Ultrafilters on measurable semigroups

TL;DR

This paper explores the structure of ultrafilters on measurable semigroups, demonstrating that the collection of ultrafilters forms a compact right topological semigroup with certain algebraic properties.

Contribution

It introduces the concept of ultrafilters on measurable semigroups and proves their structure as a compact right topological semigroup.

Findings

m^eta is a compact right topological semigroup.

Ultrafilters on measurable semigroups have specific algebraic properties.

The paper establishes foundational properties of m^eta.

Abstract

Let $(S,\cdot)$ be a semigroup and $\mathfrak{m}$ be a $\sigma$-algebra on $S$. We say $(S,\cdot,\mathfrak{m})$ is a measurable semigroup if $\pi:S\times S\longrightarrow S$ by $\pi(x,y)=x\cdot y$ is a measurable function. In this paper , we consider to $\mathfrak{m}^\beta$ as the collection of all ultrafilters on $\mathfrak{m}$. We show that $\mathfrak{m}^\beta$ is a compact right topological semigroup respect to generated topology by $\sigma$-algebra $\mathfrak{m}$ on $\mathfrak{m}^\beta$. Also we study some elementary properties of the algebraic structure of $\mathfrak{m}^\beta$.

Equations

$$\mathcal{A}=\{C\in \mathfrak{m}:thereissomeB\in pwithA\cap B\subseteq C\}.$$

$$(A\cup B)\cap (D\cap C)$$

$$\subseteq (A\cap C)\cup (B\cap D)=\varnothing .$$

$$\hat{x}=\{A\in \mathfrak{m}:x\in A\}.$$

$$\hat{A}=\{p\in {\mathfrak{m}}^{\beta }:A\in p\}.$$

$$g(p)=\{A\in {\mathfrak{m}}_2:{\phi }^{-1}(A)\in p\}.$$

$$g(e(x))$$

$$=\{A\in {\mathfrak{m}}_2:x\in \widehat{{\phi }^{-1}(A)}\}$$

$$=\{A\in {\mathfrak{m}}_2:\phi (x)\in A\}$$

$$p\otimes q=\{A\in \mathfrak{m}\otimes \mathfrak{m}:\{s\in X:A_s\in q\}\in p\}$$

$$R_q^{-1}(A\times B)$$

$$=\{s\in S:A\times B\in s\otimes q\}$$

$$=\{s\in S:\{u:\{t:(u,t)\in A\times B\}\in q\}\in s\}$$

$$=\{s\in S:\{t:(s,t)\in A\times B\}\in q\}$$

$$=\{s\in S:s\in A,B\in q\}=A\in \mathfrak{m}$$

$$\{s\in S:{\rho }_q(s)\in \hat{A}\}={\rho }_q^{-1}(A)\in p.$$

$$\{s\in S:s^{-1}A\in q\}$$

$$=\{s\in S:A\in {\rho }_q(s)\}$$

$$=\{s\in S:s\in {\rho }_q^{-1}(A)\}$$

$$\{x\in X:f(x)>0\}$$

$${\mathcal{L}}_{\circ }=\{A\in \mathcal{B}:\lambda (A)=0\}$$

$${\mathcal{L}}_k=\{A\in \mathcal{B}:\lambda (A)>k\}.$$

$${\mathcal{L}}_{\circ }^{\beta }=\{p\in {\mathcal{L}}^{\beta }:\exists A\in p\lambda (A)=0\},$$

$${\mathcal{L}}_k^{\beta }=\{p\in {\mathcal{L}}^{\beta }:\forall A\in p\lambda (A)>k\}$$

$${\mathcal{L}}_{+}^{\beta }=\{p\in {\mathcal{L}}^{\beta }:\forall A\in p\lambda (A)>0\}$$

$$\{x\in S:-x+A\in p\}\in q.$$

$$x^{\ast }=\{p\in {\mathcal{L}}^{\beta }:x\in \bigcap _{A\in p}c{l}_{S}A\}.$$

$${\infty }^{\ast }=\{p\in {\mathcal{L}}^{\beta }:\forall x>0,(x,+\infty )\in p\}.$$

$$(0,+\infty )=\bigcup _{t\in G}-t+A.$$

Full text

2010 Mathematics Subject Classification: 05D10; Secondary 22A15.

*†*Corresponding Arthur

1. Introduction

A semigroup $S$ which is also a Hausdorff topological space is called a semitopological semigroup, if for each $s\in S$, $\lambda_{s}:S\rightarrow S$ and $r_{s}:S\rightarrow S$ are continuous, where for each $x\in S$, $\lambda_{s}(x)=sx$ and $r_{s}(x)=xs$. If just $r_{s}$, for each $s\in S$, is continuous, $S$ is called a right topological semigroup.

A pair $(\psi,X)$ is a semigroup compactification of $S$ if $X$ is a Hausdorff compact right topological semigroup and $\psi:S\rightarrow X$ is continuous homomorphism with dense image such that for all $s\in S$, the mapping $x\mapsto\psi(s)x:X\rightarrow X$ is continuous. For more details see [6, Section 2].

All semigroup compactifications of a semitopological semigroup as a collection of $z$-ultrafilters or $e$-ultrafilters have been described, see [1], [2], [3] and [4] for more details and some applications. This approach gives us a new light on studying this kind of compactifications. It seems that the methods presented in [1], [2], [3] and [4] can serve as a valuable tool in the study of semigroup compactifications and also of topological compactifications. But this methods depend on concepts of topology on semigroups. For a clear example, the Stone-$\check{C}$ech compactification of real additive numbers equipped to the natural topology is not semigroup compactification.

Let $(S,\cdot)$ be a semitopological semigroup. In Preliminary for a nonempty set $X$, we define ultrafiler on a $\sigma$-algebra $\mathfrak{m}$, is called $\mathfrak{m}$-ultrafilter. Also, we study $\mathfrak{m}^{\beta}$ as a collection of all $\mathfrak{m}$-ultrafilters on $\mathfrak{m}$ respect to $\sigma$-algebra generated by $\{\widehat{A}:A\in\mathfrak{m}\}$, where $\widehat{A}=\{p\in\mathfrak{m}^{\beta}:A\in p\}$.

In Section 3, for two measurable spaces $(X,\mathfrak{m})$ and $(Y,\mathfrak{n})$ we show that if $f:(X,\mathfrak{m})\rightarrow(Y,\mathfrak{n})$ is measurable function then there exists a unique measurable extension of $f$.

In Section 4, we define measurable semigroup and we extend $"\cdot"$ naturally to $"*"$ on $\mathfrak{m}^{\beta}$. Also, some elementary algebraic properties of $(\mathfrak{m}^{\beta},*)$ as extension of measurable semigroup $(S,\cdot,\mathfrak{m})$ is stated.

In Section 5, we concentrate on the Lebesgue measurable subsets od $S=(0,+\infty)$.

