On the Čech-Completeness of the Space of τ-Smooth Idempotent Probability Measures

Abstract

For the set $\mathsf{I}\left(X\right)$ of probability measures on a compact Hausdorff space X, we propose a new way to introduce the topology by using the open subsets of the space X. Then, among other things, we give a new proof that for a compact Hausdorff space X, the space $\mathsf{I}\left(X\right)$ is also a compact Hausdorff space. For a Tychonoff space X, we consider the topological space ${\mathsf{I}}_{\tau }\left(X\right)$ of $\tau$-smooth idempotent probability measures on X and show that the space ${\mathsf{I}}_{\tau }\left(X\right)$ is Čech-complete if and only if the given space X is Čech-complete.

1. Introduction

Idempotent mathematics is a branch of mathematical sciences, rapidly developing and gaining popularity over the last four decades. An important stage of development of the subject was presented in the book “Idempotency” [1] edited by J. Gunawardena (see also [2,3]). This book arose out of the well-known international workshop that was held in Bristol, England, in October 1994. Idempotent mathematics is based on replacing the usual arithmetic operations with a new set of basic operations, i.e., on replacing numerical fields by idempotent semirings and semifields. A typical example is the so-called max-plus algebra (in fact, an idempotent semifield) ${\mathbb{R}}_{\max }=\mathbb{R}\cup \{-\infty \}$ with operation $x\oplus y=\max \{x,y\}$ and $x\odot y=x+y$ ([1,4,5,6]).

M. Zarichnyi, in 2010, in [7], investigated the categorical properties of the space of idempotent probability measures. In [8], the theory was extended to the class of metric spaces. So, the space of idempotent probability measures is a new object. But it has already been studied from different points of view in Measure Theory, Functional Analysis, Probability Theory, Topology, and Category Theory. The study of spaces of idempotent probability measures leads to the problem of its investigations on wider classes of topological classes than the class of compact Hausdorff spaces, in particular, the class of Tychonoff spaces.

T. Banakh [9], T. Banakh, and T. Radul [10,11] carried out a systematic study on probability measures on Tychonoff spaces. In their studies, they fruitfully used the linearity of probability measures. Unlike probability measures, idempotent probability measures are not linear. In papers [6,12] the theory was put forward, and in [13], some categorical properties of $\tau$-smooth weakly additive (nonlinear) functionals were established.

The results obtained in [4,5,6,7,8,14] show that, in order to establish “good” properties of the space of idempotent probability measures, methods are required that are very different from classical methods (i.e., from methods suitable for probability measures which have been productively used in [5,15] and others).

Unlike the abovementioned papers [4,5,6,7,8,14], in this paper, for a compact Hausdorff space X, we introduce the notion of idempotent measures as a set-function on the family $\mathfrak{B}\left(X\right)$ of Borel subsets of X. Note that the work [16] also have this approach. Improving their theory in the current paper, we note some types of open and closed subsets of the space of idempotent probability measures. In the set of idempotent probability measures, we introduce the base of the product topology and show that, for a compact Hausdorff space X, the topological space $\mathsf{I}\left(X\right)$ of idempotent probability measures is also a compact Hausdorff space.

Further, for a Tychonoff space X, we consider the space $I_{\tau }\left(X\right)$ of $\tau$-smooth idempotent probability measures on X. Then, we establish the Čech-completeness of the space of $\tau$-smooth idempotent probability measures for the Čech-complete Tychonoff space. (Čech-complete spaces were introduced by Eduard Čech in 1937 to prove the Baire category theorem. Another important application of the Čech-completeness appears in the metrization of a topological space by a complete metric. Note that a topological space X is Čech-complete if X is a Tychonoff space, and the remainder $\beta X\setminus X$ is an $F_{\sigma }$-set in the Stone–Čech compactification $\beta X$. Locally compact spaces are Čech-complete, but the converse is not true. The space of all irrational numbers with the topology of a subspace of the real line is an example of a Čech-complete space that is not locally compact [17].) From here, since Čech-completeness is hereditary with respect to $G_{\delta }$-subsets, and a Čech-complete Tychonoff space X is $G_{\delta }$-subset in its Stone–Čech compactification $\beta X$, we conclude that a Tychonoff space X is Čech-complete if and only if $I_{\tau }\left(X\right)$ is Čech-complete. Note that a linear (in the classical sense) version of this result was established in [9]. One can see that the methods used in [5,6,9,10,12,15] are not suitable for the present case.

2. Preliminaries

Let X be a compact Hausdorff space and $\mathfrak{B}\left(X\right)$ the family of Borel subsets of X. We denote ${\overline{\mathbb{R}}}_{+}=\left[0,+\infty \right)\cup \left\{+\infty \right\}=\left[0,+\infty \right]$. The symbol D denotes the directed set. Following [16], we enter the following notion.

Definition 1.

A set function $\mu :\mathfrak{B}\left(X\right)\to {\overline{\mathbb{R}}}_{+}$ is said to be an idempotent measure on X if the following conditions hold:

(1)

$\mu (\varnothing )=0$;

(2)

$\mu (A\cup B)=\max \left\{\mu \right(A),\mu (B\left)\right\}$ for any $A,B\in \mathfrak{B}\left(X\right)$;

(3)

$\mu \left({\displaystyle \bigcup _{\alpha \in D}}{A}_{\alpha }\right)=\underset{\alpha \in D}{sup}\left\{\mu \left(A_{\alpha }\right)\right\}$ for every increasing net $\{A_{\alpha },\alpha \in D\}\subset \mathfrak{B}\left(X\right)$ such that $\underset{\alpha \in D}{\cup }{A}_{\alpha }\in \mathfrak{B}\left(X\right)$.

Remark 1.

Every idempotent measure $\mu$ is increasing, i.e., for A, B $\in \mathfrak{B}\left(X\right)$, if $A\subset B$, then, $\mu \left(A\right)\le \mu \left(B\right)$.

The set of all idempotent measures on X will be denoted by $\mathsf{IM}\left(X\right)$. If $\mu \left(X\right)=1$, the idempotent measure $\mu$ is called an idempotent probability measure on X. We denote

$$\begin{array}{c}\mathsf{I}\left(X\right)=\{\mu \in \mathsf{IM}(X):\mu (X)=1\}.\end{array}$$

A set

$$\begin{array}{c}supp\mu ={\left[X\setminus \cup \{A\in \mathfrak{B}(X):\mu (A)=0\}\right]}_X\end{array}$$

(1)

is said to be a support of the given idempotent measure $\mu$.

The support of an idempotent measure $\mu$ can be defined by the following equality:

$$\begin{array}{c}supp\mu ={\left[\cap \{C:C\in \mathfrak{B}(X),\mu (X\setminus C)=0\}\right]}_X.\end{array}$$

(2)

Lemma 1.

For every $\mu \in \mathsf{I}\left(X\right)$, we have $\mu \in I\left(\mathrm{supp}\mu \right)$.

Proof. 

For a support, we will apply (2). It is easy to see that

$$\begin{array}{cc}\hfill \mu (X\setminus supp\mu )& =\mu \left(X\setminus {\left[\cap \{C:C\in \mathfrak{B}(X),\mu (X\setminus C)=0\}\right]}_X\right)\hfill \\ \hfill & =\mu \left(\mathrm{int}\left(X\setminus \cap \{C:C\in \mathfrak{B}(X),\mu (X\setminus C)=0\}\right)\right)\hfill \\ \hfill & =\mu \left(\mathrm{int}\left(\cup \{X\setminus C:C\in \mathfrak{B}(X),\mu (X\setminus C)=0\}\right)\right)\hfill \\ \hfill & \le \mu \left(\cup \{X\setminus C:C\in \mathfrak{B}(X),\mu (X\setminus C)=0\}\right)\hfill \\ \hfill & =\sup \left\{\mu \right(X\setminus C):C\in \mathfrak{B}(X),\mu (X\setminus C)=0\}=0.\hfill \end{array}$$

On the other hand,

$$\begin{array}{c}1=\mu \left(X\right)=\mu \left(\left(X\setminus supp\mu \right)\cup supp\mu \right)=\max \{\mu (X\setminus supp\mu ),\mu (supp\mu )\},\end{array}$$

i.e., $\mu \left(\mathrm{supp}\mu \right)=1$ and $\mu \in I\left(\mathrm{supp}\mu \right)$. □

Let X be a compact Hausdorff space, $\mathcal{B}$ a base in X, $U_i\in \mathcal{B}$, $i=1,\dots ,n$, and $\epsilon >0$. For an idempotent probability measure $\mu \in \mathsf{I}\left(X\right)$, we define a set

$$\langle \mu ;U_1,\dots ,U_n;\epsilon \rangle =\{\nu \in \mathsf{I}\left(X\right):|\nu \left(U_i\right)-\mu \left(U_i\right)|<\epsilon ,i=1,\dots ,n\}.$$

Gathering all such sets, we construct a family

$$\mathcal{B}\left(\mu \right)=\{\langle \mu ;U_1,\dots ,U_n;\epsilon \rangle :U_i\in \mathcal{B},i=1,\dots ,n;\epsilon >0\},\mu \in \mathsf{I}\left(X\right),$$

and put

$${\mathcal{B}}_{\mathsf{I}\left(X\right)}=\underset{\mu \in \mathsf{I}\left(X\right)}{\cup }\mathcal{B}\left(\mu \right).$$

Proposition 1.

