Weighted Entropy of a Flow on Non-compact Sets

Abstract

Let $(X_i,\phi _i), i=1,2,\ldots ,k$ be continuous flows on compact metric spaces, and for each $1\le i\le k-1$, $(X_{i+1},\phi _{i+1})$ be a factor of $(X_i,\phi _i)$. Let $\mathbf{a}=(a_1,a_2,\ldots ,a_k)\in {\mathbb {R}}^k$ with $a_1>0$ and $a_i\ge 0$ for $2\le i\le k$. Based on the theory of Carathéodory structure, this paper introduce the $\mathbf{a}$-weighted topological entropy of a flow on non-compact sets and the $\mathbf{a}$-weighted measure-theoretic entropy of a flow. We establish the variational principle and also investigate the relationship between the $\mathbf{a}$-weighted entropy and the classical $\mathbf{a}$-weighted entropy of time one map.

Appendix: A Weighted Version of the Brin–Katok Theorem

The main result in this Appendix is the weighted version of the Brin–Katok theorem.

Let $k\ge 2$. Assume that $(X_i,d_i),i = 1,2,\ldots ,k$, are compact metric spaces, and $(X_i,T_i)$ are topological dynamical systems. Moreover, assume that for each $1\le i\le k-1$, $(X_{i+1},T_{i+1})$ is a factor of $(X_i,T_i)$ with a factor map $\pi _i : X_i \rightarrow X_{i+1}$; in other words, $\pi _1,\pi _2,\ldots ,\pi _{k-1}$ are continuous maps so that the following diagrams commute.

For convenience, we use $\pi _0$ to denote the identity map on $X_1$. Define $\tau _i : X_1\rightarrow X_{i+1}$ by $\tau _i = \pi _i\circ \pi _{i-1}\circ \cdots \circ \pi _0$ for $i = 0, 1,\ldots k-1$. Fix $\mathbf{a} = (a_1,a_2,\ldots ,a_k)\in R^k$ with $a_1 > 0$ and $a_i\ge 0$ for $i\ge 2.$

For any $n\in {\mathbb {N}}$, the n-th $\mathbf{a}$-weighted Bowen metric $d_n^{\mathbf{a }}$ on $X_1$ is defined by

$$\begin{aligned} d_n^{\mathbf{a }}(x, y) =\max \left\{ d_i\left( T_{i}^s \tau _{i-1}x,T_{i}^s \tau _{i-1}y\right) : 0\le s \le \lceil (a_1 + \cdots + a_i)n\rceil , i = 1,\ldots ,k\right\} . \end{aligned}$$

For every $\epsilon >0$, we denote by $B^\mathbf{a }_n(x, \epsilon )$ the open ball of radius $\epsilon$ in the metric $d_n^\mathbf{a }$ around x, i.e.,

$$\begin{aligned} B^\mathbf{a }_n(x,\epsilon ) =\left\{ y \in X : \ d_n^{\mathbf{a }}(x, y) < \epsilon \right\} . \end{aligned}$$

Theorem 7

Let $\mu$ be invariant measure with $h_\mu ^\mathbf{a}(T_1)<\infty$. Then for $\mu$-a.e. $x\in X_1$

(a):

$$\begin{aligned}&\lim _{\epsilon \rightarrow 0} \liminf _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}\\&\quad =\lim _{\epsilon \rightarrow 0} \limsup _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}\overset{def}{=}h_\mu ^\mathbf{a}(T_1,x); \end{aligned}$$

(b):

$h_\mu ^\mathbf{a}(T_1,x)$ is $T_1$-invariant;

(c):

$\int _{X_1}h_\mu ^\mathbf{a}(T_1,x)d\mu (x)=h_\mu ^\mathbf{a}(T_1)$.

When $\mathbf{a}=(1,0,\ldots , 0)$, the above result reduces to the Brin–Katok theorem on local entropy [4].

The proof of Theorem 7 is based on the following weighted version of the Shannon–McMillan–Breiman theorem [5].

Proposition 2

Let $(X,{\mathcal {B}},\mu ,T)$ be a measure preserving dynamical system and $k\ge 1$. Let $\alpha _1, \ldots , \alpha _k$ be k countable measurable partitions of $(X,{\mathcal {B}},\mu )$ with $H_\mu (\alpha _i)<\infty$ for each i, and $\mathbf{a}=(a_1,\ldots ,a_k)\in {\mathbb {R}}^k$ with $a_1>0$ and $a_i\ge 0$ for $i\ge 2$. Then

$$\begin{aligned} \lim _{N\rightarrow +\infty }\frac{1}{N}I_\mu \left( \bigvee _{i=1}^k (\alpha _i)_0^{\lceil (a_1+\cdots +a_i)N\rceil -1} \right) (x)=\sum _{i=1}^k a_i{\mathbb {E}}_\mu (F_i|{\mathcal {I}}_\mu )(x) \end{aligned}$$

(9)

almost everywhere, where

$$\begin{aligned} F_i(x):=I_\mu \left( \bigvee _{j=i}^{k}\alpha _j\big |\bigvee _{n=1}^\infty T^{-n}\left( \bigvee _{j=i}^{k}\alpha _j\right) \right) (x), \quad i=1,\ldots ,k \end{aligned}$$

and ${\mathcal {I}}_\mu =\{ B\in {\mathcal {B}}:\; \mu (B\triangle T^{-1}B)=0\}$. In particular, if T is ergodic, we have

$$\begin{aligned} \lim _{N\rightarrow +\infty }\frac{1}{N}I_\mu \left( \bigvee _{i=1}^k (\alpha _i)_0^{\lceil (a_1+\cdots +a_i)N\rceil -1} \right) (x)=\sum _{i=1}^k a_ih_\mu \left( T,\bigvee _{j=i}^{k}\alpha _j\right) \end{aligned}$$

almost everywhere.

