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% Full title of the paper (Capitalized)
\Title{On \v{C}ech-completeness of the space of $\tau$-smooth
idempotent probability measures}

% MDPI internal command: Title for citation in the left column
\TitleCitation{On \v{C}ech-completeness of $I_{\tahu}(X)$}

% Author Orchid ID: enter ID or remove command
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%\newcommand{\orcidauthorB}{0000-0000-0000-000X} % Add \orcidB{} behind the author's name

% Authors, for the paper (add full first names)
\Author{Lj. D. R. Ko\v cinac
$^{1,\dagger,\ddagger}$\orcid{000-0-0002-4870-7908}}, A. A. Zaitov
$^{2}$ and M. R. Eshimbetov $^{3}$*}

%\longauthorlist{yes}

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\AuthorNames{Lj. D. R. Ko\v cinac, A. A. Zaitov, and M. R.
Eshimbetov}

% MDPI internal command: Authors, for citation in the left column
\AuthorCitation{Ko\v cinac, Lj.D.R.; Zaitov, Eshimbetov, M.R.}
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% Affiliations / Addresses (Add [1] after \address if there is only one affiliation.)
\address{%
$^{1}$ \quad University of Ni\v s, Faculty of Sciences and Mathematics, 18000 Nis, Serbia; lkocinac2gmail.com\\
$^{1}$ \quad State University of Noni Pazar, 36230 Novj Pazar,
Serbia\\
$^{2}$ \quad Tashkent Architecture and Civil Engineering University,
Yangi Shahar Str. 9, 100194 Tashkent, Uzbekistan;
adilbek\_zaitov@mail.ru\\
$^{3}$ \quad Institute of Mathematics named after V.~I.~Romanovsky,
Universitet Str. 9, 100174 Tashkent, Uzbekistan;
mr.eshimbetov@gmail.com }



% Contact information of the corresponding author
\corres{Correspondence: lkocinac@gmail.come-mail.com}}

% Current address and/or shared authorship
%\firstnote{Current address: Affiliation 3.}
\secondnote{These authors contributed equally to this work.}
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%\simplesumm{} % Simple summary

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% Abstract (Do not insert blank lines, i.e. \\)
\abstract{For the set of probability measures ${\sf I}(X)$, where
$X$ is a compact Hausdorff space, we propose a new way to introduce
the topology by using open subsets of the space $X$. Then, among
other things,  we give a new proof that for a compact Hausdorff
space $X$ the space ${\sf I}(X)$ is also a compact Hausdorff space.
For a Tychonoff space $X$, we consider the topological space ${\sf
I_{\tau}}(X)$ of $\tau$-smooth idempotent probability measures on
$X$,and show that the space ${\sf I_{\tau}}(X)$ is \v{C}ech-complete
if and only if the given space $X$ is \v{C}ech-complete. }

% Keywords
\keyword{\v{C}ech-complete space, compact space, probability
measure, $\tau$-smooth idempotent probability measure, neighbourhood
system.}

% The fields PACS, MSC, and JEL may be left empty or commented out if not applicable
%\PACS{J0101}
\MSC{28C15, 28A33, 46E27, 60B05, 54D99}
%\JEL{}

\section{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'' \cite{gu} edited by J.~Gunawardena (see also
\cite{Green, Kolokoltsov-Maslow}). 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. 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 $x\odot
y = x+y$ (\cite{gu}, \cite{rt1}, \cite{va}, \cite{za1}).

M.~Zarichnyi in 2010 in \cite{zm} investigated categorical
properties of the space of idempotent probability measures. In
\cite{zbr} the theory was extended to the class of metric spaces.
So, the space of idempotent probability measures is a new object.
But it is already 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 its investigations on wider classes of
topological classes than the class of compact Hausdorff spaces, in
particular, the class of Tychonoff spaces.

T.~Banakh \cite{ba}, T.~Banakh and T.~Radul \cite{br},
\cite{fedorchuk}, 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
\cite{aaz}, \cite{az} and \cite{za1} the theory put forward, and in
\cite{rt} some categorical properties of $\tau$-smooth weakly
additive (nonlinear) functionals were established.

The results obtained in  \cite{rt1}, \cite{va}, \cite{za1},
\cite{zk}, \cite{zm}, \cite{zbr} 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 \cite{fs}, \cite{va} and others).

Unlike above mentioned papers \cite{rt1}, \cite{va}, \cite{za1},
\cite{zk}, \cite{zm}, \cite{zbr}, in this paper for a compact
Hausdorff space $X$ we introduce the notion of idempotent measures
as a set-function on the family $\mathfrak{B}(X)$. Note that the
works \cite{pu}, \cite{es}, \cite{es1} and \cite{es2} 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 ${\sf I}(X)$ of
idempotent probability measures is also a compact Hausdorff space.

Further, for a  Tychonoff space $X$, we consider the space
$I_{\tau}(X)$ of $\tau$-smooth idempotent probability measures on
$X$. Then we establish the \v{C}ech-completeness of the space of
$\tau$-smooth idempotent probability measures for the
\v{C}ech-complete Tychonoff space. (\v{C}ech-complete spaces were
introduced by Eduard \v{C}ech in 1937 to prove the Baire category
theorem. Another important application of the \v{C}ech-completeness
appears in the metrization of a topological space by a complete
metric. Note that a topological space $X$ is \v{C}ech-complete if
$X$ is a Tychonoff space and the remainder $\beta X\setminus X$ is
an $F_{\sigma}$-set in the Stone-\v{C}ech compactification $\beta
X$. Locally compact spaces are \v{C}ech-complete, but the inverse is
not take place. The space of all irrational numbers with the
topology of a subspace of the real line is an example of a
\v{C}ech-complete space that is not locally compact \cite{er}.) From
here, since \v{C}ech-completeness is hereditary with respect to
$G_{\delta}$-subsets and a \v{C}ech-complete Tychonoff space $X$ is
$G_{\delta}$-subset in its Stone-\v{C}ech compactification $\beta
X$, we conclude that a Tychonoff space $X$ is \v{C}ech-complete if
and only if $I_{\tau}(X)$ is \v{C}ech-complete. Note that a linear
(in the classical sense) version of this result was established in
\cite{ba}. One can see that the methods used in \cite{aaz},
\cite{az}, \cite{ba}, \cite{br}, \cite{fs}, \cite{va}, \cite{za1}
are not suitable for the present case.

\section{Preliminaries}

Let $X$ be a compact Hausdorff space and $\mathfrak{B}(X)$ 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 \cite{pu}, we enter the following notion.


\begin{Definition}\rm \label{d1}
A set function $\mu\colon\, \mathfrak{B}(X)\to
\overline{\mathbb{R}}_{+}$ is said to be  an \emph{idempotent
measure} on $X$ if the following conditions hold:
\begin{enumerate}
\item[$1)$] $\mu (\emptyset) = 0$;

\item[$2)$] $\mu (A\cup B) = \max \{\mu (A),\, \mu (B)\}$ for any $A,\, B\in \mathfrak{B}(X)$;

\item[$3)$] $\mu \left(\underset{\alpha\in D}{\bigcup}A_{\alpha}\right)
= \mathop{\sup}\limits_{\alpha\in D} \{\mu (A_{\alpha})\}$ for every
increasing net $\{A_{\alpha},\, \alpha\in D\}
\subset\mathfrak{B}(X)$ such that $\underset{\alpha \in
D}{\cup}A_{\alpha} \in \mathfrak{B}(X)$.
\end{enumerate}
\end{Definition}

\begin{Remark} \label{r1}{\rm
Every idempotent measure $\mu$ is increasing, i.~e. for $A$, $B$
$\in \mathfrak{B}(X)$ if $A\subset B$, then $\mu(A)\leq\mu(B)$.}
\end{Remark}

The set of all idempotent measure on $X$ we denote by $IM(X)$. If
$\mu(X)=1$, the idempotent measure $\mu$ is called an
\emph{idempotent probability measure} on $X$. We denote
\begin{gather*}
{\sf I}(X) = \{\mu\in IM(X)\colon\, \mu(X)=1\}.
\end{gather*}


A set
\begin{gather}\label{supp1}
\operatorname{supp}\,\mu = \left[X\setminus \cup\{A\in
\mathfrak{B}(X):\, \mu(A) = 0\}\right]_{X}
\end{gather}
is said to be a \textit{support} of the given idempotent measure
$\mu$.

