A Note on the Expansive Measures of Nonautonomous Dynamical Systems

Abstract

In this paper, we introduce densely measure-expansive and measure expansive center on nonautonomous dynamical systems. In addition, we analyze a variety of properties about expansive measures and densely measure-expansive nonautonomous dynamical systems enlightened by Morales. In a sense, we extend various results of Morales to nonautonomous dynamical systems.

1 Introduction

In dynamical systems, the orbits may have two opposite behaviors: keeping separate and staying close to each other. The classical conception of expansiveness is firstly introduced by Utz [13], who studies the former behavior of orbits. Expansiveness is a very important and useful dynamical property on compact metric spaces. There exist a sea of significant results in [5, 8,9,10].

As we all know, the nonautonomous dynamical systems are governed by a sequence of maps. A sea of problems are necessarily described by nonautonomous dynamical systems. There are more and more elegant results for nonautonomous dynamical systems in [2, 3, 11]. So it is natural for us to consider the properties of expansiveness on nonautonomous dynamical systems.

In [6], Lee, Morales and Shin proved that the set of expansive measures of a homeomorphism on a compact metric space is a $G_{\delta \sigma }$ subset in the set of all Borel probability measures under the weak-$*$ topology. And they proved that if the set of expansive measures is dense in the set of all Borel probability measures, then for every $\epsilon > 0$ there is an expansive measure $\mu$ with invariant support such that the Hausdorff distance between X and the support of $\mu$ less than $\epsilon$.

These results drive us to consider that whether the conclusions still hold for nonautonomous dynamical systems or not. It seems that some properties and research methods on autonomous dynamical systems can be similarly extended to nonautonomous dynamical systems. But some issues are still being explored. As a matter of fact, we generalize some results of [6] to nonautonomous dynamical systems. In this paper, we investigate various properties of expansive measures on nonautonomous dynamical systems.

Let (X, d) be a compact metric space and $\{f_i:X\rightarrow X\}_{i=1}^{+\infty }$ a sequence of homeomorphisms. Denote by $f_{1,\infty }$ the sequence $\{f_i\}_{i=1}^{\infty }$ and the induced nonautonomous dynamical system $(X,f_{1,\infty })$. Let $\mathcal {M}(X)$ be the set of all Borel probability measures of X with respect to the weak-$*$ topology. Denote by $Supp(\mu )$ the support of $\mu$. The Hausdorff distance between A and B is denoted by $d_H(A,B)$. Denote by $G_\delta$, a subset which is the intersection of countably many open sets. Also, a $G_{\delta \sigma }$ subset of a topological space is the union of countably many $G_\delta$ subsets. The detailed definitions and notations can be found in the following sections.

Now we start to state our main results.

Theorem 1.1

The set of expansive measures of $(X,f_{1,\infty })$ is a $G_{\delta \sigma }$ subset of $\mathcal {M}(X)$.

Theorem 1.2

If $(X,f_{1,\infty })$ is densely measure-expansive, then for every $\epsilon > 0$ there is an expansive measure $\mu$ such that $d_H(X,Supp(\mu ))\le \epsilon$.

This paper is arranged as follows. In Sect. 2, some fundamental preliminaries required for the other sections are given. In Sect. 3, we study various properties of expansive measures and densely measure-expansive nonautonomous dynamical systems. In addition, we give the proofs of Theorems 1.1 and 1.2.

2 Preliminaries

In this section, we will state some necessary basic concepts required for the other sections.

Denote by $\mathbb {Z}$ and $\mathbb {N}$, the set of all integers and positive integers, respectively. We say that a nonautonomous dynamical system is a pair $(X,f_{1,\infty })$, where (X, d) is a compact metric space and $\{f_i:X\rightarrow X\}_{i=1}^{+\infty }$ a sequence of homeomorphisms. Write $f_{1,\infty }= \{f_i\}_{i=1}^{\infty }$. Let $f_i^0=id_X,$

$$\begin{aligned} f_i^n= & {} f_{i+(n-1)} \circ f_{i+(n-2)}\circ \cdots \circ f_{i+1}\circ f_i,\\ f_i^{-n}= & {} (f_i^n)^{-1}=\left( f_i\right) ^{-1}\circ \left( f_{i+1}\right) ^{-1}\circ \cdots \circ \left( f_{i+(n-1)}\right) ^{-1} \end{aligned}$$

for every $i,n \in \mathbb {N}$, where $id_X$ is the identity map on X.