The built family ${\mathcal{B}}_{\mathsf{I}\left(X\right)}$ forms a base (or a neighbourhoods system) for some topology in $\mathsf{I}\left(X\right)$.

Proof. 

Since $\mathcal{B}\ne \varnothing$, there exists an open nonempty set $U\in \mathcal{B}$. On the other hand, $U\in \mathfrak{B}\left(X\right)$, so $\mu \left(U\right)$ is well defined. That is why, for every $\epsilon >0$, we have $\langle \mu ;U;\epsilon \rangle \in \mathcal{B}\left(\mu \right)$. Consequently, $\mathcal{B}\left(\mu \right)\ne \varnothing$. Clearly, $\mu \in \langle \mu ;U_1,\dots ,U_n;\epsilon \rangle$ for each $\langle \mu ;U_1,\dots ,U_n;\epsilon \rangle \in \mathcal{B}\left(\mu \right)$.

Suppose we obtain $\mu \in \langle \nu ;V_1,\dots ,V_n;\delta \rangle \in \mathcal{B}\left(\nu \right)$. Note that $(\mu -\nu )\left(V\right)=\mu \left(V\right)-\nu \left(V\right)$, and designate

$$\begin{array}{c}a=\min \{(\nu -\mu )\left(V_1\right)+\delta ,(\mu -\nu )\left(V_1\right)+\delta ,\dots ,(\nu -\mu )\left(V_n\right)+\delta ,(\mu -\nu )\left(V_n\right)+\delta \}.\end{array}$$

Obviously, $a>0$. Then, for every $\xi \in \langle \mu ;V_1,\dots ,V_n;a\rangle$, one has

$$\begin{array}{cc}\hfill |\xi \left(V_i\right)-\nu \left(V_i\right)|& =|\xi \left(V_i\right)-\mu \left(V_i\right)+\mu \left(V_i\right)-\nu \left(V_i\right)|\hfill \\ \hfill & \le |\xi \left(V_i\right)-\mu \left(V_i\right)|+|\mu \left(V_i\right)-\nu \left(V_i\right)|\hfill \\ \hfill & <a+|\mu \left(V_i\right)-\nu \left(V_i\right)|.\hfill \end{array}$$

Two cases are possible: $\mu \left(V_i\right)-\nu \left(V_i\right)\ge 0$ or $\mu \left(V_i\right)-\nu \left(V_i\right)\le 0$. In the former case, we take into attention $a<\nu \left(V_i\right)-\mu \left(V_i\right)+\delta$, and in the latter case, $a<\mu \left(V_i\right)-\nu \left(V_i\right)+\delta$. Hence, $\xi \in \langle \nu ;V_1,\dots ,V_n;\delta \rangle$; in other words, $\langle \mu ;V_1,\dots ,V_n;a\rangle \subset \langle \nu ;V_1,\dots ,V_n;\delta \rangle$.

Finally, consider an arbitrary couple of sets $\langle \mu ;U_1,\dots ,U_n;\epsilon \rangle$, $\langle \mu ;V_1,\dots ,V_k;\delta \rangle$ belonging to $\mathcal{B}\left(\mu \right)$. Denote $\theta =\min \{\epsilon ,\delta \}$, and we have

$$\begin{array}{cc}\hfill & \langle \mu ;U_1,\dots ,U_n;\epsilon \rangle \cap \langle \mu ;V_1,\dots ,V_k;\delta \rangle \hfill \\ \hfill \supset & \langle \mu ;U_1,\dots ,U_n;\theta \rangle \cap \langle \mu ;V_1,\dots ,V_k;\theta \rangle \hfill \\ \hfill =& \langle \mu ;U_1,\dots ,U_n,V_1,\dots ,V_k;\theta \rangle \in \mathcal{B}\left(\mu \right).\hfill \end{array}$$

Proposition 1 is proved. □

Remark 2.

For an open set U, sets

$$\begin{array}{c}{\langle U;\epsilon \rangle }_{-}=\{\xi \in \mathsf{I}\left(X\right):\xi \left(U\right)<\epsilon \}\mathrm{and}{\langle U;\epsilon \rangle }_{+}=\{\xi \in \mathsf{I}\left(X\right):\xi \left(U\right)>\epsilon \}\end{array}$$

are open. Indeed, we have $\mu \left(U\right)=a<\epsilon$ for every $\mu \in {\langle U;\epsilon \rangle }_{-}$. Then, for each $\nu \in {\langle \mu ;U;\epsilon -a\rangle }_{-}$, one has $-(\epsilon -a)+\mu \left(U\right)<\nu \left(U\right)<\mu \left(U\right)+(\epsilon -a)$, $2a-\epsilon <\nu \left(U\right)<\epsilon$. Hence, $\nu \in {\langle U;\epsilon \rangle }_{-}$. The openness of the second set will be established like the openness of the first set.

Remark 3.

For a closed set F, sets

$$\begin{array}{c}{[F;\epsilon ]}_{-}=\{\xi \in \mathsf{I}\left(X\right):\xi \left(F\right)\le \epsilon \}and{[F;\epsilon ]}_{+}=\{\xi \in \mathsf{I}\left(X\right):\xi \left(F\right)\ge \epsilon \}\end{array}$$

are closed. Indeed, for each $\left\{{\xi }_{\alpha }\right\}\subset {[F;\epsilon ]}_{-}\subset \mathsf{I}\left(X\right)$, there exists ${\xi }_0\in \mathsf{I}\left(X\right)$ such that ${\xi }_0={\displaystyle \underset{\alpha }{\lim }}{\xi }_{\alpha }$ (i.e., $\xi \left(A\right)={\displaystyle \underset{\alpha }{\lim }}{\xi }_{\alpha }\left(A\right)$ for each $A\in \mathfrak{B}\left(X\right)$). Hence, ${\xi }_0\left(F\right)\le \epsilon$. Thus, ${\xi }_0\in {[F;\epsilon ]}_{-}$. In the same way, one can show that ${[F;\epsilon ]}_{+}$ is also closed.

3. Idempotent Probability Measures on Compact Hausdorff Spaces

The construction of $\mathsf{I}\left(X\right)$ gives that $\mathsf{I}\left(X\right)\subset {\mathbb{R}}^{\mathfrak{B}\left(X\right)}$. Equip the set $\mathsf{I}\left(X\right)$ with the topology generated by the above neighbourhood system. Obviously, this topology coincides with the induced topology from the product topology of ${\mathbb{R}}^{\mathfrak{B}\left(X\right)}$ to $\mathsf{I}\left(X\right)$.

Theorem 1.

For a compact Hausdorff space X, the topological space $\mathsf{I}\left(X\right)$ is also a compact Hausdorff space.

Proof. 

At first, we show that $\mathsf{I}\left(X\right)$ is a Hausdorff space. Let $\mu$, $\nu \in \mathsf{I}\left(X\right)$ be different idempotent measures. Then, there exists a set $U\in \mathcal{B}$ such that $\mu \left(U\right)\ne \nu \left(U\right)$. Put $\left|\mu \right(U)-\nu (U\left)\right|=a$. One has $a\ne 0$, $|\mu \left(U\right)-\nu \left(U\right)|>\frac{a}{2}$. Assume that there exists an idempotent probability measure $\xi$ belonging to both sets $\langle \mu ;U;\frac{a}{4}\rangle$ and $\langle \nu ;U;\frac{a}{4}\rangle$. Then,

$$\begin{array}{cc}\hfill \left|\nu \right(U)-\mu (U\left)\right|& =\left|\nu \right(U)-\xi (U)+\xi (U)-\mu (U\left)\right|\hfill \\ \hfill & \le |\nu \left(U\right)-\xi \left(U\right)|+|\xi \left(U\right)-\mu \left(U\right)|<\frac{a}{4}+\frac{a}{4}=\frac{a}{2}.\hfill \end{array}$$

The obtained contradiction shows that $\langle \mu ;U;\frac{a}{4}\rangle \cap \langle \nu ;U;\frac{a}{4}\rangle =\varnothing$.

Now, we will prove the compactness of $\mathsf{I}\left(X\right)$ with respect to the topology generated by the neighbourhood system ${\mathcal{B}}_{\mathsf{I}\left(X\right)}$.

Clearly, $\mathsf{I}\left(X\right)$ is bounded in ${\mathbb{R}}^{\mathfrak{B}\left(X\right)}$. Its closedness remains to be shown. Let $\mu \in {\mathbb{R}}^{\mathfrak{B}\left(X\right)}\setminus \mathsf{I}\left(X\right)$. Then, the following cases are possible.