2. Preliminary

Let $X$ be a nonempty set and $\mathfrak{m}$ be an infinite $\sigma$-algebra on $X$. Then $(X,\mathfrak{m})$ is called measurable space. We say that $\mathfrak{m}$ separate $X$ if for each $x,y\in X$ there exist $A,B\in\mathfrak{m}$ such that $x\in A$, $y\in B$ and $A\cap B=\emptyset$. If $\mu$ is a measure on $\mathfrak{m}$, then $(X,\mathfrak{m},\mu)$ is a measure space. The outer measure $\mu^{*}$ can be defined for every nonnegative measure $\mu$. The collection of $\mu^{*}-$measurable sets is denoted by $\mathfrak{m}_{\mu}$. Also $\mu^{*}$ is a measure on $\mathfrak{m}_{\mu}$, $(X,\mathfrak{m}_{\mu},\mu^{*})$ is a Lebesgue extension or the Lebesgue completion of the measure $\mu$, and $\mathfrak{m}\subseteq\mathfrak{m}_{\mu}$, see Theorem 1.5.6 in [7].

Let $(X,\mathfrak{m}_{1})$ and $(X,\mathfrak{m}_{2})$ be measurable spaces. The function $f:X\rightarrow Y$ is an $(\mathfrak{m}_{1},\mathfrak{m}_{2})$-measurable function if $f^{-1}(A)\in\mathfrak{m}_{1}$ for every $A\in\mathfrak{m}_{2}$.

Let $(Y,\tau)$ be a topological space. The $\sigma$-algebra generated by $\tau$ is called the Borel $\sigma$-algebra and is denoted by $B_{\tau}$. The function $f:(X,\mathfrak{m})\rightarrow(Y,\tau)$ is an $\mathfrak{m}$-measurable function if $f^{-1}(A)\in\mathfrak{m}$ for every $A\in\tau$.

Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be two measurable spaces. For $X\times Y$, we consider the collection of all sets of the from $A\times B$ , where $A\in\mathcal{A}$ and $B\in\mathcal{B}$, called measurable rectangle. The $\sigma$-algebra generated by all measurable rectangles is called product of the $\sigma$-algebras $\mathcal{A}$ and $\mathcal{B}$, and is denoted by $\mathcal{A}\bigotimes\mathcal{B}$. Let $\mu$ and $\nu$ be measures on $\mathcal{A}$ and $\mathcal{B}$, respectively. Define $\mu\times\nu(A\times B)=\mu(A)\nu(B)$ for each measurable rectangles $A\times B$.

Theorem 2.1**.**

The set function $\mu\times\nu$ extends to a countably additive measure, denoted by $\mu\bigotimes\nu$, on $\mathcal{A}\bigotimes\mathcal{B}$.

Proof: See Theorem 3.3.1 in [7].

Theorem 2.2**.**

Let $(X_{1},\mathcal{A}_{1})$ and $(X_{2},\mathcal{A}_{2})$ be measurable spaces. Then the set $A_{x_{1}}=\{x_{2}\in X_{2}:(x_{1},x_{2})\in A\}\in\mathcal{A}_{2}$ for every $A\in\mathcal{A}_{1}\otimes\mathcal{A}_{2}$ and $x_{1}\in X_{1}$. Also the set $A^{x_{2}}=\{x_{1}\in X_{1}:(x_{1},x_{2})\in A\}\in\mathcal{A}_{1}$ for every $A\in\mathcal{A}_{1}\otimes\mathcal{A}_{2}$ and $x_{2}\in X_{2}$.

Proof.

See Proposition 3.3.2 in [7]. ∎

Definition 2.3**.**

Let $X$ be a nonempty set and $\mathfrak{m}$ be an infinite $\sigma$-algebra on $X$. We say $p\subseteq\mathfrak{m}$ is an $\mathfrak{m}$-filter if

$(a)$ $\emptyset\notin p$ and $X\in p$.

$(b)$ If $A,B\in\mathfrak{m}$, $A\in p$, and $A\subseteq B$, then $B\in p$.

$(c)$ If $A,B\in p$, then $A\cap B\in p$.

An $\mathfrak{m}$-ultrafilter is an $\mathfrak{m}$-filter which is not properly contained in any other $\mathfrak{m}$-filter. Zorn’s Lemma guarantees that there exist $\mathfrak{m}^{\beta}$-ultrafilters.

Lemma 2.4**.**

Let $p$ be an $\mathfrak{m}$-filter and $A\in\mathfrak{m}$. Either

$(a)$* there is some $B\in p$ such that $A\cap B=\emptyset$ or*

$(b)$* $\{C\in\mathfrak{m}:\ there\ is\ some\ B\in p\ with\ A\cap B\subseteq C\}$ is an $\mathfrak{m}$-filter.*

Proof.

If there is $(a)$, then the statement is obvious.

If $(a)$ is false, then $A\cap B\neq\emptyset$ for every $B\in p$. Let

[TABLE]

It is obvious that $A\in\mathcal{A}$, $\emptyset\notin\mathcal{A}$, and $X\in\mathcal{A}$. If $C\in\mathcal{A}$ and $C\subseteq C^{\prime}\in\mathfrak{m}$, then $C^{\prime}\in\mathcal{A}$. Also if $C,D\in\mathcal{A}$, then there exist $B_{1},B_{2}\in p$ such that $B_{1}\cap A\subseteq C$ and $B_{2}\cap A\subseteq D$. So $(B_{1}\cap B_{2})\cap A\subseteq D\cap C$ and hence $B_{1}\cap B_{2}\in p$ Since $p$ is an $\mathfrak{m}$-filter. So $D\cap C\in\mathcal{A}$ and this implies that $\mathcal{A}$ is an $\mathfrak{m}$-filter. ∎

Theorem 2.5**.**

Let $\mathfrak{m}$ be an infinite $\sigma$-algebra on $X$ and $p\subseteq\mathfrak{m}$ be an $\mathfrak{m}$-filter. The following statements are equivalent:

$(a)$* $p$ is an $\mathfrak{m}$-ultrafilter.*

$(b)$* for all $A,B\in\mathfrak{m}$, if $A\cup B\in p$ then $A\in p$ or $B\in p$.*

$(c)$* for all $A\in\mathfrak{m}$ either $A\in p$ or $A^{c}\in p$.*

$(d)$* for each $A\in\mathfrak{m}\setminus p$ there is some $B\in p$ such that $A\cap B=\emptyset$.*

Proof.

(a) implies (b). Let $A,B\in\mathfrak{m}$ such that $A\cup B\in p$. If $B\notin p$ and $A\notin p$, then there are $C,D\in p$ such that $A\cap C=\emptyset$ and by Lemma 2.4, $B\cap D=\emptyset$. So

[TABLE]

Then $\emptyset\in p$, is contradiction.

(b) implies (c). It is obvious.

(c) implies (d). Let $A\in\mathfrak{m}$ such that $A\notin p$. Then $A^{c}\in p$ so $A\cap A^{c}=\emptyset$.

(d) implies (a). Let $p$ be an $\mathfrak{m}$-filter and let $\mathcal{A}$ be an $\mathfrak{m}$-ultrafilter such that $p\subseteq\mathcal{A}$. Pick $A\in\mathcal{A}\setminus p$. Then $A\notin p$ so there is a some $B\in p$ such that $A\cap B=\emptyset$. But $\emptyset=A\cap B\in\mathcal{A}$, a contradiction. ∎

Definition 2.6**.**

Let $X$ be a nonempty set and $\mathfrak{m}$ be an infinite $\sigma$-algebra on $X$. For $x\in X$, we define

[TABLE]

and $\hat{x}$ is called $\mathfrak{m}$-principal ultrafilter.

