Proof Let $\mathcal {V}$ be a finite open cover of X with diameter smaller than the Lebesgue number of $\mathcal {U}$ . Each element $V\in \mathcal {V}$ is contained in some element $U(V)\in \mathcal {U}$ . For any $n\in \mathbb {N}$ and any string $\mathbf {V}=\{V_{g}\}_{g\in F_{n}}\in \mathcal {W}_{F_{n}}(\mathcal {V}),$ there exists a corresponding string $\mathbf {U}(\mathbf {V})=\{U(V_{g})\}_{g\in F_{n}}\in {W}_{F_{n}}(\mathcal {U})$ . If $\Gamma \subset \mathcal {W}_{F_{n}}(\mathcal {V})$ covers a set $Z\subset X$ , then $\mathbf {U}(\Gamma )=\{\mathbf {U}(\mathbf {V}):\mathbf {V}\in \Gamma \}$ also covers Z. By definition of the scaled topological entropy, $M(Z,s,N,\{F_{n}\},\mathcal {U},\mathbf {a})\leq M(Z,s,N,\{F_{n}\},\mathcal {V},\mathbf {a})$ . Then, $E_{Z}(\{F_{n}\},\mathcal {U},\mathbf {a})\leq E_{Z}(\{F_{n}\},\mathcal {V},\mathbf {a})$ . Therefore, $E_{Z}(\{F_{n}\},\mathcal {U},\mathbf {a})\leq \liminf \limits _{|\mathcal {V}|\rightarrow 0}E_{Z}(\{F_{n}\},\mathcal {V},\mathbf {a})$ , hence

$$ \begin{align*}\limsup_{|\mathcal{U}|\rightarrow0}E_{Z}(\{F_{n}\},\mathcal{U},\mathbf{a})\leq \liminf_{|\mathcal{V}|\rightarrow0}E_{Z}(\{F_{n}\},\mathcal{V},\mathbf{a}).\end{align*} $$

This implies that

$$ \begin{align*}E_{Z}(\{F_{n}\},\mathbf{a})=\lim_{|\mathcal{U}|\rightarrow0}E_{Z}(\{F_{n}\},\mathcal{U},\mathbf{a}).\end{align*} $$

$\underline {E}_{Z}(\{F_{n}\},\mathbf {a})=\lim \limits _{|\mathcal {U}|\rightarrow 0}\underline {E}_{Z}(\{F_{n}\},\mathcal {U},\mathbf {a})$ and can be proved in a similar manner.

Proof We only give the proof of (a). (b) can be proved in a similar manner as Proposition 2.3.

Firstly, we prove $E_{Z}^{B^{1}}(\{F_{n}\},\mathbf {a})\leq E_{Z}(\{F_{n}\},\mathbf {a})$ . Let $\mathcal {U}$ be a finite open cover of X. If s satisfies $M(Z,s,\{F_{n}\},\mathcal {U},\mathbf {a})=0$ , then by definition for every $\epsilon>0$ , there exists $N^{'}\in {\mathbb N}$ such that for each $N\in {\mathbb N}$ with $N>N^{'}$ ,

$$ \begin{align*}\inf_{\Gamma}\sum_{\mathbf{U}\in\Gamma}\exp(-sa(m(\mathbf{U})))<\frac{\epsilon}{2},\end{align*} $$

where the infimum is taken over all covers $\Gamma \subset \bigcup _{n\geq N}\mathcal {W}_{F_{n}}(\mathcal {U})$ of Z. So there exists $\Gamma _{N}$ that satisfies $\Gamma _{N}\subset \bigcup _{n\geq N}\mathcal {W}_{F_{n}}(\mathcal {U})$ and $\Gamma _{N}$ covers Z, such that

$$ \begin{align*}\sum_{\mathbf{U}\in\Gamma_{N}}\exp(-sa(m(\mathbf{U})))<\epsilon.\end{align*} $$

For any $n\in {\mathbb N}$ and $\mathbf {U}\in \mathcal {W}_{F_{n}}(\mathcal {U})$ , $n^{\{F_{n}\}}_{\mathcal {U}}(X(\mathbf {U}))\geq |F_{n}|$ . Then

$$ \begin{align*}\sum_{\mathbf{U}\in\Gamma_{N}}\exp(-sa(n^{\{F_{n}\}}_{\mathcal{U}}(X(\mathbf{U}))))\leq \sum_{\mathbf{U}\in\Gamma_{N}}\exp(-sa(m(\mathbf{U})))<\epsilon\end{align*} $$

and

$$ \begin{align*}\exp(-sa(n^{\{F_{n}\}}_{\mathcal{U}}(X(\mathbf{U}))))\leq \sum_{\mathbf{U}\in\Gamma_{N}}\exp(-sa(m(\mathbf{U})))<\epsilon.\end{align*} $$

This implies that

$$ \begin{align*}M_{\{F_{n}\},\mathcal{U}}(Z,s,\mathbf{a})=0,\end{align*} $$

and hence

$$ \begin{align*}E_{Z}^{B_{1}}(\{F_{n}\},\mathbf{a})\leq E_{Z}(\{F_{n}\},\mathbf{a}).\end{align*} $$

Next, we prove $ E_{Z}(\{F_{n}\},\mathbf {a})\leq E_{Z}^{B_{1}}(\{F_{n}\},\mathbf {a})$ . Let $\mathcal {U}$ be a finite open cover of X. If s satisfies $M_{\{F_{n}\},\mathcal {U}}(Z,s,\mathbf {a})=0$ , then by definition for every $\epsilon>0$ , there exists $k^{'}\in {\mathbb N}$ such that for each $k\in {\mathbb N}$ with $k>k^{'}$ ,

$$ \begin{align*}\inf_{\mathcal{D}\in \mathcal{G}(G,\mathcal{U},Z,k)}\sum_{D\in \mathcal{D}}\exp(-sa(n^{\{F_{n}\}}_{\mathcal{U}}(D)))<\frac{\epsilon}{4}.\end{align*} $$

Thus there exists $\mathcal {D}=\{D_{i}\}_{i=1}^{\infty }\in \mathcal {G}(G,\mathcal {U},Z,k)$ such that

$$ \begin{align*}\sum_{D\in \mathcal{D}}\exp(-sa(n^{\{F_{n}\}}_{\mathcal{U}}(D)))<\frac{\epsilon}{2}.\end{align*} $$

Without loss of generality, we can assume that each $D_i\in \mathcal {D}$ is open with

$$ \begin{align*}\sum_{D\in \mathcal{D}}\exp(-sa(n^{\{F_{n}\}}_{\mathcal{U}}(D)))<\epsilon.\end{align*} $$

Indeed, if $n^{\{F_{n}\}}_{\mathcal {U}}(D_i)<\infty $ , we can take $\widehat {D}_i=X(\mathbf {U})$ , where $X(\mathbf {U})= \bigcap _{g\in F_{n}}g^{-1}U_{g}$ and for each $g\in F_n$ , $gD_i\subset U_g$ . It is easy to see that $n^{\{F_{n}\}}_{\mathcal {U}}(D_i)=n^{\{F_{n}\}}_{\mathcal {U}}(\widehat {D}_i)$ and $D_i\subset \widehat {D}_i$ . If $n^{\{F_{n}\}}_{\mathcal {U}}(D_i)=\infty $ , we can take $n\in {\mathbb N}$ sufficiently large and $\widehat {D}_i=X(\mathbf {U})$ , where $X(\mathbf {U})= \bigcap _{g\in F_{n}}g^{-1}U_{g}$ and for each $g\in F_n$ , $gD_i\subset U_g$ , so that $\exp (-sa(n^{\{F_{n}\}}_{\mathcal {U}}(\widehat {D}_i)))<\frac {\epsilon }{2^{i+1}}$ .