Proof of Theorem 7

We just adapt the proof of Brin and Katok [4] for their local entropy formula.

Let $\epsilon >0$. Let $\alpha _i$ be a finite Borel partition of $X_i$, $i=1,\ldots ,k$, with $\text {diam}(\alpha _i)<\epsilon$. Then

$$\begin{aligned} B_n^\mathbf{a}(x,\epsilon )\supseteq \bigcap _{i=1}^k \left( \tau _{i-1}^{-1}\alpha _i\right) _0^{\lceil (a_1+\cdots +a_i)n\rceil -1}(x) \end{aligned}$$

for $x\in X_1$. Hence by Proposition 2, for $\mu$-a.e $x\in X_1$ we have

$$\begin{aligned}&\limsup _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}\le \limsup _{n\rightarrow +\infty } \frac{-\log \mu \left( \bigcap _{i=1}^k (\tau _{i-1}^{-1}\alpha _i)_0^{\lceil (a_1+\cdots +a_i)n\rceil -1}(x)\right) }{n}\\&\quad =\limsup _{n\rightarrow +\infty } \frac{I_\mu \left( \bigvee _{i=1}^k (\tau _{i-1}^{-1}\alpha _i)_0^{\lceil (a_1+\cdots +a_i)n\rceil -1}\right) (x)}{n}=\sum _{i=1}^k a_i{\mathbb {E}}_\mu (F_i|{\mathcal {I}}_\mu )(x) \end{aligned}$$

where

$$\begin{aligned} F_i(x):=I_\mu \left( \bigvee _{j=i}^{k}\alpha _j\big |\bigvee _{n=1}^\infty T^{-n}\left( \bigvee _{j=i}^{k}\alpha _j\right) \right) (x), \quad i=1,\ldots ,k \end{aligned}$$

and ${\mathcal {I}}_\mu =\{ B\in {\mathcal {B}}:\; \mu (B\triangle T^{-1}B)=0\}$. Moreover,

$$\begin{aligned} \int _{X_1}\limsup _{n\rightarrow +\infty }&\frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}d\mu (x)\le \int _{X_1}\sum _{i=1}^k a_i{\mathbb {E}}_\mu (F_i|{\mathcal {I}}_\mu )(x)d\mu (x) \nonumber \\&\le \sum _{i=1}^k a_ih_{\tau _{i-1}^{-1}}(T_i)=h_\mu ^\mathbf{a}(T_1). \end{aligned}$$

(10)

By the dominant convergence theorem, one has

$$\begin{aligned} \int _{X_1}\lim _{\epsilon \rightarrow 0}\limsup _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}d\mu (x) \le h_\mu ^\mathbf{a}(T_1). \end{aligned}$$

(11)

We proceed now to estimate from below. It is sufficient to show

$$\begin{aligned} \int _{X_1}\lim _{\epsilon \rightarrow 0}\liminf _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}d\mu (x) \ge h_\mu ^\mathbf{a}(T_1). \end{aligned}$$

First as in [4] we can construct finite Borel partitions $\beta _i=\{ B^i_1,B^i_2,\ldots ,B^i_{v_i}\}$ of $X_i$ for $i=1, 2,\ldots ,k$, such that $\mu \circ \tau _{i-1}^{-1}(\partial \beta _i)=0$ and $\text{ diam }(\beta _i)$ are small enough so that

$$\begin{aligned} h_{\mu \circ \tau _{i-1}^{-1}}(T_i,\beta _i)\ge h_{\mu \circ \tau _{i-1}^{-1}}(T_i)-\delta . \end{aligned}$$

Next define the partitions $\alpha _i$ of $X_i$ recursively for $i=k, k-1, \ldots , 1$, by setting $\alpha _k=\beta _k$ and

$$\begin{aligned} \alpha _j=\beta _j\vee \pi _j^{-1}(\alpha _{j+1}) \quad \text{ for }\quad j=k-1,\ldots , 1. \end{aligned}$$

Then we have

• (1) $\alpha _i\succeq \pi _i^{-1}(\alpha _{i+1})$ for $i=1,\ldots ,k-1$.

• (2) $\sum _{i=1}^k a_ih_{\mu \circ \tau _{i-1}^{-1}}(T_i,\alpha _i)\ge h_\mu ^\mathbf{a}(T_1)-(a_1+a_2+\cdots +a_k)\delta$.

• (3) $\mu \circ \tau _{i-1}^{-1}(\partial \alpha _i)=0$ for $i=1,\ldots ,k$.

Write $\alpha _i=\{ A^i_1,A^i_2,\ldots ,A^i_{u_i}\}$ for $i=1,\ldots , k$. Let $M=\max \{u_i:\; 1\le i\le k\}$ and $\Lambda =\{1,\ldots ,M\}$. Given $m\in {\mathbb {N}}$, for $\mathbf{s}=(s_i)_{i=0}^{m-1},\mathbf{t}=(t_i)_{i=0}^{k-1}\in \Lambda ^{\{0,1,\ldots ,m-1\}}$, the Hamming distance between $\mathbf{s}$ and $\mathbf{t}$ is defined to be the following value

$$\begin{aligned} \frac{1}{m}\#\left\{ i\in \{0,1,\ldots ,m-1\}: s_i\ne t_i\right\} . \end{aligned}$$