It is obvious that $p$ is the $\mathfrak{m}$-principal ultrafilter if and only if $\bigcap p\neq\emptyset$.

Theorem 2.7**.**

Let $X$ be a nonempty set, $\mathfrak{m}$ be an infinite $\sigma$-algebra on $X$, and $\mathcal{A}\subseteq\mathfrak{m}$ has the finite intersection property. Then there is an $\mathfrak{m}$-ultrafilter $p$ such that $\mathcal{A}\subseteq p$.

Proof.

By Zorn’s Lemma is obvious. ∎

Definition 2.8**.**

Let $X$ be a nonempty set, let $\mathfrak{m}$ be an infinite $\sigma$-algebra on $X$, and let $\mathcal{R}$ be a nonempty set such that $\mathcal{R}\subseteq\mathfrak{m}$. We say that $\mathcal{R}$ is partition regular if and only if whenever $A,B\in\mathfrak{m}$ and $A\cup B\in\mathcal{R}$, there exists $C\in\mathcal{R}$ such that $C\subseteq A$ or $C\subseteq B$.

Theorem 2.9**.**

Let $X$ be a nonempty set, let $\mathfrak{m}$ be an infinite $\sigma$-algebra on $X$, and let $\mathcal{R}$ be a nonempty set such that $\mathcal{R}\subseteq\mathfrak{m}$ and assume that $\emptyset\notin\mathcal{R}$. Let $R^{\uparrow}=\{B\in\mathfrak{m}:A\subseteq B\mbox{ for some }A\in\mathcal{R}\}$. The following statements are equivalent:

$(a)$* $\mathcal{R}$ is partition regular.*

$(b)$* Whenever $\mathcal{A}\subseteq\mathfrak{m}$ has the property that every finite nonempty subfamily of $\mathcal{A}$ has an intersection which is in $R^{\uparrow}$, there is an $\mathfrak{m}$-ultrafilter $p\subseteq\mathfrak{m}$ such that $\mathcal{A}\subseteq p\subseteq R^{\uparrow}$.*

$(c)$* Whenever $A\in\mathcal{R}$, there is some $\mathfrak{m}$-ultrafilter $p$ such that $A\in p\subseteq R^{\uparrow}$.*

Proof.

(a) implies (b). Let $\mathcal{B}=\{A\in\mathfrak{m}:\forall\ B\in\mathcal{R},A\cap B\neq\emptyset\}$ and note that $\mathcal{B}\neq\emptyset$. We may assume that $\mathcal{A}\neq\emptyset$, since $\{X\}$ has the hypothesized property. Let $\mathcal{C}=\mathcal{A}\cup\mathcal{B}$. We claim that $\mathcal{C}$ has the finite intersection property. To see this it suffices (since $\mathcal{A}$ and $\mathcal{B}$ are nonempty) to let $\mathcal{F}\in P_{f}(\mathcal{A})$ and $\mathcal{G}\in P_{f}(\mathcal{B})$ and show that $\bigcap\mathcal{F}\cap\bigcap\mathcal{G}\neq\emptyset$. So suppose instead that we have such $\mathcal{F}$ and $\mathcal{G}$ with $\bigcap\mathcal{F}\cap\bigcap\mathcal{G}=\emptyset$. Pick $B\in\mathcal{R}$ such that $B\subseteq\bigcap\mathcal{F}$. Then $B\cap\bigcap\mathcal{G}=\emptyset$ and so $B=\bigcup_{A\in\mathcal{G}}(B\setminus A)$. Pick $A\in\mathcal{G}$ and $C\in\mathcal{R}$ such that $C\subseteq B\setminus A$. Then $A\cap C=\emptyset$, contradicting the fact that $A\in\mathcal{B}$.

By Theorem 2.7, there is an $\mathfrak{m}$-ultrafilter $p$ such that $\mathcal{C}\subseteq p$. Given $C\in p$, $X\setminus C\notin\mathcal{B}$ (since $C\cap(X\setminus C)=\emptyset\notin p$). So pick some $B\in\mathcal{R}$ such that $B\cap(X\setminus C)=\emptyset$. That is, $B\subseteq C$.

(b) implies (c). Let $\mathcal{A}=\{A\}$.

(c) implies (a). Let $\mathcal{F}$ be a finite set of elements of $\mathfrak{m}$ with $\bigcup\mathcal{F}\in\mathcal{R}$ and let $p$ be an $\mathfrak{m}$-ultrafilter such that $\bigcup\mathcal{F}\in p$ and for each $C\in p$ there is some $B\in\mathcal{R}$ such that $B\subseteq C$. Pick by Theorem 2.5 some $A\in\mathcal{F}\cap p$. ∎

Corollary 2.10**.**

Let $(X,\mathfrak{m})$ be a measurable space. Let $\mathcal{A}\subseteq\mathfrak{m}$ be an arbitrary family. If the intersection of every finite subfamily of $\mathcal{A}$ is infinite, then $\mathcal{A}$ is contained in an $\mathfrak{m}$-ultrafilter all of whose members are infinite. More generally, if $\kappa$ is an infinite cardinal and if the intersection of every finite subfamily of $\mathcal{A}$ has cardinality at least $\kappa$, then there exists $\mathfrak{m}$-ultrafilter $p$ such that $|A|\geq\kappa$ for every $A\in p$.

Proof.

The proof is obvious. ∎

Let $(X,\mathfrak{m})$ be a measurable space. The collection of all $\mathfrak{m}$-ultrafilters is denoted by $\mathfrak{m}^{\beta}$. For each $A\in\mathfrak{m}$, we define

[TABLE]

Lemma 2.11**.**

Let $X$ be a nonempty set, let $\mathfrak{m}$ be an infinite $\sigma$-algebra on $X$, and let $A,B\in\mathfrak{m}$.

$(a)$* $\widehat{A\cap B}=\widehat{A}\cap\widehat{B}$;*

$(b)$* $\widehat{A\cup B}=\widehat{A}\cup\widehat{B}$;*

$(c)$* $\widehat{X\setminus A}=\mathfrak{m}^{\beta}\setminus\widehat{A}$;*

$(d)$* $\widehat{A}=\emptyset$ if and only if $A=\emptyset$;*

$(e)$* $\widehat{A}=\mathfrak{m}^{\beta}$ if and only if $A=X$;*

$(f)$* $\widehat{A}=\widehat{B}$ if and only if $A=B$.*

Proof.

The proof is obvious. ∎

The collection $\{\widehat{A}:A\in\mathfrak{m}\}$ is a basis for topology on $\mathfrak{m}^{\beta}$.

Theorem 2.12**.**

Let $X$ be a nonempty set, let $\mathfrak{m}$ be an infinite $\sigma$-algebra on $X$. Then $\mathfrak{m}^{\beta}$ is a compact and Hausdorff space.

Proof.

Suppose that $p$ and $q$ are distinct elements of $\mathfrak{m}^{\beta}$. If $A\in p\setminus q$, then $X\setminus A\in q$. So $\widehat{A}$ and $\widehat{X\setminus A}$ are disjoint open subsets of $\mathfrak{m}^{\beta}$ containing $p$ and $q$, respectively. Thus $\mathfrak{m}^{\beta}$ is Hausdorff.