Next, we shall state a variety of concepts. A Borel probability measure $\mu$ is nonatomic if $\mu (\{x\})=0$ for every $x \in X$. We say that a subset $\mathcal {M}\subset \mathcal {M}(X)$ is convex if $t\mu + (1-t)\nu \in \mathcal {M}$ for every $\mu ,\nu \in \mathcal {M}$ and $0 \le t \le 1$. It is known that $\mathcal {M}(X)$ is convex and compact with the weak-$*$ topology.

We say that $A\subset X$ is an invariant set of $(X,f_{1,\infty })$, if $f_j(A)\subset A$ for all $j\in \mathbb {Z}$. Recall that the support of $\mu \in \mathcal {M}(X)$ is the set $Supp(\mu )=\{x\in X: \mu (U)>0 \}\text { for any neighborhood } U \text { of }x$. It follows that $Supp(\mu )$ is a nonempty compact subset of X.

A topological space X is a Baire space if the intersection of each countable family of open and dense subsets is dense in X. We say that $A\subset X$ is a Baire subset if A is a Baire space under the topology induced by X; define $A\subset X$ is nowhere dense if the closure of A in X has no interior in X; and $A\subset X$ is meagre if it is the union of countably many nowhere dense subsets of X.

Given $A,B\subset X$, the Hausdorff distance between A and B is defined by

$$\begin{aligned} d_H(A,B)= & {} max\left\{ \sup \limits _{a\in A}d(a,B),\sup \limits _{b\in B}d(A,b)\right\} , \end{aligned}$$

where $d(a,B)=inf\{d(a,b):b\in B\}$ and d(A, b) is similar to it.

3 Expansive Systems and Expansive Measures

3.1 Some Definitions and Properties

In the following, we firstly state some notations of nonautonomous dynamical systems. Then, we shall define and study various properties of expansive measures on nonautonomous dynamical systems.

Definition 3.1

[12] We say that $(X,f_{1,\infty })$ is an expansive system, if there is $\delta >0$ for any distinct $x, y\in X$ there is $n\in \mathbb {Z}$ such that $d\left( f_1^n(x),f_1^n(y)\right) >\delta$, where $\delta$ is called the expansive constant of $(X,f_{1,\infty })$.

Two simple but important remarks are given in the following.

Remark 3.2

Equivalently, a nonautonomous dynamical system $(X,f_{1,\infty })$ is expansive system, if there is $\delta >0$ such that $\Gamma _\delta (x)=\{x\}$ for every $x \in X$ where

$$\begin{aligned} \Gamma _\delta (x)=\left\{ y\in X:d\left( f_1^j(x),f_1^j(y)\right) \le \delta \text { for all } j \in \mathbb {Z} \right\} . \end{aligned}$$

Remark 3.3

In the above definition, if $f_i=f$ for all $i\in \mathbb {Z}$, where $f:X\rightarrow X$ is homeomorphism, then the expansiveness of the system $(X,f_{1,\infty })$ is equivalent to expansiveness of f on X in [14].

In [1], the authors introduced the definition of expansive measures for a time varying bi-measurable map on nonautonomous dynamical systems. Enlightened by it, we have the following definition.

Definition 3.4

[1] Let $\mu$ be Borel probability measure of X, we say that $\mu$ is expansive, if there is $\delta >0$ such that

$$\begin{aligned} \mu \left( \Gamma _\delta (x)\right) =0 \end{aligned}$$

for every $x \in X$, where $\delta$ is called the expansive constant of $\mu$. Denote by $\mathcal {M}_{ex}(f_{1,\infty })$ the set of expansive measures of $(X,f_{1,\infty })$. Given $\epsilon >0$, we denote by

$$\begin{aligned} \mathcal {M}_{ex}(f_{1,\infty },\epsilon )= & {} \{\mu \in \mathcal {M}_{ex}(f_{1,\infty }):\epsilon \text { is an expansive constant of } \mu \}. \end{aligned}$$

Remark 3.5

• (i) $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ is a convex set of $\mathcal {M}(X)$. Indeed, Take $\mu ,\nu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ and $0 \le t \le 1$, then

  $$\begin{aligned} \left( t\mu +(1-t)\nu \right) (\Gamma _\epsilon (x))=t\mu \left( \Gamma _\epsilon (x)\right) +(1-t)\nu (\Gamma _\epsilon (x))=0 \end{aligned}$$

  for all $x\in X$. Hence, $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ is convex.