For $\mathbf{s}\in \Lambda ^{\{0,1,\ldots ,m-1\}}$ and $0<\tau \le 1$, let $Q(\mathbf{s},\tau )$ be the total number of those $\mathbf{t}\in \Lambda ^{\{0,1,\ldots ,m-1\}}$ so that the Hamming distance between $\mathbf{s}$ and $\mathbf{t}$ does not exceed $\tau$. Clearly,

$$\begin{aligned} Q_m(\tau ):=\max \limits _{\mathbf{s}\in \Lambda ^{\{0,1,\ldots ,m-1\}}}Q(\mathbf{s},\tau )\le \left( {\begin{array}{c}m\\ \lceil m\tau \rceil \end{array}}\right) M^{\lceil m\tau \rceil }. \end{aligned}$$

By the Stirling formula, there exists a small $0<\delta _1$ and a positive constant $C:=C(\delta ,M)>0$ such that

$$\begin{aligned} \left( {\begin{array}{c}m\\ \lceil m\delta _1\rceil \end{array}}\right) M^{\lceil m\delta _1\rceil }\le e^{\delta m+C} \end{aligned}$$

(12)

for all $m\in {\mathbb {N}}$.

For $\eta >0$, set

$$\begin{aligned}&U^i_\eta (\alpha _i)=\{x\in X_1:\; B(\tau _{i-1}x,\eta )\not \subseteq \alpha _i(\tau _{i-1}x)\}, \quad i=1,\ldots ,k. \end{aligned}$$

Then $\bigcap _{\eta >0} U^i_\eta (\alpha _i)=\tau _{i-1}^{-1}(\partial \alpha _i)$, and hence $\mu (U^i_\eta (\alpha _i))\rightarrow \mu (\tau _{i-1}^{-1}(\partial \alpha _i))=0$ as $\eta \rightarrow 0$. Therefore, we can choose $\epsilon >0$ such that $\mu (U^i_\eta (\alpha _i))<\delta \delta _1$ for any $0<\eta \le \epsilon$ and $i=1,\ldots ,k$.

By the Birkhoff’s ergodic theorem,

$$\begin{aligned} \lim _{n\rightarrow +\infty }&\frac{1}{\lceil (a_1+\cdots +a_k)n \rceil }\sum \limits _{i=1}^k \sum _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}\chi _{U^i_\epsilon (\alpha _i)}(T^j_1x) \end{aligned}$$

exists for almost every $x\in X_1$, where we take the convention $a_0=0$. Moreover

$$\begin{aligned}&\int _X\lim _{n\rightarrow +\infty } \frac{1}{\lceil (a_1+\cdots +a_k)n \rceil }\sum \limits _{i=1}^k \sum _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}\chi _{U^i_\epsilon (\alpha _i)}(T^j_1x)d\mu (x)\\&\quad =\frac{1}{(a_1+\cdots +a_k)} \sum \limits _{i=1}^k a_i\mu (U^i_\epsilon (\alpha _i))<\delta \delta _1, \end{aligned}$$

Thus we can find a large natural number $\ell _0$ such that $\mu (A_\ell )>1-\delta$ for any $\ell \ge \ell _0$, where

$$\begin{aligned} A_\ell&=\left\{ x\in X_1:\;\frac{1}{\lceil (a_1+\cdots +a_k)n \rceil }\sum \limits _{i=1}^k \sum _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}\chi _{U^i_\epsilon (\alpha _i)}(T^j_1x)\right. \\&\quad \left. \le \frac{\delta _1\delta }{\delta } =\delta _1\text { for all }n\ge \ell \right\} . \end{aligned}$$

Since $\tau _{0}^{-1}\alpha _1\succeq \tau _{1}^{-1}\alpha _2\succeq \cdots \succeq \tau _{k-1}^{-1}\alpha _k$, we have

$$\begin{aligned} \lim _{n\rightarrow +\infty } \frac{-\log \mu \left( \bigvee _{i=1}^k \left( \bigvee _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{i-1}^{-1}\alpha _i \right) (x)\right) }{n} =\sum _{i=1}^k a_i{\mathbb {E}}_\mu (F_i|{\mathcal {I}}_\mu )(x) \end{aligned}$$

almost everywhere by Proposition 2 where

$$\begin{aligned} F_i(x):=I_\mu \left( \bigvee _{j=i}^{k}\alpha _j\big |\bigvee _{n=1}^\infty T^{-n}(\bigvee _{j=i}^{k}\alpha _j)\right) (x), \quad i=1,\ldots ,k \end{aligned}$$

and ${\mathcal {I}}_\mu =\{ B\in {\mathcal {B}}:\; \mu (B\triangle T^{-1}B)=0\}$. For convenience, we note

$$\begin{aligned} m(x):=\sum _{i=1}^k a_i{\mathbb {E}}_\mu (F_i|{\mathcal {I}}_\mu )(x). \end{aligned}$$

It is clear, by the construction of $\{\alpha _i\}_{i=1}^k$, that

$$\begin{aligned} \int _{X_1}m(x)d\mu (x)=\sum _{i=1}^k a_ih_{\tau _{i-1}^{-1}}(T_i,\alpha _i)\ge h_\mu ^\mathbf{a}(T_1)-(a_1+a_2+\cdots ,+a_k)\delta . \end{aligned}$$

(13)

For $N\in {\mathbb {N}}$, put $F_0^N=\{x\in X_1:m(x)\ge N\}$. By Proposition 2,

$$\begin{aligned} \lim _{N\rightarrow \infty }\int _{F_0^N}m(x)d\mu (x)=0 \end{aligned}$$

