Proof. (1) From the definition, we have $\overline {\mathrm {Per}(T)}=X$ and thus by [Reference Denker, Grillenberger and Sigmund8, Proposition 21.12], $C_T(X)=X$ .

(2) When $\sup _{\varepsilon>0}\lim \nolimits _{n\to \infty } ({M(n,\varepsilon )}/{n})=0$ , the result is directly from [Reference Kwietniak, Ła̧cka, Oprocha, Kolyada, Möller, Moree and Ward16, Theorem 17(4)]. In fact, their proof still works without the assumption that $\sup _{\varepsilon>0}\lim \nolimits _{n\to \infty } ({M(n,\varepsilon )}/{n})=0$ , and we put it here. Given a pair of non-empty open subsets U and V of X, there exist $x{\kern-1pt}\in{\kern-1pt} U$ , $y{\kern-1pt}\in{\kern-1pt} V$ , and $\varepsilon{\kern-1pt}>{\kern-1pt}0$ such that $B(x,2\varepsilon ){\kern-1pt}\subset{\kern-1pt} U$ and $B(y,2\varepsilon )\subset V$ . Let $N=M(1,\varepsilon )$ , then for every $n\geq N$ , there exists $z\in X$ such that $d(x,z)\leq \varepsilon $ and $d(y,T^nz)\leq \varepsilon $ . Hence, $f^n(U)\cap V\neq \emptyset $ . As a result, $(X,T)$ is topologically mixing.

(3) Suppose that c is a pseudo-expansive constant for $(X,T)$ , then $\mathrm {Per}_n(T)$ is $(n,c)$ -separated, and hence $h_{\mathrm {top}}(T)\geq \limsup \nolimits _{n\to \infty } ({1}/{n}) \log s(n,c)\geq \limsup \nolimits _{n\to \infty }({1}/{n})\log |\mathrm {Per}_{n}(T)|.$

(4) (i) Since $G^K\neq \emptyset $ , we can choose $y\in G^K$ . Given $\varepsilon>0$ and $\mu \in K$ , by Lemma 3.1, to show that $G^K$ is residual in X, it is enough to show that $\{x{\kern-1pt}\in{\kern-1pt} X{\kern-1pt}:{\kern-1pt}V_T(x){\kern-1pt}\cap{\kern-1pt} B(\mu ,\varepsilon ){\kern-1pt}\neq{\kern-1pt} \emptyset \}$ is dense in X. Given $x\in X$ and $0<\delta < {\varepsilon }/{2}$ , let $N=M(1,\delta )$ . For any $n\in \mathbb {N}$ , by non-uniform periodic specification property, we can find $z_n\in X$ such that $d(x,z_n)\leq \delta $ and $d(T^iy,T^{N+i}z_n)\leq \delta $ for any $0\leq i\leq n-1$ . Since X is compact, we can choose an accumulation point z of $\{z_n:n\geq 1\}$ . Then $d(x,z)\leq \delta $ and $d(T^iy,T^{N+i}z)\leq \delta $ for any $i\geq 0$ . Hence, $\rho (\delta _{T^Nz}^n,\delta _y^n)\leq \delta $ for any $n\geq 1$ . As a result, $V_T(T^Nz)\cap B(\mu ,\varepsilon )\neq \emptyset $ and thus $V_T(z)\cap B(\mu ,\varepsilon )\neq \emptyset $ . Therefore, $\{x\in X:V_T(x)\cap B(\mu ,\varepsilon )\neq \emptyset \}$ is dense in X.

(ii) When $(X,T)$ is pseudo-expansive, let c be a pseudo-expansive constant. Since $G_K\neq \emptyset $ , we can choose $y\in G_K$ . Given $x\in X$ and $0<\delta \leq c$ , let $N=M(1,\delta )$ . Following the same line in part (i), we can find $z\in X$ such that $d(z,x)\leq \delta $ and $d(T^iy,T^{N+i}z)\leq \delta $ for any $i\geq 0$ . By the pseudo-expansiveness of $(X,T)$ , $\lim \nolimits _{n\to \infty }d(T^{N+n}z,T^ny)=0$ , since $y\in G_K$ , we have $T^Nz\in G_K$ . Hence, $z\in G_K$ . As a result, $G_K$ is dense in X.