• (ii) $\mathcal {M}_{ex}(f_{1,\infty })= \bigcup \limits _{\epsilon >0}\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$.

• (iii) If $(X,f_{1,\infty })$ is expansive, then $\mathcal {M}_{ex}(f_{1,\infty })=\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ for every expansive constant $\epsilon$ of $(X,f_{1,\infty })$.

Next, we will give an example about expansive systems and expansive measures as follows.

Example 3.6

Let $X=\sum _2=\prod \nolimits _{-\infty }^\infty \{0,1\}$. A metric on X is given by $d(x,y)=1/{2^k}$, where $k=min\{|i|:x_i\ne y_i\},$and $d(x,y)=0$ when $x=y$. Then, X is compact. The two-sided shift $T:X\rightarrow X$ is defined by

$$\begin{aligned} T\left( \ldots ,x_{-1},\mathop {x_0}\limits ^*,x_1,\ldots \right) =\left( \ldots ,x_{-1},x_0,\mathop {x_1}\limits ^*,x_2,\ldots \right) , \end{aligned}$$

where the symbol $*$ occurs the 0-th coordinate of each point. Denote $\mathop 0\limits ^ - =1$ and $\mathop 1\limits ^ - =0$, let $g:X \rightarrow X$ defined by

$$\begin{aligned} g\left( \ldots ,x_{-1},x_0,x_1,\ldots \right) =\left( \ldots ,\mathop {x_{-1}}\limits ^-,\mathop {x_0}\limits ^-,\mathop {x_1}\limits ^-,\ldots \right) . \end{aligned}$$

Then, T and g are homeomorphisms of X. Let $f_{1,\infty }=\{T,g,T,g,\ldots \}$ and $\mu$ be the Bernoulli measure. Thus, for all $x=\left( \ldots ,x_{-1},\mathop {x_0}\limits ^*,x_1,\ldots \right) , y=\left( \ldots ,y_{-1},\mathop {y_0}\limits ^*,y_1,\ldots \right) \in X$ with $x, y\in X$ and $x\ne y$, there is $n\in \mathbb {Z}$ such that $x_n\ne y_n$. Hence,

$$\begin{aligned} d\left( f_1^{2n}(x),f_1^{2n}(y)\right) = d\left( T^n(x),T^n(y)\right) =1>\frac{1}{4}. \end{aligned}$$

This yields that $(X,f_{1,\infty })$ is an expansive system with expansive constant $\frac{1}{4}$ and then $\mu$ is expansive.

Enlightened by [6], the following definition is introduced naturally.

Definition 3.7

We say that a nonautonomous dynamical system $(X,f_{1,\infty })$ is densely measure-expansive, if the set of expansive measures of $(X,f_{1,\infty })$ is dense in $\mathcal {M}(X)$ .

Proposition 3.8

If $(X,f_{1,\infty })$ is densely measure-expansive, then X has no isolated points.

Proof

Since $(X,f_{1,\infty })$ is densely measure-expansive, $\mathcal {M}_{ex}(f_{1,\infty })$ is dense in $\mathcal {M}(X)$. Then,

$$\begin{aligned} A:= & {} \bigcup \limits _{\mu \in \mathcal {M}_{ex}(f_{1,\infty })}Supp(\mu ) \end{aligned}$$

is dense in X by ([6], Lemma 3.6). If X has isolated point x, then we claim that $x\in A$. If $x\notin A$, there is a sequence $\{x_k\}\subset A$ such that $x_k\rightarrow x$ as $k\rightarrow \infty$ since A is dense in X. Then, for every ball neighborhood B(x, 1/n) of x, there is $x_{n_j}\in B(x,1/n)$ for all $n\in \mathbb {N}$ and some $j\in \mathbb {N}$. This contradicts with the definition of isolated points. Hence,

$$\begin{aligned} x\in \bigcup \limits _{\mu \in \mathcal {M}_{ex}(f_{1,\infty })}Supp(\mu ). \end{aligned}$$

It follows that there is $\mu \in \mathcal {M}_{ex}(f_{1,\infty })$ such that $x\in Supp(\mu )$. Since x is an isolated point, there is $\epsilon >0$ such that

$$\begin{aligned} B(x,\epsilon )=\{x\}. \end{aligned}$$

But $\mu (B(x,\epsilon ))>0$ since $x\in Supp(\mu )$. This is contradict with the fact that $\mu$ is expansive. Consequently, X has no isolated points. $\square$

More properties of $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ are given as follows. Firstly, we will study some equivalences for the expansivity of a given measure enlightened by [9].

