# Measurable functions on a $\sigma$-algebra generated by a collection of mappings

Let $(E,\mathcal{E})$ be a measurable space, i.e. $E$ is a set and $\mathcal{E}$ is a $\sigma$-algebra on $E$. We have a collection $\mathcal{H}$ of mappings from $\Omega$ to $E$, and $\mathcal{F}$ is a $\sigma$-algebra on $\Omega$ generated by $\mathcal{H}$, i.e. $\mathcal{F}$ is the smallest $\sigma$-algebra containing the union of all of the inverse images of measurable sets on $E$ under mappings in $\mathcal{H}$, we denote it by

$$\mathcal{F}=\sigma\{\bigcup\limits_{f\in\mathcal{H}}f^{-1}(\mathcal{E})\}$$

Now suppose $\varphi$ to be an arbitrary $\mathcal{F}$-measurable function, I would like to prove that there exists a countable subcollection $\mathcal{H}_0=\{f_1,f_2,\cdots\}$ of $\mathcal{H}$ so that $\varphi$ is $\mathcal{F}_0$-measurable, where $\mathcal{F}_0$ is the $\sigma$-algebra on $\Omega$ generated by $\mathcal{H}_0$.

It seems like this exercise can be solved by the monotone class theorem and the approximation theorem of measurable functions by simple functions.

Any hint or reference would be appreciated.

• Any element of $\sigma(\mathscr R)$ belongs to $\sigma(\mathscr R_0)$ for some countable $\mathscr R_0 \subseteq \mathscr R$. Commented Jun 20, 2018 at 0:08

Consider $\cup_{H_0} \sigma (H_0)$ where the union is taken over all at most countable families $H_0$ contained in $H$. It is very easy to verify that this is a sigma algebra and the it makes each function in $H$ measurable. It follows that it contains $\mathcal F$. This means that whenever $E \in \mathcal F$, $E \in \sigma (H_0)$ for some $H_0$. In other words, $I_E$ is measurable w.r.t. $\sigma (H_0)$ for some $H_0$. It follows that any simple function measurable with respect to $\mathcal F$ is measurable w.r.t. $\sigma (H_0)$ for some $H_0$; now you can go to non-negative measurable functions and finally for all measurable functions. [ You only need the fact that countable union of countable sets is countable to verify these statements].
• Can you please sketch how to prove that $$\bigcup_{H_0} \sigma (H_0)$$ is a $\sigma$-algebra? In general the union of $\sigma$-algebras isn't… so what's different here?
• Is it clear that the complement of any set in this union is also in the union? Now take $A_n,n=1,2,...$ in this union. There exist at most countable families $H_n,n=1,2,...$ such that $A_n \in \sigma (H_n)$. Let $H_0$ be the union of $H_n,n=1,2,...$. This is an at most countable subfamily of $H$. Since $A_n \in \sigma (H_0)$ for each $n$ it follows that $\cup_n A_n \in \sigma (H_0)$ so $\cup_n A_n$ is in the union of the sigma algebras. Commented Jun 20, 2018 at 8:21