Is $\sigma((X,Y))=\sigma(X,Y)$? Let


*

*$(\Omega,\mathcal A)$ and $(\Omega_i,\mathcal A_i)$ be measurable spaces

*$X_i:\Omega\to\Omega_i$ be $(\mathcal A,\mathcal A_i)$-measurable



Are we able to show that $\sigma((X_1,X_2))=\sigma(X_1,X_2)$?

By definition, $$\sigma((X_1,X_2))=(X_1,X_2)^{-1}(\mathcal A_1\otimes\mathcal A_2)\tag1$$ and we know that this is equal to $$\sigma((X_1,X_2)^{-1}(\mathcal E_1\times\mathcal E_2),\tag2$$ if $\mathcal E_i\subseteq\mathcal A_i$ with $\sigma(\mathcal E_i)=\mathcal A_i$. On the other hand, $$\sigma(X_1,X_2)=\sigma(\sigma(X_1)\cup\sigma(X_2))=\sigma(X_1^{-1}(\mathcal A_1)\cup X_2^{-1}(\mathcal A_2))\tag3$$ and $$X_i^{-1}(\mathcal A_i)=\sigma(X_i^{-1}(\mathcal E_i)).\tag4$$ But I don't see that the claim follows from these equalities.
 A: $X_1^{-1} (A)=(X_1,X_2)^{-1}(A\times \Omega_2)$, $X_2^{-1} (A)=(X_1,X_2)^{-1}( \Omega_1 \times A)$ and $(X_1,X_2)^{-1} (A\times B)= X_1^{-1}(A) \cap X_2^{-1}(B)$. Using these identities prove that each side is contained in teh other.
A: Observe that $\Omega_1\times\Omega_2$ is equipped with the product $\sigma$-algebra which means exactly that:$$(X_1,X_2)\text{ is measurable iff }X_i=\pi_i\circ(X_1,X_2)\text{ are measurable for }i=1,2\tag1$$ where $\pi_i:\Omega_1\times\Omega_2\to\Omega_i$ denotes the projection for $i=1,2$
$\sigma((X_1,X_2))$ is by definition the smallest $\sigma$-algebra on $\Omega_1\times\Omega_2$ such that $(X_1,X_2):\Omega\to\Omega_1\times\Omega_2$ is measurable.
$\sigma(X_1,X_2)$  is by definition the smallest $\sigma$-algebra on $\Omega_1\times\Omega_2$ such that $X_i:\Omega\to\Omega_i$ is measurable for $i=1,2$.
So $(1)$ tells us directly that they are the same.

Actually we have $\mathcal A_1\otimes\mathcal A_2=\sigma\left(\pi_1^{-1}(\mathcal A_1)\cup\pi_2^{-1}(\mathcal A_2)\right)$ so that: $$\sigma((X_1,X_2))=(X_1,X_2)^{-1}(\mathcal A_1\otimes\mathcal A_2)=(X_1,X_2)^{-1}\left(\sigma\left(\pi_1^{-1}(\mathcal A_1)\cup\pi_2^{-1}(\mathcal A_2)\right)\right)=$$$$\sigma\left((X_1,X_2)^{-1}\left(\pi_1^{-1}(\mathcal A_1)\cup\pi_2^{-1}(\mathcal A_2)\right)\right)=\sigma(X_1^{-1}(\mathcal A_1)\cup X_2^{-1}(\mathcal A_2))=\sigma(X_1,X_2)$$where the third equality is based on the rule:$$f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)=\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)$$For a proof of that rule see here.
