Sub sigma algebra generated by a measurable process with a random time. We have the following settings:
Given the probability space $(\Omega,\mathscr F,P)$
$T$ is a random time, if it is a $\mathscr F$-measurable random variable with values in $[0,\infty]$
A stochastic process is called measurable if for every $A\in\mathscr B(\Bbb R^d)$, the set $\{(t,\omega);X_t(\omega)\in A\}$ belongs to the product $\sigma$-algebra $\mathscr B([0,\infty))\otimes\mathscr F$. I will be appreciate for any help.
Now I want to show that for a measurable process $X_t$(with well-defined $X_{\infty}$) and a random time $T$, the collection of all sets of the form $\{X_T\in A\}$ and $\{X_T\in A\}\cup\{T=\infty\}$,forms a sub-sigma algebra of $\mathscr F$.
I understand that we have the following mapping$$\omega\rightarrow(\omega,T(\omega))\rightarrow X(\omega,T(\omega))$$ $$\mathscr F\rightarrow\mathscr B([0,\infty])\otimes\mathscr F\rightarrow\mathscr B([0,\infty])$$.
As $X_t$ is measurable, we know that $X_T^{-1}(A)$ is somehow contained in $\mathscr B([0,\infty])\otimes\mathscr F$. As I am not sure if the preimage of $X$ can be written in the form of $C\otimes D$ and then consider $T^{-1}(D)\cap C$ in the product sigma algebra, I don't know how to proceed further to the original space $\mathscr F$
 A: By assumption, the mapping
$$([0,\infty) \times \Omega; \mathcal{B}[0,\infty] \otimes \mathcal{F}) \ni (t,\omega) \mapsto X(t,\omega) \in (\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d)) \tag{1}$$
is measurable. On the other hand, the mapping
$$(\Omega,\mathcal{F}) \ni \omega \mapsto Y(\omega):= (T(\omega),\omega) \in ([0,\infty) \times \Omega; \mathcal{B}[0,\infty] \otimes \mathcal{F}) \tag{2}$$
is also measurable. Indeed: Take $B \in \mathcal{B}[0,\infty]$ and $F \in \mathcal{F}$, then
$$\{Y^{-1}(B \times F)\} = \{\omega; \omega \in F, T(\omega) \in B) = F \cap \{T \in B\} \in \mathcal{F}.$$
Since sets of the form $B \times F$ are a generator of the product-$\sigma$-algebra, this proves that $Y$ is measurable.
Combining $(1)$ and $(2)$, we find that the mapping
$$(\Omega,\mathcal{F}) \ni \omega \mapsto X \circ Y(\omega) = X(T(\omega),\omega) \in (\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$$
is measurable as the composition of the measurable mappings $(1)$ and $(2)$. Hence, $\{X_T \in A\} \in \mathcal{F}$ for any Borel set $A$.
