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

Question

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?

Answer 1

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:

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

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

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

Answer 2

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 $\mathcal{C}$