Smallest $\sigma$-algebra with adjoined set

Question

I am in self-learning mode of measure theory, and trying to solve an exercise from the book "A Probability path", S. Resnick 1, specifically:

Suppose $\mathcal{B}$ is a $\sigma$-field of subsets from $\Omega$ and suppose that $A \notin \mathcal{B}$. Show that $\sigma(\mathcal{B} \cup \{A\})$, the smallest $\sigma$-field containing both $\mathcal{B}$ and $\{A\}$ consists of sets of the form: $(A\cup B) \cap (A^{c} \cup B^{'})$, where $B, B^{'} \in \mathcal{B}$.

Assuming only $B_{1}, B_{2} \in \mathcal{B}$, then I can form the $\sigma(\mathcal{B}) = \{\Omega, \emptyset, B_1, B_2, B^{c}_1, B^{c}_{2}, (B_1 \cup B_2), (B_1 \cap B_2), (B_1 \cap B_2)^{c}, (B_1 \cup B_2)^{c}, (B_1 \cup B^{c}_{2}), (B_2 \cup B^{c}_{1}), (B_1 \cap B^{c}_{2}), (B_2 \cap B^{c}_{1})\}$

If I adjoin $A \notin \mathcal{B}$ to $\sigma(\mathcal{B})$, I do not see how to arrive to sets of the form $(A\cup B_1) \cap (A^{c} \cup B_2)$...any trick I am missing or any mistake I am doing?

Answer 1

You have to verify that sets of the given form form a sigma algebra. Since $A=(A\cup \emptyset) \cap (A\cup \Omega)$ this class contains $A$. Taking $B'=B$ we see that every set in $\mathcal B$ is also in this class. Hence this new sigma algebra contains the sigma algebra generated by $\mathcal B$ and $A$. The other inclusion is obvious.