Proposition 3.9

Let $(X,f_{1,\infty })$ be a nonautonomous dynamical system, then a Borel probability measure $\mu$ of X is an expansive measure of $(X,f_{1,\infty })$ if and only if there is $\delta >0$ such that $\mu (\Gamma _\delta (x))=0$ for $\mu$-a.e. $x\in X$.

Proof

We only need to prove the sufficiency. Let $\delta > 0$ be such that $\mu (\Gamma _\delta (x))=0$ for $\mu$-a.e. $x\in X$. We will prove that $\delta /2$ is an expansive constant of $\mu$. Suppose by not so. Then, there is $x_0\in X$ such that

$$\begin{aligned} \mu (\Gamma _{\delta /2}(x_0)) > 0. \end{aligned}$$

Denote $A =\left\{ x\in X: \mu \left( \Gamma _\delta (x)\right) = 0\right\} ,$ then $\mu (A) = 1$. Since $\mu$ is a probability measure, we obtain

$$\begin{aligned} A\cap \Gamma _{\delta /2}(x_0)\ne \emptyset . \end{aligned}$$

It yields that there is $y_0\in A\cap \Gamma _{\delta /2}(x_0)$ such that $\mu (\Gamma _\delta (y_0))=0.$ Since $y_0\in \Gamma _{\delta /2}(x_0)$, we claim that

$$\begin{aligned} \Gamma _{\delta /2}(x_0)\subset \Gamma _\delta (y_0). \end{aligned}$$

In fact, for every $x\in \Gamma _{\delta /2}(x_0)$, we have $d\left( f_1^j(x),f_1^j(x_0)\right) \le \delta /2$ for all $j \in \mathbb {Z}$. Since $d\left( f_1^j(x_0),f_1^j(y_0)\right) \le \delta /2$ for all $j \in \mathbb {Z}$. Thus, we have

$$\begin{aligned} d\left( f_1^j(x),f_1^j(y_0)\right) \le \delta \end{aligned}$$

for all $j \in \mathbb {Z}$ by triangle inequality. From the above, it follows that $x\in \Gamma _\delta (y_0)$. This proves the claim. Therefore,

$$\begin{aligned} \mu (\Gamma _{\delta /2}(x_0))\le \mu (\Gamma _\delta (y_0))=0. \end{aligned}$$

This contradicts with $\mu (\Gamma _{\delta /2}(x_0)) > 0.$ $\square$

In particular, we have the following corollary.

Corollary 3.10

Let $(X,f_{1,\infty })$ be a nonautonomous dynamical system, then a Borel probability measure $\mu$ of X is an expansive measure of $(X,f_{1,\infty })$ if and only if there is $\delta >0$ such that $\mu (\Gamma _\delta (x))=0$ for every $x\in Supp(\mu )$.

Next, we will analyze an important property about expansive measures on nonautonomous dynamical systems stimulated by [9].

Proposition 3.11

A nonautonomous dynamical system $(X,f_{1,\infty })$ has an expansive measure if and only if there is an invariant Borel set Y of $(X,f_{1,\infty })$ such that the restriction $(Y,f_{1,\infty }|_Y)$ has an expansive measure, where $f_{1,\infty }|_Y$ is a sequence of homeomorphisms on Y defined by $\{f_i|_Y:Y\rightarrow Y\}_{i=1}^{+\infty }$.

Proof

If $(X,f_{1,\infty })$ has an expansive measure $\mu$, then $\mu \left( \Gamma _\delta ^{f_{1,\infty }}(x)\right) =0$. Since

$$\begin{aligned} \Gamma _\delta ^{f_{1,\infty }|_Y}(x)\subset \Gamma _\delta ^{f_{1,\infty }}(x), \end{aligned}$$

we have $\mu \left( \Gamma _\delta ^{f_{1,\infty }|_Y}(x)\right) \le \mu \left( \Gamma _\delta ^{f_{1,\infty }}(x)\right) =0.$ Consequently, $\mu \left( \Gamma _\delta ^{f_{1,\infty }|_Y}(x)\right) =0$.