Then we can assume that there exists an open cover $\mathcal {D}=\{D_{i}\}_{i=1}^{\infty }$ of Y, such that for every $D_{i}\in \mathcal {D}$ , there exists $\mathbf {U}_i\in \mathcal {W}_{F_{n}}(\mathcal {U})$ with $D_{i}=X(\mathbf {U}_i)$ and

$$ \begin{align*}\sum_{D\in \mathcal{D}}\exp(-sa(n^{\{F_{n}\}}_{\mathcal{U}}(D)))<\epsilon.\end{align*} $$

Then $Z\subset \bigcup _{i=1}^{\infty } D_{i}=\bigcup _{i=1}^{\infty } X(\mathbf {U}_i)$ and $\sum _{i=1}^{\infty }\exp (-sa(m(\mathbf {U})))<\epsilon .$ By the arbitrariness of $\epsilon $ , we get $M(Z,s,\{F_{n}\},\mathcal {U},\mathbf {a})=0$ and $E_{Z}(\{F_{n}\},\mathbf {a})\leq E_{Z}^{B_{1}}(\{F_{n}\},\mathbf {a}).$ So

$$ \begin{align*}E_{Z}(\{F_{n}\},\mathbf{a})=E_{Z}^{B_{1}}(\{F_{n}\},\mathbf{a}).\end{align*} $$

Let $\delta (\mathcal {U})$ be the Lebesgue number of $\mathcal {U}$ . Clearly, for every $x\in X$ and $n\in {\mathbb N}$ , if $x\in X(\mathbf {U})$ for some string $\mathbf {U}\in \mathcal {W}_{F_{n}}(\mathcal {U})$ , then $B_{F_{n}}(x,\delta (\mathcal {U}))\subset X(\mathbf {U})\subset B_{F_{n}}(x,|\mathcal {U}|)$ . By Proposition 2.1, this implies that

$$ \begin{align*}E_{Z}(\{F_{n}\},\mathbf{a})=\lim_{\delta\rightarrow0}E_{Z}^{B_{2}}(\{F_{n}\},\delta,\mathbf{a})=E_{Z}^{B_{2}}(\{F_{n}\},\mathbf{a}).\end{align*} $$

Therefore, $E_{Z}(\{F_{n}\},\mathbf {a})=E_{Z}^{B_1}(\{F_{n}\},\mathbf {a})=E_{Z}^{B_2}(\{F_{n}\},\mathbf {a})$ .

Proof We will prove the first equality, the second one can be proved in a similar fashion. Let us put

$$ \begin{align*}\underline{\beta}=\underline{E}_{Z}(\{F_{n}\},\mathcal{U},\mathbf{a}),\gamma=\liminf_{n\rightarrow\infty}\frac{1}{a(|F_{n}|)}\log\aleph\left(\bigvee_{g\in F_{n}}g^{-1}\mathcal{U},Z\right).\end{align*} $$

Given $\eta>0$ , one can choose a subsequence $\{n_i\}$ such that

$$ \begin{align*}0=\underline{r}(Z,\underline{\beta}+\eta,\{F_{n}\},\mathcal{U},\mathbf{a})=\lim_{i\rightarrow\infty}R(Z,\underline{\beta}+\eta,n_i,\{F_{n}\},\mathcal{U},\mathbf{a}).\end{align*} $$

When $i\in {\mathbb N}$ is sufficiently large, we have

$$ \begin{align*}\aleph\left(\bigvee_{g\in F_{n_i}}g^{-1}\mathcal{U},Z\right)\exp(-(\underline{\beta}+\eta)a(|F_{n_i}|))\leq1,\end{align*} $$

and hence

$$ \begin{align*}\frac{1}{a(|F_{n_i}|)}\log\aleph\left(\bigvee_{g\in F_{n_i}}g^{-1}\mathcal{U},Z\right)\leq\underline{\beta}+\eta.\end{align*} $$

Let $i\rightarrow \infty $ , we have $\gamma \leq \underline {\beta }+\eta $ , therefore, $\gamma \leq \underline {\beta }.$

Let us choose a subsequence $\{n_i\}$ such that

$$ \begin{align*}+\infty=\underline{r}(Z,\underline{\beta}-\eta,\{F_{n}\},\mathcal{U},\mathbf{a})=\lim_{i\rightarrow\infty}R(Z,\underline{\beta}-\eta,n_i,\{F_{n}\},\mathcal{U},\mathbf{a}).\end{align*} $$

When $i\in {\mathbb N}$ is sufficiently large, we have

$$ \begin{align*}\aleph\left(\bigvee_{g\in F_{n_i}}g^{-1}\mathcal{U},Z\right)\exp(-(\underline{\beta}-\eta)a(|F_{n_i}|))\geq1,\end{align*} $$

and hence

$$ \begin{align*}\frac{1}{a(|F_{n_i}|)}\log\aleph\left(\bigvee_{g\in F_{n_i}}g^{-1}\mathcal{U},Z\right)\geq\underline{\beta}-\eta.\end{align*} $$

Let $i\rightarrow \infty $ , we have $\gamma \geq \underline {\beta }-\eta $ , therefore, $\gamma \geq \underline {\beta }.$

So $\gamma =\underline {\beta }$ , i.e.,

$$ \begin{align*}\underline{E}_{Z}(\{F_{n}\},\mathcal{U},\mathbf{a})=\liminf_{n\rightarrow\infty}\frac{1}{a(|F_{n}|)}\log\aleph\left(\bigvee_{g\in F_{n}}g^{-1}\mathcal{U},Z\right).\end{align*} $$

Proof $(1)$ Since $\mathcal {U}\preceq \mathcal {V}$ , each element $V\in \mathcal {V}$ is contained in some element in $\mathcal {U}$ which we denote by $U(V)$ . Therefore, for each string $\mathbf {V}\in \mathcal {W}_{F_{n}}(\mathcal {V})$ there exists a corresponding string $\mathbf {U}(\mathbf {V})\in {W}_{F_{n}}(\mathcal {U})$ . This yields that

$$ \begin{align*} \aleph\left(\bigvee_{g\in F_{n}}g^{-1}\mathcal{U},Z\right)\leq\aleph\left(\bigvee_{g\in F_{n}}g^{-1}\mathcal{V},Z\right), \end{align*} $$

and hence

$$ \begin{align*} \underline{E}_{Z}(\{F_{n}\},\mathcal{U},\mathbf{a})\leq\underline{E}_{Z}(\{F_{n}\},\mathcal{V},\mathbf{a}) \end{align*} $$

and

Let $\Gamma \subset \mathcal {W}(\mathcal {V})$ be a collection of strings that covers Z. The corresponding collection of strings $\{\mathbf {U}(\mathbf {V}):\mathbf {V}\in \Gamma \}\subset \mathcal {W}(\mathcal {U})$ also covers Z. This implies that $M(Z,s,N,\{F_{n}\},\mathcal {U},\mathbf {a})\leq M(Z,s,N,\{F_{n}\},\mathcal {V},\mathbf {a})$ for each $s\geq 0$ and $N>0$ . Thus, $M(Z,s,\{F_{n}\},\mathcal {U},\mathbf {a})\leq M(Z,s,\{F_{n}\},\mathcal {V},\mathbf {a})$ . The first statement follows.