(5) (i) Since $|X|\geq 2$ , we can choose $\varepsilon>0$ and $x,y\in X$ with $d(x,y)>3\varepsilon $ . Given $n\in \mathbb {N}$ and an n-tuple $(z_0,z_1,\ldots ,z_{n-1})\in \{x,y\}^n$ , let $N=M(1,\varepsilon )$ . Then there exists $z\in X$ with $d(z_i,T^{iN}z)\leq \varepsilon $ for any $0\leq i\leq n-1$ . Hence, for every two different n-tuples, the points we find are $(nN,\varepsilon )$ -separated. As a result, $s(nN,\varepsilon )\geq 2^n$ . Therefore,

$$ \begin{align*} h_{\mathrm{top}}(T)\geq\limsup_{n\to\infty}\frac{1}{nN}\log s(nN,\varepsilon)\geq\lim_{n\to\infty}\frac{1}{nN}\cdot n\log2>0.\end{align*} $$

(ii) Given $\delta>0$ , by item (2), $(X,T)$ is topologically mixing. Choose a transitive point $z_\ast \in X$ and $N_0\in \mathbb {N}$ with

$$ \begin{align*}d_H(\{z_\ast,Tz_\ast,\ldots,T^{N_0-1}z_\ast\},X)<\frac{\delta}{2}.\end{align*} $$

Since $|X|\geq 2$ , we can choose $0<\varepsilon < {\delta }/{2}$ and $x,y\in X$ with $d(x,y)>3\varepsilon $ . Let ${N_1=M(1,\varepsilon )}$ and $N_2=M(N_0,\varepsilon )$ , then given $n\in \mathbb {N}$ and a $(n+1)$ -tuple $(z_0,z_1,\ldots , z_{n-1},z_\ast )$ , there exists $z\in X$ with $d(z_i,T^{iN_1}z)\leq \varepsilon $ for any $0\leq i\leq n-1$ , $d(T^jz_\ast , T^{nN_1+j}z) \leq \varepsilon $ for any $0\leq j\leq N_0-1$ and $T^{m_n}z=z$ , where $m_n=nN_1+N_0-1+N_2$ . Hence, $z\in \mathrm {Per}_{m_n,\delta }(T)$ . Since for every two different $(n+1)$ -tuples, the points we find are different, we have $|\mathrm {Per}_{m_n,\delta }(T)|\geq ({1}/{m_n}) 2^n$ . As a result,

$$ \begin{align*} \limsup_{n\to\infty}\frac{1}{n}\log|\mathrm{Per}_{n,\delta}(T)|&\geq\limsup_{n\to\infty} \frac{1}{m_n}\log|\mathrm{Per}_{m_n,\delta}(T)|\\ &\geq\limsup_{n\to\infty} \bigg(\frac{n}{m_n}\log2-\frac{1}{m_n}\log m_n\bigg)>0. \end{align*} $$

(6) By item (5), $h_{\mathrm {top}}(T)>0$ and thus by Lemma 2.14, $(X,T)$ is DC2 chaotic.

(7) Since $h_{\mathrm {top}}(T)>0$ , we have that $|X|=\infty $ . By item (2), $(X,T)$ is topological mixing and thus topological transitive. Combining with item (1), we have that $(X,T)$ is Devaney chaotic.

(8) By item (1), $\overline {\mathrm {Per}(T)}=X$ , and combining with $|X|=\infty $ , we have that ${|\mathcal {M}_T^{\mathrm {erg}}(X)|\geq 2}$ . Since $\overline {\mathcal {M}_T^{\mathrm {erg}}(X)}$ is compact, we can choose a countable subset $D=\{\mu _1,\mu _2,\ldots ,\}\subset \mathcal {M}_T^{\mathrm {erg}}(X)$ with $\overline {D}=\overline {\mathcal {M}_T^{erg}(X)}$ . By Theorem 3.2(4), $G^{\mu _i}$ is residual in X for any $i\geq 1$ , and hence, $G^{D}$ is residual in X. Since $V_T(x)$ is compact for any $x\in X$ , we have that

$$ \begin{align*}G^{\overline{\mathcal{M}_T^{\mathrm{erg}}(X)}}=G^{\overline{D}}=G^D.\end{align*} $$

As a result, $G^{\overline {\mathcal {M}_T^{\mathrm {erg}}(X)}}$ is residual in X. Since $|\mathcal {M}_T^{\mathrm {erg}}(X)|\geq 2$ , we can choose $f\in C(X)$ and $\nu _1,\nu _2\in \mathcal {M}_T^{\mathrm {erg}}(X)$ with $\int _Xf\,{d}\nu _1<\int _Xf\,{d}\nu _2$ , and hence, $\emptyset \neq G^{\overline {\mathcal {M}_T^{\mathrm {erg}}(X)}}\subset I_f(T)$ and thus $\mathcal {C}(T)\neq \emptyset $ . By Lemma 2.16, $\bigcap _{f\in \mathcal {C}(T)}I_f(T)$ is residual in X.

