Is there a characterisation of measurability in a forward way similar to the closure characterisation of continuity?


A function $f\colon X \to Y$ is continuous if equivalently:


1. If $G\subset Y$ is open, then is $f^{-1}(G)$ open too.
2. $f(\overline A) \subset \overline{f( A)}$ where $\overline A$ is the closure of $A$.


A function $f\colon X \to Y$ is measurable if equivalently:


3. If $A\subset Y$ is a measurable set, then is $f^{-1}(A)$ a measurable subset of $X$ too.
4. ?



Forward definition of measurability
If $(X,M)$ is an algebra of sets, then define a relation $\delta\_M\subseteq P(X)^2$ by setting $A\mathrel{\delta\_M}B$ if there does not exist some $S\in M$ with $A\subseteq S$, $B\subseteq S^c$. Suppose that $(X,M)$, $(Y,N)$ are algebras of sets. Let $f:X\rightarrow Y$ be a function. Then it is not too hard to prove that $f^{-1}[S]\in M$ whenever $S\in N$ precisely when $A\mathrel{\delta\_M} B\Rightarrow f[A]\mathrel{\delta\_N}f[B]$ for $A,B\subseteq X$.


**Ideals**


Suppose $A,B\subseteq X$. Let $\mathcal{I}\_{A,B}$ be the collection of all
$C\in M$ where there is some $D\in M$ where $A\cap C\subseteq D,B\cap C\subseteq D^c$. Then $\mathcal{I}\_{A,B}$ is an ideal in the Boolean algebra $M$ and $\mathcal{I}\_{A,B}$ is a proper ideal in $M$ if and only if $A\delta\_MB$.


**More context: proximity spaces**


The reader is assumed to be familiar with proximity spaces when reading this answer. The relation $\delta\_M$ defined above is a proximity relation, and by definition $A\mathrel{\delta\_M}B\Rightarrow f[A]\mathrel{\delta\_N}f[B]$ precisely when $f$ is a proximity map. We shall now establish that the category of zero-dimensional proximity spaces is isomorphic to the category of algebras of sets with measurable mappings as morphisms, and the characterization of measurable transformations as mappings where $A\mathrel{\delta\_M}B\Rightarrow f[A]\mathrel{\delta\_N}f[B]$ will immediately follow as a corollary.


A proximity space $X$ is said to be zero-dimensional if whenever $A\ll B$, there is some $S$ with $A\ll S\ll S\ll B$. Equivalently, $X$ is zero-dimensional precisely when there is some $C\subseteq X$ with $A\subseteq C,B\subseteq C,C\not\delta C^c$.


If $(X,\ll)$ is a proximity space, then let $M\_\ll\subseteq P(X)$ be the set defined by letting $A\in M\_\ll$ precisely when $A\ll A$. Then $(X,M\_\ll)$ is an algebra of sets. If $(X,M)$ is an algebra of sets, then let $\ll\_M$ be the relation where we set $A\ll\_MB$ precisely when $A\subseteq S\subseteq B$ for some $S\in M$. Then $\ll\_M$ is a proximity relation on the set $X$.


Proposition:


1. Suppose that $(X,M)$ is an algebra of sets. Then $M=M\_{\ll\_M}$.
2. If $(X,\ll)$ is a zero-dimensional proximity space, then ${\ll}={\ll\_{M\_\ll}}$.


Proof:


1. If $R\in M$, then $R\ll\_M R$, so $R\in M\_{\ll\_M}$. Similarly, if
$R\in M\_{\ll\_M}$, then $R\ll\_MR$, hence there is some $S\in M$ with $R\subseteq S\subseteq R$. But this is only possible if $R=S\in M$.
2. Suppose $A\ll B$. Then by zero-dimensionality, there is some $S$ with $A\subseteq S\ll S\subseteq B$. Therefore, $S\in M\_{\ll}$, so $S\ll\_{M\_\ll}S$. We can therefore conclude that $A\ll\_{M\_\ll}B$. Now assume that $A\ll\_{M\_\ll}B$. Then there is some $S\in M\_\ll$ with
$A\subseteq S\subseteq B$. But this means that $S\ll S$, so $A\ll B$ as well. $\square$


Observation: If $(X,M)$, $(Y,N)$ are algebras of sets, then a function $f:X\rightarrow Y$ is a measurable mapping (which means that $f^{-1}[R]\in M$ whenever $R\in N$) precisely when $f$ is a proximity map from $(X,\ll\_M)$ to $(Y,\ll\_N)$.