To show that $\mathfrak{m}^{\beta}$ is compact, we shall consider a family $\mathcal{A}$ of sets of the form $\widehat{A}$ with the finite intersection property and show that $\mathcal{A}$ has a nonempty intersection. Let $\mathcal{B}=\{A\in\mathfrak{m}:\widehat{A}\in\mathcal{A}\}$. If $F\in P_{f}(\mathcal{B})$, then there is some $p\in\bigcap_{A\in\mathcal{F}}\widehat{A}$ and so $\bigcap\mathcal{F}\in p$ and thus $\bigcap\mathcal{F}\neq\emptyset$. That is, B has the finite intersection property, so by Theorem 2.7 pick $q\in\mathfrak{m}^{\beta}$ with $\mathcal{B}\subseteq q$. Then $q\in\bigcap\mathcal{A}$. ∎

Theorem 2.13**.**

Let $X$ be a nonempty set and let $\mathfrak{m}$ be an infinite $\sigma$-algebra on $X$. Then the sets of the form $\widehat{A}$ are the clopen subsets of $\mathfrak{m}^{\beta}$.

Proof.

Each set $\widehat{A}$ is closed as well as open. Suppose that $C$ is any clopen subset of $\mathfrak{m}^{\beta}$. Let $\mathcal{A}=\{\widehat{A}:A\in\mathfrak{m}\ and\ \widehat{A}\subseteq C\}$. Since $C$ is open, $\mathcal{A}$ is an open cover of $C$. Since $C$ is closed, it is compact by Theorem 2.5(b). Now by Theorem 2.12, pick a finite subfamily $\mathcal{F}$ of $\mathfrak{m}$ such that $C=\bigcup_{A\in\mathcal{F}}\widehat{A}$. Then by Lemma 2.11, $C=\widehat{\bigcup\mathcal{F}}$. ∎

Theorem 2.14**.**

Let $X$ be a nonempty set and let $\mathfrak{m}$ be a $\sigma$-algebra on $X$ containing $P_{f}(X)$.

$(a)$* For every $A\in\mathfrak{m}$, $\widehat{A}=cl_{\mathfrak{m}^{\beta}}e(A)$ where $e:X\rightarrow\mathfrak{m}^{\beta}$ is defined by $e(x)=\hat{x}$.*

$(b)$* For any $A\in\mathfrak{m}$ and any $p\in\mathfrak{m}^{\beta}$, $p\in cl_{\mathfrak{m}^{\beta}}e(A)$ if and only if $A\in p$.*

Proof.

The proof is obvious. ∎

Theorem 2.15**.**

Let $X$ be a nonempty set and let $\mathfrak{m}$ be a $\sigma$-algebra on $X$ and $\mathfrak{m}$ separate $X$. Let $e:X\rightarrow\mathfrak{m}^{\beta}$ is defined by $e(x)=\hat{x}$.

$(a)$* The mapping $e$ is $\mathfrak{m}$-measurable.*

$(b)$* If $(X,\mathfrak{m})$ is measurable space and let $P_{f}(X)\subseteq\mathfrak{m}$, then the mapping $e$ is injective.*

$(c)$* $e(X)$ is a dense subset of $\mathfrak{m}^{\beta}$.*

Proof.

$(a)$ Since $A=e^{-1}(\widehat{A})$ for every $A\in\mathfrak{m}$, so the mapping $e$ is $\mathfrak{m}$-measurable.

The proofs $(b)$ and $(c)$ are obvious. ∎

Theorem 2.16**.**

Let $X$ be a nonempty set and let $\mathfrak{m}$ be a $\sigma$-algebra on $X$ containing $P_{f}(X)$. If $U$ is an open subset of $\mathfrak{m}^{\beta}$, $cl_{\mathfrak{m}^{\beta}}U$ is also open.

Proof.

If $U=\emptyset$, the conclusion is trivial and so we assume that $U\neq\emptyset$. Put $A=e^{-1}[U]$. We claim first that $U\subseteq cl_{\mathfrak{m}^{\beta}}e(A)$. So let $p\in U$ and let $\widehat{B}$ be a basic neighborhood of $p$. Then $U\cap\widehat{B}$ is a nonempty open set and so by Theorem 2.15 $(c)$, $U\cap\widehat{B}\cap e(X)\neq\emptyset$. So pick $b\in B$ with $e(b)\in U$. Then $e(b)\in\widehat{B}\cap e(A)$ and so $\widehat{B}\cap e(A)\neq\emptyset$. Also $e[A]\subseteq U$ and hence $U\subseteq cl_{\mathfrak{m}^{\beta}}e(A)\subseteq cl_{\mathfrak{m}^{\beta}}U$. By Theorem 2.14 $(a)$, $cl_{\mathfrak{m}^{\beta}}U=cl_{\mathfrak{m}^{\beta}}e(A)=\widehat{A}$, and so $cl_{\mathfrak{m}^{\beta}}U$ is open in $\mathfrak{m}^{\beta}$. ∎

Definition 2.17**.**

Let $X$ be a nonempty set, let $\mathfrak{m}$ be a $\sigma$-algebra on $X$ containing $P_{f}(X)$, and let $\mathcal{A}$ be an $\mathfrak{m}$-filter. We define $\widehat{\mathcal{A}}=\{p\in\mathfrak{m}^{\beta}:\mathcal{A}\subseteq p\}$.

Theorem 2.18**.**

Let $X$ be a nonempty set and let $\mathfrak{m}$ be a $\sigma$-algebra on $X$ containing $P_{f}(X)$.

$(a)$* If $\mathcal{A}$ is an $\mathfrak{m}$-filter, then $\widehat{A}$ is a closed subset of $\mathfrak{m}^{\beta}$.*

$(b)$* If $\emptyset\neq A\subseteq\mathfrak{m}^{\beta}$ and $\mathcal{A}=\bigcap A$, then $\mathcal{A}$ is an $\mathfrak{m}$-filter and $\mathcal{\widehat{A}}=cl_{\mathfrak{m}^{\beta}}A$.*

Proof.

$(a)$ Let $p\in\mathfrak{m}^{\beta}\setminus\mathcal{A}$. Pick $B\in\mathcal{A}\setminus p$. Then $\widehat{X\setminus B}$ is a neighborhood of $p$ which misses $\widehat{\mathcal{A}}$.

$(b)$ $\mathcal{A}$ is the intersection of a set of $\mathfrak{m}$-filters, so $\mathcal{A}$ is an $\mathfrak{m}$-filter. Further, for each $p\in A$, $\mathcal{A}\subseteq p$ so $A\subseteq\widehat{\mathcal{A}}$ and thus by $(a)$, $cl_{\mathfrak{m}^{\beta}}A\subseteq\mathcal{\widehat{A}}$. To see that $\mathcal{\widehat{A}}\subseteq cl_{\mathfrak{m}^{\beta}}A$, let $p\in\widehat{\mathcal{A}}$ and let $B\in p$. Suppose $\widehat{B}\cap A=\emptyset$. Then for each $q\in A$, $X\setminus B\in q$ so $X\setminus B\in\mathcal{A}\subseteq p$, a contradiction. ∎

Theorem 2.19**.**

Let $X$ be a nonempty set, let $\mathfrak{m}$ be a $\sigma$-algebra on $X$ containing $P_{f}(X)$, $p\in\mathfrak{m}^{\beta}$, and $U\subseteq\mathfrak{m}^{\beta}$. If $U$ is a neighborhood of $p$ in $\mathfrak{m}^{\beta}$, then $e^{-1}[U]\in p$.