Proof. (1) First, we show that $(X,T)$ is $\mathrm {Per}(T)$ -closable. Fix $\mu \in \mathcal {M}_T^{\mathrm {erg}}(X)$ , $x\in G_\mu $ , ${\varepsilon>0}$ , and $N>0$ . Choose $p\geq N$ large enough such that

(3.2) $$ \begin{align} \frac{M(p, {\varepsilon}/{2})}{p}<\frac{\varepsilon}{2}\quad\text{and}\quad \frac{M(1,{\varepsilon}/{2})}{p}<\frac{\varepsilon}{2}. \end{align} $$

Let $x_1=x_2=x$ , $a_1=0$ , $b_1=p-1$ , $b_2=a_2=M(p,{\varepsilon }/{2})+p-1$ , and $q=M(p, {\varepsilon }/{2}) +p-1+M(1, {\varepsilon }/{2})$ , then by the non-uniform periodic specification property, there exists $y\in \mathrm {Per}_q(T)$ such that $y\in B_p(x,\varepsilon )$ . From equation (3.2), we have $q\leq (1+\varepsilon )p$ . Therefore, x is $\mathrm {Per}(T)$ -closable. From the definition, we have that $(X,T)$ is $\mathrm {Per}(T)$ -closable.

(2) Now, we show that $\mathrm {Per}(T)$ is linkable. Fix $y_1,y_2\in \mathrm {Per}(T)$ and $\unicode{x3bb} \in [0,1]$ .

(i) $\unicode{x3bb} =0$ , fix $\varepsilon>0$ , choose $p_1\in \mathbb {N}$ with ${M(p_1, {\varepsilon }/{2})}/{p_1}<\varepsilon $ , and choose $p_2$ large enough such that ${M(p_2,{\varepsilon }/{2})}/{p_2} < \varepsilon $ and ${p_1}/({p_1+p_2})\leq \varepsilon $ . Let $x_1=y_1,x_2=y_2, a_1\ =\ 0,\ b_1\ =\ p_1\ -\ 1,\ q_1\ =\ p_1 - 1\ +\ M(p_1,\ {\varepsilon }/{2}),\ a_2\ =\ q_1,\ b_2\ =\ q_1\ +\ p_2\ -\ 1, q_2 = q_1 + p_2-1 + M(p_2, {\varepsilon }/{2})$ , then $p_j\leq q_j-q_{j-1}\leq (1+\varepsilon )p_j$ and by the non-uniform periodic specification property, there exists $z\in \mathrm {Per}_{q_2}(T)$ such that $T^{q_{j-1}}z\in B_{p_j}(y_j,\varepsilon )$ for $q_0=0$ , $j=1,2$ .

(ii) $\unicode{x3bb} =1$ , fix $\varepsilon>0$ , choose $p_2\in \mathbb {N}$ with $({M(p_2,{\varepsilon }/{2})}/{p_2})<\varepsilon $ , and choose $p_1$ large enough such that $({M(p_1,{\varepsilon }/{2})}/{p_1})<\varepsilon $ and $1-\varepsilon \leq {p_1}/({p_1+p_2})$ . Let $x_1=y_1, x_2 = y_2, a_1\ =\ 0,\ b_1\ =\ p_1\ -\ 1,\ q_1\ =\ p_1 - 1\ +\ M(p_1,\ {\varepsilon }/{2}),\ a_2\ =\ q_1,\ b_2\ =\ q_1\ +\ p_2\ -\ 1,$ $q_2=q_1 + p_2-1+M(p_2, {\varepsilon }/{2})$ , then $p_j\leq q_j-q_{j-1}\leq (1+\varepsilon )p_j$ and by the non-uniform periodic specification property, there exists $z\in \mathrm {Per}_{q_2}(T)$ such that $T^{q_{j-1}}z\in B_{p_j}(y_j,\varepsilon )$ for $q_0=0$ , $j=1,2$ .