(14)

since $h_\mu ^\mathbf{a}(T_1,\alpha )\le h_\mu ^\mathbf{a}(T_1)<\infty$. Put

$$\begin{aligned} F_m^N=\left\{ x\in X_1:m(x)\in \left[ \frac{m-1}{N},\frac{m}{N}\right) \right\} \quad \text {for}\quad m=1,2,\ldots ,N^2. \end{aligned}$$

Hence we can find a large natural number $\ell _1$ such that $\mu (B_{\ell ,m}^N)>\mu (F_m^N)-\delta$ for any $\ell \ge \ell _1$ and $m\in \{1,2,\ldots ,N^2\}$, where $B_{\ell ,m}^N$ is the set of all points $x\in F_m^N$ such that

$$\begin{aligned} \frac{-\log \mu \left( \bigvee _{i=1}^k \left( \bigvee _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{m-1}^{-1}\alpha _i \right) (x)\right) }{n}\ge \frac{m-1}{N}-\delta \end{aligned}$$

(15)

for all $n\ge \ell$.

Fix $\ell \ge \max \{ \ell _0,\ell _1\}$. Let $E_m^N=A_\ell \cap B_{\ell ,m}^N$. Then $\mu (E_m^N)>\mu (F_m^N)-2\delta$. For $x\in X_1$ and $n\in {\mathbb {N}}$, the unique element

$$\begin{aligned} C(n,x)=(C_j(n,x))_{j=0}^{\lceil (a_1+\cdots +a_k)n\rceil -1} \end{aligned}$$

in $\Lambda ^{\{0,1,\ldots ,\lceil (a_1+\cdots +a_k)n\rceil -1\}}$ satisfying that $T_1^jx\in \tau _{i-1}^{-1}(A^i_{C_j(n,x)})$ for $\lceil (a_0+\cdots +a_{i-1})n\rceil \le j\le \lceil (a_1+\cdots +a_i)n\rceil -1$, $i=1,\ldots ,k$, is called the $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of x. Since each point in one atom A of $\bigvee _{i=1}^k \left( \bigvee _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{i-1}^{-1}\alpha _i \right)$ has the same $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name, we define

$$\begin{aligned} C(n,A):=C(n,x) \end{aligned}$$

for any $x\in A$, which is called the $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of A.

Now if $y\in B_n^\mathbf{a}(x,\epsilon )$, then for $i=1,\ldots ,k$ and $\lceil (a_0+\cdots +a_{i-1})n\rceil \le j\le \lceil (a_1+\cdots +a_i)n\rceil -1$, either $T_1^jx$ and $T_1^jy$ belong to the same element of $\tau _{i-1}^{-1}\alpha _i$ or $T_1^jx\in U^i_\epsilon (\alpha _i)$. Hence if $x\in E$, $n\ge \ell$ and $y\in B_n^\mathbf{a}(x,\epsilon )$, then the Hamming distance between $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of x and y does not exceed $\delta _1$. Furthermore, $B_n^\mathbf{a}(x,\epsilon )$ is contained in the set of points y whose $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name is $\delta _1$-close to $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of x. It is clear that the total number $L_n(x)$ of such $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-names admits the following estimate:

$$\begin{aligned} L_n(x)&\le \left( {\begin{array}{c}\lceil (a_1+\cdots +a_k)n\rceil \\ \lceil \lceil (a_1+\cdots +a_k)n\rceil \delta _1\rceil \end{array}}\right) M^{\lceil \lceil (a_1+\cdots +a_k)n\rceil \delta _1\rceil }\\&\le e^{\delta \lceil (a_1+\cdots +a_k)n\rceil +C}\\&\le e^{(a_1+\cdots +a_k)\delta n+C+\delta } \end{aligned}$$

where the second inequality comes from (12). More precisely, we have shown that for any $x\in X_1$ and $n\ge \ell$,

$$\begin{aligned} \begin{aligned}&B^\mathbf{a}_n(x,\epsilon )\subseteq \{ y\in X_1: C(n,y)\text { is }\delta _1 \text {-close to } C(n,x)\}\\&\quad =\bigcup \left\{ A\in \bigvee _{i=1}^k \left( \bigvee _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{i-1}^{-1}\alpha _i \right) : C(n,A)\text { is }\delta _1 \text {-close to } C(n,x)\right\} \qquad \end{aligned} \end{aligned}$$

(16)

and

$$\begin{aligned} \begin{aligned}&\#\left\{ A\in \bigvee _{i=1}^k \left( \bigvee _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{i-1}^{-1}\alpha _i \right) : C(n,A)\text { is }\delta _1 \text {-close to } C(n,x)\right\} \\&\quad \le e^{(a_1+\cdots +a_k)\delta n+C+\delta }. \end{aligned} \end{aligned}$$

(17)

Now for $n\in {\mathbb {N}},m\in \{1,2,\ldots ,N^2\}$, let $E_{n,m}^N$ denote the set of points x in $E_m^N$ such that there exists an element A in $\bigvee _{i=1}^k \left( \bigvee _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{i-1}^{-1}\alpha _i \right)$ with

$$\begin{aligned} \mu (A)>e^{\left( -\frac{m-1}{N}+(2+a_1+\cdots +a_k)\delta \right) n} \end{aligned}$$

and the $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of A is $\delta _1$-close to the $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of x. It is clear that if $x\in E_m^N\setminus E_{n,m}^N$, then for each $A\in \bigvee _{i=1}^k \left( \bigvee _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{i-1}^{-1}\alpha _i \right)$ whose $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name is $\delta _1$-close to the $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of x, one has

$$\begin{aligned} \mu (A)\le e^{\left( -\frac{m-1}{N}+(2+a_1+\cdots +a_k)\delta \right) n}. \end{aligned}$$

In the following, we wish to estimate the measure of $E_{n,m}^N$ for $n\ge \ell$.