On the other hand, assume that $(Y,f_{1,\infty }|_Y)$ has an expansive measure $\nu$. Fix $\delta >0$, since Y is invariant, we have either

$$\begin{aligned} \Gamma _{\delta /2}^{f_{1,\infty }}(x)\cap Y=\emptyset \end{aligned}$$

or

$$\begin{aligned} \Gamma _{\delta /2}^{f_{1,\infty }}(x)\cap Y\subset \Gamma _\delta ^{f_{1,\infty }|_Y}(y) \end{aligned}$$

for all $x\in X$ and some $y\in Y$. In fact, for every $x_1,x_2\in \Gamma _{\delta /2}^{f_{1,\infty }}(x)$, we have $d\left( f_1^j(x_1),f_1^j(x_2)\right) \le \delta$ for all $j \in \mathbb {Z}$ by triangle inequality. If $\Gamma _{\delta /2}^{f_{1,\infty }}(x)\cap Y\ne \emptyset$, then

$$\begin{aligned} d\left( f_1^j(y_1),f_1^j(y)\right) \le \delta \end{aligned}$$

for all $y_1,y\in \Gamma _{\delta /2}^{f_{1,\infty }}(x)\cap Y$ and $j \in \mathbb {Z}$. Thus, $y_1\in \Gamma _\delta ^{f_{1,\infty }|_Y}(y)$, and consequently

$$\begin{aligned} \Gamma _{\delta /2}^{f_{1,\infty }}(x)\cap Y\subset \Gamma _\delta ^{f_{1,\infty }|_Y}(y). \end{aligned}$$

Therefore, either

$$\begin{aligned} \Gamma _{\delta /2}^{f_{1,\infty }}(x)\cap Y=\emptyset \end{aligned}$$

or

$$\begin{aligned}&\mu \left( \Gamma _{\delta /2}^{f_{1,\infty }}(x)\right) \le \nu \left( \Gamma _\delta ^{f_{1,\infty }|_Y}(y)\right) \end{aligned}$$

for some $y\in Y$ where $\mu$ is the Borel probability measure of X defined by $\mu (A)=\nu (A\cap Y)$. Taking $\delta$ as an expansive constant of $\nu$, then $\nu \left( \Gamma _\delta ^{f_{1,\infty }|_Y}(y)\right) =0.$ Thus, $\mu \left( \Gamma _{\delta /2}^{f_{1,\infty }}(x)\right) =0$ for every $x\in X$. If $\Gamma _{\delta /2}^{f_{1,\infty }}(x)\cap Y=\emptyset$, then $\nu \left( \Gamma _{\delta /2}^{f_{1,\infty }}(x)\cap Y\right) =\mu \left( \Gamma _{\delta /2}^{f_{1,\infty }}(x)\right) =0.$ We conclude that $\mu$ is expansive with expansive constant ${\delta /2}$. This completes the proof. $\square$

In the following, we shall give the proofs of our main results.

3.2 Proof of Theorem 1.1

To prove Theorem 1.1, we need to give a lemma. Firstly, we observe that the set of expansive measures of an expansive system coincides with the set of nonatomic Borel probability measures. Notice that the latter set is a $G_\delta$ subset of $\mathcal {M}(X)$ ([7], Theorem 2.2), hence the set of expansive measures of any expansive system is a $G_\delta$ subset of $\mathcal {M}(X)$. In particular, if $(X,f_{1,\infty })$ is an expansive system, then $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ is a $G_\delta$ subset of $\mathcal {M}(X)$. In the following, we generalize this assertion to any nonautonomous dynamical systems. More precisely, we have the following result.

Lemma 3.12

If $(X,f_{1,\infty })$ is a nonautonomous dynamical system, then $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ is a $G_\delta$ subset of $\mathcal {M}(X)$ for every $\epsilon >0$.