$(2)$ The last statement follows immediately from the definitions.

Proof $(1)$ The statements follow directly from the definitions.

$(2)$ By $(1)$ , $\mathcal {E}_{Z_{i}}(\{F_{n}\},\mathbf {a})\leq \mathcal {E}_{Z}(\{F_{n}\},\mathbf {a})$ and hence $\sup _{i\geq 1}\mathcal {E}_{Z_{i}}(\{F_{n}\},\mathbf {a})\leq \mathcal {E}_{Z}(\{F_{n}\},\mathbf {a})$ .

Now we are left to show that $E_{Z}(\{F_{n}\},\mathbf {a})\geq \sup _{i\geq 1}E_{Z_{i}}(\{F_{n}\},\mathbf {a})$ . In fact, suppose $E_{Z_{i}}(\{F_{n}\},\mathbf {a})<s(i=1,2,\cdots )$ , it follows that $M(Z_{i},s,\{F_{n}\},\mathcal {U},\mathbf {a})=0$ , and hence $M(\bigcup _{i\geq 1}Z_{i},s,\{F_{n}\},\mathcal {U},\mathbf {a})=0$ . Then $E_{Z=\bigcup _{i\geq 1}Z_{i}}(\{F_{n}\},\mathbf {a})<s$ . This implies that, $E_{Z}(\{F_{n}\},\mathbf {a})\leq \sup _{i\geq 1}E_{Z_{i}}(\{F_{n}\},\mathbf {a})$ .

Proof We shall prove the result for the scaled topological entropy $E_{Z}(\{F_{n}\},[\mathbf {a}])$ , the arguments for the lower and upper scaled topological entropies are similar.

Suppose there is $[\mathbf {a}]\in \mathcal {A}$ such that $E_{Z}(\{F_{n}\},[\mathbf {a}])$ is positive and finite. Then for each $[\mathbf {b}]\succeq [\mathbf {a}]$ ,

$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{a_{1}(|F_{n}|)}{b_{1}(|F_{n}|)}=0,\end{align*} $$

for arbitrary $\mathbf {a_{1}}=\{a_{1}(|F_{n}|)\}\in [\mathbf {a}]$ and $\mathbf {b_{1}}=\{b_{1}(|F_{n}|)\}\in [\mathbf {b}]$ . Let us fix such two scaled sequences $\mathbf {a_{1}}$ and $\mathbf {b_{1}}$ . Given a small number $\beta>0$ , for all sufficiently large $n\in {\mathbb N}$ we have that $a_{1}(|F_{n}|)<\beta b_{1}(|F_{n}|)$ and hence, $M(Z,s,\{F_{n}\},\mathcal {U},\mathbf {a_{1}})\geq M(Z,s,\{F_{n}\},\mathcal {U},\beta \mathbf {b_{1}})$ . This implies that

$$ \begin{align*}E_{Z}(\{F_{n}\},\mathbf{a_{1}})\geq E_{Z}(\{F_{n}\},\beta\mathbf{b_{1}})=\frac{1}{\beta}E_{Z}(\{F_{n}\},\mathbf{b_{1}}),\end{align*} $$

i.e., $\beta E_{Z}(\{F_{n}\},\mathbf {a_{1}})\geq E_{Z}(\{F_{n}\},\mathbf {b_{1}})$ . Since $\beta $ is arbitrary, we conclude that

$$ \begin{align*}E_{Z}(\{F_{n}\},\mathbf{b_{1}})=0,\end{align*} $$

and hence

$$ \begin{align*}E_{Z}(\{F_{n}\},[\mathbf{b}])=0.\end{align*} $$

On the other hand, if $[\mathbf {b}]\preceq [\mathbf {a}]$ , then

$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{b_{2}(|F_{n}|)}{a_{2}(|F_{n}|)}=0,\end{align*} $$

for arbitrary $\mathbf {a_{2}}=\{a_{2}(|F_{n}|)\}\in [\mathbf {a}]$ and $\mathbf {b_{2}}=\{b_{2}(|F_{n}|)\}\in [\mathbf {b}]$ . Given a small number $\beta>0$ , for all sufficiently large $n\in {\mathbb N}$ we have that $b_{2}(|F_{n}|)<\beta a_{2}(|F_{n}|)$ and hence, $M(Z,s,\{F_{n}\},\mathcal {U},\mathbf {b_{2}})\geq M(Z,s,\{F_{n}\},\mathcal {U},\beta \mathbf {a_{2}})$ . It follows that $E_{Z}(\{F_{n}\},\mathbf {b_{2}})>\frac {1}{\beta }E_{Z}(\{F_{n}\},\mathbf {a_{2}})$ . Again since $\beta $ is arbitrary,

$$ \begin{align*}E_{Z}(\{F_{n}\},\mathbf{b_{2}})=\infty,\end{align*} $$

then

$$ \begin{align*}E_{Z}(\{F_{n}\},[\mathbf{b}])=\infty.\\[-30pt]\end{align*} $$

Proof We shall prove the result for the scaled topological entropy $E_{Z}\Big(\Big [\{F_n\}\Big ]_{\mathbf {a}},\mathbf {a}\Big)$ , the arguments for the lower and upper scaled topological entropies are similar.

Suppose there is $\Big [\{F_n\}\Big ]_{\mathbf {a}}\in \mathcal {F}(G)_{\mathbf {a}}$ such that $E_{Z}\Big(\Big [\{F_n\}\Big ]_{\mathbf {a}},\mathbf {a}\Big)$ is positive and finite. Then for each $\Big [\{Q_n\}\Big ]_{\mathbf {a}}\succeq \Big [\{F_n\}\Big ]_{\mathbf {a}}$ ,

$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{a(|F_{n}^{*}|)}{a(|Q_{n}^{*}|)}=0,\end{align*} $$

for arbitrary $\{F_{n}^*\}\in \Big [\{F_n\}\Big ]_{\mathbf {a}}$ and $\{Q_{n}^*\}\in \Big [\{Q_n\}\Big ]_{\mathbf {a}}$ . Let us fix such two F $\unicode{xf8} $ lner sequences $\{F_{n}^*\}$ and $\{Q_{n}^*\}$ . Given a small number $\beta>0$ , for all sufficiently large $n\in {\mathbb N}$ we have that $a(|F_{n}^*|)<\beta a(|Q_{n}^*|)$ and hence, $M(Z,s,\{F_{n}^*\},\mathcal {U},\mathbf {a})\geq M(Z,s,\\ \{Q_{n}^*\},\mathcal {U},\beta \mathbf {a})$ . By $(2)$ of Proposition 2.7,

$$ \begin{align*}E_{Z}(\{F_{n}^*\},\mathbf{a})\geq E_{Z}(\{Q_{n}^*\},\beta\mathbf{a})=\frac{1}{\beta}E_{Z}(\{Q_{n}^*\},\mathbf{a}),\end{align*} $$

i.e., $\beta E_{Z}(\{F_{n}^*\},\mathbf {a})\geq E_{Z}(\{Q_{n}^*\},\mathbf {a})$ . Since $\beta $ is arbitrary, we conclude that

$$ \begin{align*}E_{Z}(\{Q_{n}^*\},\mathbf{a})=0,\end{align*} $$

and hence

$$ \begin{align*}E_{Z}\Big(\Big[\{Q_{n}\}\Big]_{\mathbf{a}},\mathbf{a}\Big)=0.\end{align*} $$