Let $n\ge \ell$. Put

$$\begin{aligned} {\mathcal {F}}_{n,m}^N=\left\{ A\in \bigvee _{i=1}^k \left( \bigvee _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{m-1}^{-1}\alpha _i \right) : \mu (A)>e^{\left( -\frac{m-1}{N}+(2+a_1+\cdots +a_k)\delta \right) n}\right\} . \end{aligned}$$

Obviously,

$$\begin{aligned} \#{\mathcal {F}}_{n,m}^N\le e^{\left( \frac{m-1}{N}-(2+a_1+\cdots +a_k)\delta \right) n} \end{aligned}$$

since $\mu (X_1)=1$.

Let $x\in E_{n,m}^N$. On the one hand since $x\in B_{l,m}^N$,

$$\begin{aligned} \mu \left( \bigvee _{i=1}^k \left( \bigvee _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{m-1}^{-1}\alpha _i \right) (x)\right) \le e^{\left( -\frac{m-1}{N}+\delta \right) n} \end{aligned}$$

by (15). On the other hand by the definition of $E_{n,m}^N$, there exists $A\in {\mathcal {F}}_{n,m}^N$ with the $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of A is $\delta _1$-close to the $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of x, that is the $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of A is $\delta _1$-close to the $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of

$$\begin{aligned} \left( \bigvee _{i=1}^k \bigvee _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{i-1}^{-1}\alpha _i \right) (x). \end{aligned}$$

According to this, we have

$$\begin{aligned} E_{n,m}^N\subset \bigcup \{B:B\in {\mathcal {G}}_{n,m}^N\} \end{aligned}$$

(18)

where ${\mathcal {G}}_{n,m}^N$ denotes the set of all elements B in $\bigvee _{i=1}^k \left( \bigvee _{j=\lceil (a_0+\cdots +a_{m-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{i-1}^{-1}\alpha _i \right)$ satisfying $\mu (B)\le e^{\left( -\frac{m-1}{N}+\delta \right) n}$ and the $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of B is $\delta _1$-close to the $(\{\alpha _i\}_{i=1}^k,\mathbf{a};n)$-name of A for some $A\in {\mathcal {F}}_n^N$.

Since for each $A\in {\mathcal {F}}_n^N$, the total number of B in $\bigvee _{i=1}^k \left( \bigvee _{j=\lceil (a_0+\cdots +a_{i-1})n\rceil }^{\lceil (a_1+\cdots +a_i)n\rceil -1}T_1^{-j}\tau _{i-1}^{-1}\alpha _i \right)$, whose $(\{\alpha _i\},\mathbf{a};n)$-name is $\delta _1$-close to the $(\{\alpha _i\},\mathbf{a};n)$-name of A, is upper bounded by

$$\begin{aligned} \left( {\begin{array}{c}\lceil (a_1+\cdots +a_k)n\rceil \\ \lceil \lceil (a_1+\cdots +a_k)n\rceil \delta _1\rceil \end{array}}\right) M^{\lceil \lceil (a_1+\cdots +a_k)n\rceil \delta _1\rceil }\le e^{(a_1+\cdots +a_k)\delta n+C+\delta }. \end{aligned}$$

Hence

$$\begin{aligned} \#{\mathcal {G}}_{n,m}^N\le e^{(a_1+\cdots +a_k)\delta n+C+\delta }\cdot \left( \# {\mathcal {F}}_{n,m}^N\right) \le e^{\left( \frac{m-1}{N}-2\delta \right) n+C+\delta }. \end{aligned}$$

Moreover

$$\begin{aligned} \mu (E_{n,m}^N)\le e^{\left( -\frac{m-1}{N}+\delta \right) n}\cdot \left( \#{\mathcal {G}}_{n,m}^N\right) \le e^{-\delta n+C+\delta } \end{aligned}$$

by (18) and the definition of ${\mathcal {G}}_{n,m}^N$.

Next we take $\ell _2\ge \ell$ so that $\sum _{n=\ell _2}^\infty e^{-\delta n+C+\delta }<\delta$. Then $\mu (\bigcup _{n\ge \ell _2}E_{n,m}^N)<\delta$. Let $D_m^N=E_m^N\setminus \bigcup _{n\ge \ell _2}E_{n,m}^N$. Then $\mu (D_m^N)>\mu (E_m^N)-\delta >\mu (F_m^N)-3\delta$. For $x\in D_m^N$ and $n\ge \ell _2$, one has

$$\begin{aligned} \mu (B_n^\mathbf{a}(x,\epsilon ))&\le e^{(a_1+\cdots +a_k)\delta n+C+\delta }\cdot e^{\left( -\frac{m-1}{N}+(2+a_1+\cdots +a_k)\delta \right) n}\\&=e^{\left( -\frac{m-1}{N}+2(1+a_1+\cdots +a_k)\delta \right) n+C+\delta } \end{aligned}$$

by (16), (17) and the definition of $E_{n,m}^N$. Thus for $x\in D_m^N$,

$$\begin{aligned} \liminf _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}&\ge \frac{m-1}{N}-2(1+a_1+\cdots +a_k)\delta . \end{aligned}$$