The support of an idempotent measure $\mu$ can be defined by the
following equality:
\begin{gather}\label{supp2}
\operatorname{supp}\,\mu = \left[\cap\{C\colon\, C\in
\mathfrak{B}(X),\, \mu(X\setminus C) = 0\}\right]_{X}.
\end{gather}


\begin{Lemma}\label{l2}
For every $\mu\in {\sf I}(X)$ we have $\mu\in
I(\operatorname{supp}\,\mu)$.
\end{Lemma}

\begin{proof} For a support we will apply (\ref{supp2}). It is easy to see that
\begin{align*}
\mu(X\setminus \operatorname{supp}\,\mu) &= \mu\left(X\setminus\left[\cap\{C\colon\, C\in \mathfrak{B}(X),\, \mu(X\setminus C)=0\}\right]_{X}\right)\\
&=\mu\left(\operatorname{int}\left(X\setminus\cap\{C\colon\, C\in \mathfrak{B}(X),\, \mu(X\setminus C)=0\}\right)\right)\\
&= \mu\left(\operatorname{int}\left(\cup\{X\setminus C\colon\, C\in \mathfrak{B}(X),\, \mu(X\setminus C)=0\}\right)\right)\\
&\leq \mu\left(\cup\{X\setminus C\colon\, C\in \mathfrak{B}(X),\, \mu(X\setminus C)=0\}\right)\\
&= \mathop{\sup}\{\mu(X\setminus C):\, C\in \mathfrak{B}(X),\,
\mu(X\setminus C)=0\}=0.
\end{align*}
On the other side,
\begin{gather*}
1 = \mu(X)=\mu\left(\left(X\setminus
\operatorname{supp}\,\mu\right)\cup
\operatorname{supp}\,\mu\right)=\mathop{\max}\{\mu(X\setminus
\operatorname{supp}\,\mu),\, \mu(\operatorname{supp}\,\mu)\},
\end{gather*}
i.~e. $\mu(\operatorname{supp}\,\mu)=1$ and $\mu\in
I(\operatorname{supp}\,\mu)$.
\end{proof}


Let $X$ be a compact Hausdorff space, $\mathcal{B}$ a base in $X$,
$U_{i}\in \mathcal{B}$, $i=1,\, \ldots, \, n$, and $\varepsilon >
0$. For an idempotent probability measure $\mu\in {\sf I}(X)$ we
define a set
$$
\langle\mu;\, U_1,\,\ldots,\, U_n;\, \varepsilon\rangle = \{\nu\in
{\sf I}(X)\colon\, |\nu(U_i) - \mu(U_i)| <\varepsilon,\, i=1,\,
\ldots, \, n\}.
$$

Gathering all of such sets construct a family
$$
\mathcal{B}(\mu) = \{\langle\mu;\, U_1,\,\ldots,\, U_n;\,
\varepsilon\rangle\colon\, U_{i}\in \mathcal{B},\, i=1,\, \ldots, \,
n; \varepsilon > 0\}, \quad \mu\in I(X),
$$
and put
$$
\mathcal{B}_{{\sf I}(X)} = \underset{\mu\in {\sf
I}(X)}{\cup}\mathcal{B}(\mu).
$$

\begin{Proposition}\label{p1}
The built family $\mathcal{B}_{{\sf I}(X)}$ forms a base (or, a
neighbourhoods system) for some topology in ${\sf I}(X)$.
\end{Proposition}

\begin{proof}
Since $\mathcal{B} \ne \emptyset$, there exists an open nonempty set
$U\in \mathcal{B}$. On the other side, $U\in \mathfrak{B}(X)$, so
$\mu(U)$ is well defined. That is why for every $\varepsilon > 0$ we
have $\langle\mu;\, U;\, \varepsilon\rangle \in \mathcal{B}(\mu)$.
Consequently, $\mathcal{B}(\mu) \ne \emptyset$. Clearly, $\mu\in
\langle\mu;\, U_1,\,\ldots,\, U_n;\, \varepsilon\rangle$ for each
$\langle\mu;\, U_1,\,\ldots,\, U_n;\, \varepsilon\rangle \in
\mathcal{B}(\mu)$.

Suppose  we get $\mu\in \langle\nu;\, V_1,\,\ldots,\, V_n;\,
\delta\rangle\in \mathcal{B}(\nu)$. Note that $(\mu - \nu)(V) =
\mu(V) - \nu(V)$, and designate
\begin{gather*}
a=\min\{(\nu - \mu)(V_{1}) + \delta,\, (\mu - \nu)(V_{1}) +
\delta,\,\ldots,\, (\nu - \mu)(V_{n}) + \delta,\, (\mu - \nu)(V_{n})
+ \delta\}.
\end{gather*}
Obviously, $a > 0$. Then for every $\xi\in \langle\mu;\,
V_1,\,\ldots,\, V_n;\, a\rangle$ one has
\begin{align*}
|\xi(V_{i}) - \nu(V_{i})| &= |\xi(V_{i}) - \mu(V_{i}) + \mu(V_{i}) - \nu(V_{i})|\leq\\
&\leq |\xi(V_{i}) - \mu(V_{i})| + |\mu(V_{i}) - \nu(V_{i})|<\\
&< a + |\mu(V_{i}) - \nu(V_{i})|.
\end{align*}
Two cases are possible: $\mu(V_{i}) - \nu(V_{i}) \geq 0$ or $\mu(V_{i}) - \nu(V_{i}) \leq 0$. In the front case we take into attention $a < \nu(V_{i}) - \mu(V_{i}) + \delta$, and the last case:  $a < \mu(V_{i}) - \nu(V_{i}) + \delta$. Hence, $\xi\in \langle\nu;\, V_1,\,\ldots,\, V_n;\, \delta\rangle$, in the other words, $\langle\mu;\, V_1,\,\ldots,\, V_n;\, a\rangle\subset \langle\nu;\, V_1,\,\ldots,\, V_n;\, \delta\rangle$.\\


Finally, consider an arbitrary couple of sets $\langle\mu;\,
U_1,\,\ldots,\, U_n;\, \varepsilon\rangle$, $\langle\mu;\,
V_1,\,\ldots,\, V_k;\, \delta\rangle$ belonging $\mathcal{B}(\mu)$.
Denote $\theta = \min\{\varepsilon,\, \delta\}$, and we have
\begin{align*}
&\langle\mu;\, U_1,\,\ldots,\, U_n;\, \varepsilon\rangle\cap \langle\mu;\, V_1,\,\ldots,\, V_k;\, \delta\rangle \supset\\
\supset &\langle\mu;\, U_1,\,\ldots,\, U_n;\, \theta\rangle\cap \langle\mu;\, V_1,\,\ldots,\, V_k;\, \theta\rangle = \\
= &\langle\mu;\, U_1,\,\ldots,\, U_n,\, V_1,\,\ldots,\, V_k;\,
\theta\rangle\in \mathcal{B}(\mu).
\end{align*}
Proposition \ref{p1} is proved.
\end{proof}

\begin{Remark}\label{r2} {\rm For an open set $U$ sets
\begin{gather*}
\langle U;\, \varepsilon\rangle_{-} = \{\xi\in {\sf I}(X)\colon\,
\xi(U) < \varepsilon\}\,\, \mbox{ and }\,\,  \langle U;\,
\varepsilon\rangle_{+} = \{\xi\in I(X)\colon\, \xi(U) >
\varepsilon\}
\end{gather*}
are open. Indeed, we have $\mu(U) = a < \varepsilon$ for every
$\mu\in \langle U;\, \varepsilon\rangle_{-}$. Then for each $\nu\in
\langle \mu;\, U;\, \varepsilon - a \rangle_{-}$ one has
$-(\varepsilon - a) + \mu(U) < \nu(U) < \mu(U) + (\varepsilon - a)$,
$2a-\varepsilon < \nu(U) < \varepsilon$. Hence, $\nu\in\langle U;\,
\varepsilon\rangle_{-}$. The openness of the second set will be
established like to the openness of the first set.}
\end{Remark}

\begin{Remark}\label{r3} {\rm For a closed set $F$ sets
\begin{gather*}
[ F;\, \varepsilon]_{-} = \{\xi\in {\sf I}(X)\colon\, \xi(F) \leq
\varepsilon\}\,\, \mbox{ and }\,\, [ F;\, \varepsilon]_{+} =
\{\xi\in I(X)\colon\, \xi(F) \geq \varepsilon\}
\end{gather*}
are closed. Indeed, for each $\{\xi_{\alpha}\} \subset [ F;\,
\varepsilon]_{-} \subset {\sf I}(X)$ there exists $\xi_{0}\in {\sf
I}(X)$ such that $\xi_{0} = \underset\alpha\lim\, \xi_{\alpha}$
(i.~e, $\xi(A) = \underset\alpha\lim\, \xi_{\alpha}(A)$ for each
$A\in \mathfrak{B}(X)$). Hence, $\xi_{0}(F) \leq \varepsilon$. Thus,
$\xi_{0}\in[ F;\, \varepsilon]_{-}$. In the same way one can show
that $[F;\, \varepsilon]_{+}$ is also closed.}
\end{Remark}