Case 1. $\mu (\varnothing )=a\ne 0$. In this case, we have $〈\mu ;\varnothing ;\frac{\left|a\right|}{2}〉\cap \mathsf{I}\left(X\right)=\varnothing$.

Case 2. $\mu \left(X\right)=a\ne 1$. Then, it is obvious that $〈\mu ;X;\frac{|a+1|}{2}〉\cap \mathsf{I}\left(X\right)=\varnothing$.

Case 3. $\mu (G\cup \mathrm{\Gamma })=a\ne b=\max \left\{\mu \right(G),\mu (\mathrm{\Gamma }\left)\right\}$ for some open sets G and $\mathrm{\Gamma }$, whereupon, $〈\mu ;G,\mathrm{\Gamma },G\cup \mathrm{\Gamma };\frac{|b-a|}{2}〉\cap \mathsf{I}\left(X\right)=\varnothing$.

Case 4. $\mu \left({\displaystyle \bigcup _{\alpha \in D}}{\mathrm{\Gamma }}_{\alpha }\right)=a\ne b={\displaystyle \underset{\alpha \in D}{\sup }}\left\{\mu \left({\mathrm{\Gamma }}_{\alpha }\right)\right\}$ for a net ${\left\{{\mathrm{\Gamma }}_{\alpha }\right\}}_{\alpha \in D}$ of open sets.

The fourth case has the following subcases.

Subcase 4.1. Assume $0\le a<b\le 1$. By definition, there exists such an index ${\alpha }_0$ that $\mu \left({\mathrm{\Gamma }}_{{\alpha }_0}\right)>b-\frac{b-a}{4}=\frac{a+3b}{4}$. We claim that $\langle \mu ;{\mathrm{\Gamma }}_{{\alpha }_0},\underset{\alpha \in D}{\cup }{\mathrm{\Gamma }}_{\alpha };\frac{b-a}{4}\rangle \cap \mathsf{I}\left(X\right)=\varnothing$. Presume it is not so. Then, there exists $\nu$ belonging to this intersection. From here, we obtain

$$\begin{array}{cc}\hfill \left|\nu \left(\bigcup _{\alpha \in D}{\mathrm{\Gamma }}_{\alpha }\right)-\mu \left(\bigcup _{\alpha \in D}{\mathrm{\Gamma }}_{\alpha }\right)\right|& <\frac{b-a}{4},\hfill \\ \hfill |\nu \left({\mathrm{\Gamma }}_{{\alpha }_0}\right)-\mu \left({\mathrm{\Gamma }}_{{\alpha }_0}\right)|& <\frac{b-a}{4}.\hfill \end{array}$$

These inequalities provide, correspondingly,

$$\begin{array}{c}\frac{3a-b}{4}<\nu \left(\bigcup _{\alpha \in D}{\mathrm{\Gamma }}_{\alpha }\right)<\frac{3a+b}{4},\\ \frac{a+b}{2}<\nu \left({\mathrm{\Gamma }}_{{\alpha }_0}\right)<\frac{5b-a}{4}.\end{array}$$

In the issue, one obtains

$$\begin{array}{c}\underset{\alpha \in D}{\sup }\left\{\nu \left({\mathrm{\Gamma }}_{\alpha }\right)\right\}\ge \nu \left({\mathrm{\Gamma }}_{{\alpha }_0}\right)>\frac{a+b}{2}>\frac{3a+b}{4}>\nu \left(\bigcup _{\alpha \in D}{\mathrm{\Gamma }}_{\alpha }\right),\end{array}$$

i.e., ${\displaystyle \underset{\alpha \in D}{\sup }}\left\{\nu \left({\mathrm{\Gamma }}_{\alpha }\right)\right\}>\nu \left({\displaystyle \bigcup _{\alpha \in D}}{\mathrm{\Gamma }}_{\alpha }\right)$. The last inequality goes against $\nu \in \mathsf{I}\left(X\right)$. The resulting contradiction shows that our assumption is wrong, and our statement about the emptiness of the intersection is true.

Subcase 4.2. Assume $0\le b<a\le 1$. Take any $\xi \in \langle \mu ;\underset{\alpha \in D}{\cup }{\mathrm{\Gamma }}_{\alpha };\frac{a-b}{4}\rangle$. Then,

$$\begin{array}{c}\underset{\alpha \in D}{\sup }\left\{\xi \left({\mathrm{\Gamma }}_{\alpha }\right)\right\}<b+\frac{a-b}{4}<a-\frac{a-b}{4}=\mu \left(\bigcup _{\alpha \in D}{\mathrm{\Gamma }}_{\alpha }\right)-\frac{a-b}{4}<\xi \left(\bigcup _{\alpha \in D}{\mathrm{\Gamma }}_{\alpha }\right),\end{array}$$

i.e., ${\displaystyle \underset{\alpha \in D}{\sup }}\left\{\xi \left({\mathrm{\Gamma }}_{\alpha }\right)\right\}<\xi \left({\displaystyle \bigcup _{\alpha \in D}}{\mathrm{\Gamma }}_{\alpha }\right)$. Hence, $\xi \notin \mathsf{I}\left(X\right)$. Thus, $\langle \mu ;\underset{\alpha \in D}{\cup }{\mathrm{\Gamma }}_{\alpha };\frac{a-b}{4}\rangle \cap \mathsf{I}\left(X\right)=\varnothing$.

Subcase 4.3. Assume $b<0$. Then, $\mu \left({\mathrm{\Gamma }}_{\alpha }\right)\le b$ for every $\alpha \in D$. For $\frac{\left|b\right|}{2}$, there exists an index ${\alpha }_0\in D$ such that ${\displaystyle \frac{3b}{2}}=b-\frac{\left|b\right|}{2}<\mu \left({\mathrm{\Gamma }}_{{\alpha }_0}\right)<b+\frac{\left|b\right|}{2}={\displaystyle \frac{b}{2}}$. Evidently, $\langle \mu ;{\mathrm{\Gamma }}_{{\alpha }_0};\frac{\left|b\right|}{2}\rangle \cap \mathsf{I}\left(X\right)=\varnothing$.

Subcase 4.4. Assume $a<0$. Then, $\langle \mu ;\underset{\alpha \in D}{\cup }{\mathrm{\Gamma }}_{\alpha };\frac{\left|a\right|}{2}\rangle \cap \mathsf{I}\left(X\right)=\varnothing$.

Subcase 4.5. Assume $b>1$. Then, there exists an index ${\alpha }_0\in D$ such that $\frac{b+1}{2}=b-\frac{b-1}{2}<\mu \left({\mathrm{\Gamma }}_{{\alpha }_0}\right)<b+\frac{b-1}{2}=\frac{3b-1}{2}$. Hence, $\langle \mu ;{\mathrm{\Gamma }}_{{\alpha }_0};\frac{b-1}{2}\rangle \cap \mathsf{I}|\left(X\right)=\varnothing$.

Subcase 4.6. Assume $a>1$. Then, $\langle \mu ;\underset{\alpha \in D}{\cup }{\mathrm{\Gamma }}_{\alpha };\frac{a-1}{2}\rangle \cap \mathsf{I}\left(X\right)=\varnothing$.

On this, all possible cases have been exhausted. Consequently, $\mu$ is an interior point of ${\mathbb{R}}^{\mathfrak{B}\left(X\right)}\setminus \mathsf{I}\left(X\right)$, which immediately implies that $\mathsf{I}\left(X\right)$ is closed in ${\mathbb{R}}^{\mathfrak{B}\left(X\right)}$, which implies its compactness. The proof of Theorem 1 is completed. □

For a mapping $f:X\to Y$ of compact Hausdorff spaces X and Y, we define a mapping

$$\begin{array}{c}\mathsf{I}\left(f\right):\mathsf{I}\left(X\right)\to \mathsf{I}\left(Y\right)\end{array}$$

by the rule

$$\begin{array}{c}\mathsf{I}\left(f\right)\left(\mu \right)\left(B\right)=\mu \left(f^{\leftarrow }\left(B\right)\right),B\in \mathfrak{B}\left(Y\right).\end{array}$$

Proposition 2.

For every pair of compact Hausdorff spaces X and Y and any continuous mapping $f:X\to Y$, the mapping $\mathsf{I}\left(f\right):\mathsf{I}\left(X\right)\to \mathsf{I}\left(Y\right)$ is continuous.

Proof. 