On the other hand, if $\Big [\{F_n\}\Big ]_{\mathbf {a}}\succeq \Big [\{Q_n\}\Big ]_{\mathbf {a}}$ , then

$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{a(|Q_{n}^{**}|)}{a(|F_{n}^{**}|)}=0\end{align*} $$

for arbitrary $\{F_{n}^{**}\}\in \Big [\{F_n\}\Big ]_{\mathbf {a}}$ and $\{Q_{n}^{**}\}\in \Big [\{Q_n\}\Big ]_{\mathbf {a}}$ . Given a small number $\beta>0$ , for all sufficiently large $n\in {\mathbb N}$ we have that $a(|Q_{n}^{**}|)<\beta a(|F_{n}^{**}|)$ and hence, $M(Z,s,\{Q_{n}^{**}\},\mathcal {U},\mathbf {a})\geq M(Z,s,\{F_{n}^{**}\},\mathcal {U},\beta \mathbf {a})$ . It follows that $E_{Z}(\{Q_{n}^{**}\},\mathbf {a})>\frac {1}{\beta }E_{Z}(\{F_{n}^{**}\},\mathbf {a})$ . Again since $\beta $ is arbitrary,

$$ \begin{align*}E_{Z}(\{Q_{n}^{**}\},\mathbf{a})=\infty,\end{align*} $$

then

$$ \begin{align*}E_{Z}\left(\Big[\{Q_{n}\}\Big]_{\mathbf{a}},\mathbf{a}\right)=\infty.\\[-37pt]\end{align*} $$

Proof Suppose there is $[\mathbf {a}]\in \mathcal {A}$ such that $E_{\mu }(\{F_{n}\},[\mathbf {a}])$ is positive and finite. Then for each $[\mathbf {b}]\succeq [\mathbf {a}]$ ,

$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{a_{1}(|F_{n}|)}{b_{1}(|F_{n}|)}=0\end{align*} $$

for arbitrary $\mathbf {a_{1}}=\{a_{1}(|F_{n}|)\}\in [\mathbf {a}]$ and $\mathbf {b_{1}}=\{b_{1}(|F_{n}|)\}\in [\mathbf {b}]$ . Let us fix such two scaled sequences $\mathbf {a_{1}}$ and $\mathbf {b_{1}}$ . Given a small number $\beta>0$ , for all sufficiently large $n\in {\mathbb N}$ we have that $a_{1}(|F_{n}|)<\beta b_{1}(|F_{n}|)$ . By Proposition 3.4, we have that

$$ \begin{align*}E_{\mu}(\{F_{n}\},\mathbf{a_{1}})\geq E_{\mu}(\{F_{n}\},\beta\mathbf{b_{1}})=\frac{1}{\beta}E_{\mu}(\{F_{n}\},\mathbf{b_{1}}),\end{align*} $$

i.e., $\beta E_{\mu }(\{F_{n}\},\mathbf {a_{1}})\geq E_{\mu }(\{F_{n}\},\mathbf {b_{1}})$ . Since $\beta $ is arbitrary, we conclude that

$$ \begin{align*}E_{\mu}(\{F_{n}\},\mathbf{b_{1}})=0,\end{align*} $$

and hence

$$ \begin{align*}E_{\mu}(\{F_{n}\},[\mathbf{b}])=0.\end{align*} $$

On the other hand, if $[\mathbf {b}]\preceq [\mathbf {a}]$ then

$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{b_{2}(|F_{n}|)}{a_{2}(|F_{n}|)}=0\end{align*} $$

for arbitrary $\mathbf {a_{2}}=\{a_{2}(|F_{n}|)\}\in [\mathbf {a}]$ and $\mathbf {b_{2}}=\{b_{2}(|F_{n}|)\}\in [\mathbf {b}]$ . Given a small number $\beta>0$ , for all sufficiently large $n\in {\mathbb N}$ we have that $b_{2}(|F_{n}|)<\beta a_{2}(|F_{n}|)$ . It follows that $E_{\mu }(\{F_{n}\},\mathbf {b_{2}})>\frac {1}{\beta }E_{\mu }(\{F_{n}\},\mathbf {a_{2}})$ . Again since $\beta $ is arbitrary,

$$ \begin{align*}E_{\mu}(\{F_{n}\},\mathbf{b_{2}})=\infty,\end{align*} $$

and hence

$$ \begin{align*}E_{\mu}(\{F_{n}\},[\mathbf{b}])=\infty.\\[-30pt]\end{align*} $$

Proof Suppose there is $\Big [\{F_n\}\Big ]_{\mathbf {a}}\in \mathcal {F}(G)_{\mathbf {a}}$ such that $E_{\mu }\Big(\Big [\{F_n\}\Big ]_{\mathbf {a}},\mathbf {a}\Big)$ is positive and finite. Then for each $\Big [\{Q_n\}\Big ]_{\mathbf {a}}\succeq \Big [\{F_n\}\Big ]_{\mathbf {a}}$ ,

$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{a(|F_{n}^{*}|)}{a(|Q_{n}^{*}|)}=0\end{align*} $$

for arbitrary $\{F_{n}^*\}\in \Big [\{F_n\}\Big ]_{\mathbf {a}}$ and $\{Q_{n}^*\}\in \Big [\{Q_n\}\Big ]_{\mathbf {a}}$ . Let us fix such two F $\unicode{xf8} $ lner sequences $\{F_{n}^*\}$ and $\{Q_{n}^*\}$ . Given a small number $\beta>0$ , for all sufficiently large $n\in {\mathbb N,}$ we have that $a(|F_{n}^*|)<\beta a(|Q_{n}^*|)$ . By $(2)$ of Proposition 3.4, we have that

$$ \begin{align*}E_{\mu}(\{F_{n}^*\},\mathbf{a})\geq E_{\mu}(\{Q_{n}^*\},\beta\mathbf{a})=\frac{1}{\beta}E_{\mu}(\{Q_{n}^*\},\mathbf{a}),\end{align*} $$

i.e., $\beta E_{\mu }(\{F_{n}^*\},\mathbf {a})\geq E_{\mu }(\{Q_{n}^*\},\mathbf {a})$ . Since $\beta $ is arbitrary, we conclude that

$$ \begin{align*}E_{\mu}(\{Q_{n}^*\},\mathbf{a})=0,\end{align*} $$

and hence

$$ \begin{align*}E_{\mu}\Big(\Big[\{Q_{n}\}\Big]_{\mathbf{a}},\mathbf{a}\Big)=0.\end{align*} $$

On the other hand, if $\Big [\{F_n\}\Big ]_{\mathbf {a}}\succeq \Big [\{Q_n\}\Big ]_{\mathbf {a}}$ then

$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{a(|Q_{n}^{**}|)}{a(|F_{n}^{**}|)}=0\end{align*} $$

for arbitrary $\{F_{n}^{**}\}\in \Big [\{F_n\}\Big ]_{\mathbf {a}}$ and $\{Q_{n}^{**}\}\in \Big [\{Q_n\}\Big ]_{\mathbf {a}}$ . Given a small number $\beta>0$ , for all sufficiently large $n\in {\mathbb N}$ we have that $a(|Q_{n}^{**}|)<\beta a(|F_{n}^{**}|)$ . It follows that $E_{\mu }(\{Q_{n}^{**}\},\mathbf {a})>\frac {1}{\beta }E_{\mu }(\{F_{n}^{**}\},\mathbf {a})$ . Again since $\beta $ is arbitrary,

$$ \begin{align*}E_{\mu}(\{Q_{n}^{**}\},\mathbf{a})=\infty,\end{align*} $$

then

$$ \begin{align*}E_{\mu}\Big(\Big[\{Q_{n}\}\Big]_{\mathbf{a}},\mathbf{a}\Big)=\infty.\\[-30pt]\end{align*} $$

Proof $(1)$ For a fixed $r>0$ and $k\in \mathbb {N}$ , let

$$ \begin{align*}L_{k}=\left\{x\in L:\liminf_{n\rightarrow\infty}\frac{-\log\mu(B_{F_{n}}(x,\epsilon))}{a(|F_{n}|)}<s+r\ \text{for all}~\epsilon\in (0,\frac{1}{k})\right\}.\end{align*} $$

Then we have $L=\bigcup _{k=1}^{\infty }L_{k}$ , since $\underline {h}_\mu (\{F_{n}\},\mathbf {a},x)\leq s$ for all $x\in L$ .