Proof:


$\rightarrow$ Suppose that $f$ is a measurable mapping. Then whenever
$A,B\subseteq Y$ and $A\ll\_N B$, there is some $S\in N$ with $A\subseteq S\subseteq B$. Therefore, $f^{-1}[S]\in M$ and $f^{-1}[A]\subseteq f^{-1}[S]\subseteq f^{-1}[B]$. Therefore, $f^{-1}[A]\ll\_M f^{-1}[B]$, so $f$ is a proximity map.


$\leftarrow$ Suppose that $f$ is a proximity map. Then whenever $S\in N$, we have $S\ll\_NS$, so $f^{-1}[S]\ll\_Mf^{-1}[S]$, hence $f^{-1}[S]\in M$ as well, so $f$ is a measurable mapping. $\square$


Proposition: If $(X,M)$ is a separating algebra of sets, then the Smirnov compactification of $(X,\ll\_M)$ is simply the Stone space of $M$.


Proof: Let $S(M)$ denote the Stone space of $M$. Let $\iota:X\rightarrow S(M)$ be the mapping where $\iota(x\_0)=\{A\in M:x\_0\in A\}$. Then the space $S(M)$ is a compactification of $\iota[X]$. Suppose now that $A,B\subseteq X$. We now need to show that $A\mathrel{\not\delta\_M} B$ if and only if $\overline{\iota[A]}\cap\overline{\iota[B]}=\emptyset$.


If $A\mathrel{\not\delta\_M}B$, then there is some $S\in M$ with $A\subseteq S$, $B\subseteq S^c$. Let $C\subseteq S(M)$ be the clopen set consisting of all ultrafilters $\mathcal{U}\subseteq M$ with $S\in\mathcal{U}$. Then $C^c$ is the set of all ultrafilters $\mathcal{U}\subseteq M$ with $S^c\in\mathcal{U}$. If $a\in A$, then $a\in S$, so $S\in\iota(a)$, hence $\iota(a)\in C$. Therefore, $\iota[A]\subseteq C$. Similarly, if $b\in B$, then $b\in S^c$, so $S^c\in\iota(b)$, hence $\iota(b)\in C^c$. Therefore, $\iota[B]\subseteq C^c$. In particular, $\overline{\iota[A]}\cap\overline{\iota[B]}\subseteq C\cap C^c=\emptyset$.


Suppose now that $\overline{\iota[A]}\cap\overline{\iota[B]}=\emptyset$. Then there is some clopen set $C\subseteq S(M)$ with $\iota[A]\subseteq C$, $\iota[B]\subseteq C^c$. But by Stone duality, there is some $S\in M$ where $C=\{\mathcal{U}\in S(M):S\in\mathcal{U}\}$. Therefore, if $a\in A$, then $\iota(a)\in C$, so $S\in\iota(a)$, hence $a\in S$. Thus $A\subseteq S$. Similarly, if $b\in B$, then $\iota(b)\in C^c$, so $S^c\in\iota(b)$, hence $b\in S^c$. Therefore, $B\subseteq S^c$. We conclude that
$A\mathrel{\not\delta\_M}B$.


We may therefore conclude that the Stone space $S(M)$ is the Smirnov compactification of $X$.


$\square$


Proposition: A separated proximity space $(X,\ll)$ is zero-dimensional if and only if the Smirnov compactification $K$ is zero-dimensional.


Proof: $\DeclareMathOperator\Cl{Cl}$By the above proposition, we know that if $(X,\ll)$ is zero-dimensional, then $K=S(M\_\ll)$ which is zero-dimensional. Suppose that $K$ is zero-dimensional. Let $A,B\subseteq X$ and suppose that $A\mathrel{\not\delta} B$. Then $\Cl\_K(A)\cap\Cl\_K(B)=\emptyset$. Therefore, there is some clopen subset $C\subseteq K$ where $\Cl\_K(A)\subseteq C$, $\Cl\_K(B)\subseteq K\setminus C$. In this case,
$\Cl\_K(C\cap X)\cap\Cl\_K(X\setminus C)=\emptyset$, so
$A\subseteq C\cap X$, $B\subseteq X\setminus C$ and $(C\cap X)\mathrel{\not\delta}(X\setminus C)$. Therefore, $\delta$ is zero-dimensional. $\square$


