Characterisation of $\sigma$-algebra generated by a family of functions by the product of those functions I have seen a special case of the following characterisation of the $\sigma$-algebra generated by a family of functions, and I thought it generalised to the claim below. I haven't seen a proof of the special case I'm referring to (where the family of spaces/functions is finite), so I can't compare.

Let $(X_\alpha, \mathcal{E}_\alpha)_{\alpha \in A}$ be an arbitrary family of measurable spaces, and let $(X, \mathcal{E})$ be the product space, i.e. let $X = \prod_{\alpha \in A} X_\alpha$ and $\mathcal{E} = \bigotimes_{\alpha \in A} \mathcal{E}_\alpha$. That is, $\mathcal{E}$ is the $\sigma$-algebra generated by the projections $\pi_\alpha$.
Consider a set $Y$ and a family $\mathcal{A} = (f_\alpha)_{\alpha \in A}$ of functions $f_\alpha \colon Y \to X_\alpha$, and equip $Y$ with the $\sigma$-algebra $\mathcal{F} = \sigma(\mathcal{A})$ generated by $\mathcal{A}$. Let $f \colon Y \to X$ be the product of the $f_\alpha$, i.e. the unique measurable function such that $\pi_\alpha \circ f$ is measurable for all $\alpha \in A$. Then I claim that $\mathcal{F} = f^{-1}(\mathcal{E})$. That is, the $\sigma$-algebra generated by the $f_\alpha$ is the same as the one generated by $f$ alone.

For the first inclusion, let $B \in \mathcal{E}_\alpha$ for some $\alpha$, and consider the set $f_\alpha^{-1}(B)$, since $\mathcal{F}$ is generated by sets on this form. But
$$ f_\alpha^{-1}(B) = (\pi_\alpha \circ f)^{-1}(B) = f^{-1}(\pi_\alpha^{-1}(B)) \in f^{-1}(\mathcal{E}), $$
since $f_\alpha^{-1}(B) \in \mathcal{E}$ by definition of $\mathcal{E}$.
For the second inclusion, let $\mathcal{B}$ be the sets on the form $\pi_\alpha^{-1}(B)$ for $B \in \mathcal{E}_\alpha$. Then $\mathcal{E} = \sigma(\mathcal{B})$, so $f^{-1}(\mathcal{E}) = f^{-1}(\sigma(\mathcal{B})) = \sigma(f^{-1}(\mathcal{B}))$, and so it is enough to show that $f^{-1}(\mathcal{B}) \subseteq \mathcal{F}$. And for $B \in \mathcal{E}_\alpha$ we have
$$ f^{-1}(\pi_\alpha^{-1}(B)) = (\pi_\alpha \circ f)^{-1}(B) = f_\alpha^{-1}(B) \in \mathcal{F}. $$
Is my argument correct? If not, what goes wrong? I don't believe I've seen this explicitly mentioned anywhere, and I couldn't find it by searching, so I was worried that the claim might not be true!
 A: I figured I would go back and clear this up since no one else has commented.
The product $\sigma$-algebra on $X$ is the initial $\sigma$-algebra induced by the projections $\pi_\alpha$. Let $\mathcal F$ be the initial $\sigma$-algebra on $Y$ induced by $f \colon Y \to X$. Then $\mathcal F$ is also induced by the maps $\pi_\alpha \circ f = f_\alpha$:
First we have $\mathcal F = \sigma ( f^{-1}(\mathcal E) )$, and it is easy to show that we may replace $\mathcal E$ by any collection of sets $\mathcal D$ that generates $\mathcal E$ (it suffices to show that measurability of $f$ only depends on preimages of sets in $\mathcal D$). So pick $\mathcal D = \bigcup_{\alpha \in A} \pi_\alpha^{-1}(\mathcal E_\alpha)$. It follows that
$$ \mathcal F
= \sigma \biggl( f^{-1} \Bigr(\bigcup_{\alpha \in A} \pi_\alpha^{-1}(\mathcal E_\alpha) \Bigl) \biggr)
= \sigma \biggl( \bigcup_{\alpha \in A} (\pi_\alpha \circ f)^{-1}(\mathcal E_\alpha) \biggr), $$
which proves the claim. Indeed, this argument generalises verbatim to the case where $\mathcal F$ and $\mathcal E$ are general initial $\sigma$-algebras. (And incidentally, the proof works just as well for initial topologies, see e.g. Willard Exercise 8H.)