Proof

Fix $\epsilon >0$, define $C(\rho )=\{\mu \in \mathcal {M}(X):\mu (\Gamma _\epsilon (x))\ge \rho \text { for some } x\in X\}$. Then, we can get

$$\begin{aligned} \mathcal {M}_{ex}(f_{1,\infty },\epsilon )=\bigcap \limits _{m=1}^\infty (\mathcal {M}(X)\backslash C(1/m)). \end{aligned}$$

Thus, it suffices to show that $C(\rho )$ is closed in $\mathcal {M}(X)$ for every $\rho >0$. In fact, take $\rho > 0$ together with a sequence $\{\mu _n\}\subset C(\rho )$ such that $\mu _n\rightarrow \mu$ for some $\mu \in \mathcal {M}(X)$. Note that $\{\mu _n\}\subset C(\rho )$, we can choose a sequence $\{x_n\}\subset X$ such that

$$\begin{aligned} \mu _n(\Gamma _\epsilon (x_n))\ge \rho \end{aligned}$$

for each $n \in \mathbb {N}$. Since X is compact, we can assume that $x_n\rightarrow x$ for some $x \in X$. Fix a compact neighborhood C of $\Gamma _\epsilon (x)$. Denote by $O = int(C)$ the interior of C. Hence, $\Gamma _\epsilon (x)\subset O$. Let us prove that

$$\begin{aligned} \Gamma _\epsilon (x_n)\subset C \end{aligned}$$

for all n large. Otherwise, there is a subsequence $n_k \rightarrow \infty$ such that $\Gamma _\epsilon (x_{n_k})\nsubseteq O$ for all $k\in \mathbb {N}$. Then, we can select a sequence $\{z_k\}\in \Gamma _\epsilon (x_{n_k})\backslash O$. Again by the compactness of X, we can assume that $z_k\rightarrow z$ for some $z\in X$. Clearly, $z\notin O$. However, since $z_k\in \Gamma _\epsilon (x_{n_k})$ and it implies that

$$\begin{aligned} d\left( f_1^j(z_k),f_1^j(x_{n_k})\right) \le \epsilon \text { for all } j \in \mathbb {Z}. \end{aligned}$$

Fixing j and letting $k \rightarrow \infty$, we get

$$\begin{aligned} d\left( f_1^j(z),f_1^j(x)\right) \le \epsilon \text { for all } j \in \mathbb {Z}. \end{aligned}$$

Consequently, $z\in \Gamma _\epsilon (x)$. We have $z\in O$ as $\Gamma _\epsilon (x)\subset O$, which is a contradiction. Then, $\Gamma _\epsilon (x_n)\subset C$ for all n large. It shows that $\mu _n\left( \Gamma _\epsilon (x_n)\right) \le \mu _n(C)$ for all n large. Since $\mu _n\rightarrow \mu$, and by ([14], p.149) we obtain

$$\begin{aligned} \rho \le \limsup \limits _{n\rightarrow \infty }\mu _n\left( \Gamma _\epsilon (x_n)\right) \le \limsup \limits _{n\rightarrow \infty }\mu _n(C)\le \mu (C). \end{aligned}$$

Thus, $\mu (C)\ge \rho$ for every compact neighborhood C of $\Gamma _\epsilon (x)$. Consequently, $\mu \left( \Gamma _\epsilon (x)\right) \ge \rho$. This proves that $\mu \in C(\rho )$ and so $C(\rho )$ is closed for any $\rho >0$. Thus, the claim follows, it implies that $\mathcal {M}(X)\backslash C(1/m)$ is open for every $m\in \mathbb {N}$ and so $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ is a $G_\delta$ subset. $\square$

With the above lemma, we can give the proof of Theorem 1.1 as follows.

Proof of Theorem 1.1

Since

$$\begin{aligned} \mathcal {M}_{ex}(f_{1,\infty })= \bigcup \limits _{n=1}^\infty \mathcal {M}_{ex}(f_{1,\infty },1/n). \end{aligned}$$

By Lemma 3.12, $\mathcal {M}_{ex}(f_{1,\infty },1/n)$ is a $G_\delta$ subset of $\mathcal {M}(X)$. Consequently, $\mathcal {M}_{ex}(f_{1,\infty })$ is a $G_{\delta \sigma }$ subset of $\mathcal {M}(X)$. $\square$

We obtain immediately the following corollary of Lemma 3.12.

Corollary 3.13

If $(X,f_{1,\infty })$ is a nonautonomous dynamical system, then we have that $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ is a Baire subspace of $\mathcal {M}(X)$ for every $\epsilon >0$.

Proof

Since X is compact, thus $\mathcal {M}(X)$ is compact (hence complete) under the weak-$*$ topology. On the other hand, by Alexandroff’s Theorem ([4], Theorem 10.19, p. 712), every $G_\delta$ subset of a complete metric space is topologically complete. Then, it follows from Lemma 3.12 that $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ is topologically complete. Therefore, the result holds because every topologically complete subset is Baire (see for instance [4], p. 716). $\square$

3.3 Proof of Theorem 1.2

In order to prove Theorem 1.2, we need to prove several lemmas in the following.

