Proving that a class of good sets forms a $\sigma$-field. Let $\mathcal{C}$ be a class of subsets of $\Omega $, $A\subset \Omega $, $\mathcal{F}=\sigma(\mathcal{C})$, and $\mathcal{G}=\{ B \in \mathcal{F}:B\cap A\in \sigma_A(\mathcal{C}\cap A)\}$.
How can I prove that $\mathcal{G}$ is a sigma field? 
For example: $B\in \mathcal{G}\iff (B\cap A \in \sigma_A(\mathcal{C}\cap A)) \land (B\in \mathcal{F}) \Rightarrow (B^c\cup A^c \in \sigma_A(\mathcal{C}\cap A)) \land (B^c\in \mathcal{F}) $, but what I would want is $(B^c\cap A \in \sigma_A(\mathcal{C}\cap A)) \land (B^c\in \mathcal{F}) $
I'm trying to use the good sets principle to prove that $\sigma_A(\mathcal{C}\cap A)=\mathcal{F}\cap A$.
Any help would be appreciated.
 A: HINT: Let $\mathcal{B} = \{ B \in \mathcal{F} : B \cap A \in \sigma_A(\mathcal{C} \cap A)\}$, and show that $\mathcal B = \mathcal F$.

Note first that $\mathcal{C} \subset \mathcal{B}$.  Since $\mathcal{F}$ and $\sigma_A(\mathcal{C} \cap A)$ are both $\sigma$-fields, it's straightforward to show that $\mathcal B$ is a $\sigma$-field.   But since $\mathcal{B}$ is a $\sigma$-field with $\mathcal C \subset\mathcal{B}$, we must have $\mathcal{F} = \sigma(C) \subset \mathcal{B}$; we also have that $\mathcal B \subset \mathcal F$ by definition, thereby showing that $\mathcal B = \mathcal F$.  Finally, by definition of $\mathcal B$, we have that $\mathcal B \cap A \subset \sigma_A(\mathcal C \cap A)$.  Since $\mathcal B = \mathcal F$, this is shows $\mathcal F \cap A \subset \sigma_A(\mathcal C \cap A)$.  
The reverse inclusion is immediate: since $\mathcal C \subset \mathcal F$, we have $\sigma_A(\mathcal C \cap A) \subset \sigma_A(\mathcal F \cap A) = \mathcal F \cap A$.

EDIT: How to show that $\mathcal{B}$ is a $\sigma$-field:


*

*Since $\Omega \in \mathcal{F}$ and $\Omega \cap A = A \in \sigma_A(\mathcal{C} \cap A)$, we also have $\Omega \in \mathcal{B}$.

*Let $E \in \mathcal{B}$.  Then $E \in \mathcal{F} \implies \Omega \setminus E \in \mathcal{F}$.  Similarly, $E \cap A \in \sigma_A(\mathcal{C} \cap A) \implies A \setminus (E \cap A) \in \sigma_A(\mathcal{C} \cap A)$.  But $A \setminus (E \cap A) = \{x \in A : x \notin E\} = (\Omega \setminus E) \cap A$.  Thus $\Omega \setminus E \in \mathcal{B}$.

*Let $\{E_i\}_{i = 1}^\infty \subset \mathcal{B}$.  Then $$\{E_i\}_{i=1}^\infty \subset \mathcal{F} \implies \bigcup\limits_{i} E_i \in \mathcal{F}.$$
Similarly, we have $$\{E_i \cap A\}_{i = 1}^\infty \subset \sigma_A(\mathcal{C} \cap A) \implies \bigcup\limits_{i} (E_i \cap A) \in \sigma_A(\mathcal{C} \cap A).$$ Since $\bigcup_i (E_i \cap A) = (\bigcup_i E_i) \cap A$, this implies that $\bigcup_i E_i \in \mathcal{B}$, thereby completing the proof.
