# Showing a collection generates a sigma algebra

Let $$X = X_1 \times X_2$$, let $$\pi_i : X \to X_i$$ ($$i=1,2$$) be coordinate maps. Let $$\mathcal{M}_i$$ be $$\sigma-algebra$$ on $$X_i$$. Let $$\mathcal{F} = \{ \pi_i^{-1}(E_i) : E_i \in \mathcal{M}_i \}$$. $$\sigma( \mathcal{F} )$$ is $$\sigma-$$algebra on the product $$X$$.

# attempt:

Lets take an element from $$\mathcal{F}$$(which also belongs to $$\sigma( \mathcal{F} )$$), say $$\pi_1^{-1}(E_1)$$. We know

$$\pi_1^{-1}(E_1) = \{ (x_1,x_2) : x_1 \in E_1, \; \; x_2 \in X_2 \}$$

If we have that $$E_2 = X_2$$, then it must be the case that $$\pi_1^{-1}(E_1) = E_1 \times E_2$$, and so $$\pi_1^{-1}(E_1) \in \mathcal{E}$$. Hence, $$\sigma( \mathcal{F} ) \subset \mathcal{E}$$. But we know $$\mathcal{E} \subset \sigma( \mathcal{E} )$$. In particular, we have $$\boxed{ \sigma(\mathcal{F}) \subset \sigma(\mathcal{E}) }$$

For the other direction, we take one element from $$\mathcal{E}$$, and notice

$$E_1 \times E_2 = \{ (x_1,x_2) : x_i \in E_i \} = \{ (x_1,x_2) : x_1 \in E_1, x_2 \in X_2 \} \cap \{ (x_1,x_2) : x_1 \in X_1 , x_2 \in E_2 \} = \pi_1^{-1}(E_1) \cap \pi_2^{-1}(E_2) \in \mathcal{F} \subset \sigma( \mathcal{F} )$$

and so we obtain that $$\mathcal{E} \subset \sigma( \mathcal{F} )$$ which implies that $$\sigma( \mathcal{E} ) \subset \sigma( \mathcal{F} )$$ which in turn gives us that

$$\sigma( \mathcal{E} ) = \sigma( \mathcal{F} )$$

as was to be shown.

Is this a correct proof ? Can we generalize this to any indexed collection?

This can only be generalized to a countable indexed collection, by the result of your earlier question. The first part is ok. About the second part, it is not true that $\pi_1^{-1}(E_1) \cap \pi_2^{-1}(E_2) \in \mathcal{F}$, in general.
We want to prove that $\sigma({\cal F}) \subset \sigma({\cal E})$: take $E_1 \in {\cal M}_1$ and $E_2 \in {\cal M}_2$. Then $$\pi_1^{-1}(E_1) = E_1 \times X_2 \in {\cal E}, \quad \pi_2^{-1}(E_2) = X_1 \times E_2 \in {\cal E}.$$Since my $E_i$ were arbitrary, we get: $${\cal F} \subseteq {\cal E}\subseteq \sigma({\cal E}) \implies \sigma({\cal F})\subseteq \sigma({\cal E}).$$On the other hand, take $E_1 \times E_2 \in {\cal E}$, arbitrary: we have: $$E_1 \times E_2 = \pi_1^{-1}(E_1) \cap \pi_2^{-1}(E_2) \in \sigma({\cal F}),$$because the RHS is a countable (two!) intersection of measurables. So: $${\cal E} \subseteq \sigma({\cal F}) \implies \sigma({\cal E})\subseteq \sigma({\cal F}),$$and we conclude that $\sigma({\cal E})=\sigma({\cal F})$.