Measurability via generators (checking preimages on a generating class)

This note records a standard “workhorse” criterion: to check measurability, it suffices to check preimages on a generating family.

If you need background on “generated σ-algebras”, see Generated σ-algebras: what it means (and common confusions). For π/λ-systems and the monotone class theorem, see σ-algebras, π/λ-systems, and the monotone class theorem.

Setup

Let $(E,\Sigma_1)$ and $(F,\Sigma_2)$ be measurable spaces, and let $f:E\to F$ be a function.

Let $\mathcal F_0\subseteq \Sigma_2$ be a generating family for $\Sigma_2$, meaning

$$ \Sigma_2=\sigma(\mathcal F_0). $$

Set-theoretic preliminaries: functions and preimages

Definition (function / mapping)

A function (mapping) $f:E\to F$ assigns to each $x\in E$ a unique value $f(x)\in F$.

Definition (preimage)

For any set $B\subseteq F$, the preimage of $B$ under $f$ is

$$ f^{-1}(B):=\{x\in E:\ f(x)\in B\}\subseteq E. $$

Lemma (basic preimage identities)

For any function $f:E\to F$ and any sets $B,B_1,B_2,\dots\subseteq F$:

• Whole space / empty set:
  • $f^{-1}(F)=E$
  • $f^{-1}(\emptyset)=\emptyset$
• Complement:
  • $f^{-1}(F\setminus B)=E\setminus f^{-1}(B)$
• Countable unions:
  • $f^{-1}\!\left(\bigcup_{n\ge1} B_n\right)=\bigcup_{n\ge1} f^{-1}(B_n)$

These are purely set-theoretic facts; no measurability assumptions are needed.

Definition: “measurable relative to $(\Sigma_1,\Sigma_2)$”

Let $(E,\Sigma_1)$ and $(F,\Sigma_2)$ be measurable spaces. A function $f:E\to F$ is $(\Sigma_1,\Sigma_2)$-measurable (a.k.a. measurable relative to $\Sigma_1,\Sigma_2$) if

$$ \forall B\in\Sigma_2,\quad f^{-1}(B)\in\Sigma_1. $$

Proposition (measurability via a generator)

The following are equivalent:

1. $f$ is $(\Sigma_1,\Sigma_2)$-measurable, i.e. $f^{-1}(B)\in\Sigma_1$ for every $B\in\Sigma_2$.
2. $f^{-1}(B)\in\Sigma_1$ for every $B\in\mathcal F_0$.

Necessity

If $f$ is measurable, then by definition $f^{-1}(B)\in\Sigma_1$ for all $B\in\Sigma_2$. Since $\mathcal F_0\subseteq \Sigma_2$, the condition holds in particular for all $B\in\mathcal F_0$.

Sufficiency (the “$\mathcal F_1$ trick”)

Assume $f^{-1}(B)\in\Sigma_1$ for all $B\in\mathcal F_0$.

Define

$$ \mathcal F_1:=\{B\subseteq F:\ f^{-1}(B)\in\Sigma_1\}. $$

Claim 1: $\mathcal F_1$ is a σ-algebra on $F$.

• Whole space: because $f:E\to F$, we have the set-theoretic identity

  $$ f^{-1}(F)=E. $$

  Since $E\in\Sigma_1$, it follows that $F\in\mathcal F_1$.
  Note: this uses only the definition of preimage + “$f$ maps into $F$”; no measurability is assumed here.

• Complement: if $B\in\mathcal F_1$, then $f^{-1}(B)\in\Sigma_1$. Since $\Sigma_1$ is a σ-algebra, $E\setminus f^{-1}(B)\in\Sigma_1$. Using the identity

  $$ f^{-1}(F\setminus B)=E\setminus f^{-1}(B), $$

  we get $F\setminus B\in\mathcal F_1$.

• Countable union: if $B_1,B_2,\dots\in\mathcal F_1$, then $f^{-1}(B_n)\in\Sigma_1$ for all $n$, so $\bigcup_{n\ge1} f^{-1}(B_n)\in\Sigma_1$. Using

  $$ f^{-1}\!\left(\bigcup_{n\ge1}B_n\right)=\bigcup_{n\ge1} f^{-1}(B_n), $$

  we get $\bigcup_{n\ge1}B_n\in\mathcal F_1$.

So $\mathcal F_1$ is a σ-algebra.

Claim 2: $\mathcal F_0\subseteq \mathcal F_1$.

This is exactly the assumption: for each $B\in\mathcal F_0$, we have $f^{-1}(B)\in\Sigma_1$, hence $B\in\mathcal F_1$.

Now since $\mathcal F_1$ is a σ-algebra containing $\mathcal F_0$, it must contain the σ-algebra generated by $\mathcal F_0$:

$$ \sigma(\mathcal F_0)\subseteq \mathcal F_1. $$

But $\sigma(\mathcal F_0)=\Sigma_2$, so $\Sigma_2\subseteq \mathcal F_1$. Unpacking the definition of $\mathcal F_1$, this means:

$$ \forall B\in\Sigma_2,\quad f^{-1}(B)\in\Sigma_1, $$

which is exactly measurability of $f$.

Example: Borel measurability into $\mathbb R$

If $(F,\Sigma_2)=(\mathbb R,\mathcal B(\mathbb R))$, a common generator is the family of open intervals $\{(a,b):a

So to show $f:E\to\mathbb R$ is Borel-measurable, it often suffices to check that $f^{-1}((-\infty,a))\in\Sigma_1$ for all rational $a$.