Now fix $k\geq 1$ and $0<\epsilon <\frac {1}{5k}$ . For each $x\in L_{k}$ , there exists a strictly increasing sequence $\{n_{j}\}_{j=1}^{\infty }$ (depending on the point x) such that

$$ \begin{align*}\mu(B_{F_{n_{j}}}(x,\epsilon))\geq \exp(-(s+r)a(|F_{n_{j}}|))\ \text{for all}~j\geq1 .\end{align*} $$

For any $N\in \mathbb {N}$ , the set $L_{k}$ is contained in the union of the sets in the family

$$ \begin{align*}\mathcal{F}=\{B_{F_{n_{j}}}(x,\epsilon):x\in L_{k}, n_{j}\geq N\}.\end{align*} $$

By Lemma 4.1, there exists a finite or countable subfamily $\mathcal {B}=\{B_{F_{n_{i}}}(x_i,\epsilon )\}_{i\in \mathcal {I}}\subset \mathcal {F}$ of pairwise disjoint balls such that

$$ \begin{align*}L_{k}\subset\bigcup_{i\in I}B_{F_{n_{i}}}(x_{i},5\epsilon).\end{align*} $$

The subfamily is at most countable since $\mu $ is a probability measure and the elements in $\mathcal {B}$ are pairwise disjoint and have positive $\mu $ -measure. Note that

$$ \begin{align*}\mu(B_{F_{n_{i}}}(x_{i},\epsilon))\geq \exp(-(s+r)a(|F_{n_{i}}|))\ \text{for all}~i\in \mathcal{I}.\end{align*} $$

The disjointness of $\{B_{F_{n_{i}}}(x_i,\epsilon )\}_{i\in \mathcal {I}}$ yields that

$$ \begin{align*}M(L_{k},s+r,N,\{F_{n}\},5\epsilon,\mathbf{a})\leq\sum_{i\in \mathcal{I}}\exp(-(s+r)a(|F_{n_{i}}|))\leq\sum_{i\in \mathcal{I}}\mu(B_{F_{n_{i}}}(x_{i},\epsilon))\leq1.\end{align*} $$

It follows that

$$ \begin{align*}M(L_{k},s+r,\{F_{n}\},5\epsilon,\mathbf{a})=\lim_{N\rightarrow\infty}M(L_{k},s+r,N,\{F_{n}\},5\epsilon,\mathbf{a})\leq1.\end{align*} $$

Hence,

$$ \begin{align*}E_{L_{k}}(\{F_{n}\},5\epsilon,\mathbf{a})\leq s+r,\end{align*} $$

which implies that

$$ \begin{align*}E_{L_{k}}(\{F_{n}\},\mathbf{a})\leq s+r\ \text{for all}~k\geq1.\end{align*} $$

Hence,

$$ \begin{align*}E_{L}(\{F_{n}\},\mathbf{a})=E_{\bigcup_{k=1}^{\infty}L_{k}}(\{F_{n}\},\mathbf{a})=\sup_{k\geq1}E_{L_{k}}(\{F_{n}\},\mathbf{a})\leq s+r.\end{align*} $$

Since r can be arbitrary, this implies that $E_{L}(\{F_{n}\},\mathbf {a})\leq s.$

$(2)$ For a fixed $r>0$ and $k\in \mathbb {N}$ , let

$$ \begin{align*}L_{k}=\left\{x\in L:\liminf_{n\rightarrow\infty}\frac{-\log\mu(B_{F_{n}}(x,\epsilon))}{a(|F_{n}|)}>s-r\ \text{for all}~\epsilon\in (0,\frac{1}{k})\right\}.\end{align*} $$

Since $\underline {h}_\mu (\{F_{n}\},\mathbf {a},x)\geq s$ for all $x\in L$ , we have that $L_{k}\subset L_{k+1}$ and $L=\bigcup _{k=1}^{\infty }L_{k}$ . Fix a sufficiently large $k\geq 1$ with $\mu (L_{k})>\frac {1}{2}\mu (L)>0$ . For each $N\in \mathbb {N}$ , set

$$ \begin{align*}L_{k,N}=\left\{x\in L_{k}:\frac{-\log\mu(B_{F_{n}}(x,\epsilon))}{a(|F_{n}|)}>s-r\ \text{for all}~n\geq N,\epsilon\in (0,\frac{1}{k})\right\}.\end{align*} $$

It is easy to see that $L_{k,N}\subset L_{k,N+1}$ and $L_{k}=\bigcup _{N=1}^{\infty }L_{k,N}$ . Thus we can pick $N^{\ast }\geq 1$ such that $\mu (L_{k,N^{\ast }})>\frac {1}{2}\mu (L_{k})>0$ . For simplicity of notation, let $L^{\ast }=L_{k,N^{\ast }}$ and $\epsilon ^{\ast }=\frac {1}{k}$ . By the choice of $L^{\ast }$ , we have that

$$ \begin{align*}\mu(B_{F_{n}}(x,\epsilon))\leq \exp(-(s-r)a(|F_{n}|))\ \text{for all}~x\in L^{\ast},0<\epsilon<\epsilon^{\ast},n\geq N^{\ast} .\end{align*} $$

Fix a sufficiently large $N>N^{\ast }$ . For each cover $\mathcal {F}=\{B_{F_{n_{i}}}(y_{i},\frac {\epsilon }{2})\}_{i\geq 1}$ of $L^{\ast }$ with $0<\epsilon <\epsilon ^{\ast }$ and $n_{i}\geq N\geq N^{\ast }$ for each $i\geq 1$ . Without loss of generality, assume that $L^{\ast }\bigcap B_{F_{n_{i}}}(y_{i},\frac {\epsilon }{2})\neq \emptyset $ for all i. Thus, for each $i\geq 1$ pick a point $x_{i}\in L^{\ast }\bigcap B_{F_{n_{i}}}(y_{i},\frac {\epsilon }{2})$ so that

$$ \begin{align*}B_{F_{n_{i}}}\left(y_{i},\frac{\epsilon}{2}\right)\subset B_{F_{n_{i}}}(x_{i},\epsilon).\end{align*} $$

It follows that

$$ \begin{align*}\sum_{i\geq1}\exp(-(s-r)a(|F_{n_{i}}|))\geq\sum_{i\geq1}\mu(B_{F_{n_{i}}}(x_{i},\epsilon))\geq\mu(L^{\ast}).\end{align*} $$

Therefore,

$$ \begin{align*}M(L^{\ast},s-r,N,\{F_{n}\},\frac{\epsilon}{2},\mathbf{a})\geq \mu(L^{\ast})>0.\end{align*} $$

Consequently,

$$ \begin{align*}M(L^{\ast},s-r,\{F_{n}\},\frac{\epsilon}{2},\mathbf{a})=\lim_{N\rightarrow\infty}M(L^{\ast},s-r,N,\{F_{n}\},\frac{\epsilon}{2},\mathbf{a})\geq \mu(L^{\ast})>0,\end{align*} $$

which implies that $E_{L^{\ast }}(\{F_{n}\},\mathbf {a})\geq s-r.$

It follows that

$$ \begin{align*}E_{L}(\{F_{n}\},\mathbf{a})\geq E_{L^{\ast}}(\{F_{n}\},\mathbf{a})\geq s-r.\end{align*} $$

Since r can be arbitrary, this implies that $E_{L}(\{F_{n}\},\mathbf {a})\geq s$ completing the proof of the theorem.