(iii) $0<\unicode{x3bb} <1$ , fix $0<\varepsilon <\min \{\unicode{x3bb} ,1-\unicode{x3bb} \}$ and choose $q\in \mathbb {N}$ with

(3.3) $$ \begin{align} \frac{1}{q}\leq\min \bigg\{\frac{1}{\unicode{x3bb}+\varepsilon}-1,\frac{1}{2({1}/({\unicode{x3bb}+\varepsilon})-1)}\bigg\}. \end{align} $$

Choose $p\in \mathbb {N}$ large enough such that

$$ \begin{align*} \bigg(\frac{1}{\unicode{x3bb}+\varepsilon}-1\bigg)\bigg\lceil\frac{p}{q}\bigg\rceil+1\leq \bigg(\frac{1}{\unicode{x3bb}-\varepsilon}-1\bigg)\bigg\lceil\frac{p}{q}\bigg\rceil, \end{align*} $$

$$ \begin{align*} \bigg(\frac{1}{\unicode{x3bb}+\varepsilon}-1\bigg)\frac{p}{q}+\frac{1}{\unicode{x3bb}+\varepsilon}\leq p\quad\text{and}\quad\frac{M(p,{\varepsilon}/{2})}{p}<\frac{\varepsilon}{q^2}. \end{align*} $$

Let $p_1=\lceil {p}/{q}\rceil \leq p$ , then ${M(p_1,{\varepsilon }/{2})}/{p_1}<\varepsilon $ . Let $p_2=\lceil ({1}/({\unicode{x3bb} +\varepsilon })-1)\lceil {p}/{q}\rceil \rceil $ , then $p_2\geq ({1}/({\unicode{x3bb} +\varepsilon })-1)({p}/{q})\geq {p}/{q^2}$ and

$$ \begin{align*} p_2\leq \bigg(\frac{1}{\unicode{x3bb}+\varepsilon}-1\bigg) \bigg\lceil\frac{p}{q}\bigg\rceil+1\leq \bigg(\frac{1}{\unicode{x3bb}+\varepsilon}-1\bigg)\bigg(\frac{p}{q}+1\bigg)+1\leq p. \end{align*} $$

Hence, ${M(p_2,{\varepsilon }/{2})}/{p_2}<\varepsilon $ . Since

$$ \begin{align*} \frac{1}{\unicode{x3bb}+\varepsilon}-1\leq \frac{p_2}{p_1}\leq\frac{1}{p_1}\bigg[\bigg(\frac{1}{\unicode{x3bb}+\varepsilon}-1\bigg) p_1+1\bigg]\leq\frac{1}{\unicode{x3bb}-\varepsilon}-1, \end{align*} $$

we have $\unicode{x3bb} -\varepsilon \leq {p_1}/({p_1+p_2}) \leq \unicode{x3bb} +\varepsilon $ . Let $x_1=y_1,x_2=y_2,a_1=0,b_1=p_1-1, q_1=p_1-1+M(p_1,{\varepsilon }/{2}),a_2=q_1,b_2= q_1+ p_2-1,q_2\kern1.8pt{=}\kern1.8pt q_1\kern1.7pt{+}\kern1.7pt p_2\kern1.8pt{-}\kern1.8pt 1+ M(p_2,{\varepsilon }/{2})$ , then $p_j\leq q_j-q_{j-1}\leq (1+\varepsilon )p_j$ and by the non-uniform periodic specification property, there exists $z\in \mathrm {Per}_{q_2}(T)$ such that $T^{q_{j-1}}z\in B_{p_j}(y_j,\varepsilon )$ for $q_0=0$ , $j=1,2$ .

Combining parts (i), (ii), and (iii), we have that $\mathrm {Per}(T)$ is linkable.

Proof. By Lemma 3.3, $(X,T)$ is $\mathrm {Per}(T)$ -closable and $\mathrm {Per}(T)$ is linkable. By Lemma 2.8, $\mathcal {M}_T^{\mathrm {co}}(X)$ is dense in $\mathcal {M}_T^{\mathrm {erg}}(X)$ . Since $C_T(X)=X$ , by Lemma 2.10, we have:

1. (1) $\mathcal {M}_T^{\mathrm {co}}(X)$ is dense in $\mathcal {M}_T(X)$ ;

2. (2) for any non-empty compact connected subset $V\subset \mathcal {M}_T(X)$ , $G_V$ is dense in X.

Proof. When $\sup _{\varepsilon>0}\lim \nolimits _{n\to \infty }({M(n,\varepsilon )}/{n})=0$ , the result is directly from [Reference Kwietniak, Ła̧cka, Oprocha, Kolyada, Möller, Moree and Ward16, Remark 29]. In fact, when we only assume that $\sup _{\varepsilon>0}\liminf \nolimits _{n\to \infty } ({M(n,\varepsilon )}/{n})=0$ , the result still holds. Here is the proof.