\section{Idempotent probability measures on compact Hausdorff spaces}

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

\begin{Theorem}\label{t1}
For a compact Hausdorff space $X$ the topological space ${\sf I}(X)$
is also a compact Hausdorff space.
\end{Theorem}
\begin{proof}
At first we show that ${\sf I|}(X)$ is a Hausdorff space. Let $\mu$,
$\nu\in {\sf I}(X)$ be different idempotent measures. Then there
exists a set $U\in\mathcal{B}$ such that $\mu(U) \neq \nu(U)$. Put
$|\mu(U) - \nu(U)| = a$. One has $a\ne 0$, $|\mu(U) - \nu(U)| >
\tfrac{a}{2}$. Assume that there exists an idempotent probability
measure $\xi$ belonging to both sets $\langle\mu;\, U;\,
\tfrac{a}{4}\rangle$ and $\langle\nu;\, U;\, \tfrac{a}{4}\rangle$.
Then
\begin{align*}
|\nu(U) - \mu(U)| &= |\nu(U) - \xi(U) + \xi(U) - \mu(U)| \leq\\
& \leq|\nu(U) - \xi(U)| + |\xi(U) - \mu(U)| < \tfrac{a}{4} +
\tfrac{a}{4} = \tfrac{a}{2}.
\end{align*}
The obtained contradiction shows that $\langle\mu;\, U;\,
\tfrac{a}{4}\rangle \cap \langle\nu;\, U;\, \tfrac{a}{4}\rangle
=\emptyset$.


Now we will prove the compactness of $I(X)$ with respect to the
topology generated by the  neighbourhoods system $\mathcal{B}_{{\sf
I}(X)}$.

Clearly, ${\sf I}(X)$ is bounded in $\mathbb{R}^{\mathfrak{B}(X)}$.
It remains to show its closedness. Let $\mu\in
\mathbb{R}^{\mathfrak{B}(X)}\setminus I(X)$. Then the following
cases are possible.

\emph{Case 1}. $\mu(\emptyset) = a \ne 0$. In this case we have
$\langle \mu;\, \emptyset;\, \tfrac{|a|}{2}\rangle \cap {\sf I}(X) =
\emptyset$.

\emph{Case 2}. $\mu(X) = a \ne 1$. Then it os obvious that $\langle
\mu;\, X;\, \tfrac{|a+1|}{2}\rangle \cap {\sf I}(X) = \emptyset$.
\\

\emph{Case 3}. $\mu(G\cup \Gamma) = a \ne b = \max\{\mu(G),\,
\mu(\Gamma)\}$ for some open sets $G$ and $\Gamma$. Whereupon,
$\langle \mu;\, G,\, \Gamma,\, G\cup \Gamma;\, \tfrac{|b -
a|}{2}\rangle \cap I(X) = \emptyset$.

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

The fourth case has the following subcases.

\emph{Subcase 4.1}. Assume $0\leq a < b \leq 1$. By the definition
there exist such an index $\alpha_{0}$ that
$\mu(\Gamma_{\alpha_{0}}) > b - \tfrac{b - a}{4} = \tfrac{a +
3b}{4}$. We claim that $\langle \mu;\, \Gamma_{\alpha_{0}},\,
\underset{\alpha\in D}{\cup} \Gamma_{\alpha};\, \tfrac{b -
a}{4}\rangle \cap {\sf I}(X) = \emptyset$. Presume, it is not so.
Then there exists $\nu$ belonging to this intersection. From here we
get
\begin{align*}
\left|\nu\left(\underset{\alpha\in D}{\bigcup} \Gamma_{\alpha}\right) - \mu\left(\underset{\alpha\in D}{\bigcup} \Gamma_{\alpha}\right)\right| &< \tfrac{b - a}{4},\\
|\nu(\Gamma_{\alpha_{0}}) - \mu(\Gamma_{\alpha_{0}})| &< \tfrac{b -
a}{4}.
\end{align*}
These inequalities provide, correspondingly,
\begin{gather*}
\tfrac{3a - b}{4} < \nu\left(\underset{\alpha\in D}{\bigcup} \Gamma_{\alpha}\right) < \tfrac{3a + b}{4},\\
\tfrac{a + b}{2} < \nu(\Gamma_{\alpha_{0}}) < \tfrac{5b - a}{4}.
\end{gather*}
In the issue one obtains
\begin{gather*}
\underset{\alpha\in D}{\sup}\{\nu(\Gamma_{\alpha})\} \geq
\nu(\Gamma_{\alpha_{0}}) > \tfrac{a + b}{2} > \tfrac{3a + b}{4} >
\nu\left(\underset{\alpha\in D}{\bigcup} \Gamma_{\alpha}\right),
\end{gather*}
i.e. $\underset{\alpha\in D}{\sup}\{\nu(\Gamma_{\alpha})\} >
\nu\left(\underset{\alpha\in D}{\bigcup} \Gamma_{\alpha}\right)$.
The last inequality goes against $\nu\in {\sf I}(X)$. The resulting
contradiction shows that our assumption is wrong, and our statement
about the emptiness of the intersection is true.

\emph{Subcase 4.2}. Assume $0\leq b < a \leq 1$. Take an any
$\xi\in\langle \mu;\, \underset{\alpha\in D}{\cup}
\Gamma_{\alpha};\, \tfrac{a - b}{4}\rangle$. Then
\begin{gather*}
\underset{\alpha\in D}{\sup}\{\xi(\Gamma_{\alpha})\}  < b + \tfrac{a
- b}{4} < a - \tfrac{a - b}{4} = \mu\left(\underset{\alpha\in
D}{\bigcup} \Gamma_{\alpha}\right) - \tfrac{a - b}{4} <
\xi\left(\underset{\alpha\in D}{\bigcup} \Gamma_{\alpha}\right),
\end{gather*}
i.~e. $\underset{\alpha\in D}{\sup}\{\xi(\Gamma_{\alpha})\} <
\xi\left(\underset{\alpha\in D}{\bigcup} \Gamma_{\alpha}\right)$.
Hence, $\xi\not \in I(X)$. Thus, $\langle \mu;\, \underset{\alpha\in
D}{\cup} \Gamma_{\alpha};\, \tfrac{a - b}{4}\rangle\cap {\sf I}(X) =
\emptyset$.

\emph{Subcase 4.3}. Assume $b < 0$. Then $\mu(\Gamma_{\alpha})\leq
b$ for every $\alpha\in D$. For $\tfrac{|b|}{2}$ there exists an
index $\alpha_{0}\in D$ such that $\frac{3b}{2} = b - \tfrac{|b|}{2}
< \mu(\Gamma_{\alpha_{0}}) < b + \tfrac{|b|}{2} = \frac{b}{2}$.
Evidently, $\langle \mu;\, \Gamma_{\alpha_{0}};\,
\tfrac{|b|}{2}\rangle \cap {\sf I}(X) = \emptyset$.

\emph{Subcase 4.4}. Assume $a < 0$. Then $\langle \mu;\,
\underset{\alpha\in D}{\cup} \Gamma_{\alpha};\,
\tfrac{|a|}{2}\rangle \cap {\sf I}(X) = \emptyset$.

