This yields that

Indeed, let $\psi \colon X \to \mathbb {R}$ be a bounded continuous potential. Observe that, for $\tau \in \mathcal {E}_{K, A_n}$ , $(\phi ^n)^{\tau _n}_{*} = (\phi _{*}^{\tau })|_{A_n^G}$ . Moreover, for every $B \in \mathcal {B}$ , since $\tau _n^{-1}(A^G_n) = A^G_n$ and , we have that . Then, we obtain

where $\psi ^n = \psi |_{A_n^G}$ .

Furthermore, Lemma 5.12 guarantees that the sequence of induced measures is tight, and we can apply Prokhorov’s theorem to guarantee the existence of a limit point for some subsequence $\{\mu _{n_k}\}_k$ , which we denote by . Now, we are going to prove that is a conformal measure for $\phi $ . For that, consider a bounded continuous potential $\psi \colon X \to \mathbb {R}$ , $A \in \mathcal {F}(\mathbb {N})$ , $K \in \mathcal {F}(G)$ , and $\tau \in \mathcal {E}_{K, A}$ . Then,

where the fourth equality follows from the fact that for k large enough, $A \subseteq A_{n_k}$ , and the last equality follows from weak convergence and the fact that $\psi \mathop {\textrm {exp}}\nolimits \phi _{*}^{\tau }$ is a continuous and bounded function. Indeed, first notice that $\phi _{*}^{\tau }$ is a uniform limit of continuous functions that are bounded from above, since $\phi $ is exp-summable. Therefore, the same holds for $\phi _{*}^{\tau }$ , so that $\mathop {\textrm {exp}}\nolimits (\phi _{*}^{\tau })$ is continuous and bounded (from above and below). Since A, K, and $\tau $ are arbitrary, this proves that is conformal for $\phi $ and, therefore, DLR for $\phi $ .

It remains to show that is G-invariant. For that, notice that, due to the weak convergence, for any $B \in \mathcal {B}$ ,

where we have used that, for each $k \in \mathbb {N}$ , is G-invariant due to G-invariance of $A_n^G$ and to the fact that $\mu _{n_k}$ is G-invariant.

Proof. Begin by noticing that, since $\mu $ is a Bowen-Gibbs measure for $\phi $ , we have that, in particular, for $\epsilon = 1$ , there exist $K \in \mathcal {F}(G)$ , $\delta> 0$ , and a $(K,\delta )$ -invariant set $F \in \mathcal {F}(G)$ with $1_G \in F$ , such that, for every $x \in X$ , it holds that