Thus

$$\begin{aligned}&\int _{X_1}\liminf _{n\rightarrow +\infty } \frac{-\log \mu \left( B_n^\mathbf{a}(x,\epsilon )\right) }{n}d\mu (x)\ge \sum _{m=1}^{N^2}\int _{D_i}\liminf _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}d\mu (x)\\&\quad \ge \sum _{m=1}^{N^2}\left( \frac{m-1}{N}-2(1+a_1+\cdots +a_k)\delta \right) \cdot \mu (D_m^N)\\&\quad \ge \sum _{m=1}^{N^2}\frac{m}{N}\cdot (\mu (F_m^N)-3\delta )-2(1+a_1+\cdots +a_k)\delta -\frac{1}{N}\\&\quad >\sum _{m=1}^{N^2}\int _{F_m^N}m(x)d\mu (x)-2(1+a_1+\cdots +a_k)\delta -\frac{1}{N}-\sum _{m=1}^{N^2}\frac{3m}{N}\delta \\&\quad =\int _{X_1}m(x)d\mu (x)-\int _{F_0^N}m(x)d\mu (x)-2(1+a_1+\cdots +a_k)\delta -\frac{1}{N}-\frac{3}{2}(N^3+N)\delta \\&\quad \ge h_\mu ^\mathbf{a}(T_1)-\int _{F_0^N}m(x)d\mu (x)-3(1+a_1+\cdots +a_k)\delta -\frac{1}{N}-\frac{3}{2}(N^3+N)\delta . \end{aligned}$$

Since $\liminf _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}$ increase as $\epsilon$ decrease, then

$$\begin{aligned}&\int _{X_1}\lim _{\epsilon \rightarrow 0}\liminf _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}d\mu (x)\ge h_\mu ^\mathbf{a}(T_1)\\&\quad -\int _{F_0^N}m(x)d\mu (x)-3(1+a_1+\cdots +a_k)\delta -\frac{1}{N}-\frac{3}{2}(N^3+N)\delta . \end{aligned}$$

Let $\delta \rightarrow 0$, $N\rightarrow \infty$ successively. We have by (14)

$$\begin{aligned} \int _{X_1}\lim _{\epsilon \rightarrow 0}\liminf _{n\rightarrow +\infty } \frac{-\log \mu \left( B_n^\mathbf{a}(x,\epsilon )\right) }{n}d\mu (x)\ge h_\mu ^\mathbf{a}(T_1). \end{aligned}$$

(19)

Combining (19) with (11), one has

$$\begin{aligned} \lim _{\epsilon \rightarrow 0}&\liminf _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}d\mu (x)\nonumber \\&=\lim _{\epsilon \rightarrow 0}\limsup _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}d\mu (x) \overset{def}{=}h_\mu ^\mathbf{a}(T_1,x) \end{aligned}$$

(20)

for $\mu$-a.e. $x\in X_1$ since $h_\mu ^\mathbf{a}(T_1)<\infty$. (a) and (c) is immediately from (11),(19) and (20). Since

$$\begin{aligned} B_n^\mathbf{a}(T_1x,\delta )\supseteq B_{n+1}^\mathbf{a}(x,\delta ). \end{aligned}$$

It follows from (a) that

$$\begin{aligned} h_\mu ^\mathbf{a}(T_1,x)\ge h_\mu ^\mathbf{a}(T_1,T_1x). \end{aligned}$$

The last inequality together with (c) gives (b). This finishes the proof. $\square$

Since ${\mathcal {M}}_{T_1}$ is a compact convex set we can use the Choquet representation theorem to express each member of ${\mathcal {M}}_{T_1}$ in terms of the ergodic members of ${\mathcal {M}}_{{T_1}}$. For each $\mu \in {\mathcal {M}}_{T_1}$, there is a unique measure $\nu _\mu$ on the Borel subsets of the compact metrisable space ${\mathcal {M}}_{T_1}$ such that $\nu _\mu ({\mathcal {E}}_{{T_1}})=1$ and $\forall f\in C(X_1)$

$$\begin{aligned} \int _{X_1}f(x)d\mu _(x)=\int _{{\mathcal {E}}_{{T_1}}}\left( \int _{X_1}f(x)dm(x)\right) d\nu _\mu (m). \end{aligned}$$

We write $\mu =\int _{{\mathcal {E}}_{{T_1}}}md\nu _\mu (m)$ and call this the ergodic decomposition of $\mu$ (see [7]). Using the ergodic decomposition, we can define

$$\begin{aligned} h^\mathbf{a,u}_\mu ({T_1}):=\sup \{s\in {\mathbb {R}}:\nu _\mu \left( \{m\in {\mathcal {E}}_{{T_1}}:h^\mathbf{a}_m({T_1})>s\right) \}>0\}. \end{aligned}$$

We have the following Corollary.

Corollary 1

Let $\mu$ be invariant probability measure with $h_\mu ^\mathbf{a}(T_1)<\infty$. Then

$$\begin{aligned} {\overline{ess}}\{h_\mu ^\mathbf{a}(T_1,x):x\in X_1\}=h_\mu ^\mathbf{a,u}(T_1) \end{aligned}$$

where ${\overline{ess}}\{h_\mu ^\mathbf{a}(T_1,x):x\in X_1\}$ is the upper bound of $s\in {\mathbb {R}}$ such that

$$\begin{aligned} \mu \{x\in X_1:h_\mu ^\mathbf{a}(T_1,x)>s\}>0. \end{aligned}$$

Proof

First we are to show that $h_\mu ^\mathbf{a}(T_1,x)\le h_\mu ^\mathbf{a,u}(T_1)$ for $\mu$-a.e. $x\in X_1$. We can assume that $h_\mu ^\mathbf{a,u}(T_1)<\infty ,$ otherwise we have nothing to prove. Given positive number $\gamma >0$, put

$$\begin{aligned} E_\gamma =\left\{ x:\lim _{\epsilon \rightarrow 0} \limsup _{n\rightarrow +\infty } \frac{-\log \mu (B_n^\mathbf{a}(x,\epsilon ))}{n}>h_\mu ^\mathbf{a,u}(T_1)+\gamma \right\} . \end{aligned}$$