Proof Let $Z\subset X,s\geq 0,\epsilon ,\delta>0$ , set $f=\chi _ Z,c_i\equiv 1$ in the definition of scaled weighted topological entropy, we have

$$ \begin{align*}W(Z,s,N,\epsilon,\{F_{n}\},\mathbf{a})\leq M(Z,s,N,\{F_{n}\},\epsilon,\mathbf{a}) ~for~every~ N\in \mathbb{N}.\end{align*} $$

Next, we prove $M(Z,s+\delta ,N,\{F_{n}\},6\epsilon ,\mathbf {a})\leq W(Z,s,N,\epsilon ,\{F_{n}\},\mathbf {a})$ for large enough $N\in \mathbb {N}$ .

There are $\xi>0$ and $N_{1}$ such that $a(|F_{n}|)\geq n\xi $ for all $n\geq N_{1}$ . Let $N>\max \{N_{1},2\}$ such that $n^{2}e^{-n\delta \xi }\leq 1$ for all $n\geq N$ . Let $\{(B_{F_{n_{i}}}(x_i,\epsilon ),c_i)\}_{i\in \mathcal {I}}$ be a family so that $\mathcal {I}\subset \mathbb {N},x_i\in X,0\leq c_i<\infty ,n_i\geq N$ and

$$ \begin{align*} \sum_i c_i\chi_{B_i}\geq\chi_Z, \end{align*} $$

where $B_i:=B_{F_{n_{i}}}(x_i,\epsilon )$ . We claim that

(4.1) $$ \begin{align} M(Z,s+\delta,N,\{F_{n}\},6\epsilon,\mathbf{a})\leq\sum_{i\in I}c_{i}\exp(-sa(|F_{n_{i}}|)) \end{align} $$

and hence $M(Z,s+\delta ,N,\{F_{n}\},6\epsilon ,\mathbf {a})\leq W(Z,s,N,\epsilon ,\{F_{n}\},\mathbf {a})$ .

We denote

$$ \begin{align*}\mathcal{I}_n=\{i\in \mathcal{I}:n_i=n\},\end{align*} $$

and

$$ \begin{align*}\mathcal{I}_{n,k}=\{i\in \mathcal{I}_n:i\leq k\}\end{align*} $$

for $n\geq N,k\in \mathbb {N}.$ We write $B_i:=B_{F_{n_{i}}}(x_i,\epsilon ),5B_i:=B_{F_{n_{i}}}(x_i,5\epsilon )$ for $i\in \mathcal {I}$ . Obviously, we may assume $B_i\neq B_j$ for $i\neq j$ . For $t>0$ , set

$$ \begin{align*} Z_{n,t}=\left\{x\in Z:\sum _{i\in \mathcal{I}_n}c_i\chi_{B_i}(x)>t\right\}\ \ \end{align*} $$

and

$$ \begin{align*} Z_{n,t,k}=\left\{x\in Z:\sum _{i\in \mathcal{I}_{n,k}}c_i\chi_{B_i}(x)>t\right\}. \end{align*} $$

We divide the proof of (4.1) into the following three steps.

Step 1. For each $n\geq N, k\in \mathbb {N}$ and $t>0$ , there exists a finite set $\mathcal {J}_{n,k,t}\subset \mathcal {I}_{n,k}$ such that the ball $B_i(i\in \mathcal {J}_{n,k,t})$ are pairwise disjoint, $Z_{n,t,k}\subset \cup _{i\in \mathcal {J}_{n,k,t}}5B_i$ , and

$$ \begin{align*} \aleph(\mathcal{J}_{n,k,t})\exp(-sa(|F_{n_{i}}|))\leq\frac{1}{t}\sum_{i\in \mathcal{I}_{n,k}}c_{i}\exp(-sa(|F_{n_{i}}|)). \end{align*} $$

We will use the method of Federer [Reference Federer10, 2.10.24] and Mattila [Reference Mattila23, Lemma 8.16] for amenable group actions. Since $\mathcal {I}_{n,k}$ is finite, by approximating the $c_i$ ’s from above, we may assume that each $c_i$ is a positive rational, and then by multiplying with a common denominator we may assume that each $c_i$ is a positive integer. Let m be the least integer with $m\geq t$ . Denote $\mathcal {B}=\{B_i,i\in \mathcal {I}_{n,k}\}$ , and define ${u:\mathcal {B}\rightarrow \mathbb {Z}}$ , by $u(B_i)=c_i$ . Since $B_i\neq B_j$ for $i\neq j$ , so u is well defined. We define by introduction integer-valued functions $v_0,v_1,\cdots ,v_m$ on $\mathcal {B}$ and sub-families $\mathcal {B}_1,\mathcal {B}_{2},\ldots ,\mathcal {B}_m$ of $\mathcal {B}$ starting with $v_0=u$ . Using Lemma 4.1 repeatedly, we define inductively for $j=1,\cdots ,m$ , disjoint subfamilies $\mathcal {B}_{i}$ of $\mathcal {B}$ such that

$$ \begin{align*} \mathcal{B}_j\subset\{B\in\mathcal{B}:v_{j-1}(B)\geq1\}, \end{align*} $$

$$ \begin{align*} Z_{n,k,t}\subset\cup_{B\in\mathcal{B}_j}5B, \end{align*} $$

and the functions $v_j$ such that

$$ \begin{align*}v_j(B)=\left\{\begin{array}{ll} v_{j-1}(B)-1,& \text{for}\ \ B\in\mathcal{B}_j, \\ v_{j-1}(B),& \text{for}\ \ B\in{\mathcal{B}\backslash\mathcal{B}_j\text{.}}\end{array}\right .\end{align*} $$

This is possible for $j<m$ ,

$$ \begin{align*}Z_{n,k,t}\subset\left\{x:\sum_{B\in\mathcal{B}:B\ni x}v_j(B)\geq m-j\right\},\end{align*} $$

whence every $x\in Z_{n,k,t}$ belongs to some ball $B\in \mathcal {B}$ with $v_j(B)\geq 1$ . Thus

$$ \begin{align*} \sum_{j=1}^{m}\aleph(\mathcal{B}_{j})\exp(-sa(|F_{n}|))&=\sum_{j=1}^{m}{\sum_{B\in{\mathcal{B}}_{j}}(v_{j-1}(B)-v_{j}(B))\exp(-sa(|F_{n}|))}\\ &\leq \sum_{B\in \mathcal{B}}{\sum_{j=1}^{m}(v_{j-1}(B)-v_{j}(B))\exp(-sa(|F_{n}|))} \\ &\leq \sum_{B\in \mathcal{B}}u(B)\exp(-sa(|F_{n}|))\\&=\sum_{i\in \mathcal{I}_{n,k}}c_i\exp(-sa(|F_{n}|)). \end{align*} $$

Choose $j_0\in \{1,\cdots ,m\}$ so that $\aleph (\mathcal {B}_{j_0})$ is the smallest. Then

$$ \begin{align*} \aleph(\mathcal{B}_{j_0})\exp(-sa(|F_{n}|))&\leq&\frac{1}{m}\sum_{i\in \mathcal{I}_{n,k}}c_i\exp(-sa(|F_{n}|))\leq\frac{1}{t}\sum_{i\in \mathcal{I}_{n,k}}c_i\exp(-sa(|F_{n}|)). \end{align*} $$

So $\mathcal {J}_{n,k,t}=\{i\in \mathcal {I}:B_i\in \mathcal {B}_{j_0}\}$ is desired.

