$\sigma(\mathcal{C}\cap A)=\sigma(\mathcal{C})\cap A$ 
Suppose $\mathcal{C}$ is a class of subsets of $\Omega$. Let $A$ be a non-empty subset of $\Omega$. Show that
$$\sigma(\mathcal{C}\cap A)=\sigma(\mathcal{C})\cap A,\text{ a }\sigma\text{-field of subsets of }A$$
where, $\mathcal{C}\cap A:=\{X\cap A:X\in\mathcal{C}\}$.

Proving

*

*$\sigma(\mathcal{C})\cap A$ is a $\sigma$-field of subsets of $A$

*$\sigma(\mathcal{C})\cap A$ contains $\mathcal{C}\cap A$
is easy. I also need to prove


*For any field containing $\mathcal{C}\cap A, F(\mathcal{C}\cap A)$ (say), $F(\mathcal{C}\cap A)\supseteq\sigma(\mathcal{C})\cap A$.

However for any $B\in\sigma(\mathcal{C})\cap A$, I am not getting anywhere. Hints please?
 A: Important: Note that in order to prove $\sigma(\mathcal{C}\cap A)=\sigma(\mathcal{C})\cap A$, we must consider $\sigma(\mathcal{C}\cap A)$ as the $\sigma$-field generated in $A$, that means, a $\sigma$-field of subsets of $A$.
You already know:

*

*$\sigma(\mathcal{C})\cap A$ is a $\sigma$-field of subsets of $A$

*$\sigma(\mathcal{C})\cap A$ contains $\mathcal{C}\cap A$
From 1 and 2, it follows immediately that $\sigma(\mathcal{C}\cap A) \subseteq\sigma(\mathcal{C})\cap A$, because $\sigma(\mathcal{C}\cap A)$ is the intersection of all $\sigma$-fields containing $\mathcal{C}\cap A$.
Now, we must prove that $\sigma(\mathcal{C})\cap A \subseteq \sigma(\mathcal{C}\cap A)$. Let us define
$$\mathcal{K}=\{S + T : S \in \sigma(\mathcal{C}\cap A) \text{ and }  T \in \sigma(\mathcal{C}\cap A^c)\} $$
where $+$ indicates disjoint union. It is to see that $\mathcal{K}$ is a $\sigma$-field of subsets of $\Omega$  and $\mathcal{C} \subseteq \mathcal{K}$.
It follows immediately that $\sigma(\mathcal{C}) \subseteq \mathcal{K}$. So we have
$$ \sigma(\mathcal{C}) \cap A \subseteq \mathcal{K} \cap A = \sigma(\mathcal{C}\cap A)$$
So we have proved that $\sigma(\mathcal{C})\cap A \subseteq \sigma(\mathcal{C}\cap A)$. So, we have $\sigma(\mathcal{C}\cap A)=\sigma(\mathcal{C})\cap A$.
Remark: Another way to prove that $\sigma(\mathcal{C})\cap A \subseteq \sigma(\mathcal{C}\cap A)$ is to consider
$$ \mathcal{H} = \{ S \subseteq \Omega : S \cap A \in \sigma(\mathcal{C}\cap A)\}$$
It is easy to prove that $\mathcal{H}$ is a $\sigma$-field of subsets of $\Omega$  and $\mathcal{C} \subseteq \mathcal{H}$. It follows immediately that $\sigma(\mathcal{C}) \subseteq \mathcal{H}$. So we have
$$ \sigma(\mathcal{C}) \cap A \subseteq \mathcal{H} \cap A \subseteq \sigma(\mathcal{C}\cap A)$$
So we have proved that $\sigma(\mathcal{C})\cap A \subseteq \sigma(\mathcal{C}\cap A)$.
A: In general if $\mathcal B$ is a $\sigma$-algebra on $A$ then it can be proved that the collection $\{B\in\mathcal P(\Omega)\mid B\cap A\in\mathcal B\}$ is a $\sigma$-algebra on $\Omega$.
I leave it to you to prove this.
Applying that to $\mathcal B=\sigma(\mathcal C\cap A)$ we arrive at a $\sigma$-algebra on $\Omega$ that moreover contains $\mathcal C$. Then we are allowed to conclude that it also contains $\sigma(\mathcal C)$ which means exactly that $\sigma(\mathcal C)\cap A\subseteq\sigma(\mathcal C\cap A)$.
