Let $B_t$ be a $\mathcal F_t$ measurable Brownian motion.
During the lecture we proved that $f:(t,\omega)\to(t,B_t(\omega))$ is $\mathcal B [0,T]\otimes \mathcal F_t \to \mathcal B(\mathbb R^+)\otimes B(\mathbb R)$ measurable. for doing so we took a set $A\times B\in \mathcal B(\mathbb R^+)\times B(\mathbb R)$ then $$f^{-1}(A\times B) = [(A \cap[0,T]\times \Omega]\cap g^{-1}(B)\in \mathcal B [0,T]\otimes \mathcal F_t$$ where $g$ is the function: $(t,\omega)\to B_t(\omega)$
What is not clear to me is: $1)$ why do we take $A\times B\in \mathcal B(\mathbb R^+)\times B(\mathbb R)$? I know that $\mathcal B(\mathbb R^+)\times B(\mathbb R)$ generate $\mathcal B(\mathbb R^+)\otimes B(\mathbb R)$ but still is not clear to me.
$2)$ could someone explain why $f^{-1}(A\times B) = [(A \cap[0,T]\times \Omega]\cap g^{-1}(B)$?
To prove that a function $g$ from a measure space $(\Omega_1,\mathcal F_1,\mu)$ to $\mathbb R^{2}$ is measurable w.r.t. the Borel sigma algebra on $\mathbb R^{2}$ it is enough to show that $g^{-1}(A\times B)$ belongs to $\mathcal F_1$ for any two Borel sets $A$ and $B$. [And a similar result holds if $\mathbb R^{2}$ is replaced by $\mathbb R^{+} \times \mathbb R$].
To see this you have to note two things:
a) The Borel sigma algebra of $\mathbb R^{2}$ coincides with the product of the Borel sigma algebra of $\mathbb R$ with itself.
b) The collection of all Borel set $C$ in $\mathbb R^{2}$ such that $g^{-1}(C) \in \mathcal F_1$ is a sigma algebra. Since it contains the product set $A \times B$ it contains all Borel sets in $\mathbb R^{2}$ finishing the proof.
The last part is straighforward. Show that $(t,\omega)$ belongs to LHS iff it belongs to RHS.