Why is $\mathcal{C}=\{\{ X_T \in A \} \text{ or } \{ X_T \in A \} \cup \{T=\infty \},A \in \mathcal{B}(R) \}$ closed under complementation Why is $\mathcal{C}=\{\{ X_T \in A \} \text{ or } \{ X_T  \in A\} \cup \{T=\infty \},A \in \mathcal{B}(R) \}$ closed under complementation where  X is a random measurable process and T a random time? 
This is (Problem 1.17) in Karatzas and Shreve. 
My problem: 
If I choose $D=\{ X_T \in A \}$  then clearly $D^c=\{ X_T \in A^c \} \in C$ but when D is of the form $D=\{ X_T \in A \} \cup \{T=\infty \}$ then $D^c=\{ X_T \in A^c\} \cap \{T< \infty \}$ which I cant show is in $\mathcal{C}$
Can you please help?
 A: OK so we want to examine the complement of $D = \{X_T\in A\}\cup \{T = \infty\}$, which is
\begin{align*}
D^C = \{X_T\in A\}^C\cap \{T = \infty\}^C \\
= \{X_T\not\in A\}\cap \{T < \infty\}\\
= \{X_T\in A^C\}\cap \{T <\infty\}\\
= \{X_T\in A^C\} = \{X_T\in A'\}
\end{align*}
where we let $A' = A^C$. The last part here should mean we are done. We know that $\mathcal{B}(\mathbb{R})$ is closed under complementation, so $A^C\in\mathcal{B}(\mathbb{R})$, meaning that $D^C = \{X_T\in A'\}$ is an element of $\mathcal{C}$
A: The statement of the problem is: 

Prove that 
  $\{\{X_T\in A\}, A\in \mathcal B(\mathbb R)\}\cup\{\{X_T\in A\}\cup \{T=\infty\}, A\in \mathcal B(\mathbb R)\}$ is a sub-$\sigma$-field.

In the book, $X_T$ is defined only on $\{T<\infty\}$, so that $\{X_T\in A\}$ should be understood as $\{w\in\{T<\infty\}, X_T(w)\in A \}$.


*

*$\Omega$ is in the collection: simply note that $\{X_T\in \mathbb R\}\cup \{T=\infty\} = \Omega$

*The collection is closed under complementation: 


*

*$\{X_T\in A\}^c = \{T=\infty\}\cup \{X_T\in A^c\}$

*$(\{X_T\in A\} \cup \{T=\infty\})^c =  \{X_T\in A\}^c\cap \{T<\infty\} = \{X_T\in A^c\}$


*The collection is closed under countable union: I leave it to you.