Step 2. For each $n\in \mathbb {N}$ and $t>0$ , we have

(4.2) $$ \begin{align} M(Z_{n,t},s+\delta,N,\{F_{n}\},6\epsilon,\mathbf{a})\leq\frac{\exp(-\delta a(|F_{n}|))}{t}\sum_{i\in \mathcal{I}_n}c_i\exp(-sa(|F_{n}|)). \end{align} $$

Assume $Z_{n,t}\neq \emptyset $ , otherwise (4.2) is obvious. Since $Z_{n,k,t}\uparrow Z_{n,t}$ , $Z_{n,k,t}\neq \emptyset $ for large enough $k\in {\mathbb N}$ . Let $\mathcal {J}_{n,k,t}$ be the sets constructed in Step 1. Then $\mathcal {J}_{n,k,t}\neq \emptyset $ for large enough $k\in {\mathbb N}$ . Set $E_{n,k,t}=\{x_i:i\in \mathcal {J}_{n,k,t}\}$ . Note that the family of all non-empty compact subsets of X is compact with respect to Hausdorff distance(Federer [Reference Federer10, 2.10.21]). It follows that there is a subsequence $(k_j)$ of natural numbers and a non-empty compact set $E_{n,t}\subset X$ such that $E_{n,k_j,t}$ converges to $E_{n,t}$ in the Hausdorff distance as $j\rightarrow \infty $ . Since any two points in $E_{n,k,t}$ have a distance (with respect to $d_{F_{n}}$ ) not less than $\epsilon $ , so do the points in $E_{n,t}$ . Thus $E_{n,t}$ is a finite set, moreover, $\aleph (E_{n,k_j,t})=\aleph (E_{n,t})$ when $j\in {\mathbb N}$ is large enough.

Hence

$$ \begin{align*}\bigcup_{x\in E_{n,t}}B_{F_{n}}(x,5.5\epsilon)\supset \bigcup_{x\in E_{n,k_j,t}}B_{F_{n}}(x,5\epsilon)=\bigcup_{i\in\mathcal{J}_{n,k_j,t}}5B_i\supset Z_{n,k_j,t}\end{align*} $$

when $j\in {\mathbb N}$ is large enough, and thus $\bigcup _{x\in E_{n,t}}B_{F_{n}}(x,6\epsilon )\supset Z_{n,t}$ . By the way, since $\aleph {(E_{n,k_j,t})}=\aleph (E_{n,t})$ when $j\in {\mathbb N}$ is large enough, we have

$$ \begin{align*} \aleph(E_{n,t})\exp(-sa(|F_{n}|))\leq \frac{1}{t}\sum_{i\in \mathcal{I}_n}c_i\exp(-sa(|F_{n}|)). \end{align*} $$

Therefore,

$$ \begin{align*} M(Z_{n,t},s+\delta,N,\{F_{n}\},6\epsilon,\mathbf{a}) &\leq\aleph(E_{n,t})\exp(-(s+\delta)a(|F_{n}|))\\ &\leq\frac{\exp(-\delta a(|F_{n}|))}{t}\sum_{i\in \mathcal{I}_n}c_i\exp(-sa(|F_{n}|))\\ &\leq\frac{\exp(-\delta n\xi )}{t}\sum_{i\in \mathcal{I}_n}c_i\exp(-sa(|F_{n}|))\\ &\leq\frac{1}{n^{2}t}\sum_{i\in \mathcal{I}_n}c_i\exp(-sa(|F_{n}|)).\\ \end{align*} $$

Step 3. For any $t\in (0,1)$ , we have

$$ \begin{align*} M(Z,s+\delta,N,\{F_{n}\},6\epsilon,\mathbf{a})\leq\frac{1}{t}\sum_{i\in \mathcal{I}}c_i\exp(-sa(|F_{n_{i}}|)), \end{align*} $$

which implies (4.1). In fact, fix $t\in (0,1)$ . Note that $\sum _{n=N}^{\infty }n^{-2}<1$ . Then $Z\subset \bigcup _{n=N}^\infty Z_{n,n^{-2}t}$ . Also note that $M(Z,s,N,\{F_{n}\},\epsilon ,\mathbf {a})$ is an outer measure of X, so we get

$$ \begin{align*} M(Z,s+\delta,N,\{F_{n}\},6\epsilon,\mathbf{a})&\leq\sum_{n=N}^{\infty}M(Z_{n,n^{-2}t},s+\delta,N,\{F_{n}\},6\epsilon,\mathbf{a})\\& \leq\sum_{n=N}^{\infty}\frac{1}{t}{\sum_{i\in\mathcal{I}_n}c_i\exp(-sa(|F_{n}|))}=\frac{1}{t}\sum_{i\in\mathcal{I}}c_i\exp(-sa(|F_{n_{i}}|)). \end{align*} $$

Proof Clearly $c<\infty $ . We define a function p on the space $C(X)$ of continuous real-valued functions on X by

$$ \begin{align*} p(f)=\frac{1}{c}\text{W}(\chi_K\cdot f,s,N,\epsilon,\{F_{n}\},\mathbf{a}). \end{align*} $$

Let $\mathbf {1}\in C(X)$ denote the constant function $\mathbf {1}(x)\equiv 1$ . It is easy to verify that:

1. (1) $p(tf)=tp(f)$ for $f\in C(X)$ and $t\geq 0$ ,

2. (2) $p(f+g)\leq p(f)+p(g)$ for $f,g\in C(X)$ ,

3. (3) $p(\mathbf {1})=1,0\leq P(f)\leq \parallel f\parallel _{\infty }$ for $f\in C(X)$ , and $p(g)=0$ for $g\in C(X),g\leq 0.$

By the Hahn–Banach theorem, we can extend the linear functional ${t\rightarrow tp(1),t\in \mathbb {R}}$ , from the subspace of constant functions to a linear functional $L:C(X)\rightarrow \mathbb {R}$ satisfying

$$ \begin{align*}L(\mathbf{1})=p(\mathbf{1})=1 \text{ and }-p(-f)\le L(f)\le p(f) \text{ for any }f\in C(X).\end{align*} $$

If $f\in C(X)$ with $f\geq 0$ , then $p(-f)=0$ and so $L(f)\geq 0$ . Hence we can use the Riesz representation theorem to find a Borel probability measure $\mu $ on X such that $L(f)=\int f d\mu $ for $f\in C(X)$ .

Next, we prove $\mu (K)=1$ . For any compact set $E\subset X\backslash K$ , by Urysohn lemma there exists $f\in C(X)$ such that $0\leq f\leq 1,f(x)=1$ for $x\in E$ and $f(x)=0$ for ${x\in K}$ . Then ${f\cdot \chi _K=0}$ and thus $p(f)=0$ . Hence $\mu (E)\leq L(f)\leq p(f)=0$ . This shows ${\mu (X\backslash K)=0,}$ that is $\mu (K)=1$ .