Fix $\varepsilon>0$ , $\delta _1>0$ , and $\delta _2>0$ . Since $\liminf \nolimits _{n\rightarrow \infty } ({M(n,\varepsilon )}/{n})=0$ , we can choose $L_1$ large enough such that

$$ \begin{align*} \frac{M(L_1,\varepsilon)-1}{L_1+M(L_1,\varepsilon)-1}\leq\delta_1. \end{align*} $$

Let $L_2=L_1+M(L_1,\varepsilon )-1$ and $L_3=\max _{1\leq i\leq L_2}M(i,\varepsilon )=M(L_2,\varepsilon )$ . Choose ${N=N(\varepsilon ,\delta _1,\delta _2)}$ large enough such that ${L_3}/{N}\leq \delta _2$ . Now, we fix $n\geq N$ and a sequence $\{x_i\}_{i=1}^\infty $ . Suppose that $n-1=kL_2+p$ , where $0\leq p\leq L_2-1$ . For $i\in \mathbb {N}$ and ${0\leq j\leq k}$ , denote $y_{i,j}=T^{jL_2}x_i$ . Then for any $m\in \mathbb {N}$ , consider the orbit segments:

$$ \begin{align*} \langle y_{1,0}, Ty_{1,0}, &\ldots,T^{L_1-1}y_{1,0}\rangle, \langle y_{1,1}, Ty_{1,1}, \ldots,T^{L_1-1}y_{1,1}\rangle,\ldots,\\ &\langle y_{1,k-1},Ty_{1,k-1},\ldots, T^{L_1-1}y_{1,k-1}\rangle, \langle y_{1,k},Ty_{1,k},\ldots,T^{p}y_{1,k}\rangle; \end{align*} $$

$$ \begin{align*} \langle y_{2,0}, Ty_{2,0}, &\ldots,T^{L_1-1}y_{2,0}\rangle, \langle y_{2,1}, Ty_{2,1}, \ldots,T^{L_1-1}y_{2,1}\rangle,\ldots,\\ &\langle y_{2,k-1},Ty_{2,k-1},\ldots, T^{L_1-1}y_{2,k-1}\rangle, \langle y_{2,k},Ty_{2,k},\ldots,T^{p}y_{2,k}\rangle; \end{align*} $$

$$ \begin{align*}\cdots\end{align*} $$

$$ \begin{align*}\langle y_{m,0}, Ty_{m,0}, &\ldots,T^{L_1-1}y_{m,0}\rangle, \langle y_{m,1}, Ty_{m,1}, \ldots,T^{L_1-1}y_{m,1}\rangle ,\ldots,\\ &\langle y_{m,k-1},Ty_{m,k-1},\ldots, T^{L_1-1}y_{m,k-1}\rangle, \langle y_{m,k},Ty_{m,k},\ldots,T^{p}y_{m,k}\rangle. \end{align*} $$

By the non-uniform specification property, there exists $z_m\in X$ such that for $1\leq i\leq m$ , $0\leq j\leq k-1$ , and $0\leq l\leq L_1-1$ , we have

$$ \begin{align*} d(T^{jL_2+(i-1)(n-1+M(p+1,\varepsilon))+l}z_m,T^ly_{i,j})\leq\varepsilon \end{align*} $$

and for $1\leq i\leq m$ and $0\leq l\leq p$ , we have

$$ \begin{align*} d(T^{kL_2+(i-1)(n-1+M(p+1,\varepsilon))+l}z_m,T^ly_{i,k})\leq\varepsilon. \end{align*} $$

For $i\in \mathbb {N}$ , denote $h_i=(i-1)(n-1+M(p+1,\varepsilon ))$ , then $h_1=0$ and

$$ \begin{align*}n\leq h_{i+1}-h_{i}=n-1+M(p+1,\varepsilon)\leq n-1+L_3\leq n(1+\delta_2).\end{align*} $$

Since X is compact, we can choose z as an accumulation point of $\{z_m:m\geq 1\}$ . Then for any $i\geq 1$ ,

$$ \begin{align*} &|\{0\leq j\leq n-1: d(T^{h_i+j}z,T^jx_i)>\varepsilon\}|\leq k(M(L_1,\varepsilon)-1)=\frac{k(M(L_1,\varepsilon)-1)}{n}n\\&\quad =\frac{k(M(L_1,\varepsilon)-1)}{k(L_1+M(L_1,\varepsilon)-1)+p+1}n\leq \frac{M(L_1,\varepsilon)-1)}{L_1+M(L_1,\varepsilon)-1}n\leq\delta_1n. \end{align*} $$

Therefore, $(X,T)$ satisfies the approximate product property.

Proof. Combining with Lemmas 3.5 and 2.22 completes the proof.