Lemma 3.14

If $(X,f_{1,\infty })$ is a nonautonomous dynamical system and $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )\ne \emptyset$ for given $\epsilon >0$, then there is a meagre subset $\mathcal {D}_\epsilon$ of $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ such that

$$\begin{aligned} Supp(\mu )=\bigcup \limits _{\nu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )}Supp(\nu ) \end{aligned}$$

for every $\mu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )\backslash \mathcal {D}_\epsilon$.

Proof

Let $2_c^X$ denote the set of compact subsets of X. This set is a compact metric space if endowed with the Hausdorff distance. Define

$$\begin{aligned} \Psi :\mathcal {M}_{ex}(f_{1,\infty },\epsilon )\rightarrow 2_c^X \end{aligned}$$

by $\Psi (\mu )=Supp(\mu )$. Denote by $\mathcal {D}_\epsilon$ the set of discontinuity points of $\Psi$. Next, we prove that $\mathcal {D}_\epsilon$ satisfies the desired property. First of all, since $\Psi$ is lower-semicontinuous, it follows from ([4], Corollary 1, p. 71) that $\mathcal {D}_\epsilon$ is meagre. Next, take $\mu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )\backslash \mathcal {D}_\epsilon$ and $\nu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )$. Define $\mu _t = (1-t)\mu + t\nu$ for $0 \le t \le 1$. Since $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ is convex by Remark 3.5(i), it shows that $\mu _t\in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ for every $0 \le t \le 1$. Moreover, $\mu _t\rightarrow \mu$ as $t\rightarrow 0$. Since $\mu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )\backslash \mathcal {D}_\epsilon$, it implies that $\Psi$ is continuous at $\mu$. Hence,

$$\begin{aligned} \Psi (\mu _t) = Supp(\mu _t) = Supp(\mu )\cup Supp(\nu ) \end{aligned}$$

converges to $\Psi (\mu ) = Supp(\mu )$. From this, we get $Supp(\nu ) \subset Supp(\mu ).$ Since $\nu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ is arbitrary, we conclude that

$$\begin{aligned} \bigcup \limits _{\nu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )}Supp(\nu )\subset Supp(\mu ). \end{aligned}$$

Since $\mu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )$, we have

$$\begin{aligned} Supp(\mu )\subset \bigcup \limits _{\nu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )}Supp(\nu ). \end{aligned}$$

Hence,

$$\begin{aligned} Supp(\mu )=\bigcup \limits _{\nu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )}Supp(\nu ) \end{aligned}$$

for every $\mu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )\backslash \mathcal {D}_\epsilon$. $\square$

In light of the definition of measure center in [6, 15], we have the following concept.

Definition 3.15

Given $\epsilon >0$, the $\epsilon$-measure expansive center of $(X,f_{1,\infty })$ is the union of the support of all the expansive measures with $\epsilon$ as an expansivity constant, that is

$$\begin{aligned} E(f_{1,\infty },\epsilon )=\bigcup \limits _{\nu \in \mathcal {M}_{ex}(f_{1,\infty },\epsilon )}Supp(\nu ). \end{aligned}$$

Lemma 3.16

Let $(X,f_{1,\infty })$ be a nonautonomous dynamical system and $\epsilon >0$, if $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )\ne \emptyset$, then there is a dense subset $\mathcal {R}_\epsilon$ of $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ such that

$$\begin{aligned} Supp(\mu )=E(f_{1,\infty },\epsilon ) \end{aligned}$$

for every $\mu \in \mathcal {R}_\epsilon$.

Proof

Let $\mathcal {D}_\epsilon$ be the meagre subset of $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ given by Lemma 3.14. By Corollary 3.13, we have that $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$ is a Baire space. Therefore, $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )\backslash \mathcal {D}_\epsilon$ is dense in $\mathcal {M}_{ex}(f_{1,\infty },\epsilon )$. Then, Lemma 3.14 implies the result by taking $\mathcal {R}_\epsilon =\mathcal {M}_{ex}(f_{1,\infty },\epsilon )\backslash \mathcal {D}_\epsilon$. $\square$

Next, lemma is a simple property about $\epsilon$-measure expansive center of $(X,f_{1,\infty })$.

