I'm reading "Limit Thoerem for Stochastic Processes" and finding it hard to calculate the Stochastic interval.For example :
In proposition 2.10,$T$ is a stopping time:
If $A\in\mathcal F_0$,I need to prove that $[0,T_A[=[0,T[\cup(A^c\times\mathbb R)$
If $A=B\cap\{t<T\},B\in\mathcal F_t$,I need to prove that $[0,T_{A^c}[=[0,T[\cup(A\times(t,\infty))$
Remark:$T_A=T1_{\omega\in A}+\infty 1_{\omega\notin A}$
For the 1st one:If $\omega\in A^c$,all $t$ satisfy the condition, so $(A^c\times \mathbb R)$,but when $\omega\notin A$,then $t<T$,I think it is not $[0,T[$,It should have some restrictions on $A$.
Same questions for the 2rd one.I think both the 2 parts in $\cup$ is incorrect.
No, the proof in the book is correct.
By definition,
$$[0,T_A[ = \{(\omega,t); 0 \leq t < T_A(\omega)\}.$$
Since $T_A(\omega) = \infty$ for $\omega \notin A$ and $T_A(\omega)=T(\omega)$ for $\omega \in A$, we get
$$[0,T_A[ = \{(\omega,t); \omega \in A, 0 \leq t< T(\omega)\} \cup \{(\omega,t); \omega \notin A, t \geq 0\}.$$
Since
$$\{(\omega,t); \omega \notin A, t \geq 0\} \supseteq \{(\omega,t); \omega \notin A, 0 \leq t < T(\omega)\}$$
this implies
$$[0,T_A[ = \{(\omega,t); \omega \in \Omega, 0 \leq t< T(\omega)\} \cup \{(\omega,t); \omega \notin A, t \geq 0\}$$
This proves $$[0,T_A[ = [0,T[ \cup (A^c \times [0,\infty)).$$ The proof of the second one is very similar, I leave it to you.