Let $\: \langle X,\delta\rangle \: $ be a separated proximity space.

Let $\: \mu^* \: : \: 2^{X} \: \to \: [0,+\infty] \: $ be a proximal outer measure.

Let $U$ be an open subset of $X$.

Does it follow that $U$ is Caratheodory measurable by $\mu^\ast$?


I shall first give an example of an open set in a proximity space that is not measurable. Let $X$ be an uncountable discrete space with the proximity $\delta$ induced by the one-point compactification of $X$. In this case, if $A\overline{\delta}B$ if and only if $A\cap B=\emptyset$ and either $A$ or $B$ is finite. Let $\mu^{*}$ be the outer measure on $X$ where $\mu^{*}(A)=1$ if $A$ is uncountable and $\mu^{*}(A)=0$ otherwise. Clearly $\mu^{*}$ is an outer measure, and $\mu^{*}$ is proximal. If $A\overline{\delta} B$, then we can assume without loss of generality that $A$ is finite, so $\mu(A\cup B)=\mu(B)=\mu(A)+\mu(B)$. However, if $U\subseteq X$ is a subset where both $U$ and $U^{c}$ are uncountable, then $\mu^{*}$ cannot be outer measurable since $\mu^{*}(X)=1$, but $\mu^{*}(U)+\mu^{*}(U^{c})=1+1=2$. Therefore $U$ is an open set in $X$ that is not measurable.

While there may be open sets that are not measurable, every proximally cozero set is measurable. In many common spaces the proximally cozero sets coincide with the open sets, so it is sometimes the case that every open set is measurable.

If $(X,\delta)$ is a proximity space, then a proximally zero set is a set of the form $f^{-1}(0)$ for some proximity map $f:(X,\delta)\rightarrow[0,1]$. A proximally cozero set is a set of the form $f^{-1}(0,1]$ for some proximity map $f:(X,\delta)\rightarrow[0,1]$. In other words, a proximally cozero set is a complement of a proximally zero set. It can be shown that the proximally cozero sets are precisely the sets of the form $\bigcup_{n}A_{n}$ for some sequence $A_{1}\prec A_{2}\prec...\prec A_{n}\prec...$. The $\sigma$-algebra of proximally Baire sets is the $\sigma$-algebra generated by the proximally zero sets. See my paper http://arxiv.org/abs/1202.3127 for more information on proximally zero sets and the $\sigma$-algebra of proximally Baire sets.

$\textbf{Theorem}$: Let $(X,\delta)$ be a proximity space, and let $\mu^{*}$ be an outer measure on $X$ such that if $A\overline{\delta}B$, then $\mu^{*}(A\cup B)=\mu^{*}(A)+\mu^{*}(B)$.

  1. If $A_{n}\prec A_{n+1}$ for all $n$ and $E\subseteq X$, then $\mu^{*}(\bigcup_{n}(A_{n}\cap E))=^{\lim}_{n\rightarrow\infty}\mu^{*}(A_{n}\cap E)$.

  2. Every proximally Baire set in $(X,\delta)$ is measurable.


  1. Let $x=^{\lim}_{n\rightarrow\infty}\mu^{*}(A_{n}\cap E)$. If $x=\infty$, then clearly $\mu^{*}(\bigcup_{n}(A_{n}\cap E))=\infty=^{\lim}_{n\rightarrow\infty}\mu^{*}(A_{n}\cap E)$. Now assume that $x<\infty$. Then for each $\epsilon>0$ there is some $N$ where $x-\mu^{*}(A_{N}\cap E)<\epsilon$. Let $B_{n}=E\cap(A_{n+1}\setminus A_{n})$ for all $n$. Then for all positive integers $L$, the sets $A_{N}\cap E,B_{N+1},B_{N+3},...,B_{N+2L+1}$ are pairwise separated, and the sets $A_{N}\cap E,B_{N+2},B_{N+4},...,B_{N+2L}$ are pairwise separated as well. Therefore, we have $$\mu^{*}(A_{N}\cap E)+\mu^{*}(B_{N+1})+\mu^{*}(B_{N+3})+...+\mu^{*}(B_{N+2L+1})$$ $$=\mu^{*}((A_{N}\cap E)\cup B_{N+1}\cup...\cup B_{N+2L+1}) \leq\mu^{*}(A_{N+2L+2}\cap E)\leq x,$$ so $$\mu^{*}(B_{N+1})+\mu^{*}(B_{N+3})+...+\mu^{*}(B_{N+2L+1})\leq x-\mu^{*}(A_{N}\cap E)<\epsilon.$$ Similarly, we have $$\mu^{*}(A_{N}\cap E)+\mu^{*}(B_{N+2})+...+\mu^{*}(B_{N+2L})$$ $$=\mu^{*}((A_{N}\cap E)\cup B_{N+2}\cup...\cup B_{N+2L})$$ $$\leq\mu^{*}(A_{N+2L+1})\leq x.$$ Therefore, $\mu^{*}(B_{N+2})+\mu^{*}(B_{N+4})+...+\mu^{*}(B_{N+2L})\leq x-\mu^{*}(A_{N}\cap E)<\epsilon$. We therefore conclude that $\mu^{*}(B_{N+1})+\mu^{*}(B_{N+2})+...+\mu^{*}(B_{N+2L+1})<2\epsilon$. Since $L$ is arbitrary, we have $\sum_{k=1}^{\infty}\mu^{*}(B_{N+k})\leq 2\epsilon$. Therefore, we conclude that $$\mu^{*}(\bigcup_{n}(A_{n}\cap E))\leq\mu^{*}(A_{N+1}\cap E)+\sum_{k=1}^{\infty}\mu^{*}(B_{N+k})\leq x+2\epsilon.$$

Since $\epsilon$ is arbitrary, we have $\mu^{*}(\bigcup_{n}(A_{n}\cap E))\leq x=^{\lim}_{n\rightarrow\infty}\mu^{*}(A_{n}\cap E)\leq \mu^{*}(\bigcup_{n}(A_{n}\cap E))$. Therefore $\mu^{*}(\bigcup_{n}(A_{n}\cap E))=^{\lim}_{n\rightarrow\infty}\mu^{*}(A_{n}\cap E)$.

  1. It suffices to show that every proximally cozero set is measurable. Let $U$ be a proximally cozero set. Then $U=\bigcup_{n}U_{n}$ for some sequence $U_{1}\prec U_{2}\prec...$. Therefore if $E\subseteq X$, then $$\mu^{*}(E\cap U)+\mu^{*}(E\cap U^{c})=^{\lim}_{n\rightarrow\infty}\mu^{*}(E\cap U_{n})+\mu^{*}(E\cap U^{c})$$ $$=^{\lim}_{n\rightarrow\infty}\mu^{*}((E\cap U_{n})\cup(E\cap U^{c}))\leq\mu^{*}(E).$$ QED.

In particular, if $(\mathbb{R},\rho)$ is a proximity space compatible with the euclidean topology on $\mathbb{R}$, then every proximity map $f:(X,\delta)\rightarrow(\mathbb{R},\rho)$ is measurable.

In every Lindelof proximity space where every open subspace is Lindelof, every open set is a proximally cozero set, so in a Lindelof proximity space, every open set is measurable. Also, if our proximity is induced by a metric, then every closed set is a proximally zero set, so every closed set is measurable.


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