I was watching some lecture when I came across the following statement:
Consider the scenario where $\mathcal{F} = \sigma(X,Y)$ and
$X: (\Omega, \mathcal{F}) \rightarrow (\mathbb{R},\mathcal{B})$,
$Y: (\Omega, \mathcal{F}) \rightarrow (\mathbb{R},\mathcal{B})$.
The claim is now that, for each $A \in \mathcal{F} \: \exists B \in \mathcal{B} $ s.t. either $X^{-1}(B) = A$ or $Y^{-1}(B) = A$.
Isn't this statement wrong? It is not necessarily true that $\sigma(X) \cup \sigma(Y) $ is a $\sigma$-field.
If one considers $\Omega = [0,10]$,$X= \chi_{[0,5]} $ and $Y= \chi_{[1,6]} $, then $\sigma(X) = \{ \Omega , \emptyset, [0,5], (5,10] \}$, $\sigma(Y) = \{ \Omega , \emptyset, [1,6], [0,1) \cup (6,10] \}$. Therefor $[1,5] \in \sigma(X,Y)$ (as intersection of $[0,5]$ and $[1,6]$), for which both $X$ and $Y$ cannot produce this as pre-image. Am I missing something here? Is my counter-example wrong?
If my counter example is correct, wouldn't a proper statement be something like:
$A \in \mathcal{F} \: \exists B \in \mathcal{B} $ s.t. either $X^{-1}(B) = A$ or $Y^{-1}(B) = A$ iff $\sigma(X) \cup \sigma(Y)$ is $\sigma$-field ?