Take any $\mu \in \mathsf{I}\left(X\right)$, and let $\nu =\mathsf{I}\left(f\right)\left(\mu \right)\in \mathsf{I}\left(Y\right)$. Consider an arbitrary neighbourhood $\langle \nu ;V_1,\dots ,V_k;\delta \rangle$ of $\nu$. Then, for any $\xi \in \langle \mu ;f^{\leftarrow }\left(V_1\right),\dots ,f^{\leftarrow }\left(V_k\right);\delta \rangle$, we have

$$\begin{array}{c}|\mathsf{I}\left(f\right)\left(\xi \right)\left(V_i\right)-\nu \left(V_i\right)|=|\xi \left(f^{\leftarrow }\left(V_i\right)\right)-\mu \left(f^{\leftarrow }\left(V_i\right)\right)|<\delta ,\end{array}$$

which yields $\mathsf{I}\left(f\right)\left(\xi \right)\in \langle \nu ;V_1,\dots ,V_k;\delta \rangle$. By virtue of the arbitrariness of $\xi$, we obtain that $\mathsf{I}\left(f\right)\left(\langle \mu ;f^{\leftarrow }\left(V_1\right),\dots ,f^{\leftarrow }\left(V_k\right);\delta \rangle \right)\subset \langle \nu ;V_1,\dots ,V_k;\delta \rangle$. □

4. $\mathbf{\tau }$-Smooth Idempotent Probability Measures

To continue our investigation, we need the notion of an outer idempotent measure.

An outer idempotent measure of an arbitrary subset $A\subset X$ is defined by

$$\begin{array}{c}{\mu }^{*}\left(A\right)=\inf \{\mu \left(B\right):B\supset A,B\in \mathfrak{B}\left(X\right)\}.\end{array}$$

So, we obtained an extension ${\mu }^{*}$ of $\mu$. Now, Remark 1 may be improved as follows.

Remark 4.

For every idempotent measure $\mu$, its extension ${\mu }^{*}$ is increasing, i.e., if $A\subset B$, then, ${\mu }^{*}\left(A\right)\le {\mu }^{*}\left(B\right)$.

Lemma 2.

For any pair of A, $B\subset X$ and every idempotent measure, we have

$$\begin{array}{c}{\mu }^{*}(A\cup B)=\max \{{\mu }^{*}\left(A\right),{\mu }^{*}\left(B\right)\}.\end{array}$$

Proof. 

It is easy to see that ${\mu }^{*}(A\cup B)\ge \max \{{\mu }^{*}\left(A\right),{\mu }^{*}\left(B\right)\}$. Let us show the inverse inclusion, i.e., ${\mu }^{*}(A\cup B)\le \max \{{\mu }^{*}\left(A\right),{\mu }^{*}\left(B\right)\}$. By the definition of the outer idempotent measure, one has

$${\mu }^{*}(A\cup B)=\inf \{\mu \left(E\right):E\in \mathfrak{B}\left(X\right),E\supset A\cup B\}.$$

Clearly,

$$\{E:E\in \mathfrak{B}(X),E\supset A\cup B\}\supset \{C\cup D:C,D\in \mathfrak{B}(X),C\supset A\mathrm{and}D\supset B\}.$$

Then,

$$\begin{array}{cc}\hfill {\mu }^{*}(A\cup B)& =\inf \left\{\mu \right(E):E\in \mathfrak{B}(X),E\supset A\cup B\}\hfill \\ \hfill & \le \inf \left\{\mu \right(C\cup D):C,D\in \mathfrak{B}(X),C\supset A\mathrm{and}D\supset B\}\hfill \\ \hfill & =\inf \left\{\max \right\{\mu \left(C\right),\mu \left(D\right)\}:C,D\in \mathfrak{B}(X),C\supset A\mathrm{and}D\supset B\}\hfill \\ \hfill & =\max \left\{\inf \left\{\mu \right(C):C\in \mathfrak{B}(X),C\supset A\},\inf \left\{\mu \right(D):D\in \mathfrak{B}(X),D\supset B\}\right\}\hfill \\ \hfill & =\max \{{\mu }^{*}\left(A\right),{\mu }^{*}\left(B\right)\}\hfill \end{array}$$

Thus, ${\mu }^{*}(A\cup B)=\max \{{\mu }^{*}\left(A\right),{\mu }^{*}\left(B\right)\}$. The proof of Lemma 2 is completed. □

Let X be a Tychonoff space, and $\beta X$ the Stone–Čech compactification of X. We determine the following set:

$$\begin{array}{c}{\mathsf{I}}_{\tau }\left(X\right)=\{\mu \in I\left(\beta X\right):\mu \left(F\right)=0\mathrm{for}\mathrm{every}F\in \mathfrak{B}\left(\beta X\right),F\subset \beta X\setminus X\}.\end{array}$$

(3)

It is easy to see that $\mu \in I_{\tau }\left(X\right)$ implies ${\mu }^{*}\left(X\right)=1$.

From the definition, we have

$$\begin{array}{cc}\hfill {\mathsf{I}}_{\tau }(\varnothing )& =\{\mu \in \mathsf{I}(\beta X):\mu (F)=0,\forall F\in \mathfrak{B}(\beta X),F\subset \beta X\}\hfill \\ \hfill & =\{\mu \in \mathsf{I}(\beta X):(\mu \left(\beta X\right)=1)\wedge (\mu \left(\beta X\right)=0\left)\right\}=\varnothing ,\hfill \end{array}$$

i.e.,

$$\begin{array}{c}{\mathsf{I}}_{\tau }(\varnothing )=\varnothing .\end{array}$$

(4)

Elements of ${\mathsf{I}}_{\tau }\left(X\right)$ are said to be $\tau$-smooth idempotent probability measures.

For each $\mu \in {\mathsf{I}}_{\tau }\left(X\right)$, we define a set function $\tilde{\mu }:\mathfrak{B}\left(X\right)\to \mathbb{R}$ on the family $\mathfrak{B}\left(X\right)$ of all Borel subsets of X by the formula

$$\begin{array}{c}\tilde{\mu }\left(A\right)={\mu }^{*}\left(A\right)=\inf \{\mu \left(B\right):B\in \mathfrak{B}\left(\beta X\right),B\supset A\},A\in \mathfrak{B}\left(X\right).\end{array}$$

Lemma 3.

$\tilde{\mu }$ is an idempotent probability measure on X.

Proof. 

Evidently, $\tilde{\mu }(\varnothing )=0$. Equality (2) in Definition 1 holds because of Lemma 2. We should show that equality (3) in Definition 1 is also true.

Let $\{A_{\alpha },\alpha \in D\}\subset \mathfrak{B}\left(X\right)$ be an increasing net such that $\underset{\alpha \in D}{\cup }{A}_{\alpha }\in \mathfrak{B}\left(X\right)$. For every $\alpha$, there exists $M_{\alpha }\in \mathfrak{B}\left(\beta X\right)$ such that $M_{\alpha }\cap X=A_{\alpha }$ and $\underset{\alpha \in D}{\cup }{M}_{\alpha }\in \mathfrak{B}\left(\beta X\right)$. One has

$$\begin{array}{c}\left\{M\in \mathfrak{B}\left(\beta X\right):M\supset \underset{\alpha \in D}{\cup }{A}_{\alpha }\right\}\supset \left\{\underset{\alpha \in D}{\cup }{M}_{\alpha }\in \mathfrak{B}\left(\beta X\right):\mathfrak{B}\left(\beta X\right)\ni M_{\alpha }\supset A_{\alpha },\alpha \in D\right\}.\end{array}$$

Then,

$$\begin{array}{cc}\hfill \tilde{\mu }\left(\bigcup _{\alpha \in D}{A}_{\alpha }\right)& =\inf \left\{\mu \left(M\right):M\in \mathfrak{B}\left(\beta X\right),M\supset \underset{\alpha \in D}{\cup }{A}_{\alpha }\right\}\hfill \\ \hfill & \le \inf \left\{\mu \left(\bigcup _{\alpha \in D}{M}_{\alpha }\right):\underset{\alpha \in D}{\cup }{M}_{\alpha }\in \mathfrak{B}\left(\beta X\right),\mathfrak{B}\left(\beta X\right)\ni M_{\alpha }\supset A_{\alpha },\alpha \in D\right\}\hfill \\ \hfill & =\underset{\begin{array}{c}{\displaystyle \bigcup _{\alpha \in D}}{M}_{\alpha }\in \mathfrak{B}\left(\beta X\right),\\ M_{\alpha }\supset A_{\alpha },\alpha \in D\end{array}}{\inf }\left\{\mu \left(\bigcup _{\alpha \in D}{M}_{\alpha }\right)\right\}=\underset{\begin{array}{c}{M}_{\alpha }\in \mathfrak{B}\left(\beta X\right),\\ M_{\alpha }\supset A_{\alpha },\alpha \in D\end{array}}{\inf }\left\{\underset{\alpha \in D}{\sup }\left\{\mu \left(M_{\alpha }\right)\right\}\right\}\hfill \\ \hfill & =\underset{\alpha \in D}{\sup }\left\{\underset{\begin{array}{c}{M}_{\alpha }\in \mathfrak{B}\left(\beta X\right),\\ M_{\alpha }\supset A_{\alpha }\end{array}}{\inf }\left\{\mu \left(M_{\alpha }\right)\right\}\right\}=\underset{\alpha \in D}{sup}\left\{\tilde{\mu }\left(A_{\alpha }\right)\right\},\hfill \end{array}$$

i.e., $\tilde{\mu }\left(\underset{\alpha \in D}{\cup }{A}_{\alpha }\right)\le {\displaystyle \underset{\alpha \in D}{\sup }}\left\{\tilde{\mu }\left(A_{\alpha }\right)\right\}$. Remark 4 implies the inverse inequality. The proof of Lemma 3 is completed. □

It is easy to see that the idempotent measure $\tilde{\mu }$ is $\tau$-smooth. Conversely, each $\tau$-smooth probability measure $\tilde{\mu }$ on X defines a measure $\mu \in {\mathsf{I}}_{\tau }\left(X\right)$, by means of the formula $\mu \left(A\right)=\tilde{\mu }(A\cap X)$, $A\in \mathfrak{B}\left(\beta X\right)$.