Lemma 3.17

$E\left( f_{1,\infty },n^{-1}\right)$ is an increasing sequence of sets.

Proof

Since

$$\begin{aligned} E\left( f_{1,\infty },n^{-1}\right) =\bigcup \limits _{\nu \in \mathcal {M}_{ex}\left( f_{1,\infty },n^{-1}\right) }Supp(\nu ) \end{aligned}$$

and we have

$$\begin{aligned} \nu \left( \Gamma _\frac{1}{n}(x)\right) =0 \end{aligned}$$

for every $\nu \in \mathcal {M}_{ex}\left( f_{1,\infty },n^{-1}\right)$. In addition,

$$\begin{aligned} \Gamma _\frac{1}{n}(x)=\left\{ y:d\left( f_1^j(x),f_1^j(y)\right) \le \frac{1}{n}\text { for all } j \in \mathbb {Z}\right\} . \end{aligned}$$

From this, we see that

$$\begin{aligned} \Gamma _\frac{1}{n+1}(x)=\left\{ y:d\left( f_1^j(x),f_1^j(y)\right) \le \frac{1}{n+1}<\frac{1}{n}\text { for all } j \in \mathbb {Z}\right\} . \end{aligned}$$

Thus,

$$\begin{aligned} \Gamma _\frac{1}{n+1}(x)\subset \Gamma _\frac{1}{n}(x) \end{aligned}$$

for all $n\in \mathbb {N}$. It implies that

$$\begin{aligned} 0\le \nu \left( \Gamma _\frac{1}{n+1}(x)\right) \le \nu \left( \Gamma _\frac{1}{n}(x)\right) =0 \end{aligned}$$

and then

$$\begin{aligned} \nu \left( \Gamma _\frac{1}{n+1}(x)\right) =0. \end{aligned}$$

It yields that $\nu \in \mathcal {M}_{ex}\left( f_{1,\infty },(n+1)^{-1}\right)$. Hence,

$$\begin{aligned} \mathcal {M}_{ex}\left( f_{1,\infty },n^{-1}\right) \subset \mathcal {M}_{ex}\left( f_{1,\infty },(n+1)^{-1}\right) . \end{aligned}$$

Consequently,

$$\begin{aligned} E\left( f_{1,\infty },n^{-1}\right) \subset E\left( f_{1,\infty },(n+1)^{-1}\right) . \end{aligned}$$

This completes the proof. $\square$

Next, we start to prove Theorem 1.2 based on the above lemmas.

Proof of Theorem 1.2

Since $(X,f_{1,\infty })$ is densely measure-expansive, which implies that $\mathcal {M}_{ex}(f_{1,\infty })$ is dense in $\mathcal {M}(X)$. Then, we have that

$$\begin{aligned} \bigcup \limits _{\nu \in \mathcal {M}_{ex}(f_{1,\infty })}Supp(\nu ) \end{aligned}$$

is dense in X by ([6], Lemma 3.6). Since

$$\begin{aligned} \bigcup \limits _{\nu \in \mathcal {M}_{ex}\left( f_{1,\infty }\right) }Supp(\nu )= & {} \bigcup \limits _{n=1}^\infty \\&\left( \bigcup \limits _{\nu \in \mathcal {M}_{ex}\left( f_{1,\infty },n^{-1}\right) }Supp(\nu )\right) =\bigcup \limits _{n=1}^\infty E\left( f_{1,\infty },n^{-1}\right) . \end{aligned}$$

It follows that

$$\begin{aligned} \bigcup \limits _{n=1}^\infty E\left( f_{1,\infty },n^{-1}\right) \end{aligned}$$

is dense in X. By Lemma 3.17, $E\left( f_{1,\infty },n^{-1}\right)$ is an increasing sequence of sets. Then,

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }d_H\left( X,E\left( f_{1,\infty },n^{-1}\right) \right) =d_H\left( X,\bigcup \limits _{n=1}^\infty E(f_{1,\infty },n^{-1})\right) =0. \end{aligned}$$

Now Lemma 3.16 provides a sequence of expansive measures $\mu _n$ such that $Supp(\mu _n) = E\left( f_{1,\infty },n^{-1}\right)$. Therefore,

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }d_H(X,Supp(\mu _n))=0. \end{aligned}$$

Hence, $d_H(X,Supp(\mu _n))\le \epsilon$ for every $\epsilon >0$. This proves the result. $\square$