\emph{Subcase 4.5}. Assume $b > 1$. Then there exists an index
$\alpha_{0}\in D$ such that $\tfrac{b + 1}{2} = b - \tfrac{b - 1}{2}
< \mu(\Gamma_{\alpha_{0}}) < b + \tfrac{b - 1}{2} = \tfrac{3b -
1}{2}$. Hence, $\langle \mu;\, \Gamma_{\alpha_{0}};\, \tfrac{b -
1}{2}\rangle \cap {\sf I|}(X) = \emptyset$.

\emph{Subcase 4.6}. Assume $a > 1$. Then $\langle \mu;\,
\underset{\alpha\in D}{\cup} \Gamma_{\alpha};\, \tfrac{a -
1}{2}\rangle \cap {\sf I}(X) = \emptyset$.

On this, all possible cases have been exhausted. Consequently, $\mu$
is an interior point of $\mathbb{R}^{\mathfrak{B}(X)}\setminus {\sf
I}(X)$ which immediately implies that ${\sf I}(X)$ is closed in
$\mathbb{R}^{\mathfrak{B}(X)}$ which means its compactness. The
proof of Theorem \ref{t1} is completed.
\end{proof}

For a mapping $f\colon\, X\to Y$ of compact Hausdorff spaces $X$ and
$Y$ we define a mapping
\begin{gather*}
{\sf I}(f)\colon\, {\sf I}(X)\to {\sf I}(Y)
\end{gather*}
by the rule
\begin{gather*}
{\sf I}(f)(\mu)(B) = \mu(f^{\gets}(B)),\qquad B\in \mathfrak{B}(Y).
\end{gather*}

\begin{Proposition}\label{p2}
For every pair of compact Hausdorff spaces $X$ and $Y$ and any
continuous mapping $f\colon\, X\to Y$ the mapping ${\sf
I}(f)\colon\, {\sf I}(X)\to {\sf I}(Y)$ is continuous.
\end{Proposition}

\begin{proof}
Take any $\mu\in {\sf I}(X)$, and let $\nu = {\sf I}(f)(\mu)\in {\sf
I}(Y)$. Consider an arbitrary neighbourhood $\langle\nu;\, V_{1},\,
\ldots,\, V_{k}; \delta\rangle$ of $\nu$. Then for any $\xi\in
\langle\mu;\, f^{\gets}(V_{1}),\, \ldots,\, f^{\gets}(V_{k});
\delta\rangle$ we have
\begin{gather*}
|{\sf I}(f)(\xi)(V_{i})) - \nu(V_{i}))| = |\xi(f^{\gets}(V_{i})) -
\mu(f^{\gets}(V_{i}))| < \delta,
\end{gather*}
which yields ${\sf I}(f)(\xi)\in \langle\nu;\, V_{1},\, \ldots,\,
V_{k}; \delta\rangle$. By virtue of arbitrariness of $\xi$ we obtain
that ${\sf I}(f)\left(\langle\mu;\, f^{\gets}(V_{1}),\, \ldots,\,
f^{\gets}(V_{k}); \delta\rangle\right)\subset  \langle\nu;\,
V_{1},\, \ldots,\, V_{k}; \delta\rangle$.
\end{proof}


\section{$\tau$-smooth idempotent probability measures}

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

An \emph{outer idempotent measure} of an arbitrary subset $A\subset
X$ defines as
\begin{gather*}
\mu^{\ast}(A) = \mathop{\inf }\{\mu(B):\, B \supset A,\, B\in
\mathfrak{B}(X)\}.
\end{gather*}

So, we got an extension $\mu^{\ast}$ of $\mu$. Now Remark \ref{r1}
may be improved as follows

\begin{Remark} \label{r4}{\rm
For every idempotent measure $\mu$ its extension $\mu^{\ast}$ is
increasing, i.~e. if $A\subset B$, then
$\mu^{\ast}(A)\leq\mu^{\ast}(B)$.}
\end{Remark}

\begin{Lemma}\label{l3}
For any pair of $A$, $B\subset X$ and every idempotent measure we
have
\begin{gather*}
\mu^{\ast}(A\cup B)=\mathop{\max}\{\mu^{\ast}(A),\, \mu^{\ast}(B)\}.
\end{gather*}
\end{Lemma}

\begin{proof}
It is easy to see that $\mu^{\ast}(A\cup B)\geq
\mathop{\max}\{\mu^{\ast}(A),\, \mu^{\ast}(B)\}$. Let us show the
inverse inclusion, i.~e. $\mu^{\ast}(A\cup B)\leq
\mathop{\max}\{\mu^{\ast}(A),\, \mu^{\ast}(B)\}$. By the definition
of the outer idempotent measure one has
\[
\mu^{\ast}(A\cup B) = \mathop{\inf}\{\mu(E)\colon\, E\in
\mathfrak{B}(X),\, E\supset A\cup B\}.
\]
Clearly,
\[
\{E\colon\, E\in \mathfrak{B}(X),\, E\supset A\cup B\} \supset
\{C\cup D\colon\, C,\, D\in \mathfrak{B}(X),\, C\supset A\, \mbox{
and }\, D\supset B\}.
\]
Then
\begin{align*}
\mu^{\ast}(A\cup B) &= \mathop{\inf}\{\mu(E)\colon\, E\in \mathfrak{B}(X),\, E\supset A\cup B\}\leq\\
&\leq \mathop{\inf}\{\mu(C\cup D)\colon\, C,\, D\in \mathfrak{B}(X),\ C\supset A\, \mbox{ and }\, D\supset B\}=\\
&=\mathop{\inf}\{\mathop{\max}\{\mu(C),\, \mu(D)\}\colon\, C,\, D\in \mathfrak{B}(X),\ C\supset A\, \mbox{ and }\, D\supset B\}=\\
&=\mathop{\max}\left\{\mathop{\inf}\{\mu(C)\colon\, C\in \mathfrak{B}(X),\, C\supset A\},\ \mathop{\inf}\{\mu(D)\colon\, D\in \mathfrak{B}(X),\, D\supset B\}\right\}=\\
&=\mathop{\max}\{\mu^{\ast}(A),\, \mu^{\ast}(B)\}
\end{align*}
Thus, $\mu^{\ast}(A\cup B)=\mathop{\max}\{\mu^{\ast}(A),\,
\mu^{\ast}(B)\}$. The proof of Lemma \ref{l3} is completed.
\end{proof}

Let $X$ be a Tychonoff space, $\beta\, X$ the Stone-\v{C}ech
compactification of $X$. We determine the following set:
\begin{gather}\label{itaux}
{\sf I_{\tau}}(X) = \{\mu\in I(\beta\, X)\colon\,  \mu(F)=0\, \mbox{
for every } F\in \mathfrak{B}(\beta\, X), F\subset \beta\,
X\setminus X\}.
\end{gather}

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

From the definition we have
\begin{align*}
{\sf I_{\tau}}(\emptyset) &= \{\mu\in {\sf I}(\beta X)\colon\, \mu(F)=0,\, \forall F\in \mathfrak{B}(\beta\, X),\, F\subset \beta X\}=\\
&=\{\mu\in {\sf I}(\beta X)\colon\, (\mu(\beta X)=1) \wedge
(\mu(\beta X)=0)\}=\emptyset,
\end{align*}
i.~e.
\begin{gather}\label{emptyset}
{\sf I_{\tau}}(\emptyset) = \emptyset.
\end{gather}
Elements of ${\sf I_{\tau}}(X)$ are said to be  $\tau$-\emph{smooth
idempotent probability measures}.

For each $\mu\in {\sf I_{\tau}}(X)$ we define a set function
$\tilde{\mu}\colon\, \mathfrak{B}(X)\to \mathbb{R}$ on the family
$\mathfrak{B}(X)$ of all Borel subsets of $X$ by the formula
\begin{gather*}
\tilde{\mu}(A)=\mu^{\ast}(A)=\mathop{\inf}\{\mu(B)\colon\,
B\in\mathfrak{B}(\beta\, X),\, B\supset A\},\qquad
A\in\mathfrak{B}(X).
\end{gather*}

\begin{Lemma}\label{l4}
$\tilde{\mu}$ is an idempotent probability measure on $X$.
\end{Lemma}

\begin{proof}
Evidently, that $\tilde{\mu}(\emptyset) = 0$. Equality 2) in
Definition \ref{d1} holds because of Lemma \ref{l3}. We should show
that equality 3) in Definition \ref{d1} is also true.