Lemma 4.

Let X be a Tychonoff space. If $\mu \in {\mathsf{I}}_{\tau }\left(X\right)$, then, $\mu \left(A\right)=\mu \left(B\right)$ for any two Borel subsets $A,B\subset \beta X$ such that $A\cap X=B\cap X$.

Proof. 

An arbitrary set $A\subset \beta X$ can be expressed in the form $A=(A\cap X)\cup (A\setminus X)$. We will show that

$$\begin{array}{c}\mu \left(A\right)={\mu }^{*}((A\cap X)\cup (A\setminus X))=\max \{{\mu }^{*}(A\cap X),{\mu }^{*}(A\setminus X)\}={\mu }^{*}(A\cap X).\end{array}$$

The outer measures of sets $A\cap X$ and $A\setminus X$ are

$$\begin{array}{c}{\mu }^{*}(A\cap X)=\inf \{\mu (A\cap C):C\in \mathfrak{B}\left(\beta X\right),C\supset X\}\end{array}$$

and

$$\begin{array}{c}{\mu }^{*}(A\setminus X)=\inf \{\mu \left(D\right):D\in \mathfrak{B}\left(\beta X\right),D\supset A\setminus X\}.\end{array}$$

From the relations $D\supset A\setminus X$, $A\supset A\setminus X$, the outer measure of the set $A\setminus X$ can be rewritten as

$${\mu }^{*}(A\setminus X)=\inf \{\mu (A\cap D):D\in \mathfrak{B}\left(\beta X\right),D\supset A\setminus X\}.$$

Since $\mu \in {\mathsf{I}}_{\tau }\left(X\right)$, for any sets $D\in \mathfrak{B}\left(\beta X\right)$, $D\supset A\setminus X$ and $C\in \mathfrak{B}\left(\beta X\right)$, $C\supset X$, we have $\mu \left(C\right)=1=\mu \left(\beta X\right)\ge \mu \left(D\right)$. Then, $\mu (A\cap C)\ge \mu (A\cap D)$ for each $A\in \mathfrak{B}\left(\beta X\right)$. Since the sets C and D are arbitrary, we have the inequality ${\mu }^{*}(A\cap X)\ge {\mu }^{*}(A\setminus X)$. So, $\mu \left(A\right)={\mu }^{*}(A\cap X)$.

Like this, we can obtain the equality $\mu \left(B\right)={\mu }^{*}(B\cap X)$ for a set B as well. By the data, $A\cap X=B\cap X$, which implies $\mu \left(A\right)=\mu \left(B\right)$. □

Let X and Y be Tychonoff spaces, $f:X\to Y$ a continuous mapping, and $\beta f:\beta X\to \beta Y$ the Stone–Čech compactification of f.

Theorem 2.

For Tychonoff spaces X, Y and a continuous mapping $f:X\to Y$, we have

$$=I\left(\beta f\right)\left({\mathsf{I}}_{\tau }\left(X\right)\right)\subset {\mathsf{I}}_{\tau }\left(Y\right).$$

Proof. 

Take an arbitrary $\mu \in \mathsf{I}\left(\beta X\right)$ and suppose $\mathsf{I}\left(\beta f\right)\left(\mu \right)\notin {\mathsf{I}}_{\tau }\left(Y\right)$. Then, we have $\mathsf{I}\left(\beta f\right)\left(\mu \right)\left(\mathrm{\Gamma }\right)=a>0$ for some $\mathrm{\Gamma }\in \mathfrak{B}\left(\beta Y\right)$, $\mathrm{\Gamma }\cap Y=\varnothing$. From the equality ${\left(\beta f\right)}^{\leftarrow }(\beta Y\setminus Y)=\beta X\setminus X$, we obtain ${\left(\beta f\right)}^{\leftarrow }\left(\mathrm{\Gamma }\right)\cap X=\varnothing$, and on the other hand, ${\left(\beta f\right)}^{\leftarrow }\left(\mathrm{\Gamma }\right)\in \mathfrak{B}\left(\beta X\right)$, whereupon, $\mu \left({\left(\beta f\right)}^{\leftarrow }\left(\mathrm{\Gamma }\right)\right)=a>0$. Hence, $\mu \notin {\mathsf{I}}_{\tau }\left(X\right)$. Thus, the required inclusion and Theorem 2 are established. □

The above proven Theorem 2 gives us an opportunity to determine the following mapping:

$$\begin{array}{c}{\mathsf{I}}_{\tau }{\left(f\right)=\mathsf{I}\left(\beta f\right)|}_{{\mathsf{I}}_{\tau }\left(X\right)}:{\mathsf{I}}_{\tau }\left(X\right)\to {\mathsf{I}}_{\tau }\left(Y\right).\end{array}$$

Let us recall that a mapping $f:X\to Y$ between topological spaces is called perfect if it is closed and the preimage $f^{\leftarrow }\left(y\right)$ of every point $y\in Y$ is compact.

Theorem 3.

The operation ${\mathsf{I}}_{\tau }$ putting the mapping ${\mathsf{I}}_{\tau }\left(f\right):{\mathsf{I}}_{\tau }\left(X\right)\to {\mathsf{I}}_{\tau }\left(Y\right)$ in correspondence with the mapping $f:X\to Y$ preserves the class of perfect mappings.

Proof. 

Let $f:X\to Y$ be a perfect mapping of Tychonoff spaces. Then, for the extension $\beta f:\beta X\to \beta Y$ of f, one has $\beta f(\beta X\setminus X)\subset \beta Y\setminus Y$ [9]. We consider a mapping $I\left(\beta f\right):I\left(\beta X\right)\to I\left(\beta Y\right)$ and prove that the relation $\mathsf{I}\left(\beta f\right)(\mathsf{I}\left(\beta X\right)\setminus {\mathsf{I}}_{\tau }\left(X\right))\subset \mathsf{I}\left(\beta Y\right)\setminus {\mathsf{I}}_{\tau }\left(Y\right)$ holds as well.

Let us take an arbitrary element $\mu \in \mathsf{I}\left(\beta Y\right)\setminus {\mathsf{I}}_{\tau }\left(Y\right)$. We should show that $\mathsf{I}{\left(\beta f\right)}^{\leftarrow }\left(\mu \right)\subset \mathsf{I}\left(\beta X\right)\setminus {\mathsf{I}}_{\tau }\left(X\right)$. Since $\mu \notin {\mathsf{I}}_{\tau }\left(Y\right)$, we have $\mu \left(\mathrm{\Gamma }\right)=a>0$ for some $\mathrm{\Gamma }\in \mathfrak{B}\left(\beta Y\right)$, $\mathrm{\Gamma }\cap Y=\varnothing$. By definition, for every $\nu \in I{\left(\beta f\right)}^{\leftarrow }\left(\mu \right)$, we have $I\left(\beta f\right)\left(\nu \right)=\mu$. From the equality ${\left(\beta f\right)}^{\leftarrow }(\beta Y\setminus Y)=\beta X\setminus X$, we obtain ${\left(\beta f\right)}^{\leftarrow }\left(\mathrm{\Gamma }\right)\cap X=\varnothing$, and on the other hand, ${\left(\beta f\right)}^{\leftarrow }\left(\mathrm{\Gamma }\right)\in \mathfrak{B}\left(\beta X\right)$, whereupon, $\mu \left(\mathrm{\Gamma }\right)=\nu \left({\left(\beta f\right)}^{\leftarrow }\left(\mathrm{\Gamma }\right)\right)=a>0$. Hence, $\nu \in \mathsf{I}\left(\beta X\right)\setminus {\mathsf{I}}_{\tau }\left(X\right)$. By virtue of the arbitrariness of $\nu \in I{\left(\beta f\right)}^{\leftarrow }\left(\mu \right)$, we conclude that $\mathsf{I}{\left(\beta f\right)}^{\leftarrow }\left(\mu \right)\subset \mathsf{I}\left(\beta X\right)\setminus {\mathsf{I}}_{\tau }\left(X\right)$.