Proof.

If $U$ is a neighborhood of $p$, there is a basic open subset $\widehat{A}$ of $\mathfrak{m}^{\beta}$ for which $p\in\widehat{A}\subseteq U$. This implies that $A\in p$ and so $e^{-1}[U]\in p$, because $A\subseteq e^{-1}[U]$. ∎

3. **More Topology of $\mathfrak{m}^{\beta}$ **

Now let $(X,\mathfrak{m}_{1})$ and $(Y,\mathfrak{m}_{2})$ be two measurable spaces such that $P_{f}(X)\subseteq\mathfrak{m}_{1}$ and $P_{f}(Y)\subseteq\mathfrak{m}_{2}$. Also let $\varphi:X\rightarrow Y$ be an $(\mathfrak{m}_{1},\mathfrak{m}_{2})$-measurable function. In Lemma 3.1, we show that $\varphi$ has a unique extension.

Lemma 3.1**.**

Let $(X,\mathfrak{m}_{1})$ and $(Y,\mathfrak{m}_{2})$ be measurable spaces such that $P_{f}(X)\subseteq\mathfrak{m}_{1}$ and $P_{f}(Y)\subseteq\mathfrak{m}_{2}$. Let $\varphi:(X,\mathfrak{m}_{1})\rightarrow(Y,\mathfrak{m}_{2})$ be an $(\mathfrak{m}_{1},\mathfrak{m}_{2})$-measurable function. Then there exists a continuous function $\varphi_{\beta}:\mathfrak{m}_{1}^{\beta}\rightarrow\mathfrak{m}_{2}^{\beta}$ such that

$(a)$* $\varphi_{\beta}(p)=\{A\in\mathfrak{m}_{2}:\varphi^{-1}(A)\in p\}$.*

$(b)$* $e_{Y}\circ\varphi$ = $\varphi_{\beta}\circ e_{X}$.*

$(c)$* If $A\in p\in\mathfrak{m}_{1}^{\beta}$, then $\varphi(A)\in\varphi_{\beta}(p)$ and if $B\in\varphi_{\beta}(p)$, then $\varphi^{-1}(B)\in p$.*

Proof.

$(a)$ It is obvious that $\{A\in\mathfrak{m}_{2}:\varphi^{-1}(A)\in p\}$ is an $\mathfrak{m}_{2}$-ultrafilter. For every $p\in\mathfrak{m}_{1}^{\beta}$, let

[TABLE]

Then for every $x\in X$,

[TABLE]

So $g(e(x))=\varphi(x)$.

To see that $g$ is continuous, let $A\in\mathfrak{m}_{2}$. Then $\widehat{(g\circ e)^{-1}(A)}=\widehat{\varphi^{-1}(A)}$. Since $g$ is a continuous extension of $\varphi$, we have $\widetilde{\varphi}=g$.

The proofs $(b)$ and $(c)$ are obvious. ∎

Theorem 3.2**.**

Let $(X,\mathfrak{m})$ be a measurable space, such that $P_{f}(X)\subseteq\mathfrak{m}$. Then every $G_{\delta}$-subset of $X^{*}$ has nonempty interior in $X^{*}$.

Proof.

Choose $p\in\bigcap_{n\in\mathbb{N}}(U_{n}\cap X^{*})$, where $U_{n}$ is an open subset of $\mathfrak{m}^{\beta}$ for each $n\in\mathbb{N}$. Choose a subset $A_{n}$ of $X$ such that $p\in A^{*}_{n}\subseteq U_{n}$ and $A_{n+1}\subseteq A_{n}$ for each $n\in\mathbb{N}$. Now choose an infinite sequence $\{a_{n}\}_{n=1}^{\infty}$ of distinct points of $X$ such that $a_{n}\in A_{n}$. Let $A=\{a_{n}:n\in\mathbb{N}\}$. It is obvious that $A^{*}$ is an open subset of $X^{*}$ and $A^{*}\subseteq U_{n}$ for each $n\in\mathbb{N}$. ∎

Theorem 3.3**.**

Let $(X,\mathfrak{m})$ be a measurable space, such that $P_{f}(X)\subseteq\mathfrak{m}$. Then every countable union of nowhere dense subsets of $X^{*}$ is nowhere dense in $X^{*}$.

Proof.

Let $A_{n}$ be a nowhere dense subset of $X^{*}$ for each $n\in\mathbb{N}$. We show that $B=\bigcup_{n\in\mathbb{N}}clA_{n}$ is nowhere dense. Suppose instead that $U=int_{X^{*}}clB\neq\emptyset$. By the Baire Category Theorem $X^{*}\setminus B$ is dense in $X^{*}$ so $U\setminus B\neq\emptyset$. Thus $U\setminus B$ is a nonempty $G_{\delta}$-set which thus by Theorem 3.2, has nonempty interior. This is a contradiction. ∎

4. ** Measurable Semigroups**

In this section we extend the operation of a measurable semigroup.

Definition 4.1**.**

(a) A right measurable semigroup is a triple $(S,\cdot,\mathfrak{m})$ where $(S,\cdot)$ is a semigroup, $(S,\mathfrak{m})$ is a measurable space, and $\rho_{x}:S\rightarrow S$ is $(\mathfrak{m},\mathfrak{m})$-measurable function for each $x\in S$ where $\rho_{x}(y)=yx$ for each $y\in S$.

(b) A left measurable semigroup is a triple $(S,\cdot,\mathfrak{m})$ where $(S,\cdot)$ is a semigroup, $(S,\mathfrak{m})$ is a measurable space, and $\lambda_{x}:S\rightarrow S$ is $(\mathfrak{m},\mathfrak{m})$-measurable function for each $x\in S$ where $\lambda_{x}(y)=xy$ for each $y\in S$.

(c) A measurable semigroup is a triple $(S,\cdot,\mathfrak{m})$ where $(S,\cdot)$ is a semigroup, $(S,\mathfrak{m})$ is a measurable space, and $\pi:S\times S\rightarrow S$ is $(\mathfrak{m}\otimes\mathfrak{m},\mathfrak{m})$-measurable function where $\pi(x,y)=xy$.

In this section we assume that $P_{f}(S)\subseteq\mathfrak{m}$, so $\{s\}$ belongs to $\mathfrak{m}$ for each $s\in S$.

Definition 4.2**.**

Let $(X,\mathfrak{m})$ be a measurable space. The tensor product of two $\mathfrak{m}$-ultrafilters $p$ and $q$ in $\mathfrak{m}^{\beta}$ is denoted by $p\otimes q$ and is defined as following

[TABLE]

where $A_{s}=\{t\in S:(s,t)\in A\}$.

Lemma 4.3**.**

If $p,q\in\mathfrak{m}^{\beta}$, then $p\otimes q\in(\mathfrak{m}\otimes\mathfrak{m})^{\beta}$.

Proof.

The proof is obvious. ∎

Lemma 4.4**.**

The following statements hold:

$(a)$* $S\otimes S$ is dense in $(\mathfrak{m}\otimes\mathfrak{m})^{\beta}$.*

$(b)$* $\lim_{s\rightarrow p}\lim_{t\rightarrow q}s\otimes t=p\otimes q$ for every $p,q\in\mathfrak{m}^{\beta}$.*

Proof.