Let $\{A_{\alpha},\, \alpha \in D\} \subset\mathfrak{B}(X)$ be an
increasing net such that $\underset{\alpha \in D}{\cup}A_{\alpha}
\in \mathfrak{B}(X)$. For every $\alpha$ there exists $M_{\alpha}
\in \mathfrak{B}(\beta\, X)$ such that $M_{\alpha} \cap X =
A_{\alpha}$ and $\underset{\alpha \in D}{\cup}M_{\alpha}\in
\mathfrak{B}(\beta\, X)$. One has
\begin{gather*}
\left\{M\in \mathfrak{B}(\beta\, X)\colon\, M\supset
\underset{\alpha \in D}{\cup}A_{\alpha}\right\} \supset
\left\{\underset{\alpha \in D}{\cup}M_{\alpha}\in
\mathfrak{B}(\beta\, X)\colon\, \mathfrak{B}(\beta\, X)\ni
M_{\alpha}\supset A_{\alpha},\, \alpha \in D \right\}.
\end{gather*}
Then
\begin{align*}
\tilde{\mu} \left(\underset{\alpha \in D}{\bigcup}A_{\alpha}\right)
& =
\inf\left\{\mu(M)\colon\, M\in \mathfrak{B}(\beta\, X),\, M\supset \underset{\alpha \in D}{\cup}A_{\alpha}\right\} \leq\\
&\leq \inf\left\{\mu\left(\underset{\alpha \in
D}{\bigcup}M_{\alpha}\right)\colon\, \underset{\alpha \in
D}{\cup}M_{\alpha}\in
\mathfrak{B}(\beta\, X),\, \mathfrak{B}(\beta\, X)\ni M_{\alpha}\supset A_{\alpha},\, \alpha \in D \right\}=\\
&= \inf_{\substack{\underset{\alpha \in D}{\bigcup}M_{\alpha}\in
\mathfrak{B}(\beta\, X),\\ M_{\alpha}\supset A_{\alpha},\, \alpha
\in D}} \left\{\mu\left(\underset{\alpha \in
D}{\bigcup}M_{\alpha}\right) \right\} =
\inf_{\substack{M_{\alpha}\in \mathfrak{B}(\beta\, X),\\
M_{\alpha}\supset A_{\alpha},\, \alpha \in D}}
\left\{\underset{\alpha \in D}{\sup}\{\mu(M_{\alpha})\}\right\}=\\
&= \underset{\alpha \in D}{\sup} \left\{
\inf_{\substack{M_{\alpha}\in \mathfrak{B}(\beta\, X),\\
M_{\alpha}\supset A_{\alpha}}} \{\mu(M_{\alpha})\}\right\}=
\mathop{\sup}\limits_{\alpha \in D} \{\tilde{\mu} (A_{\alpha})\},
\end{align*}
i.~e. $\tilde{\mu} (\underset{\alpha \in D}{\cup}A_{\alpha}) \leq
\underset{\alpha \in D}{\sup}\{\tilde{\mu} (A_{\alpha})\}$. Remark
\ref{r4} implies the inverse inequality. The proof of Lemma \ref{l4}
is finished.
\end{proof}

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 $\mu\in {\sf I_{\tau}}(X)$,
by means of the formula $\mu(A)=\tilde{\mu}(A\cap X)$,  $A\in
\mathfrak{B}(\beta\, X)$.

\begin{Lemma} \label{l5}
Let $X$ be a Tychonoff space. If $\mu\in {\sf I_{\tau}}(X)$, then
$\mu(A)=\mu(B)$ for any two Borel subsets $A,\, B\subset \beta\, X$
such that $A\cap X=B\cap X$.
\end{Lemma}

\begin{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{gather*}
\mu(A)=\mu^{\ast}((A\cap X) \cup(A\setminus X)) = \mathop{\max
}\{\mu^{\ast}(A\cap X), \, \mu^{\ast}(A\setminus X)\} =
\mu^{\ast}(A\cap X).
\end{gather*}
Outer measures of sets $A\cap X$ and $A\setminus X$ are
\begin{gather*}
\mu^{\ast}(A\cap X)=\mathop{\inf }\{\mu(A\cap C)\colon\, C\in
\mathfrak{B}(\beta\, X), \, C\supset X\}
\end{gather*}
and

\begin{gather*}
\mu^{\ast}(A\setminus X)=\mathop{\inf }\{\mu(D)\colon\, D\in
\mathfrak{B}(\beta\, X), \, D\supset A\setminus X\}.
\end{gather*}

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

Since $\mu\in {\sf I_{\tau}}(X)$, for any sets $D\in
\mathfrak{B}(\beta\, X)$, $D\supset A\setminus X$  and $C\in
\mathfrak{B}(\beta\, X)$, $C\supset X$, we have
$\mu(C)=1=\mu(\beta\, X)\geq \mu(D)$. Then $\mu(A\cap C)\geq
\mu(A\cap D)$ for each $A\in \mathfrak{B}(\beta\, X)$. Since the
sets $C$ and $D$ are arbitrary, we have the inequality
$\mu^{\ast}(A\cap X)\geq \mu^{\ast}(A\setminus X)$. So,
$\mu(A)=\mu^{\ast}(A\cap X)$.

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

Let $X$ and $Y$ be Tychonoff spaces, $f\colon\, X\to Y$ a continuous
mapping, and $\beta\, f\colon\, \beta\, X\to \beta\, Y$ the
Stone-\v{C}ech compactification of $f$.

\begin{Theorem} \label{t2} For Tychonoff spaces $X$, $Y$ and a continuous mapping $f\colon\, X\to
Y$ we have  $I(\beta\, f)({\sf I_{\tau}}(X))\subset {\sf
I_{\tau}}(Y)$.
\end{Theorem}

\begin{proof}
Take an arbitrary $\mu\in {\sf I}(\beta\,X)$ and suppose ${\sf
I}(\beta\, f)(\mu)\not\in {\sf I}_{\tau}(Y)$. Then we have ${\sf
I}(\beta\, f)(\mu)(\Gamma) = a > 0$ for some $\Gamma\in
\mathfrak{B}(\beta\,Y)$, $\Gamma\cap Y = \emptyset$. From the
equality $(\beta\, f)^{\gets} (\beta\,Y\setminus Y) =
\beta\,X\setminus X$, we get $(\beta\, f)^{\gets} (\Gamma) \cap X =
\emptyset$, and on the other side, $(\beta\, f)^{\gets} (\Gamma)\in
\mathfrak{B}(\beta\,X)$. Whereupon, $\mu((\beta\, f)^{\gets}
(\Gamma)) = a > 0$. Hence, $\mu\not\in {\sf I_{\tau}}(X)$. Thus, the
required inclusion, and Theorem \ref{t2} are established.
\end{proof}

Just proved Theorem \ref{t2} gives us an opportunity to determine
the following mapping
\begin{gather*}
{\sf I_{\tau}}(f) = {\sf I}(\beta\, f)|_{{\sf I_{\tau}}(X)}\colon\,
{\sf I_{\tau}}(X)\to {\sf I_{\tau}}(Y).
\end{gather*}

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


\begin{Theorem} \label{t3}
The operation ${\sf I_{\tau}}$ putting the mapping ${\sf
I_{\tau}}(f)\colon\, {\sf I_{\tau}}(X)\to {\sf I_{\tau}}(Y)$ in
correspondence with the mapping $f\colon\, X\to Y$, preserves the
class of perfect mappings.
\end{Theorem}

\begin{proof}
Let $f\colon\, X\to Y$ be a perfect mapping of Tychonoff spaces.
Then for the extension $\beta\, f\colon\, \beta\, X\to \beta\, Y$ of
$f$ one has $\beta\, f(\beta\, X\setminus X)\subset \beta\,
Y\setminus Y$ [3]. We consider a mapping $I(\beta\, f)\colon\,
I(\beta\, X)\to I(\beta\, Y)$, and prove that the relation ${\sf
I}(\beta\, f)({\sf I}(\beta\, X)\setminus {\sf I_{\tau}}(X))\subset
{\sf I}(\beta\, Y)\setminus {\sf I_{\tau}}(Y)$ holds as well.