Thus, $\mathsf{I}\left(\beta f\right)(\mathsf{I}\left(\beta X\right)\setminus {\mathsf{I}}_{\tau }\left(X\right))\subset \mathsf{I}\left(\beta Y\right)\setminus {\mathsf{I}}_{\tau }\left(Y\right)$. Since $\mathsf{I}\left(\beta f\right):\mathsf{I}\left(\beta X\right)\to \mathsf{I}\left(\beta Y\right)$ is a mapping between compact Hausdorff spaces, the mapping ${\mathsf{I}}_{\tau }{\left(f\right)=\mathsf{I}\left(\beta f\right)|}_{{\mathsf{I}}_{\tau }\left(X\right)}:{\mathsf{I}}_{\tau }\left(X\right)\to {\mathsf{I}}_{\tau }\left(Y\right)$ is perfect. Theorem 3 is proved. □

Theorem 4.

The operation ${\mathsf{I}}_{\tau }$ putting the mapping ${\mathsf{I}}_{\tau }\left(f\right):{\mathsf{I}}_{\tau }\left(X\right)\to {\mathsf{I}}_{\tau }\left(Y\right)$ in correspondence with the mapping $f:X\to Y$ preserves the class of embeddings.

Proof. 

Let $f:X\to Y$ be an embedding of topological spaces and $\beta f:\beta X\to \beta Y$ its Stone–Čech compact extension.

As we have mentioned above, $\beta f(\beta X\setminus X)\subset \beta Y\setminus f\left(X\right)$. Put $A={\mathsf{I}}_{\tau }\left(f\left(X\right)\right)$. One can prove the relation $\mathsf{I}\left(\beta f\right)(\mathsf{I}\left(\beta X\right)\setminus {\mathsf{I}}_{\tau }\left(X\right))\subset \mathsf{I}\left(\beta Y\right)\setminus A$ similarly to the proof of Theorem 3.

Theorem 2 gives $\mathsf{I}\left(\beta f\right)\left({\mathsf{I}}_{\tau }\left(X\right)\right)\subset A$. Therefore, ${\mathsf{I}}_{\tau }{\left(f\right)=\mathsf{I}\left(\beta f\right)|}_{{\mathsf{I}}_{\tau }\left(X\right)}:{\mathsf{I}}_{\tau }\left(X\right)\to A$ is a proper mapping.

We will show that ${\mathsf{I}}_{\tau }\left(f\right):{\mathsf{I}}_{\tau }\left(X\right)\to {\mathsf{I}}_{\tau }\left(Y\right)$ is an injective mapping. Then, it follows that ${\mathsf{I}}_{\tau }\left(f\right)$ is an embedding. Let $\mu ,\eta \in {\mathsf{I}}_{\tau }\left(X\right)$ be two different measures. Then, there exists a Borel set $Z\subset \beta X$ such that $\mu \left(Z\right)\ne \eta \left(Z\right)$. We should prove that ${\mathsf{I}}_{\tau }\left(f\right)\left(\mu \right)\left(\beta f\left(Z\right)\right)\ne {\mathsf{I}}_{\tau }\left(f\right)\left(\eta \right)\left(\beta f\left(Z\right)\right)$.

Put $Z^{\prime }={\left(\beta f\right)}^{\leftarrow }\left(\beta f\left(Z\right)\right)$. By definition of ${\mathsf{I}}_{\tau }\left(f\right)\left(\mu \right)$, we have ${\mathsf{I}}_{\tau }\left(f\right)\left(\mu \right)\left(\beta f\left(Z\right)\right)=\mu \left(Z^{\prime }\right)$ and ${\mathsf{I}}_{\tau }\left(f\right)\left(\eta \right)\left(\beta f\left(Z\right)\right)=\eta \left(Z^{\prime }\right)$. Since f is an embedding, $Z^{\prime }\cap X=Z\cap X$ holds. In reality, since $Z=(Z\cap X)\cup (Z\setminus X)$, one has

$$\begin{array}{cc}\hfill Z^{\prime }\cap X& =\left({\left(\beta f\right)}^{\leftarrow }\left(\beta f\right)((Z\cap X)\cup (Z\setminus X))\right)\cap X\hfill \\ \hfill & =\left({\left(\beta f\right)}^{\leftarrow }(\beta f(Z\cap X)\cup \beta f(Z\setminus X))\right)\cap X\hfill \\ \hfill & =(\left({\left(\beta f\right)}^{\leftarrow }\left(\beta f(Z\cap X)\right)\right)\cup \left({\left(\beta f\right)}^{\leftarrow }\left(\beta f(Z\setminus X)\right)\right))\cap X\hfill \\ \hfill & ={\left(\beta f\right)}^{\leftarrow }\left(\beta f(Z\cap X)\right)=Z\cap X.\hfill \end{array}$$

Then, by Lemma 4, we have

$$\begin{array}{c}{\mathsf{I}}_{\tau }\left(f\right)\left(\mu \right)\left(\beta f\left(Z\right)\right)=\mu \left(Z^{\prime }\right)=\mu \left(Z\right)\ne \eta \left(Z\right)=\eta \left(Z^{\prime }\right)={\mathsf{I}}_{\tau }\left(f\right)\left(\eta \right)\left(\beta f\left(Z\right)\right).\end{array}$$

Theorem 4 is proved. □

Theorem 5.

The operation ${\mathsf{I}}_{\tau }$ preserves the preimages of Borel sets, i.e., for every continuous mapping $f:X\to Y$ of Tychonoff spaces and any Borel subset $A\subset Y$, the equality ${\mathsf{I}}_{\tau }{\left(f\right)}^{\leftarrow }\left({\mathsf{I}}_{\tau }\left(A\right)\right)={\mathsf{I}}_{\tau }\left(f^{\leftarrow }\left(A\right)\right)$ holds.

Proof. 

We state that ${\mathsf{I}}_{\tau }\left(f^{\leftarrow }\left(A\right)\right)\subset {\mathsf{I}}_{\tau }{\left(f\right)}^{\leftarrow }\left({\mathsf{I}}_{\tau }\left(A\right)\right)$ for any (not necessary Borel) subset $A\subset Y$. Take an arbitrary $\mu \in \mathsf{I}\left(\beta X\right)$, and suppose ${\mathsf{I}}_{\tau }\left(f\right)\left(\mu \right)\notin {\mathsf{I}}_{\tau }\left(A\right)$. Then, we have ${\mathsf{I}}_{\tau }\left(f\right)\left(\mu \right)\left(\mathrm{\Gamma }\right)=a>0$ for some $\mathrm{\Gamma }\in \mathfrak{B}\left(\beta Y\right)$, $\mathrm{\Gamma }\cap A=\varnothing$. Obviously, $f^{\leftarrow }\left(\mathrm{\Gamma }\right)\cap f^{\leftarrow }\left(A\right)=f^{\leftarrow }(\mathrm{\Gamma }\cap A)=\varnothing$, and on the other hand, $f^{\leftarrow }\left(\mathrm{\Gamma }\right)\in \mathfrak{B}\left(\beta X\right)$, whereupon, $\mu \left(f^{\leftarrow }\left(\mathrm{\Gamma }\right)\right)={\mathsf{I}}_{\tau }\left(f\right)\left(\mu \right)\left(\mathrm{\Gamma }\right)=a>0$. Hence, $\mu \notin {\mathsf{I}}_{\tau }\left(f^{\leftarrow }\left(A\right)\right)$.

Let us show that the inverse inclusion ${\mathsf{I}}_{\tau }{\left(f\right)}^{\leftarrow }\left({\mathsf{I}}_{\tau }\left(A\right)\right)\subset {\mathsf{I}}_{\tau }\left(f^{\leftarrow }\left(A\right)\right)$ holds for any Borel subset $A\subset Y$. Take an arbitrary element $\mu \in {\mathsf{I}}_{\tau }{\left(f\right)}^{\leftarrow }\left({\mathsf{I}}_{\tau }\left(A\right)\right)$. Then, ${\mathsf{I}}_{\tau }\left(f\right)\left(\mu \right)\in {\mathsf{I}}_{\tau }\left(A\right)$, and we have

$$\begin{array}{c}\mu (X\setminus f^{\leftarrow }\left(A\right))=\mu \left(f^{\leftarrow }(Y\setminus A)\right)={\mathsf{I}}_{\tau }\left(f\right)\left(\mu \right)(Y\setminus A)=0.\end{array}$$

Consequently, $\mu \in {\mathsf{I}}_{\tau }\left(f^{\leftarrow }\left(A\right)\right)$. □

The authors do not know the answer to the next question.

Question 1.

Let $f:X\to Y$ be a continuous mapping of Tychonoff spaces, $A\subset Y$ an arbitrary set, $B\subset X$ a Borel set containing $f^{\leftarrow }\left(A\right)$. Does there exist a Borel set C in Y such that $f^{\leftarrow }\left(A\right)\subset f^{\leftarrow }\left(C\right)\subset B$?