By Theorem 7(b), (c), it is clear that $\mu (E_\gamma \triangle T_1^{-1}E_\gamma )=0$ and $\mu (E_\gamma ^c)>0$. If $\mu (E_\gamma )>0$, define measure $\mu _1$ and $\mu _2$ by

$$\begin{aligned} \mu _1(B)=\frac{\mu (E_\gamma \cap B)}{\mu (E_\gamma )}\text { and } \mu _2(B)=\frac{\mu (E_\gamma ^c\cap B)}{\mu (E_\gamma ^c)},\ \ B\in {\mathcal {B}}(X_1). \end{aligned}$$

Note that $\mu _1,\mu _2\in {\mathcal {M}}_{T_1}$, we have the unique ergodic decomposition

$$\begin{aligned} \mu _1=\int _{{\mathcal {E}}_{T_1}}md\nu _{\mu _1}(m)\text { and } \mu _2=\int _{{\mathcal {E}}_{T_1}}md\nu _{\mu _2}(m). \end{aligned}$$

Notice that $\mu =\mu (E_\gamma )\mu _1+(1-\mu (E_\gamma ))\mu _2$, we have

$$\begin{aligned} \mu&=\mu (E_\gamma )\int _{{\mathcal {E}}_{{T_1}}}md\nu _{\mu _1}(m)+\mu (E_\gamma ^c)\int _{{\mathcal {E}}_{T_1}}md\nu _{\mu _2}(m)\\&=\int _{{\mathcal {E}}_{T_1}}md\left( \mu (E_\gamma )\nu _{\mu _1}+\mu (E_\gamma ^c)\nu _{\mu _2}\right) (m). \end{aligned}$$

Suppose the ergodic decomposition of $\mu$ is

$$\begin{aligned} \mu =\int _{{\mathcal {E}}_{T_1}}md\nu _{\mu }(m). \end{aligned}$$

(21)

By the uniqueness of ergodic decomposition, we have

$$\begin{aligned} \nu _{\mu }=\mu (E_\gamma )\nu _{\mu _1}+\mu (E_\gamma ^c)\nu _{\mu _2}. \end{aligned}$$

Hence $\nu _{\mu _1}\ll \nu _{\mu }$. Therefore

$$\begin{aligned} h_{\mu _1}^\mathbf{a}(T_1)\le h_{\mu _1}^\mathbf{a,u}(T_1)\le h_{\mu }^\mathbf{a,u}(T_1). \end{aligned}$$

(22)

However,

$$\begin{aligned} h_{\mu _1}^\mathbf{a}(T_1)&=\int _{X_1}\lim _{\epsilon \rightarrow 0} \limsup _{n\rightarrow +\infty }\frac{-\log \mu _1\left( B_n^\mathbf{a}(x,\epsilon )\right) }{n}d\mu _1(x)\\&\ge \int _{X_1}\lim _{\epsilon \rightarrow 0} \limsup _{n\rightarrow +\infty }\frac{-\log \frac{1}{\mu (E_\gamma )}\mu \left( B_n^\mathbf{a}(x,\epsilon )\right) }{n}d\mu _1(x)\\&=\int _{X_1}\lim _{\epsilon \rightarrow 0} \limsup _{n\rightarrow +\infty }\frac{-\log \mu \left( B_n^\mathbf{a}(x,\epsilon )\right) }{n}d\mu _1(x)\ge h_{\mu }^\mathbf{a,u}+\gamma . \end{aligned}$$

It would be in contradiction with (22). Hence $h_\mu ^\mathbf{a}(T_1,x)\le h_\mu ^\mathbf{a,u}(T_1)+\gamma$. We have $h_\mu ^\mathbf{a}(T_1,x)\le h_\mu ^\mathbf{a,u}(T_1)$ by taking $\gamma \searrow 0$.

Next, we are to show ${\overline{ess}}\{h_\mu ^\mathbf{a}(T_1,x):x\in X_1\}\ge h_\mu ^\mathbf{a,u}(T_1)$. As in (21), we have the ergodic decomposition $\nu _\mu$ of $\mu$. If $h_m^\mathbf{a}(T_1)=h_\mu ^\mathbf{a,u}(T_1)$, for $\nu _\mu$-a.e. $m\in {\mathcal {E}}_{T_1}$, we have

$$\begin{aligned} h_\mu ^\mathbf{a,u}(T_1)=h_\mu ^\mathbf{a}(T_1)=\int _{X_1}h_\mu ^\mathbf{a}(T_1,x)d\mu (x)<\infty . \end{aligned}$$

Since $h_\mu ^\mathbf{a}(T_1,x)\le h_\mu ^\mathbf{a,u}(T_1)$ for $\mu$-a.e. $x\in X_1$, it is clear that $h_\mu ^\mathbf{a}(T_1,x)=h_\mu ^\mathbf{a,u}(T_1)$ for $\mu$-a.e. $x\in X_1$.