Proof. We have shown that

$$ \begin{align*} \{\mu\in\mathcal{M}_T(X):G_{\mu}\neq\emptyset\}\subseteq\overline{\mathcal{M}_T^{\mathrm{co}}(X)} \end{align*} $$

in the proof of Theorem A (see equation (4.6)) and thus

$$ \begin{align*} \mathcal{U}\subseteq\{\mu\in\mathcal{M}_T(X):G_{\mu}=\emptyset\}. \end{align*} $$

It suffices to show that

(5.16) $$ \begin{align} \overline{\mathcal{M}_T^{\mathrm{co}}(X)}\subseteq\{\mu\in\mathcal{M}_T(X):G_{\mu}\neq\emptyset\}. \end{align} $$

Arbitrarily given $\mu \in \overline {\mathcal {M}_T^{\mathrm {co}}(X)}$ , we wish to show that $G_{\mu }\neq \emptyset $ . We only need consider the case that $\mu \in \overline {\mathcal {M}_T^{\mathrm {co}}(X)}\setminus \mathcal {M}_T^{\mathrm {co}}(X)$ .

Since $\mathrm {Per}_1(T)$ is closed, we can choose $y^{(n)}\in X$ with minimal period $p_n\geq 2$ such that $\nu _n=\delta _{y^{(n)}}^{p_n}\rightarrow \mu $ . Let $d_n=\rho (\nu _n,\mu )$ . Denote by $B_n$ the closed ball in $\mathcal {M}_T(X)$ with center $\nu _n$ and radius $r_n=d_n+2^{-n}$ .

It is easy to check that the sequence $\{B_n:n\in \mathbb {N}\}$ satisfies the following:

1. (a) $B_n\cap B_{n+1}\cap \{\mu \}\neq \emptyset $ ;

2. (b) $\bigcap \limits \nolimits _{k\geq 1}\bigcup \nolimits _{n\geq k}B_n=\{\mu \}$ ;

3. (c) $\lim \nolimits _{n\rightarrow \infty }r_n=0$ .

Without loss of generality, for every $n\in \mathbb {N}$ , we assume that

$$ \begin{align*} y^{(n)}=(1^{s_1(n)}0^{t_1(n)}\cdots 1^{s_{k(n)}(n)}0^{t_{k(n)}(n)})^{\infty}, \end{align*} $$

where $s_i(n),t_i(n)>0$ and $cs_i(n)\leq t_i(n)$ for $1\leq i\leq k(n)$ . Define

$$ \begin{align*} w^{(n)}=1^{s_1(n)}0^{t_1(n)}\cdots1^{s_{k(n)}(n)}0^{t_{k(n)}(n)}\in\mathcal{L}_{p_n}(X). \end{align*} $$

Let $M_j$ be sufficiently large such that $\mathrm {diam}[w]<2^{-j-2}$ for all $w\in \mathcal {L}_{M_j}(X)$ . Choose $N_j$ sufficiently large such that

(5.17) $$ \begin{align} N_{j+1}>4^{j+2}\bigg(\sum_{i=1}^j(p_iN_i+2M_i+p_{i+1})\bigg) \quad\text{and}\quad\frac{M_j+p_{j+1}}{N_j}<4^{-j-1} \end{align} $$

for $j\in \mathbb {N}$ . Let $0\leq R_j<p_{j+1}$ such that $p_{j+1}$ divides $\sum \nolimits _{i=1}^j(p_iN_i+2M_i+R_i)$ and define

(5.18) $$ \begin{align} x=(w^{(1)})^{N_1}y^{(1)}_0\cdots y^{(1)}_{M_1-1}0^{M_1+R_1} (w^{(2)})^{N_2}y^{(2)}_0\cdots y^{(2)}_{M_2-1}0^{M_2+R_2}\cdots\!. \end{align} $$

Clearly, $x\in X$ . We introduce x in another point if view. Let

$$ \begin{align*} & a_1=0 & & b_1=p_1N_1 &\\ & a_2=p_1N_1+2M_1+R_1 & & b_2=a_2+p_2N_2 &\\ & \cdots &\\ & a_{n+1}=\sum_{j=1}^n(p_jN_j+2M_j+R_j) & & b_{n+1}=a_{n+1}+p_{n+1}N_{n+1}. & \end{align*} $$

It is easy to see that $a_{n+1}=b_n+2M_n+R_n$ . Hence, by equation (5.17), we can easily see that

(5.19) $$ \begin{align} \frac{a_{n+1}-b_n+p_{n+1}}{N_n} \leq\frac{a_{n+1}-b_n+a_n}{N_n}\leq 4^{-n} \quad\text{and}\quad\frac{a_n}{b_n}\leq4^{-n-1}. \end{align} $$