An affirmative answer to Question 1 makes it possible to obtain a more rigorous result in Theorem 5 by getting rid of the condition on one of the sets to be Borel.

Theorem 6.

Let X be a Tychonoff space and A, $B\subset X$ any subsets such that at least one of them is Borel. Then, the equality ${\mathsf{I}}_{\tau }(A\cap B)={\mathsf{I}}_{\tau }\left(A\right)\cap {\mathsf{I}}_{\tau }\left(B\right)$ holds.

Proof. 

Assume $A\cap B=\varnothing$. Then, (4) implies ${\mathsf{I}}_{\tau }(A\cap B)=\varnothing$.

Now, we will show that ${\mathsf{I}}_{\tau }\left(A\right)\cap {\mathsf{I}}_{\tau }\left(B\right)=\varnothing$. Suppose A is a Borel set. For an arbitrary $\mu \in {\mathsf{I}}_{\tau }\left(A\right)$, we have $\mu (\beta X\setminus A)=0$. Consequently, ${\mu }^{*}\left(B\right)=0$, which provides $\mu \notin {\mathsf{I}}_{\tau }\left(B\right)$. On the other hand, for every $\mu \in {\mathsf{I}}_{\tau }\left(B\right)$, we have $\mu \left(A\right)=0$ and $\mu \notin {\mathsf{I}}_{\tau }\left(A\right)$. So, $I_{\tau }\left(A\right)\cap {\mathsf{I}}_{\tau }\left(B\right)=\varnothing$. Hence, we obtain $I_{\tau }(A\cap B)={\mathsf{I}}_{\tau }\left(A\right)\cap {\mathsf{I}}_{\tau }\left(B\right)$ in this case.

Consider the case $A\cap B\ne \varnothing$. The inclusion ${\mathsf{I}}_{\tau }(A\cap B)\subset {\mathsf{I}}_{\tau }\left(A\right)\cap {\mathsf{I}}_{\tau }\left(B\right)$ directly follows from Theorem 4.

Let us show the inverse inclusion, i.e., ${\mathsf{I}}_{\tau }(A\cap B)\supset {\mathsf{I}}_{\tau }\left(A\right)\cap {\mathsf{I}}_{\tau }\left(B\right)$. Suppose that $\mu \in {\mathsf{I}}_{\tau }\left(A\right)\cap {\mathsf{I}}_{\tau }\left(B\right)\subset {\mathsf{I}}_{\tau }\left(X\right)$, and A is a Borel set. Since $\mu \in {\mathsf{I}}_{\tau }\left(A\right)$, we obtain $\mu (\beta X\setminus A)=0$. Then, the inclusion $\mu \in I_{\tau }\left(B\right)$ implies ${\mu }^{*}(B\setminus A)={\mu }^{*}(B\cap (\beta X\setminus A))=0$ and ${\mu }^{*}(B\cap A)=1$. We should prove $\mu \left(\mathrm{\Phi }\right)=0$ for every $\mathrm{\Phi }\in \mathfrak{B}\left(\beta X\right)$ with $\mathrm{\Phi }\cap (A\cap B)=\varnothing$. We decompose $\mathrm{\Phi }={\mathrm{\Phi }}_A\cup {\mathrm{\Phi }}_B$ setting ${\mathrm{\Phi }}_A=\mathrm{\Phi }\setminus A$ and ${\mathrm{\Phi }}_B=\mathrm{\Phi }\cap A$. Then, $\mu \left({\mathrm{\Phi }}_A\right)=0$ by definition. And $\mu \left({\mathrm{\Phi }}_B\right)=0$ because ${\mathrm{\Phi }}_B\cap B=\varnothing$. Consequently, $\mu \left(\mathrm{\Phi }\right)=0$. So, $\mu \in {\mathsf{I}}_{\tau }(A\cap B)$.

Thus, Theorem 6 is completely proved. □

Remark 5.

The assertion of Theorem 6 cannot be improved, i.e., the requirement that at least one of the given sets be a Borel set cannot be omitted. Indeed, consider the segment $[0;1]$. We will say $x\sim y$ if and only if $y-x\in \mathbb{Q}$; here, $\mathbb{Q}$ is the set of all rational numbers. The relation ∼ is an equivalence relation, (i.e., reflexive, symmetric, and transitive). $\left[x\right]$ denotes an equivalence class containing $x\in [0;1]$, whereupon we obtain a partition of $[0;1]$ into disjoint equivalence classes. Each class $\left[x\right]$ has many countable elements. By $\mathcal{A}$, we denote a system of all mutually distinct equivalence classes. Since $[0;1]=\underset{\left[x\right]\in \mathcal{A}}{\cup }\left[x\right]$, the system $\mathcal{A}$ has a power of the continuum. On the other hand, one can see that each class is everywhere dense in $[0;1]$. From each class (using the Choice Axiom), we take one point so that the resulting set A is everywhere dense in $[0;1]$. Then, from each class (using the Choice Axiom again), we choose a point that differs from the one taken before and construct a set B which is also everywhere dense in $[0;1]$. Neither A nor B are Borel sets, and moreover, $A\cap B=\varnothing$ by virtue of the construction. It is clear there exists no couples of Borel sets U and V in $[0;1]$ such that $A\subset U$, $B\subset V$ and $U\cap V=\varnothing$. Note, Borel sets in $[0;1]$ which can contain A and B are only $(0;1)$, $[0;1)$, $(0;1]$ or $[0;1]$.

Suppose the only Borel set containing the set A is $[0;1)$. Two cases are possible: $B\subset (0;1)$ or $B\subset (0;1]$. Consequently, ${\mathsf{I}}_{\tau }\left(A\right)={\mathsf{I}}_{\tau }\left([0;1)\right)$, and ${\mathsf{I}}_{\tau }\left(B\right)={\mathsf{I}}_{\tau }\left((0;1)\right)$ or ${\mathsf{I}}_{\tau }\left(B\right)={\mathsf{I}}_{\tau }\left((0;1]\right)$. Since $[0;1)$, $(0;1)$ and $(0;1]$ are Borel sets, we have

$${\mathsf{I}}_{\tau }\left(A\right)\cap {\mathsf{I}}_{\tau }\left(B\right)={\mathsf{I}}_{\tau }\left((0;1)\right)\ne {\mathsf{I}}_{\tau }(\varnothing )={\mathsf{I}}_{\tau }(A\cap B).$$

In the other cases ($A\subset (0;1)$, $A\subset (0;1]$), in exactly the same way as the above, one can establish that the conclusion of Theorem 6 is not true for the sets A and B.

One can select sets A and B such that $A\cap B\subset \left[\frac{1}{3};\frac{2}{3}\right]$ and the intersection is dense in $\left(\frac{1}{3};\frac{2}{3}\right)$. Then, the only Borel set containing the intersection is $\left(\frac{1}{3};\frac{2}{3}\right)$. Evidently,

$${\mathsf{I}}_{\tau }\left(A\right)\cap {\mathsf{I}}_{\tau }\left(B\right)={\mathsf{I}}_{\tau }\left((0;1)\right)\ne {\mathsf{I}}_{\tau }\left(\left(\frac{1}{3};\frac{2}{3}\right)\right)={\mathsf{I}}_{\tau }(A\cap B).$$

The set of type A (or B) considered in Remark 5 was found by Giuseppe Vitali in 1905. Therefore, such a set is called the Vitali set.

5. Idempotent Probability Measures Spaces on Čech-Complete Tychonoff Spaces

We begin our investigation of the spaces of idempotent probability measures on a Čech-complete Tychonoff space by stating the following result.

Theorem 7.

Let X be a Tychonoff space and $\{X_{\alpha }:\alpha \in D\}$ any family of Borel subsets of X such that $\underset{\alpha \in D}{\cap }{X}_{\alpha }\in \mathfrak{B}\left(\beta X\right)$. Then, the equality ${\mathsf{I}}_{\tau }\left(\underset{\alpha \in D}{\cap }{X}_{\alpha }\right)=\underset{\alpha \in D}{\cap }{\mathsf{I}}_{\tau }\left(X_{\alpha }\right)$ holds.

Proof. 

It is easy to see that ${\mathsf{I}}_{\tau }\left({\displaystyle \bigcap _{\alpha \in D}}{X}_{\alpha }\right)\subset \underset{\alpha \in D}{\cap }{\mathsf{I}}_{\tau }\left(X_{\alpha }\right)$. Indeed, it is clear that $\underset{\alpha \in D}{\cap }{X}_{\alpha }\subset X_{\alpha }$ for every $\alpha \in D$. Since $\mu \in {\mathsf{I}}_{\tau }\left({\displaystyle \bigcap _{\alpha \in D}}{X}_{\alpha }\right)$, we have $\mu \left(\beta X\setminus \left({\displaystyle \bigcap _{\alpha \in D}}{X}_{\alpha }\right)\right)=0$. Hence, $\mu (\beta X\setminus X_{\alpha })=0$ for arbitrary $\alpha \in D$, i.e., $\mu \in {\displaystyle \bigcap _{\alpha \in D}}{\mathsf{I}}_{\tau }\left(X_{\alpha }\right)$.