In the end, we prove $\mu (B_{F_n}(x,\epsilon ))\leq \frac {1}{c}\exp (-sa(|F_{n}|))$ for any $x\in X,n\geq N$ . In fact, for any compact set $E\subset B_{F_n}(x,\epsilon )$ , by Urysohn lemma again, there is $f\in C(X)$ , such that $0\leq f\leq 1, f(y)=1$ for $y\in E$ and $f(y)=0$ for $y\in X\backslash B_{F_n}(x,\epsilon )$ . Then $\mu (E)\leq L(f)\leq p(f)$ . Since $\chi _K\cdot f\leq \chi _{B_{F_n}(x,\epsilon )}$ and $n\geq N$ , we get $W(\chi _K\cdot f,s,N,\epsilon ,\{F_{n}\},\mathbf {a})\leq \exp (-sa(|F_{n}|))$ and hence $p(f)\leq \frac {1}{c}\exp (-sa(|F_{n}|))$ . Therefore, we have $\mu (E)\leq \frac {1}{c}\exp (-sa(|F_{n}|))$ . It follows that

$$ \begin{align*} \mu(B_{F_n}(x,\epsilon))=\sup\{\mu(E):E\text{ is a compact subset of}~ B_{F_n}(x,\epsilon)\}\leq\frac{1}{c}\exp(-sa(|F_{n}|)). \end{align*} $$

Proof Firstly, we prove $E_{K}^{B_{2}}(\{F_{n}\},\mathbf {a})\geq \underline {h}_\mu (\{F_{n}\},\mathbf {a})$ , for any $\mu \in \mathcal {M}(X),\mu (K)=1$ . We set

$$ \begin{align*} \underline{h}_\mu(\{F_{n}\},\mathbf{a},x,\epsilon)=\liminf_{n\rightarrow\infty}-\frac{1}{a(|F_{n}|)}\log\mu(B_{F_{n}}(x,\epsilon)) \end{align*} $$

for $x\in X,n\in \mathbb {N},\epsilon>0$ . It’s easy to see that $\underline {h}_\mu (\{F_{n}\},\mathbf {a},x,\epsilon )$ is nonnegative and increases as $\epsilon $ decreases. By the monotone convergence theorem, we get

$$ \begin{align*} \lim\limits_{\epsilon\rightarrow0}\int\underline{h}_\mu(\{F_{n}\},\mathbf{a},x,\epsilon)d\mu(x)=\int\underline{h}_\mu(\{F_{n}\},\mathbf{a},x)d\mu(x)=\underline{h}_\mu(\{F_{n}\},\mathbf{a}). \end{align*} $$

Thus to show $E_{K}^{B_{2}}(\{F_{n}\},\mathbf {a})\geq \underline {h}_\mu (\{F_{n}\},\mathbf {a})$ , we only to show

$$ \begin{align*}E_{K}^{B_{2}}(\{F_{n}\},\mathbf{a})\geq\int\underline{h}_\mu(\{F_{n}\},\mathbf{a},x,\epsilon)d\mu(x) \ \text{for any}~\epsilon>0.\end{align*} $$

Now we fix $\epsilon>0, l\in \mathbb {N}$ , set $u_{l}=\min \{l,\int \underline {h}_\mu (\{F_{n}\},\mathbf {a},x,\epsilon )d\mu (x)-\frac {1}{l}\}$ , then exist a Borel set $A_{l}\subset X,\mu (A_{l})>0,N\in \mathbb {N}$ such that

(4.3) $$ \begin{align} \mu(B_{F_{n}}(x,\epsilon))\leq \exp(-u_la(|F_{n}|)), \ \text{for all}~x\in A_{l}, n\geq N. \end{align} $$

Let $\{B_{F_{n_{i}}}(x_i,\epsilon /2)\}$ ba a finite or countable family such that $x_i\in X,n_i\geq N$ and $K\cap A_{l}\subset \bigcup _iB_{F_{n_{i}}}(x_i,\epsilon /2)$ . We may as well assume that for each $i\in {\mathbb N}$ , $B_{F_{n_{i}}}(x_i,\epsilon /2)\bigcap (K\cap A_{l})\neq \emptyset $ , and select $y_i\in B_{F_{n_{i}}}(x_i,\epsilon /2)\bigcap (K\cap A_{l})$ . Then by (4.3), we have

$$ \begin{align*} \sum_i\exp(-u_{l}a(|F_{n_{i}}|))&\geq \sum_i\mu(B_{F_{n_{i}}}(y_i,\epsilon))\\&\geq \sum_i\mu(B_{F_{n_{i}}}(x_i,\epsilon/2)) \\&\geq\mu(K\cap A_{l})=\mu(A_{l})>0. \end{align*} $$

So, we get

$$ \begin{align*} M\left(K,u_{l},\{F_{n}\},\frac{\epsilon}{2},\mathbf{a}\right)&\geq M(K,u_{l},N,\{F_{n}\},\frac{\epsilon}{2},\mathbf{a})\\&\geq M(K\cap A_{l},u_{l},N,\{F_{n}\},\frac{\epsilon}{2},\mathbf{a}) \\&\geq\mu(A_{l})>0. \end{align*} $$

Therefore, $E_{K}^{B_{2}}(\{F_{n}\},\mathbf {a})\geq u_{l}$ . Letting $l\rightarrow \infty $ , we get

$$ \begin{align*}E_{K}^{B_{2}}(\{F_{n}\},\mathbf{a})\geq\int\underline{h}_\mu(\{F_{n}\},\mathbf{a},x,\epsilon)d\mu(x).\end{align*} $$

Thus $E_{K}^{B_{2}}(\{F_{n}\},\mathbf {a})\geq \underline {h}_\mu (\{F_{n}\},\mathbf {a})$ .

We next prove $E_{K}^{B_{2}}(\{F_{n}\},\mathbf {a})\leq \{\underline {h}_\mu (\{F_{n}\},\mathbf {a}):\mu \in \mathcal {M}(X),\mu (K)=1\}$ . We may as well assume $E_{K}^{B_{2}}(\{F_{n}\},\mathbf {a})>0$ , otherwise the conclusion is obvious. By Proposition 4.1, $E_{K}^{B_{2}}(\{F_{n}\},\mathbf {a})=h_{top}^W(K,\{F_{n}\},\mathbf {a})$ . Suppose $0<s<h_{top}^W(K,\{F_{n}\},\mathbf {a})$ , then there exists $\epsilon>0$ and $N\in \mathbb {N}$ , such that $c=W(K,s,N,\epsilon ,\{F_{n}\},\mathbf {a})>0$ . By Lemma 4.2, there exists $\mu \in \mathcal {M}(X),\mu (K)=1$ , such that

$$ \begin{align*}\mu(B_{F_{n}}(x,\epsilon))\leq \frac{1}{c}\exp(-sa(|F_{n}|))\end{align*} $$

for any $x\in X,n\geq N$ . And then $\underline {h}_\mu (\{F_{n}\},\mathbf {a},x)\geq s$ for each $x\in X$ . Therefore, $\underline {h}_\mu (\{F_{n}\},\mathbf {a})\geq \int \underline {h}_\mu (\{F_{n}\},\mathbf {a},x)d\mu (x)\geq s.$ By Proposition 2.2, the proof is completed.