Let

$$ \begin{align*} \begin{cases} x_n=y^{(j)}_{n} & \text{if } a_j\leq n\leq b_j+M_j-1\text{ for some } j\in\mathbb{N},\\ x_n=0 & \text{if } b_j+M_j\leq n<a_{j+1}\text{ for some } j\in\mathbb{N}, \end{cases} \end{align*} $$

and define $x=x_0x_1\cdots $ . It is easy to check that the definition above is equivalent to equation (5.18). Note that

(5.20) $$ \begin{align} d(T^nx,T^ny^{(j)})<2^{-j-2}\quad\text{for } a_j\leq n\leq b_j-1. \end{align} $$

Now we will show that $\delta _x^k\rightarrow \mu $ . Let $i=i(k)$ be the positive integer such that ${b_i\leq k<b_{i+1}}$ . Define

$$ \begin{align*} \alpha_k= \begin{cases} \displaystyle\frac{b_i-a_i}{b_i-a_i+k-r-a_{i+1}} & \text{if } k>a_{i+1},\\ 1 & \text{if } k\leq a_{i+1}, \end{cases} \end{align*} $$

where $0\leq r<p_{i+1}$ is an integer such that $p_{i+1}$ divides $k-r-a_{i+1}$ when $k>a_{i+1}$ . Define $\gamma _k=\alpha _k\nu _i+(1-\alpha _k)\nu _{i+1}$ . Arbitrarily given $f\in C(X)$ with $\|f\|\leq 1$ , we claim that

$$ \begin{align*} \lim_{k\rightarrow\infty}\bigg|{\int_X}f\,{d}\gamma_k-\int_Xf\,{d}\delta_x^k\bigg| = 0. \end{align*} $$

For any $\varepsilon{\kern-1.7pt}>{\kern-1.7pt}0$ , where we denote by $\omega_f(\varepsilon)$ the oscillation ${\max\{|f(y{\kern-1pt}){\kern-1.5pt}-{\kern-1.5pt}f(z{\kern-1pt})|{\kern-1pt}:{\kern-1pt}d(y,{\kern-1pt}z){\kern-1pt}\leq{\kern-1pt}\varepsilon\}}$ , choose a positive integer K sufficiently large such that for any $k\geq K$ , $i=i(k)$ , the following hold:

• • $4^{-i+1}<\varepsilon $ ;

• • $\omega _f(2^{-i-3})\leq \omega _f(2^{-i-2})< {\varepsilon }/{4}$ ;

• • ${M_i}/{N_i}< {\varepsilon }/{16}$ .

By equation (5.20), we obtain that

$$ \begin{align*} \bigg|\frac{1}{b_i-a_i}\sum_{j=a_i}^{b_i-1}f(T^jy^{(i)}) -\frac{1}{b_i-a_i}\sum_{j=a_i}^{b_i-1}f(T^jx)\bigg| \leq\omega_f(2^{-i-2})<\frac{\varepsilon}{4}. \end{align*} $$

This implies

(5.21) $$ \begin{align} \bigg|{\int_X}f\,{d}\nu_i- \frac{1}{b_i-a_i}\sum_{j=a_i}^{b_i-1}f(T^jx)\bigg|<\frac{\varepsilon}{4}. \end{align} $$

If $k\leq a_{i+1}$ , then $\gamma _k=\nu _i$ . Since $\|f\|\leq 1$ , by equations (5.19), (5.21), and Lemma 2.3, we conclude that

$$ \begin{align*} \bigg|{\int_X}f\,{d}\delta_x^k-\int_Xf\,{d}\gamma_k\bigg| &\leq\bigg|{\int_X}f\,{d}\delta_x^k-\frac{1}{b_i-a_i}\sum_{j=a_i}^{b_i-1}f(T^jx)\bigg|\\ &\quad +\bigg|\frac{1}{b_i-a_i}\sum_{j=a_i}^{b_i-1}f(T^jx)-\int_Xf\,{d}\nu_i\bigg|\\ &<\bigg|\frac{1}{k}\!\sum_{j=0}^{k-1}\! f(T^jx)-\frac{1}{b_i-a_i}\!\sum_{j=a_i}^{b_i-1}\!f(T^jx)\bigg| +\frac{\varepsilon}{4} \leq\frac{\varepsilon}{4}+\frac{2(k-b_i+a_i)}k\\ &\leq\frac{\varepsilon}{4}+\frac{2(a_{i+1}-b_i+a_i)}{N_i} <\frac{\varepsilon}{4}+2\cdot 4^{-i}<\varepsilon. \end{align*} $$

If $k>a_{i+1}$ , then one has

(5.22) $$ \begin{align} \int_Xf\,{d}\nu_{i+1}=\frac{1}{k-r-a_{i+1}}\sum_{j=a_{i+1}}^{k-r-1}f(T^jy^{(i+1)}). \end{align} $$