Now, we suppose

$$\begin{aligned} \nu _\mu \left\{ m\in {\mathcal {E}}_{T_1}:h_m^\mathbf{a}(T_1)<h_\mu ^\mathbf{a,u}(T_1)\right\} >0. \end{aligned}$$

Clearly, in this case, there exists $0<s_0<h_\mu ^\mathbf{a,u}(T_1)$ such that

$$\begin{aligned} \nu _\mu \left( \left\{ m\in {\mathcal {E}}_{T_1}:h_m^\mathbf{a}(T_1)<s_0\right\} \right) >0. \end{aligned}$$

For $s_0<s<h_\mu ^\mathbf{a,u}(T_1)$. Put ${\mathcal {E}}=\{m\in {\mathcal {E}}_{T_1}:h_m^\mathbf{a}(T_1)\le s\}$ and define measure $\rho _1$ and $\rho _2$ by

$$\begin{aligned} \rho _1=\frac{\int _{{\mathcal {E}}}md\nu _{\mu }(m)}{\nu _\mu ({\mathcal {E}})}\text { and } \rho _2=\frac{\int _{{\mathcal {E}}^c}md\nu _{\mu }(m)}{\nu _\mu ({\mathcal {E}}^c)}. \end{aligned}$$

$\rho _1$ and $\rho _2$ is well defined since both $\nu _\mu ({\mathcal {E}})$ and $\nu _\mu ({\mathcal {E}}^c)$ are positive. It is clear that

$$\begin{aligned} \mu =\nu _\mu ({\mathcal {E}})\rho _1+\nu _\mu ({\mathcal {E}}^c)\rho _2. \end{aligned}$$

On the one hand

$$\begin{aligned} h_{\mu }^\mathbf{a}(T_1)= \nu _\mu ({\mathcal {E}})h_{\rho _1}^\mathbf{a}(T_1)+\nu _\mu ({\mathcal {E}}^c)h_{\rho _2}^\mathbf{a}(T_1). \end{aligned}$$

(23)

On the other hand

$$\begin{aligned} h_{\mu }^\mathbf{a}(T_1)&= \int _{X_1}h_\mu ^\mathbf{a}(T_1,x)d\mu (x)=\int _{X_1}h_\mu ^\mathbf{a}(T_1,x)d\left( \nu _\mu ({\mathcal {E}})\rho _1+\nu _\mu ({\mathcal {E}}^c)\rho _2\right) (x)\\&=\nu _\mu ({\mathcal {E}})\int _{X_1}h_\mu ^\mathbf{a}(T_1,x)d\rho _1(x)+\nu _\mu ({\mathcal {E}}^c)\int _{X_1}h_\mu ^\mathbf{a}(T_1,x)d\rho _2(x). \end{aligned}$$

For $\rho _1$-a.e. $x\in X_1$

$$\begin{aligned} h_\mu ^\mathbf{a}(T_1,x)&=\lim _{\epsilon \rightarrow 0} \limsup _{n\rightarrow +\infty }\frac{-\log \mu \left( B_n^\mathbf{a}(x,\epsilon )\right) }{n}\\&=\lim _{\epsilon \rightarrow 0} \limsup _{n\rightarrow +\infty }\frac{-\log \left( \left( \nu _\mu ({\mathcal {E}})\rho _1+\nu _\mu ({\mathcal {E}}^c)\rho _2\right) \left( B_n^\mathbf{a}(x,\epsilon )\right) \right) }{n}\\&\le \lim _{\epsilon \rightarrow 0} \limsup _{n\rightarrow +\infty }\frac{-\log \left( \nu _\mu ({\mathcal {E}})\rho _1\left( B_n^\mathbf{a}(x,\epsilon )\right) \right) }{n}\\&=h_{\rho _1}^\mathbf{a}(T_1,x). \end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned} h_{\mu }^\mathbf{a}(T_1)&=\nu _\mu ({\mathcal {E}})\int _{X_1}h_\mu ^\mathbf{a}(T_1,x)d\rho _1(x)+\nu _\mu ({\mathcal {E}}^c)\int _{X_1}h_\mu ^\mathbf{a}(T_1,x)d\rho _2(x)\\&\le \nu _\mu ({\mathcal {E}})\int _{X_1}h_{\rho _1}^\mathbf{a}(T_1,x)d\rho _1(x)+\nu _\mu ({\mathcal {E}}^c)\int _{X_1}h_\mu ^\mathbf{a}(T_1,x)d\rho _2(x)\\&=\nu _\mu ({\mathcal {E}})h_{\rho _1}^\mathbf{a}(T_1)+\nu _\mu ({\mathcal {E}}^c)\int _{X_1}h_\mu ^\mathbf{a}(T_1,x)d\rho _2(x). \end{aligned} \end{aligned}$$

(24)

Combining (24) and (23), one has

$$\begin{aligned} \int _{X_1}h_\mu ^\mathbf{a}(T_1,x)d\rho _2(x)\ge h_{\rho _2}^\mathbf{a}(T_1)>s. \end{aligned}$$

Hence

$$\begin{aligned} \rho _2\left( \{x\in X_1:h_\mu ^\mathbf{a}(T_1,x)>s\}\right) >0 \end{aligned}$$

which implies

$$\begin{aligned} \mu \left( \{x\in X_1:h_\mu ^\mathbf{a}(T_1,x)>s\}\right) \ge \nu _\mu ({\mathcal {E}}^c)\rho _2\left( \left\{ x\in X_1:h_\mu ^\mathbf{a}(T_1,x)>s\right\} \right) >0. \end{aligned}$$

Hence ${\overline{ess}}\{h_\mu ^\mathbf{a}(T_1,x):x\in X_1\}>s.$ We have

$$\begin{aligned} {\overline{ess}}\{h_\mu ^\mathbf{a}(T_1,x):x\in X_1\}\ge h_\mu ^\mathbf{a,u}(T_1) \end{aligned}$$

by taking $s\rightarrow h_\mu ^\mathbf{a,u}(T_1)$. This completes the proof. $\square$