The proof is obvious. ∎

Lemma 4.5**.**

The mapping $R_{q}:S\rightarrow(\mathfrak{m}\otimes\mathfrak{m})^{\beta}$ by $R_{q}(s)=s\otimes q$ is a $(\mathfrak{m},\mathfrak{m}\otimes\mathfrak{m})$-measurable function where $q\in\mathfrak{m}^{\beta}$.

Proof.

Since $\mathfrak{m}\times\mathfrak{m}$ generates $\mathfrak{m}\otimes\mathfrak{m}$, then

[TABLE]

where $A,B\in\mathfrak{m}$. ∎

Lemma 4.6**.**

Let $s\in S$ and $q\in\mathfrak{m}^{\beta}$. Then $\pi^{\beta}\circ R_{q}(s)=\pi^{\beta}(s\otimes q)=s\ast q$

Proof.

The proof is obvious. ∎

Theorem 4.7**.**

Let $(S,\cdot,\mathfrak{m})$ be a measurable semigroup. Then there is a unique binary operation $\ast:\mathfrak{m}^{\beta}\times\mathfrak{m}^{\beta}\rightarrow\mathfrak{m}^{\beta}$ satisfying the following three conditions:

$(a)$* For every $s,t\in S$, $s\ast t=s\cdot t$.*

$(b)$* For each $q\in\mathfrak{m}^{\beta}$, the function $\rho_{q}:\mathfrak{m}^{\beta}\rightarrow\mathfrak{m}^{\beta}$ is continuous, where $\rho_{q}(p)=p\ast q$.*

$(c)$* For each $s\in S$, the function $\lambda_{s}:\mathfrak{m}^{\beta}\rightarrow\mathfrak{m}^{\beta}$ is continuous, where $\lambda_{s}(q)=s\ast q$.*

Proof.

Given any $s\in S$, define $l_{s}:S\rightarrow S$ by $l_{s}(t)=st$, then $l_{s}$ is a measurable function. So there is a unique continuous function $\lambda_{s}:\mathfrak{m}^{\beta}\rightarrow\mathfrak{m}^{\beta}$ such that $\lambda_{s}(e(t))=l_{s}(t)$ for each $t\in S$. If $s\in S$ and $q\in\mathfrak{m}^{\beta}$, we define $s\ast q=\lambda_{s}(q)$. Then $(c)$ holds and so does $(a)$, because $\lambda_{s}$ extends $l_{s}$.

Now we extend $\ast$ to the rest of $\mathfrak{m}^{\beta}\times\mathfrak{m}^{\beta}$. Given $q\in\mathfrak{m}^{\beta}$, define $r_{q}=\pi^{\beta}\circ R_{q}:S\rightarrow\mathfrak{m}^{\beta}$. The mapping $r_{q}$ is measurable, so there is a unique continuous extension $\rho_{q}:\mathfrak{m}^{\beta}\rightarrow\mathfrak{m}^{\beta}$. If $p,q\in\mathfrak{m}^{\beta}$, we define $p\ast q=\rho_{q}(p)$. This is the only possible definition which satisfies the required conditions. ∎

Theorem 4.8**.**

Let $(S,\cdot,\mathfrak{m})$ be a measurable semigroup. Then the extended operation on $\mathfrak{m}^{\beta}$ is associative.

Proof.

The proof is obvious. ∎

Theorem 4.9**.**

Let $(S,\cdot,\mathfrak{m})$ be a measurable semigroup. Then $(\mathfrak{m}^{\beta},\ast)$ is a compact right semigroup.

Proof.

The proof is obvious. ∎

Theorem 4.10**.**

Let $(S,\cdot,\mathfrak{m})$ be a measurable semigroup and $A\in\mathfrak{m}$.

$(a)$* For any $s\in S$ and $q\in\mathfrak{m}^{\beta}$, $A\in s\cdot q$ if and only if $s^{-1}A\in q$.*

$(b)$* For any $p,q\in\mathfrak{m}^{\beta}$, $A\in p\cdot q$ if and only if $\{s\in S:s^{-1}A\in q\}\in p$.*

Proof.

$(a)$ Necessity. Let $A\in s\cdot q$. Then $A\in\lambda_{s}(q)$ and hence by Lemma 3.1, $s^{-1}A=\lambda_{s}^{-1}(A)\in q$.

Sufficiency. Assume $s^{-1}A\in q$ and suppose that $A\notin s\cdot q$. Then $S\setminus A\in s\cdot q$. So, by the already established necessity, $s^{-1}(S\setminus A)\in q$. This is a contradiction since $s^{-1}A\cap s^{-1}(S\setminus A)=\emptyset$.

$(b)$ Necessity. Let $A\in p\cdot q$. Then by Lemma 3.1,

[TABLE]

So $\{s\in S:A\in s\cdot q\}\in p$. Hence $\{s\in S:s^{-1}A\in q\}\in p$.

Sufficiency. Let $\{s\in S:s^{-1}A\in q\}\in p$. Since

[TABLE]

$\rho_{q}^{-1}(A)\in p$. So $A\in p\cdot q$. ∎

Definition 4.11**.**

Let $(S,\cdot,\mathfrak{m})$ be a measurable semigroup, then $\mathfrak{m}^{\beta}$ is called measurable semigroup compactification of $(S,\cdot,\mathfrak{m})$.

Now we state some algebraic properties of semigroup extension of measurable semigroup $(S,\cdot,\mathfrak{m})$.

Theorem 4.12**.**

Let $(S,\cdot,\mathfrak{m})$ be a measurable semigroup, then $\mathfrak{m}^{\beta}$ has an idempotent.

Proof.

See Theorem 2.5 in [9]. ∎

Example 4.13**.**

a) Let $\mathcal{L}$ denote the collection of all Lebesgue measurable subsets od real numbers. Then $(\mathcal{L}^{\beta},+)$ and $(\mathcal{L}^{\beta},\cdot)$ are measurable semigroup compactification of $(\mathbb{R},\mathcal{L})$.

b)Every vector space has a Hamel basis, i.e. a maximal linearly independent subset. Since Real numbers as a vector space on rational numbers has infinite dimension, therefore there exists an uncountable maximal linearly independent $I\subseteq[0,1]$. Now choice a countable subset $A\subseteq I$. Then the generated vector space by $I\setminus A$ is not closed. Define $i(x)=x$ when $x\in A$ and $i(x)=0$ for $x\in I\setminus A$. We denote $i$ as to extension $i$ to $\mathbb{R}$ as a linear transformation. It is obvious that $i$ is Borel measurable but it is not continuous.

Now define $*_{i}:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ by $*_{i}(x,y)=i(x+y)$ for each $x,y\in\mathbb{R}$. Then $(\mathbb{R},*_{i})$ is measurable semigroup, but it is not semitopological semigroup.

Theorem 4.14**.**

Let $(S,\cdot,\mathfrak{m})$ be a measurable semigroup. Then $S^{*}=\mathfrak{m}^{\beta}\setminus S$ is a subsemigroup of $\mathfrak{m}^{\beta}$ if and only if for any $A\in P_{f}(S)$ and for any infinite subset $B$ of $\mathfrak{m}$ there exists $F\in P_{f}(B)$ such that $\bigcap_{x\in F}x^{-1}A$ is finite.

