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.
The result does not need the Monotone Class Theorem, just some basic manipulations with $\sigma$-fields. The crucial lemma is this:
Lemma: If $f:\Omega\to E$ is a map, and $\cal C$ a collection of subsets of $E$, then $\sigma f^{-1}{\cal C}=f^{-1}\sigma\, {\cal C}.$
Proof. ($\Rightarrow$) Since ${\cal C}\subseteq \sigma {\cal C}$ we have $$f^{-1}{\cal C}\subseteq f^{-1}\sigma {\cal C}.\tag1$$ Since $f^{-1}\sigma {\cal C}$ is a $\sigma$-field (check this!) generating $\sigma$-fields on both sides of (1) gives $\sigma f^{-1}{\cal C}\subseteq f^{-1}\sigma {\cal C}.$
($\Leftarrow$) Since $f^{-1}{\cal C}\subseteq \sigma f^{-1}{\cal C}$ we have $${\cal C}\subseteq {\cal C}^\prime\tag2$$ where ${\cal C}^\prime=\{B\subseteq E: f^{-1}B\in \sigma f^{-1}{\cal C}\}$. Now ${\cal C}^\prime$ is $\sigma$-field (check this!), so when we generate $\sigma$-fields on both sides of (2) we get $\sigma {\cal C}\subseteq {\cal C}^\prime$, and hence $f^{-1}\sigma {\cal C}\subseteq \sigma f^{-1}{\cal C}.$
For fixed $i\in I$ we have ${\cal N}\supseteq f_i^{-1}{\cal N}_i.$ Generating $\sigma$-fields on both sides and applying the lemma gives $$\sigma({\cal N})\supseteq \sigma f^{-1}_i{\cal N}_i =f_i^{-1}\sigma {\cal N}_i=f^{-1}_i{\cal E}_i=\sigma(f_i).$$
This is true for every $i$, so $$\sigma({\cal N})\supseteq \sigma(f_i, i\in I).$$ The other inclusion is trivial (check this!) and so $$\sigma({\cal N}) =\sigma(f_i, i\in I).$$