Let us take an arbitrary element $\mu\in {\sf I}(\beta\, Y)\setminus
{\sf I_{\tau}}(Y)$. We should show that ${\sf I}(\beta\,
f)^{\gets}(\mu) \subset {\sf I}(\beta\, X)\setminus {\sf
I_{\tau}}(X)$. Since $\mu\not\in {\sf I_{\tau}}(Y)$, we have
$\mu(\Gamma) = a > 0$ for some $\Gamma\in \mathfrak{B}(\beta\,Y)$,
$\Gamma\cap Y = \emptyset$. By definition, for every $\nu\in
I(\beta\, f)^{\gets}(\mu)$ we have $I(\beta f)(\nu)=\mu$. From the
equality $(\beta\, f)^{\gets} (\beta\,Y\setminus Y) =
\beta\,X\setminus X$, we get $(\beta\, f)^{\gets} (\Gamma) \cap X =
\emptyset$, and on the other side, $(\beta\, f)^{\gets} (\Gamma)\in
\mathfrak{B}(\beta\,X)$. Whereupon, $\mu(\Gamma) = \nu((\beta\,
f)^{\gets} (\Gamma)) = a > 0$. Hence, $\nu\in {\sf I}(\beta\,
X)\setminus {\sf I_{\tau}}(X)$. By virtue of arbitrariness of
$\nu\in I(\beta\, f)^{\gets}(\mu)$, we conclude that ${\sf
I}(\beta\, f)^{\gets}(\mu)\subset {\sf I}(\beta\, X)\setminus {\sf
I_{\tau}}(X)$.

Thus, ${\sf I}(\beta\, f)({\sf I}(\beta\, X)\setminus {\sf
I_{\tau}}(X))\subset {\sf I}(\beta\, Y)\setminus {\sf I_{\tau}}(Y)$.
Since ${\sf I}(\beta\, f)\colon\, {\sf I}(\beta\, X)\to {\sf
I}(\beta\, Y)$ is a mapping between compact Hausdorff spaces, the
mapping ${\sf I_{\tau}}(f)={\sf I}(\beta\, f)|_{{\sf
I_{\tau}}(X)}\colon\, {\sf I_{\tau}}(X)\to {\sf I_{\tau}}(Y)$ is
perfect. Theorem \ref{t3} is proved.
\end{proof}

\begin{Theorem}\label{t4}
The operation ${\sf I_{\tau}}$ putting the mapping ${\sf
I_{\tau}}(f)\colon\, {\sf I_{\tau}}(X)\to {\sf I_{\tau}}(Y)$ in
correspondence with the mapping $f\colon\, X\to Y$, preserves the
class of embeddings.
\end{Theorem}

\begin{proof}
Let $f\colon\, X\to Y$ be an embedding of topological spaces and
$\beta\, f\colon\, \beta\, X\to \beta\, Y$ its Stone-\v{C}ech
compact extension.

As we reminded above, $\beta\, f(\beta\, X\setminus X)\subset
\beta\, Y\setminus f(X)$.  Put $A = {\sf I_{\tau}}(f(X))$.  One can
prove the relation ${\sf I}(\beta\, f)({\sf I}(\beta\, X)\setminus
{\sf I_{\tau}}(X))\subset {\sf I}(\beta\, Y)\setminus A$ similarly
to the proof of Theorem \ref{t3}.

Theorem \ref{t2} gives ${\sf I}(\beta\, f)({\sf I_{\tau}}(X))\subset
A$. Therefore, ${\sf I_{\tau}}(f)= {\sf I}(\beta\, f)|_{{\sf
I_{\tau}}(X)}\colon\, {\sf I_{\tau}}(X)\to A$ is a proper mapping.

We will show that  ${\sf I_{\tau}}(f)\colon\, {\sf I_{\tau}}(X)\to
{\sf I_{\tau}}(Y)$ is an injective mapping. Then it follows that
${\sf I_{\tau}}(f)$ is an embedding. Let $\mu,\, \eta\in {\sf
I_{\tau}}(X)$ are two different measures. Then there exists a Borel
set $Z\subset \beta\, X$ such that $\mu(Z)\neq\eta(Z)$. We should
prove that ${\sf I_{\tau}}(f)(\mu)(\beta\, f(Z))\neq {\sf
I_{\tau}}(f)(\eta)(\beta\, f(Z))$.

Put $Z^{\prime}=(\beta\, f)^{\gets}(\beta\, f(Z))$. By definition of
${\sf I_{\tau}}(f)(\mu)$ we have ${\sf I_{\tau}}(f)(\mu)(\beta\,
f(Z))=\mu(Z^{\prime})$ and ${\sf I_{\tau}}(f)(\eta)(\beta\,
f(Z))=\eta(Z^{\prime})$. Since, $f$ is an embedding, $Z^{\prime}\cap
X=Z\cap X$ holds. Really, since $Z=(Z\cap X)\cup(Z\setminus X)$ one
has
\begin{align*}
Z^{\prime}\cap X&=((\beta\, f)^{\gets}(\beta\, f)((Z\cap X)\cup(Z\setminus X)))\cap X=\\
&=((\beta\, f)^{\gets}(\beta\, f(Z\cap X)\cup \beta\, f(Z\setminus X)))\cap X=\\
&=(((\beta\, f)^{\gets}(\beta\, f(Z\cap X)))\cup ((\beta\, f)^{\gets}(\beta\, f(Z\setminus X))))\cap X=\\
&=(\beta\, f)^{\gets}(\beta\, f(Z\cap X))=Z\cap X.
\end{align*}

Then, by Lemma \ref{l5}, we have
\begin{gather*}
{\sf I_{\tau}}(f)(\mu)(\beta\,
f(Z))=\mu(Z^{\prime})=\mu(Z)\neq\eta(Z)=\eta(Z^{\prime})= {\sf
I_{\tau}}(f)(\eta)(\beta\, f(Z)).
\end{gather*}
Theorem \ref{t4} is proved.
\end{proof}


\begin{Theorem} \label{t5}
The operation ${\sf I_{\tau}}$  preserves preimages of Borel sets,
i.~e. for every continuous mapping $f\colon\, X\to Y$ of Tychonoff
spaces and any Borel subset $A\subset Y$ the equality ${\sf
I_{\tau}}(f)^{\gets}({\sf I_{\tau}}(A))= {\sf
I_{\tau}}(f^{\gets}(A))$ holds.
\end{Theorem}

\begin{proof}
We state that ${\sf I_{\tau}}(f^{\gets}(A))\subset {\sf
I_{\tau}}(f)^{\gets}({\sf I_{\tau}}(A))$ for any (not necessary
Borel) subset $A\subset Y$. Take an arbitrary $\mu\in {\sf
I}(\beta\,X)$ and suppose ${\sf I_{\tau}}(f)(\mu)\not\in {\sf
I_{\tau}}(A)$. Then we have ${\sf I_{\tau}}(f)(\mu)(\Gamma) = a > 0$
for some $\Gamma\in \mathfrak{B}(\beta\,Y)$, $\Gamma\cap A =
\emptyset$. Obviously, $f^{\gets} (\Gamma) \cap f^{\gets}(A) =
f^{\gets} (\Gamma\cap A) = \emptyset$, and on the other side,
$f^{\gets} (\Gamma)\in \mathfrak{B}(\beta\,X)$. Whereupon,
$\mu(f^{\gets} (\Gamma)) = {\sf I_{\tau}}(f)(\mu)(\Gamma) = a > 0$.
Hence, $\mu\not\in {\sf I_{\tau}}(f^{\gets}(A))$.

