On proving a consequence of the monotone class theorem. In the book of Mark Yor, Continuous Martingales and Brownian Motion I read this theorem that is left without proof: 

I suspect this should follow from the monotone class theorem but on what family of sets do I apply it?
 A: I agree that that section in the book makes it sound like Theorem 2.3 is a consequence of the "monotone class theorem" 2.1 (which I know as the "$\pi$-$\lambda$ theorem"). But I think all that is happening here is that the authors are stating the setup in which they will later be applying Theorems 2.1 and 2.2, and then the authors state a relevant fact about this setup in Theorem 2.3. (Although it is a bit odd, as Theorem 2.4 does seem to be an application of Theorem 2.2 - but then Theorem 2.4 itself looks like it's missing a condition, as the space of all $\sigma(f_i,i\in I)$-simple functions fulfils the conditions of Theorem 2.4.)
In fact, Theorem 2.3 does not even rely on $\mathcal{N}_i$ being closed under finite intersections. [I expect the main point about the monotone class theorem, in the context of this setup, is to be able to prove later on that the law of a stochastic process $(f_i)_{i\in I}$ is determined by the probabilities of events of the form $\{f_i(\omega) \in A_i \ \forall \, i \in J \}$ with $J \subset I$ finite and $A_i \in \mathcal{N}_i$. This does rely on the given generator $\mathcal{N}_i$ of $\mathcal{E}_i$ being closed under finite intersections.]
The proof of Theorem 2.3 is as follows:
It is clear that $\mathcal{N} \subset \sigma(f_i:i\in I)$, and so $\sigma(\mathcal{N}) \subset \sigma(f_i:i\in I)$. For the other direction: For each $i \in I$, since $\mathcal{E}_i = \sigma(\mathcal{N}_i)$, we have that $\{f_i^{-1}(A):A \in \mathcal{E}_i\} = \sigma(f_i^{-1}(A):A \in \mathcal{N}_i)$.$^\dagger$ Hence
\begin{align*}
\sigma(f_i:i \in I) \ &\subset \ \sigma\!\left( \bigcup_{i \in I} \ \{f_i^{-1}(A):A \in \mathcal{N}_i\} \right) \\
&\subset \ \sigma(\mathcal{N}) \hspace{6mm} \text{(by just considering singletons $J=\{i\}$).}
\end{align*}
QED
$^\dagger$This is a very important and fundamental (but not immediately obvious) fact. You can try to prove it yourself, otherwise you can probably find a proof by looking up the proof that "continuous functions are measurable".