Proof.

Necessity. Let $A\in P_{f}(S)$ and an infinite subset $B\in\mathfrak{m}$ be given. Suppose that for each $F\in P_{f}(B)$, $\bigcap_{x\in F}x^{-1}A$ is infinite. Then $\{x^{-1}A:x\in B\}$ has the property that all of its finite intersections are infinite so by Corollary 2.10 we may pick $p\in S^{*}$ such that $\{x^{-1}A:x\in B\}\subseteq p$. Pick $q\in S^{*}$ such that $B\in q$. Then $A\in q\cdot p$ and $A$ is finite so $q\cdot p\in S$, a contradiction.

Sufficiency. Let $p,q\in S^{*}$ be given and suppose that $q\cdot p=y\in S$, (that is, precisely, that $q\cdot p$ is the $\mathfrak{m}$-principal ultrafilter generated by y). Let $A=\{y\}$ and let $B=\{x\in S:x^{-1}A\in p\}$. Then $B\in p$ while for each $F\in P_{f}(B)$, one has $\bigcap_{x\in F}x^{-1}A\in p$ so that $\bigcap_{x\in F}x^{-1}A$ is infinite, a contradiction. ∎

Definition 4.15**.**

Let $S$ be a measurable semigroup. We say that S is weakly left cancellative if and only if $\lambda_{s}^{-1}(\{y\})$ is finite for every $x,y\in S$.

Theorem 4.16**.**

Let $(S,\cdot,\mathfrak{m})$ be a measurable semigroup. Then $S^{*}$ is a left ideal of $\mathfrak{m}^{\beta}$ if and only if $S$ is weakly left cancellative.

Proof.

Necessity. Let $x,y\in S$ be given, let $A=\lambda_{s}^{-1}(\{y\})$ and suppose that $A$ is infinite. Pick $p\in S^{*}\cap\widehat{A}$. Then $y\cdot p=x$, a contradiction.

Sufficiency. Since $S$ is infinite, $S^{*}\neq\emptyset$. Let $p\in S^{*}$, let $q\in\mathfrak{m}^{\beta}$ and suppose that $q\cdot p=x\in S$. Then $\{x\}\in q\cdot p$ so $\{y\in S:y^{-1}\{x\}\in p\}\in q$ and is hence nonempty. So pick $y\in S$ such that $y^{-1}\{x\}\in p$. But $y^{-1}\{x\}=\lambda_{y}^{-1}(\{x\})$ so $\lambda_{y}^{-1}(\{x\})$ is infinite, a contradiction. ∎

Theorem 4.17**.**

Let $(S,\cdot,\mathfrak{m})$ be a measurable semigroup. The following statements are equivalent:

$(a)$* $S^{*}$ is a right ideal of $\mathfrak{m}^{\beta}$.*

$(b)$* Given any finite subset $A\in\mathfrak{m}$, any sequence $\langle z_{n}\rangle_{n=1}^{\infty}$ in $S$, and any one-to-one sequence $\langle x_{n}\rangle_{n=1}^{\infty}$ in $S$, there exist $n<m$ in $\mathbb{N}$ such that $x_{n}\cdot z_{m}\notin A$.*

$(c)$* Given any $a\in S$, any sequence $\langle z_{n}\rangle_{n=1}^{\infty}$ in $S$, and any one-to-one sequence $\langle x_{n}\rangle_{n=1}^{\infty}$ in $S$, there exist $n<m$ in $\mathbb{N}$ such that $x_{n}\cdot z_{m}\neq a$.*

Proof.

See Theorem 4.32 in [9]. ∎

5. Applications

Recall a $\sigma$-algebra on a topological space $X$ is the Borel $\sigma$-algebra generated by all open sets; it is denoted by $\mathcal{B}(X)$. A set in $\mathcal{B}(X)$ is called the Borel set in the space X.

A $\sigma$-algebra on a topological space X is generated by all sets of the form

[TABLE]

where $f$ is a real valued continuous function on X, is called Baire $\sigma$-algebra and denoted by $Ba(X)$. So $A\in Ba(X)$ is called a Baire set. By Corollary 5.3.5 in [7], when $X$ is a metric space then $A\subseteq X$ is Borel set if and only if $A$ is a Baire set.

Let $\mathcal{L}$ denote the collection of all Lebesgue measurable sets on $S=(0,+\infty)$, and let $\lambda$ be the Lebesgue measure on real numbers. It is obvious that $\mathcal{B}(S)\subseteq\mathcal{L}$. For each $p,q\in\mathcal{L}^{\beta}$, $p+q$ and $p\cdot q$ are well define, see Theorem 4.10. So $\mathcal{L}^{\beta}$ is a multiplicative and additive semigroup.

Now define

[TABLE]

and for $k\in(0,1]$ define

[TABLE]

We define $\mathcal{L}_{+}=\bigcup_{k>\circ}\mathcal{L}_{k}$. It is obvious that $\mathcal{L}_{0}$, $\mathcal{L}_{k}$ and $\mathcal{L}_{k}$ are partition regular, and also $\mathcal{L}_{0}$, $\mathcal{L}_{k}$ and $\mathcal{L}_{+}$ are invariant under translate, where $A+x$ is the translate of $A$ by $x\in S$. Therefore the below collections

[TABLE]

[TABLE]

and

[TABLE]

are non-empty sets, by Theorem 2.9.

Let $\mathfrak{m}$ be a $\sigma$-algebra on $X$ and let $A\in\mathfrak{m}$, then $\mathfrak{m}_{A}=\{T\cap A:T\in\mathfrak{m}\}$ is a $\sigma$-algebra and we can write $\mathfrak{m}_{A}^{\beta}\subseteq\mathfrak{m}^{\beta}$, because for each $p_{A}\in\mathfrak{m}_{A}^{\beta}$ there exists a unique $p\in\mathfrak{m}$ such that $p_{A}\subseteq p$.

Lemma 5.1**.**

*a) $\mathcal{L}_{k}^{\beta}$ and $\mathcal{L}_{+}^{\beta}$ are left ideals of $(\mathcal{L}^{\beta},+)$. Also $\mathcal{L}_{+}^{\beta}$ is left ideal of $(\mathcal{L}^{\beta},\cdot)$.

b) $\mathcal{L}_{\circ}^{\beta}$ is a multiplicative and additive subsemigroup of $\mathcal{L}^{\beta}$.*

Proof.

a) Pick $p\in\mathcal{L}_{k}^{\beta}$ and $q\in\mathcal{L}^{\beta}$. Let $A\in q+p$, so

[TABLE]

Since $p\in\mathcal{L}_{k}^{\beta}$ so $\lambda(A)=\lambda(-x+A)>k$. This implies $q+p\in\mathcal{L}_{k}^{\beta}$. By similar way, $\mathcal{L}_{+}^{\beta}$ is left ideal of $(\mathcal{L}^{\beta},+)$.