Let us show that the inverse inclusion ${\sf
I_{\tau}}(f)^{\gets}({\sf I_{\tau}}(A))\subset {\sf
I_{\tau}}(f^{\gets}(A))$ holds for any Borel subset $A\subset Y$.
Take an arbitrary element $\mu\in {\sf I_{\tau}}(f)^{\gets}({\sf
I_{\tau}}(A))$. Then ${\sf I_{\tau}}(f)(\mu)\in {\sf I_{\tau}}(A)$,
and we have
\begin{gather*}
\mu(X\setminus f^{\gets}(A)) = \mu (f^{\gets}(Y\setminus A)) = {\sf
I_{\tau}}(f)(\mu)(Y\setminus A) = 0.
\end{gather*}
Consequently, $\mu\in {\sf I_{\tau}}(f^{\gets}(A))$.
\end{proof}

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

\begin{Question}\label{q1}
Let $f\colon\, X\to Y$ be a continuous mapping of Tychonoff spaces,
$A\subset Y$ an arbitrary set, $B\subset X$ a Borel set containing
$f^{\gets}(A)$. Does there exist a Borel set $C$ in $Y$ such that
$f^{\gets}(A)\subset f^{\gets}(C)\subset B$?
\end{Question}

An affirmative answer to Question \ref{q1} makes it possible to
obtain a more rigorous result in Theorem \ref{t5} by getting rid of
the condition on one of the sets to be Borel.

\begin{Theorem} \label{t6}
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 ${\sf
I_{\tau}}(A\cap B)={\sf I_{\tau}}(A)\cap {\sf I_{\tau}}(B)$ holds.
\end{Theorem}

\begin{proof}
Assume $A\cap B=\emptyset$. Then (\ref{emptyset}) implies ${\sf
I_{\tau}}(A\cap B)=\emptyset$.

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

Consider the case $A\cap B\neq\emptyset$. The inclusion ${\sf
I_{\tau}}(A\cap B) \subset {\sf I_{\tau}}(A)\cap {\sf I_{\tau}}(B)$
directly follows from Theorem \ref{t4}.

Let us show the inverse inclusion, i.~e. ${\sf I_{\tau}}(A\cap
B)\supset {\sf I_{\tau}}(A)\cap {\sf I_{\tau}}(B)$. Suppose, $\mu\in
{\sf I_{\tau}}(A)\cap {\sf I_{\tau}}(B)\subset {\sf I_{\tau}}(X)$,
and $A$ is a Borel set. Since $\mu\in {\sf I_{\tau}}(A)$, we get
$\mu(\beta\, X\setminus A)=0$. Then the inclusion $\mu\in
I_{\tau}(B)$ implies $\mu^{\ast}(B\setminus A) = \mu^{\ast}(B\cap
(\beta\, X\setminus A)) =0$ and $\mu^{\ast}(B\cap A)=1$. We should
proof $\mu(\Phi)=0$ for every $\Phi\in\mathfrak{B}(\beta\, X)$ with
$\Phi\cap(A\cap B)=\emptyset$. We decompose $\Phi =
\Phi_{A}\cup\Phi_{B}$ setting $\Phi_{A} = \Phi\setminus A$ and
$\Phi_{B} = \Phi\cap A$. Then $\mu(\Phi_{A}) = 0$ by definition. And
$\mu(\Phi_{B}) = 0$ because $\Phi_{B}\cap B = \emptyset$.
Consequently, $\mu(\Phi)=0$. So, $\mu\in {\sf I_{\tau}}(A\cap B)$.

Thus, Theorem \ref{t6} is completely proved.
\end{proof}

\begin{Remark}\label{r5} \rm
The assertion of Theorem \ref{t6} cannot be improved, i.~e. the
requirement that at least one of the given sets be Borel set cannot
be omitted. Really, consider segment $[0;\, 1]$. We will say
$x\thicksim y$ if and only if $y-x\in \mathbb{Q}$, here $\mathbb{Q}$
is the set of all rational numbers. The relation $\thicksim$ is an
equivalence relation, (i.~e. reflexive, symmetric and transitive).
Denote by $[x]$ an equivalence class containing $x\in [0;\, 1]$.
Whereupon, we got a partition of $[0;\, 1]$ into disjoint
equivalence classes. Each class $[x]$ has countable many elements.
By $\mathcal{A}$ we denote a system of all mutually distinct
equivalence classes. Since $[0;\, 1] = \underset{[x]\in
\mathcal{A}}{\cup}[x]$, the system $\mathcal{A}$ has a power of the
continuum. On the other side, 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 $A\cap B =
\emptyset$ by virtue of the construction. It is clear, there exists
no couple of Borel sets $U$ and $V$ in $[0;\, 1]$ such that
$A\subset U$, $B\subset V$ and $U\cap V = \emptyset$. 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, ${\sf I_{\tau}} (A) = {\sf I_{\tau}}([0;\, 1))$, and
${\sf I_{\tau}} (B) = {\sf I_{\tau}}((0;\, 1))$ or ${\sf I_{\tau}}
(B) = {\sf I_{\tau}}((0;\, 1])$. Since $[0;\, 1)$,  $(0;\, 1)$ and
$(0;\, 1]$ are Borel sets we have
$$
{\sf I_{\tau}} (A) \cap {\sf I_{\tau}} (B) = {\sf I_{\tau}}((0;\,
1)) \ne {\sf I_{\tau}}(\emptyset) = {\sf I_{\tau}} (A\cap B).
$$


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 \ref{t6} is not true for the sets $A$ and $B$.

One can select sets $A$ and $B$ such that $A\cap B \subset
\left[\tfrac{1}{3};\, \tfrac{2}{3}\right] $ and the intersection is
dense in $\left(\tfrac{1}{3};\, \tfrac{2}{3}\right)$. Then the only
Borel set containing the intersection is $\left(\tfrac{1}{3};\,
\tfrac{2}{3}\right)$. Evidently,
$$
{\sf I_{\tau}} (A) \cap {\sf I_{\tau}} (B) = {\sf
I_{\tau}}\left((0;\, 1)) \ne {\sf I_{\tau}}(\left(\tfrac{1}{3};\,
\tfrac{2}{3}\right)\right) = {\sf I_{\tau}} (A\cap B).
$$
\end{Remark}

The set of type $A$ (or $B$) considered in Remark \ref{r5} was found
by Giuseppe Vitali in 1905. Therefore, such a set is called the
Vitali set.

\section{Idempotent probability measures spaces on \v{C}ech-complete Tychonoff spaces}

We begin our investigation of spaces of idempotent probability
measures on a \v{C}ech-complete Tychonoff space by stating the
following result.

\begin{Theorem}\label{t7}
Let $X$ be a Tychonoff space and $\{X_{\alpha}:\, \alpha\in D\}$ any
family of Borel subsets of $X$ such that $\underset{\alpha\in D}\cap
X_{\alpha}\in \mathfrak{B}(\beta\, X)$. Then the equality ${\sf
I_{\tau}}\left(\underset{\alpha\in D}\cap
X_{\alpha}\right)=\underset{\alpha\in D}\cap {\sf
I_{\tau}}(X_{\alpha})$ holds.
\end{Theorem}

\begin{proof}
It is easy to see that ${\sf I_{\tau}}\left(\underset{\alpha\in
D}\bigcap X_{\alpha}\right)\subset\underset{\alpha\in D}\cap {\sf
I_{\tau}}(X_{\alpha})$. Really, it is clear that
$\underset{\alpha\in D}\cap X_{\alpha}\subset X_{\alpha}$ for every
$\alpha\in D$. Since $\mu\in {\sf I_{\tau}}\left(\underset{\alpha\in
D}\bigcap X_{\alpha}\right)$, we have $\mu\left(\beta
X\setminus\left(\underset{\alpha\in D}\bigcap
X_{\alpha}\right)\right)=0$. Hence $\mu(\beta X\setminus
X_{\alpha})=0$ for arbitrary $\alpha\in D$, i.~e. $\mu\in
\underset{\alpha\in D}\bigcap {\sf I_{\tau}}(X_{\alpha})$.

