If $\mathcal{E}$ are subsets of $X$ and $A \subset X$. Show that $\mathcal{A}\mathcal{(E \cap}$ A)=$\mathcal{A}\mathcal{(E)} \cap$ A If $\mathcal{E}$ is collections of subsets of a set $X$ and let $A \subset X$ be a subset. Show that the generated $\sigma$-algebra of $\mathcal{(E \cap}$ A) =the generated $\sigma$-algebra $\mathcal{E}$ $ \cap A$, i.e., 
$\mathcal{A}\mathcal{(E \cap}$ A)=$\mathcal{A}\mathcal{(E)}  \cap$ A
 A: First define
$$
M = \{B \in \sigma(\mathcal{E}) \mid B \in \sigma(\mathcal{E} \cap A)\}. 
$$
Show that this is a $\sigma$ algebra containing $\mathcal{E}$, proving one inclusion. 
This proof technique (showing that all sets having a certain property form a $\sigma$ algebra and contain a given set if generators, hence the generated $\sigma$ algebra) is sometimes known as the good set principle. You should remember it. 
For the other, note that $\{E \cap A \mid E \in \sigma (\mathcal{E} ) \}$ is contained in the $\sigma$ algebra (!) (over A) $\sigma(\mathcal{E}) \cap A$. 
A: Notation 


*

*$\sigma_X(\mathcal E)$ is the sigma algebra generated by $\mathcal E$ in $X$

*$A^c$ is the complement of the subset $A$ in $X$

*$\mathcal C\cap A = \{C\cap A| C\in\mathcal C\}$ for a collection of subsets $\mathcal C$

*$2^C$ is the collection of all subsets of $C$


To me the crux of the proof is the following 
Lemma If $A\subseteq X$, $\mathcal B$ is a sigma algebra on $A$ and $\mathcal C$ is a sigma algebra on $A^c$ then
$$
    \sigma_X(\mathcal B\cup\mathcal C) \subseteq \{B\cup C| B\in\mathcal B, C\in\mathcal C\}
$$
Proof: The collection on the right clearly contains $\mathcal B,\mathcal C$. Using DeMorgan laws and the distribution of the intersection over the union you can prove that it is a sigma algebra on X. Q.E.D.
Now the main result:
Theorem Given $A\subseteq X$ and $\mathcal E\subseteq 2^X$ we have
$$
    \sigma_A(\mathcal E\cap A) = \sigma_X(\mathcal E)\cap A
$$
Proof: It's easy to see that $\sigma_X(\mathcal E)\cap A$ is a sigma algebra on $A$ and contains $\mathcal E\cap A$ so $\sigma_A(\mathcal E\cap A)\subseteq \sigma_X(\mathcal E)\cap A$.
To prove the other inclusion, call $\mathcal B = \sigma_A(\mathcal E\cap A)$, $\mathcal C = \sigma_{A^c}(\mathcal E\cap A^c)$ and consider $\sigma_X(\mathcal B\cup\mathcal C)$.
It's a sigma algebra on $X$ and contains $\mathcal E$, so it contains $\sigma_X(\mathcal E)$. It also contains $A$.
Then a generic $D\cap A \in\sigma_X(\mathcal E)\cap A\subseteq\sigma_X(\mathcal B\cup\mathcal C)$ being a subset of $A$ is in $\mathcal B$ by the lemma (it can be represented as the union of a set from $\mathcal B$ and a set from $\mathcal C$, but its intersection with $A^c$ is empty so it is in $\mathcal B$), which proves $\sigma_X(\mathcal E)\cap A\subseteq \sigma_A(\mathcal E\cap A)$. Q.E.D.
