Let $\Omega \neq \emptyset$ be a nonempty set and let $(\Omega_1, \mathcal{A}_1),...,(\Omega_n, \mathcal{A}_n)$ be a finite number of measurable spaces. Furthermore, let $X_i : \Omega \rightarrow \Omega_i$ be maps and $X : \Omega \rightarrow \Omega_1 \ \times ... \times \ \Omega_n $ defined as $X(w) := (X_1(w),...,X_n(w))$.
We denote the sigma-operator with $\sigma$, so that for maps to measurable spaces $X_1,...,X_n$, $\sigma(X_1,...,X_n)$ is the smallest $\sigma$-algebra so that all $X_i$ are measurable. I have the suspicion that the following equality holds:
$$ \sigma(X_1,...,X_n) = X^{-1}(\mathcal{A}_1 \ \otimes...\otimes \ \mathcal{A}_n) $$
Is this always the case?
Kind regards and thanks for any help,
Joker