Let us show the inverse inclusion, i.~e. ${\sf
I_{\tau}}\left(\underset{\alpha\in D}\bigcap
X_{\alpha}\right)\supset\underset{\alpha\in D}\cap {\sf
I_{\tau}}(X_{\alpha})$. Suppose, $\mu\in\underset{\alpha\in D}\cap
{\sf I_{\tau}}(X_{\alpha})$. Then $\mu\in {\sf
I_{\tau}}(X_{\alpha})$ for every $\alpha\in D$. Hence for any
$\alpha\in D$ one has $\mu(\beta X\setminus X_{\alpha})=0$.
Consequently,
\begin{align*}
\mu\left(\beta X\setminus\left(\underset{\alpha\in D}\bigcap
X_{\alpha}\right)\right)=\mu\left(\underset{\alpha\in
D}\bigcup\left(\beta X\setminus
X_{\alpha}\right)\right)=\mathop{\sup}\limits_{\alpha\in D}
\{\mu(\beta X\setminus X_{\alpha})\}=0.
\end{align*}

Thus, $\mu\in {\sf I_{\tau}}\left(\underset{\alpha\in D}\bigcap
X_{\alpha}\right)$. Theorem \ref{t7} is proved.
\end{proof}

\begin{Corollary}\label{c1}
Let $X$ be a Tychonoff space and $\{X_{\alpha}:\, \alpha\in D\}$ any
family of closed subsets of $X$. Then ${\sf
I_{\tau}}\left(\underset{\alpha\in D}\bigcap
X_{\alpha}\right)=\underset{\alpha\in D}\cap {\sf
I_{\tau}}(X_{\alpha})$.
\end{Corollary}

\begin{Corollary}\label{c2}
Let $X$ be a Tychonoff space and $\{X_{\alpha}\colon\, \alpha\in
D\}$ any family of open subsets of $X$ such that
$\underset{\alpha\in D}\bigcap X_{\alpha}\in \mathfrak{B}(\beta\,
X)$. Then ${\sf I_{\tau}}\left(\underset{\alpha\in D}\bigcap
X_{\alpha}\right)=\underset{\alpha\in D}\cap {\sf
I_{\tau}}(X_{\alpha})$.
\end{Corollary}

Note that the following assertion ensures that every $X_{\alpha}$ is
closed in Corollary \ref{c1}.

\begin{Proposition}\label{p3}
For any closed subset $F\subset X$ of a Tychonoff space $X$ the set
${\sf I_{\tau}}(F)$ is closed in ${\sf I_{\tau}}(X)$.
\end{Proposition}

\begin{proof}
Take any $\mu\in {\sf I_{\tau}}(X)\setminus {\sf I_{\tau}}(F)$. Then
there exists an open subset $V\subset \beta X$ such that $F\cap V =
\emptyset$ and $\mu(V) = a > 0$. Consider an open neighbourhood
$\langle\mu;\, V;\, \tfrac{a}{2}\rangle$ of $\mu$ and claim that
$\langle\mu;\, V;\, \tfrac{a}{2}\rangle \cap {\sf I_{\tau}}(F) =
\emptyset$. Really, for every $\xi\in I_{\tau}(F)$ one has $|\xi(V)
- \mu(V)| = |0 - a| = a > \tfrac{a}{2}$, and $\xi \notin
\langle\mu;\, V;\, \tfrac{a}{2}\rangle$. Or, for each
$\zeta\in\langle\mu;\, V;\, \tfrac{a}{2}\rangle$ one has $|\zeta(V)
- \mu(V)| = |\zeta(V) - a| < \tfrac{a}{2}$, which gives $\zeta(V) >
\tfrac{a}{2} \ne 0$. Hence, $\zeta\notin {\sf I_{\tau}}(F)$. Thus,
${\sf I_{\tau}}(F)$ is closed in ${\sf I_{\tau}}(X)$.
\end{proof}

Now we are ready to formulate the following result. Note that a
linear (in the classical sense) version of this result was
established in \cite{ba}

\begin{Theorem} \label{t8}
The operation ${\sf I_{\tau}}$ preserves the \v{C}ech-completeness
of Tychonoff spaces, in the other words, if $X$ is a
\v{C}ech-complete Tychonoff space, then ${\sf I_{\tau}}(X)$ is also
a \v{C}ech-complete Tychonoff space.
\end{Theorem}

\begin{proof}
Let $X$ be a \v{C}ech-complete Tychonoff space. Then $X$ is a
$G_{\delta}$-set in $\beta X$, i.~e. there are countable many open
sets $U_{n}\subset \beta X$, such that
$X=\underset{n=1}{\overset{\infty}{\cap}}U_{n}$.  Then
\begin{align*}
{\sf I_{\tau}}(X) &=  {\sf
I_{\tau}}\left(\underset{n=1}{\overset{\infty}{\cap}}U_{n}\right) =
\left\{\mu\in
{\sf I}(\beta X)\colon\, \mu\left(\beta X\setminus \underset{n=1}{\overset{\infty}{\cap}}U_{n}\right) = 0\right\} =\\
& = \underset{n=1}{\overset{\infty}{\cap}} \left\{\mu\in I(\beta
X)\colon\, \mu(\beta X\setminus U_{n}) < \tfrac{1}{n}\right\}.
\end{align*}
Applying Remark \ref{r2}, by \cite{er} we get that ${\sf
I_{\tau}}(X)$ is a \v{C}ech-complete Tychonoff space. Theorem
\ref{t8} is proved.
\end{proof}

Since \v{C}ech-completeness is hereditary with respect to
$G_{\delta}$-subsets (Theorem 3.9.6, \cite{er}), one obtains the
following result.

\begin{Corollary} \label{c3}
For a Tychonoff space $X$  the space ${\sf I_{\tau}}(X)$
\v{C}ech-complete if and only if $X$ is \v{C}ech-complete.
\end{Corollary}

From Theorem \ref{t8} and Corollary \ref{l2} one can immediately
extract that, in contrast to closed sets, for an open set $U$, the
set ${\sf I_{\tau}}(U)$ is a $G_{\delta}$-set in $I(\beta X\setminus
X)$ for a \v{C}ech-complete Tychonoff space $X$.

\begin{Proposition}\label{p4}
For an open set $U\subset X$ of a \v{C}ech-complete Tychonoff space
$X$ the set ${\sf I_{\tau}}(U)$ is a $G_{\delta}$-set.
\end{Proposition}

\begin{proof}
Since $X$ is a \v{C}ech-complete Tychonoff space, it is a
$G_{\delta}$-set in $\beta X$, and let
$X=\underset{n=1}{\overset{\infty}{\cap}}U_{n}$, where $U_{n}$ are
open subsets of $\beta X$. Then $X\in \mathfrak{B}(\beta X)$, and $U
\in \mathfrak{B}(\beta X)$ for each open $U\subset X$. An open set
$U\subset X$ admits a representation $U =
\underset{n=1}{\overset{\infty}{\cap}}(U\cap U_{n})$. Finally,
\begin{align*}
{\sf I_{\tau}}(U) &=  {\sf
I_{\tau}}\left(\underset{n=1}{\overset{\infty}{\cap}}(U\cap
U_{n})\right) =
\left\{\mu\in {\sf I}(\beta X)\colon\, \mu\left(\beta X\setminus \underset{n=1}{\overset{\infty}{\cap}}(U\cap U_{n})\right) = 0\right\} =\\
& = \underset{n=1}{\overset{\infty}{\cap}} \left\{\mu\in {\sf
I}(\beta X)\colon\, \mu(\beta X\setminus (U\cap U_{n})) <
\tfrac{1}{n}\right\}.
\end{align*}
Proposition \ref{p4} is proved.
\end{proof}\

\section{Conclusions}

We provided a manner to define topology $T$ on the set ${\sf I(X)R$
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 $(I(X), T)$ is compact Hausdorff if $XR$ is so.
The topology $T$ (on $\beta X$) is used to define topology on the
set ${\sf I_{\tau}(X)$ of $\tau$-smooth idempotent probability
measures on a Tychonoff space $X$. Some properties of ${\sf
I}_{\tau$ are established and it is shown that the space ${\sf
I}_{\tau}(X)$ is \v Cech-complete if and only if $X$ is \v
Cech-complete. It is natural to continue investigation of properties
of ${\sf I}_[\tau}(X)$ for some other topological properties which
are  not studied in the literature.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\authorcontributions{Conceptualization, Lj. K., A. Z. and M.E.; investigation, Lj. K/, A. Z. and M.E.; resources, Lj. K. A. Z.
All authors have read and agreed to the published version of the
manuscript.}

\funding{This research received no external funding.}E.;

\dataavailability{Not applicable.]}

\conflictsofinterest{The author declares no conflict of interest.}


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