If $Y_1=(X_1,...,X_l)$ then $\sigma(Y_1)=\sigma\left(\cup_{i=1}^l \sigma(X_i)\right)$. $(\Omega, \mathcal A, P)$ probability space, $(\Omega_j, \mathcal A_j)$ measure space, $X_j : \Omega \rightarrow \Omega_j, ~ j =1, \ldots, n$ random variables. Furthermore
$\mathcal B_1 := \bigotimes_{j=1}^l \mathcal A_j, ~ \mathcal B_2 := \bigotimes_{j=l+1}^n \mathcal A_j, ~ \overline{\Omega_1} := \times_{j=1}^l \Omega_j, ~ \overline{\Omega_2} := \times_{j=l+1}^n \Omega_j, ~ \\ Y_1 := (X_1, \ldots, X_l), ~ Y_2 := (X_{l+1}, \ldots, X_n), ~ l \in \{1, \ldots, n-1 \}$.
I want to show that for independent $X_1, \ldots, X_n$ $Y_1$ and $Y_2$ are independent, too. Can somebody tell me why $\sigma(Y_1) := Y_1^{-1} (\mathcal B_1)$ is the same as $\sigma(\bigcup_{j=1}^l \sigma(X_j))$? I've already shown that the latter is independent from $\sigma(\bigcup_{j=l+1}^n \sigma(X_j))$, so it's just this step missing. 
Thanks in advance!
 A: Note that $\mathcal{B}_1$ is generated by the set 
$$\mathcal{G}=\{A_1\times \cdots \times A_j | A_i \in \mathcal{A}_i \text{ for } i\in\{1,..,l\}\}$$
therefore we have that $Y_1^{-1} (\mathcal{B}_1)=\sigma (Y_1^{-1}(\mathcal{G}))$ (see why here). 
Now For any $G\in \mathcal{G}$, there exists $A_1\in\mathcal{A}_1,...,A_l\in\mathcal{A}_l$ such that $G=A_1\times \cdots \times A_l$. We see that 
$$Y^{-1}_1(G)=\bigcap_{i=1}^lX_i^{-1}(A_i)\in \sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right),$$
since $X_i^{-1}(A_i)\in \sigma(X_i)$ for each $i\in\{1,...,l\}$. We have now shown that
$$
Y^{-1}_1(\mathcal{G}) \subset \sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right) \implies Y_1^{-1} (\mathcal{B}_1)=\sigma (Y_1^{-1}(G)) \subset\sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right) 
$$
For the converse we note that the bundled map $Y_1=(X_1,...,X_l)$ is $Y_1^{-1} (\mathcal{B}_1)/\mathcal{A}_1\otimes\cdots\otimes \mathcal{A}_l $-measurable, and this happens if and only if each coordinate map $X_i$ is $Y_1^{-1} (\mathcal{B}_1)/\mathcal{A}_i$ measurable for $i\in\{1,...,l\}$ (see why here). Thus we conclude that $Y_1^{-1} (\mathcal{B}_1)$ also makes all $X_1,...,X_l$ measurable. But since $\sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right) $ is the smallest sigma-algebra making $X_1,...,X_l$ measurable we have that
$$
\sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right)  \subset Y_1^{-1} (\mathcal{B}_1).
$$
We conclude that
$$
\sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right)  = Y_1^{-1} (\mathcal{B}_1).
$$
