Consider a family $f_i$, $i \in I$, of mappings of a set $\Omega$ into measurable spaces $(E_i,\mathcal{E}_i)$. We assume that for each $i \in I$ there is a subclass $\mathcal{N}_i$ of $\mathcal{E}_i$ , closed under finite intersections and such that $\sigma(\mathcal{N}_i) = \mathcal{E}_i$. Let $$ \mathcal{N} := \bigg\{ \bigcap_{j \in J} f_j^{-1}(A_j) \colon A_j \in \mathcal{N}_j, J \subseteq I \text{ finite}\bigg\}, $$ then by the Monotone Class Theorem we have $\sigma(\mathcal{N}) = \sigma(f_i \colon i \in I)$. $\qquad$ $(\ast)$
I don't see why $(\ast)$ is an immediate consequence of the Monotone Class Theorem, which states the following:
Monotone Class Theorem. Let $\mathcal{M}$ be a collection of subsets of a set $\Omega$ such that
- $\Omega \in \mathcal{M}$,
- if $A,B \in \mathcal{M}$ and $A \subseteq B$, then $B \setminus A \in \mathcal{M}$,
- if $\{A_n\}_{n \geq 1}$ is an increasing sequence of elements of $\mathcal{M}$, then $\bigcup_{n \geq 1} A_n \in \mathcal{M}$.
If $\mathcal{F} \subseteq \mathcal{M}$, where $\mathcal{F}$ is closed under finite intersections, then $\sigma(\mathcal{F}) \subseteq \mathcal{M}$.
My attempt of proof of $(\ast)$: I know that $\sigma(f_i \colon i \in I) = \sigma(\mathcal{G})$, where $$ \mathcal{G} = \bigcup_{i \in I} f_i^{-1}(\mathcal{N}_i). $$ At this point we only need that $\sigma(\mathcal{N}_i) = \mathcal{E}_i$ but we don't use that $\mathcal{N}_i$ is closed under finite intersections. Now we want to show $\sigma(\mathcal{N}) = \sigma(\mathcal{G})$. It is clear that $\mathcal{N} \subseteq \sigma(\mathcal{G})$ and hence $\sigma(\mathcal{N}) \subseteq \sigma(\mathcal{G})$. Thus it remains to show that $\mathcal{G} \subseteq \sigma(\mathcal{N})$. At this point I got stuck and I don't see how to use the Monotone Class Theorem.