Forward definition of measurability

Question

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. ?

Answer 1

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. $\square$

Answer 2

I completed the elementary part of @Joseph Van Name's answer:

Sets $U \subset X$ and $V \subset X$ in a measure space $(X, \mathcal{A})$ are called $\mathcal{A}$-proximal, written $U \mathrel{\delta_\mathcal{A}} V$, if there is no $A \in \mathcal{A}$ such that $U \subset A$ and $V \subset A^c$.

A map $f\colon X \to Y$ between measure spaces $(X, \mathcal{A})$ and $(Y, \mathcal{B})$ is a called proximal map if $U \mathrel{\delta_\mathcal{A}} V \Rightarrow f[U] \mathrel{\delta_\mathcal{B}} f[V]$ for all $U, V \subset X$.

Statement:

• Then $f$ is $\mathcal{A}$–$\mathcal{B}$-measurable if and only if $f$ is a proximal map.

Proof:

"If." Take $B \in \mathcal{B}$. From rules for images and preimages, $f[f^{-1}[B]] \subset B$, $f[(f^{-1}[B])^c] = f[f^{-1}[B^c]] \subset B^c$. To avoid contradiction with proximality of $f$, there is $A \in \mathcal{A}$ such that $f^{-1}[B] \subset A$ as well as $(f^{-1}[B])^c \subset A^c$ or equivalently $f^{-1}[B]\supset A$. Thus $f^{-1}[B] \in \mathcal{A}$.

"Only if." Let $f$ be $\mathcal{A}$–$\mathcal{B}$ measurable and take $U, V \subset X$ with $U \mathrel{\delta_\mathcal{A}} V$ and assume that there is $B\in \mathcal{B}$ such that $f[U] \subset B$, $f[V] \subset B^c$. By measurability of $f$, $ f^{-1}[B] \in \mathcal{A}$.
This time $ f^{-1}[B] \supset f^{-1}[f[U]] \supset U$ and $ f^{-1}[B] \supset f^{-1}[f[V]] \supset V$. This contradicts $U \mathrel{\delta_\mathcal{A}} V$, therefore $f[U] \mathrel{\delta_\mathcal{B}} f[V]$.