Consider a family of measurable functions $\mathcal{F}:= (f_{ij} \mid (i,j)\in I\times J)$ indexed over some non-empty sets $I,J$.
I was wondering if, for $\sigma(\mathcal{F})$ the $\sigma$-algebra generated by $\mathcal{F}$, it then holds that
$$\sigma(\mathcal{F}) = \sigma\left(\bigcup_{j\in J}\sigma(f_{ij}\mid i\in I)\right)\ ?$$
(If somehow required, it may be assumed that $I$ and $J$ are both countable.)
This is true for arbritrary $I, J$. Suppose the $f_{ij}$ take values in a measurable space $(G, \mathcal G)$.
Note that $\sigma(f_{ij}\mid i\in I) = \sigma(\{f_{ij} \in A\} \mid i\in I, A \in \mathcal G) \subset \sigma(\{f_{ij} \in A\} \mid (i,j)\in I\times J, A \in \mathcal G) \ \forall j \in J$, so that $\bigcup_{j\in J}\sigma(f_{ij}\mid i\in I) \subset \sigma(\mathcal F)$. Hence, $\sigma\left(\bigcup_{j\in J}\sigma(f_{ij}\mid i\in I)\right) \subset \sigma(\mathcal{F})$.
Also, $\{f_{ij} \in A\} \in \bigcup_{j\in J}\sigma(f_{ij}\mid i\in I) \ \forall\ i \in I, j \in J, A \in \mathcal G$, and so $\sigma(\mathcal F) = \sigma(\{f_{ij} \in A\} \mid (i,j)\in I\times J, A \in \mathcal G) \subset \sigma\left(\bigcup_{j\in J}\sigma(f_{ij}\mid i\in I)\right) $