Let us show the inverse inclusion, i.e., ${\mathsf{I}}_{\tau }\left({\displaystyle \bigcap _{\alpha \in D}}{X}_{\alpha }\right)\supset {\displaystyle \underset{\alpha \in D}{\cap }}{\mathsf{I}}_{\tau }\left(X_{\alpha }\right)$. Suppose that $\mu \in {\displaystyle \underset{\alpha \in D}{\cap }}{\mathsf{I}}_{\tau }\left(X_{\alpha }\right)$. Then, $\mu \in {\mathsf{I}}_{\tau }\left(X_{\alpha }\right)$ for every $\alpha \in D$. Hence, for any $\alpha \in D$, one has $\mu (\beta X\setminus X_{\alpha })=0$. Consequently,

$$\begin{array}{c}\hfill \mu \left(\beta X\setminus \left(\bigcap _{\alpha \in D}{X}_{\alpha }\right)\right)=\mu \left(\bigcup _{\alpha \in D}\left(\beta X\setminus X_{\alpha }\right)\right)=\underset{\alpha \in D}{sup}\left\{\mu (\beta X\setminus X_{\alpha })\right\}=0.\end{array}$$

Thus, $\mu \in {\mathsf{I}}_{\tau }\left({\displaystyle \bigcap _{\alpha \in D}}{X}_{\alpha }\right)$. Theorem 7 is proved. □

Corollary 1.

Let X be a Tychonoff space and $\{X_{\alpha }:\alpha \in D\}$ any family of closed subsets of X. Then, ${\mathsf{I}}_{\tau }\left({\displaystyle \bigcap _{\alpha \in D}}{X}_{\alpha }\right)=\underset{\alpha \in D}{\cap }{\mathsf{I}}_{\tau }\left(X_{\alpha }\right)$.

Corollary 2.

Let X be a Tychonoff space and $\{X_{\alpha }:\alpha \in D\}$ any family of open subsets of X such that ${\displaystyle \bigcap _{\alpha \in D}}{X}_{\alpha }\in \mathfrak{B}\left(\beta X\right)$. Then, ${\mathsf{I}}_{\tau }\left({\displaystyle \bigcap _{\alpha \in D}}{X}_{\alpha }\right)={\displaystyle \underset{\alpha \in D}{\cap }}{\mathsf{I}}_{\tau }\left(X_{\alpha }\right)$.

Note that the following assertion ensures that every $X_{\alpha }$ is closed in Corollary 1.

Proposition 3.

For any closed subset $F\subset X$ of a Tychonoff space X, the set ${\mathsf{I}}_{\tau }\left(F\right)$ is closed in ${\mathsf{I}}_{\tau }\left(X\right)$.

Proof. 

Take any $\mu \in {\mathsf{I}}_{\tau }\left(X\right)\setminus {\mathsf{I}}_{\tau }\left(F\right)$. Then, there exists an open subset $V\subset \beta X$ such that $F\cap V=\varnothing$ and $\mu \left(V\right)=a>0$. Consider an open neighbourhood $\langle \mu ;V;\frac{a}{2}\rangle$ of $\mu$ and claim that $\langle \mu ;V;\frac{a}{2}\rangle \cap {\mathsf{I}}_{\tau }\left(F\right)=\varnothing$. Indeed, for every $\xi \in I_{\tau }\left(F\right)$, one has $|\xi \left(V\right)-\mu \left(V\right)|=|0-a|=a>\frac{a}{2}$, and $\xi \notin \langle \mu ;V;\frac{a}{2}\rangle$. Or for each $\zeta \in \langle \mu ;V;\frac{a}{2}\rangle$, one has $|\zeta \left(V\right)-\mu \left(V\right)|=|\zeta \left(V\right)-a|<\frac{a}{2}$, which gives $\zeta \left(V\right)>\frac{a}{2}\ne 0$. Hence, $\zeta \notin {\mathsf{I}}_{\tau }\left(F\right)$. Thus, ${\mathsf{I}}_{\tau }\left(F\right)$ is closed in ${\mathsf{I}}_{\tau }\left(X\right)$. □

Now, we are ready to formulate the following result. Note that a linear (in the classical sense) version of this result was established in [9].

Theorem 8.

The operation ${\mathsf{I}}_{\tau }$ preserves the Čech-completeness of Tychonoff spaces; in other words, if X is a Čech-complete Tychonoff space, then, ${\mathsf{I}}_{\tau }\left(X\right)$ is also a Čech-complete Tychonoff space.

Proof. 

Let X be a Čech-complete Tychonoff space. Then, X is a $G_{\delta }$-set in $\beta X$, i.e., there are many countable open sets $U_n\subset \beta X$, such that $X=\underset{n=1}{\stackrel{\infty }{\cap }}{U}_n$. Then,

$$\begin{array}{cc}\hfill {\mathsf{I}}_{\tau }\left(X\right)& ={\mathsf{I}}_{\tau }\left(\underset{n=1}{\stackrel{\infty }{\cap }}{U}_n\right)=\left\{\mu \in \mathsf{I}\left(\beta X\right):\mu \left(\beta X\setminus \underset{n=1}{\stackrel{\infty }{\cap }}{U}_n\right)=0\right\}\hfill \\ \hfill & =\underset{n=1}{\stackrel{\infty }{\cap }}\left\{\mu \in I\left(\beta X\right):\mu (\beta X\setminus U_n)<\frac{1}{n}\right\}.\hfill \end{array}$$

Applying Remark 2, by [17], we find that ${\mathsf{I}}_{\tau }\left(X\right)$ is a Čech-complete Tychonoff space. Theorem 8 is proved. □

Since Čech-completeness is hereditary with respect to $G_{\delta }$-subsets (Theorem 3.9.6, [17]), one obtains the following result.

Corollary 3.

For a Tychonoff space X, the space ${\mathsf{I}}_{\tau }\left(X\right)$ is Čech-complete if and only if X is Čech-complete.

From Theorem 8 and Corollary 1, one can immediately extract that, in contrast to closed sets, for an open set U, the set ${\mathsf{I}}_{\tau }\left(U\right)$ is a $G_{\delta }$-set in $I(\beta X\setminus X)$ for a Čech-complete Tychonoff space X.

Proposition 4.

For an open set $U\subset X$ of a Čech-complete Tychonoff space X, the set ${\mathsf{I}}_{\tau }\left(U\right)$ is a $G_{\delta }$-set.

Proof. 

Since X is a Čech-complete Tychonoff space, it is a $G_{\delta }$-set in $\beta X$, and let $X=\underset{n=1}{\stackrel{\infty }{\cap }}{U}_n$, where $U_n$ are open subsets of $\beta X$. Then, $X\in \mathfrak{B}\left(\beta X\right)$, and $U\in \mathfrak{B}\left(\beta X\right)$ for each open $U\subset X$. An open set $U\subset X$ admits a representation $U=\underset{n=1}{\stackrel{\infty }{\cap }}(U\cap U_n)$. Finally,

$$\begin{array}{cc}\hfill {\mathsf{I}}_{\tau }\left(U\right)& ={\mathsf{I}}_{\tau }\left(\underset{n=1}{\stackrel{\infty }{\cap }}(U\cap U_n)\right)=\left\{\mu \in \mathsf{I}\left(\beta X\right):\mu \left(\beta X\setminus \underset{n=1}{\stackrel{\infty }{\cap }}(U\cap U_n)\right)=0\right\}\hfill \\ \hfill & =\underset{n=1}{\stackrel{\infty }{\cap }}\left\{\mu \in \mathsf{I}\left(\beta X\right):\mu (\beta X\setminus (U\cap U_n))<\frac{1}{n}\right\}.\hfill \end{array}$$

Proposition 4 is proved. □

6. Conclusions

We have provided a way to define topology T on the set $\mathsf{I}\left(X\right)$ of probability measures on a compact Hausdorff space X and described some closed and some open sets in that topology. It is shown that the space $\left(\mathsf{I}\right(X),T)$ is compact Hausdorff if X is so. The topology T (on $\beta X$) is used to define topology on the set ${\mathsf{I}}_{\tau }\left(X\right)$ of $\tau$-smooth idempotent probability measures on a Tychonoff space X. Some properties of ${\mathsf{I}}_{\tau }$ are established, and it is shown that the space ${\mathsf{I}}_{\tau }\left(X\right)$ is Čech-complete if and only if X is Čech-complete. It is natural to continue the investigation of properties of ${\mathsf{I}}_{\tau }\left(X\right)$ for some other topological properties which are not studied in the literature.