b) Pick $p,q\in\mathcal{L}_{\circ}^{\beta}$, so there exist $A_{p}\in p$ and $A_{q}\in q$ such that $A_{p}\cap A_{q}=\emptyset$ and $\lambda(A_{p})=\lambda(A_{q})=0$. Now Let $A=A_{p}\cup A_{q}$, and define $p_{\circ}=\{T\cap A:T\in p\}$ and $q_{\circ}=\{T\cap A:T\in q\}$. It is obvious that $p_{\circ}$ and $q_{\circ}$ are $\mathcal{L}$-ultrafilters on $A$, $p_{\circ}\subseteq p$ and $q_{\circ}\subseteq q$. It is obvious that $p_{\circ}+q_{\circ}\subseteq p+q$. By similar way, we can show that $\mathcal{L}_{\circ}^{\beta}$ is a multiplicative subsemigroup. ∎

For $x\in S$, we define

[TABLE]

It is obvious that $x^{*}=\{p\in\mathcal{L}^{\beta}:\forall y>0,\,\,\,(x-y,x+y)\in p\}$, and $\widehat{x}\in x^{*}$. We say $p\in x^{*}$ is a near point to $x$. We define $B(S)=\cup_{x\in S}x^{*}$ and $\infty^{*}=\mathcal{L}^{\beta}-B(S)$. An element $p\in B(S)$ is called a bounded ultrafilter and $p\in\infty^{*}$ is called an unbounded ultrafilter. It is obvious that

[TABLE]

Lemma 5.2**.**

*For each $x,y\in S$, then

a) $x^{*}$ is a non empty and compact subset of $\mathcal{L}^{\beta}$,

b) $x^{*}+y^{*}\subseteq(x+y)^{*}$, and

c) $x^{*}y^{*}\subseteq(xy)^{*}$.*

Proof.

See Lemma 2.3(i) in [5]. ∎

By above Lemma, it is obvious that $0^{*}$ is a compact multiplicative and additive subsemigroup of $\mathcal{L}^{\beta}$.

Lemma 5.3**.**

$\infty^{*}$* is a multiplicative and additive subsemigroup of $\mathcal{L}^{\beta}$.*

Proof.

Let $p,q\in\infty^{*}$ and let $pq\in x^{*}$ for some $x\in S$. Therefore $(a,b)\in pq$ for some $a,b\in S$, and so for some $t\in S$, $t^{-1}(a,b)\in q$ is a contradiction. So $\infty^{*}$ is a multiplicative subsemigroup. By similar way, $\infty^{*}$ is an additive semigroup. ∎

We say that $A\subseteq S$ is meager if and only if $A$ is the countable union of nowhere dense sets. Now we state a relation between Baire and meager sets.

Definition 5.4**.**

a) A Lebesgue measurable set $A\subseteq S$ is called Baire large at $\infty$ if and only if for each $x>0$, $A\cap(x,+\infty)$ is not meager. The collection of all Baire large sets at $\infty$ is denoted by $BL(\infty)$.

b) A Lebesgue measurable set $A\subseteq S$ is called Baire large at [math] if and only if for each $x>0$, $A\cap(0,x)$ is not meager. The collection of all Baire large sets at [math] is denoted by $BL(0)$.

Lemma 5.5**.**

*a) $BL(\infty)$ and $BL(0)$ are partition regular.

b) $\mathcal{B}\mathcal{L}(0)=\{p\in\mathcal{L}^{\beta}:\forall A\in p,A\in BL(0)\}$ is left ideal in $(0^{*},\cdot)$.

c) $\mathcal{B}\mathcal{L}(\infty)=\{p\in\mathcal{L}^{\beta}:\forall A\in p,A\in BL(\infty)\}$ is left ideal in $(\infty^{*},\cdot)$.

d) $\mathcal{B}\mathcal{L}(\infty)$ is left ideal in $(\infty^{*},+)$.*

Proof.

a) It is obvious.

b) By a) and Theorem 2.9 implies $BL(0)$ is non empty. Now similar Lemma 17.39 in [9], let $p\in BL(0)$, $q\in 0^{*}$ and let $A\in qp$. Pick $x\in(0,1)$ such that $x^{-1}A\in p$. Let $x^{-1}A\cap(0,\epsilon)$ is not meager for some $\epsilon>0$, so $x(x^{-1}A\cap(0,\epsilon))$ is not meager and $x(x^{-1}A\cap(0,\epsilon))\subseteq A\cap(0,\epsilon)$ .

c) and d). The proof is similar to b). ∎

Definition 5.6**.**

a) $A\subseteq(0,+\infty)$ is called syndetic if and only if there exists some $G\in P_{f}(0,+\infty)$ such that

[TABLE]

b) $A\subseteq(0,+\infty)$ is called thick if and only if for each $F\in P_{f}((0,+\infty))$ there exists $x\in A$ such that $x+F\subseteq A$.

c) $A\subseteq(0,+\infty)$ is called piecewise syndetic if and only if $\bigcup_{t\in G}-t+A$ is thick for some $G\in P_{f}(0,+\infty)$.

Theorem 5.7**.**

a) Let $p\in\mathcal{L}^{\beta}$. Then $p\in K(\mathcal{L}^{\beta},+)$ if and only if $\{x>0:-x+A\in p\}$ is syndetic for each $A\in p$.

b) Let $A\subseteq(0,+\infty)$. Then $K(\mathcal{L}^{\beta},+)\cap\overline{A}\neq\emptyset$ if and only if $A$ is piecewise syndetic.

Proof.

The proof is similar to the proof of Theorem 4.39 and 4.40 in [9]. ∎

The following Theorem has been stated as Theorem 1.3 in [8].

Theorem 5.8**.**

The closure of the minimal ideal $K(\mathcal{L}^{\beta},+)$ is a left ideal of $(\mathcal{L}^{\beta},\cdot)$. In particular there is a multiplicative idempotent in $cl_{\mathcal{L}^{\beta}}K(\mathcal{L}^{\beta})$.

Proof.

Let $p\in cl_{\mathcal{L}^{\beta}}K(\mathcal{L}^{\beta},+)$ and $q\in\mathcal{L}^{\beta}$. We show that $qp\in cl_{\mathcal{L}^{\beta}}K(\mathcal{L}^{\beta},+)$. So let $A\in qp$. Then $\{x>0:x^{-1}A\in p\}\in q$. Pick $x>0$ such that $x^{-1}A\in p$. Since $p\in cl_{\mathcal{L}^{\beta}}K(\mathcal{L}^{\beta})$, so $x^{-1}A$ is piecewise syndetic. Since $\overline{x^{-1}A}$ is a neighborhood of $p$, so there exists an $\eta\in\overline{x^{-1}A}\cap K(\mathcal{L}^{\beta},+)$. This implies that $x^{-1}A\in\eta$, and hence $A\in x\eta$. Now Let $l_{x}:(0,\infty)\rightarrow(0,+\infty)$ is defined by $l_{x}(y)=xy$. It is obvious that $l_{x}$ is an additive isomorphism. So $l_{x}$ has a unique continuous isomorphism extension $l^{\beta}_{x}:\mathcal{L}^{\beta}\rightarrow\mathcal{L}^{\beta}$. Therefore $xK(\mathcal{L}^{\beta},+)=K(\mathcal{L}^{\beta},+)$ and so $A\in x\eta$ is piecewise syndetic. This implies that $\overline{A}\cap K(\mathcal{L}^{\beta},+)\neq\emptyset$ for each $A\in qp$. So $qp\in cl_{\mathcal{L}^{\beta}}K(\mathcal{L}^{\beta},+)$. ∎