By equation (5.20), we obtain that

(5.23) $$ \begin{align} \bigg|\frac{1}{k-r-a_{i+1}}\sum_{j=a_{i+1}}^{k-r-1}f(T^jy^{(i+1)}) -\frac{1}{k-r-a_{i+1}}\sum_{j=a_{i+1}}^{k-r-1}f(T^jx)\bigg| \leq\omega_f(2^{-i-3})<\frac{\varepsilon}{4}. \ \end{align} $$

By equations (5.22) and (5.23), we obtain that

(5.24) $$ \begin{align} \bigg|{\int_X}f\,{d}\nu_{i+1}-\frac{1}{k-r-a_{i+1}}\sum_{j=a_{i+1}}^{k-r-1}f(T^jx)\bigg| <\frac{\varepsilon}{4}. \end{align} $$

Define $S\kern1.4pt{=}\kern1.4pt([a_i,b_i-1]\kern1.3pt{\cup}\kern1.3pt [a_{i+1},k-r-1])\kern1.3pt{\cap}\kern1.3pt \mathbb {Z}$ . Clearly, ${|S|\kern1.4pt{=}\kern1.4ptk-r-a_{i+1}+b_i-a_i}$ . Note that

(5.25) $$ \begin{align} \frac{1}{|S|}\sum_{j\in S}f(T^jx) =\frac{\alpha_k}{b_i-a_i}\sum_{j=a_i}^{b_i-1}f(T^jy^{(i)}) +\frac{1-\alpha_k}{k-r-a_{i+1}}\sum_{j=a_{i+1}}^{k-r-1}f(T^jy^{(i+1 )}). \end{align} $$

Since $\|f\|\leq 1$ , by equation (5.24) and Lemma 2.3, we conclude that

$$ \begin{align*} \bigg|{\int_{X}}f\,{d}\delta_x^k-\frac{1}{|S|}\sum_{j\in S}f(T^jx)\bigg| &=\bigg|\frac{1}{k}\sum_{j=0}^{k-1}f(T^jx)-\frac{1}{|S|}\sum_{j\in S}f(T^jx)\bigg|\\ &\leq\frac{2(k-|S|)}{k}=\frac{2(a_{i+1}+r-b_i+a_i)}{k}\\ &\leq\frac{2(a_{i+1}-b_i+p_{i+1})}{N_i}+\frac{a_i}{b_i}. \end{align*} $$

Hence, we obtain by equation (5.19) that

(5.26) $$ \begin{align} \bigg|{\int_{X}}f\,{d}\delta_x^k-\frac{1}{|S|}\sum_{j\in S}f(T^jx)\bigg| <\frac34 \varepsilon. \end{align} $$

Combining equations (5.21), (5.24), (5.25), and (5.26), we conclude that

$$ \begin{align*} \bigg|{\int_X}f\,{d}\delta_x^k-\int_Xf\,{d}\gamma_k\bigg| &\leq\,\bigg|{\int_{X}}f\,{d}\delta_x^k-\frac{1}{|S|}\sum_{j\in S}f(T^jx)\bigg|\\ &\quad +\alpha_k\bigg|\frac{1}{b_i-a_i}\sum_{j=a_i}^{b_i-1}f(T^jx)-\int_Xf\,{d}\nu_i\bigg|\\ &\quad +(1-\alpha_k)\bigg|\frac{1}{k-r-a_{i+1}}\sum_{j=a_{i+1}}^{k-r-1}f(T^jx) -\int_Xf\,{d}\nu_{i+1}\bigg|\\ &<\frac{3\varepsilon}{4}+\alpha_k\frac{\varepsilon}{4}+(1-\alpha_k)\frac{\varepsilon}{4}=\varepsilon. \end{align*} $$

Therefore, for any $k\geq K$ , one has

$$ \begin{align*} \bigg|{\int_X} f\,{d}\delta_x^k-\int_Xf\,{d}\gamma_k\bigg|<\varepsilon. \end{align*} $$

Hence,

$$ \begin{align*} \lim_{k\rightarrow\infty}\bigg|{\int_X}f\,{d}\gamma_k-\int_Xf\,{d}\delta_x^k\bigg|=0, \end{align*} $$

and thus

$$ \begin{align*} \mu=\lim_{k\rightarrow\infty}\gamma_k=\lim_{k\rightarrow\infty}\delta_x^k. \end{align*} $$

We have shown that $x\in G_{\mu }$ . This implies equation (5.16) and completes the proof of Proposition 5.1.