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Keywords:
real measurable function; lattice-ordered ring; realcompact measurable space; real Riesz map; free ideal
real measurable function; lattice-ordered ring; realcompact measurable space; real Riesz map; free ideal
Summary:
Let $ M(X, \mathscr{A})$ ($M^{*}(X, \mathscr{A})$) be the $f$-ring of all (bounded) real-measurable functions on a $T$-measurable space $(X, \mathscr{A})$, let $M_{K}(X, \mathscr{A})$ be the family of all $f\in M(X, \mathscr{A})$ such that ${{\,\mathrm{coz}}}(f)$ is compact, and let $M_{\infty }(X, \mathscr{A})$ be all $f\in M(X, \mathscr{A})$ that $\lbrace x\in X: |f(x)|\ge \frac{1}{n}\rbrace $ is compact for any $n\in \mathbb{N}$. We introduce realcompact subrings of $M(X, \mathscr{A})$, we show that $M^{*}(X, \mathscr{A})$ is a realcompact subring of $M(X, \mathscr{A})$, and also $M(X, \mathscr{A})$ is a realcompact if and only if $(X, \mathscr{A})$ is a compact measurable space. For every nonzero real Riesz map $\varphi : M(X, \mathscr{A})\rightarrow \mathbb{R}$, we prove that there is an element $x_0\in X$ such that $\varphi (f) =f(x_0)$ for every $f\in M(X, \mathscr{A})$ if $(X, \mathscr{A})$ is a compact measurable space. We confirm that $M_{\infty }(X, \mathscr{A})$ is equal to the intersection of all free maximal ideals of $M^{*}(X, \mathscr{A})$, and also $M_{K}(X, \mathscr{A})$ is equal to the intersection of all free ideals of $M(X, \mathscr{A})$ (or $M^{*}(X, \mathscr{A})$). We show that $M_{\infty }(X, \mathscr{A})$ and $M_{K}(X, \mathscr{A})$ do not have free ideal.
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