Subsets and Sigma Algebras: Proving the Equality of Generated Sigma Algebras 

Note that a $\sigma$-algebra on a set $\Omega$ if it contains $\Omega$, and it is closed under complement and countable unions. By definition 
\[F=\left\{ E\cap Y:\,E\in \sigma (\mathcal{E}_{1})\right\}.\]
First we show that $F$ is a $\sigma$-algebra. 
(1) Since $X\in B$, we have $Y=Y\cap X=Y \in F$. 
(2) Let $S=E \cap Y \in F$. Then $E^c \in \sigma (\mathcal{E}_{1})$, and hence 
\[S^c=E^c \cap Y \in F,\]
hence $S$ is closed under compliment. Note that $F$ is a collection of subsets of $Y$ and the compliments should be taken with respect to the set $Y$. 

(3) Suppose $S_i=E_i \cap Y \in F$, $E_i =\sigma (\mathcal{E}_{1})$. Then
\[\cup S_i= \cup (E_i \cap Y)= (\cup E_i )\cap Y \in F,\]
since $(\cup E_i ) \in \sigma (\mathcal{E}_{1})$. This means that $F$ is closed under countable union. 

Therefore $F$ is a $\sigma$-algebra. 

Now we are ready to prove what you want. 


I) We show that $\sigma(\mathcal{E}_{2}) \subset F$. Since $\mathcal{E}_{2}  \subset F$, we have


\[\sigma(\mathcal{E}_{2}) \subset F,\]
as by definition $\sigma(\mathcal{E}_{2})$ is the smallest $\sigma$-algebra containing $\mathcal{E}_{2}$. 

II) Next we show $F \subset \sigma(\mathcal{E}_{2})$.
 
We show that the set \[\mathcal{G}=\{A:A\cap Y \in\sigma(\mathcal{E_1}\cap Y)=\sigma (\mathcal{E_2})\}\] is a $\sigma$-algebra. The first requirement holds since $X \in\mathcal{G}$. If $A\in\mathcal{G}$ then $A\cap Y\in\sigma(\mathcal{E}_1\cap Y)$, implying that
 
\[A^c\cap Y=Y\setminus (A\cap Y)\in\sigma(\mathcal{E}_1\cap Y)\] as well (the set difference of two sets in a $\sigma$ algebra is in the $\sigma$ algebra as well). It follows that $A^c\in\mathcal{G}$, showing that the it is closed under compliment. Next note that if $A_n\in\mathcal{G}$ for $n\in\mathbb{N}$ then
 
\[(\cup_n A_n)\cap Y = \cup_n (A_n\cap Y)\in\sigma(\mathcal{E}_1\cap Y)\]
 
(since $A_n\cap Y\in\sigma(\mathcal{E}_1\cap Y)$) and thus $\cup_n A_n\in\mathcal{G}$. Hence $\mathcal{G}$ is closed under countable union. 
 
Therefore $\mathcal{G}$ is a $\sigma$-algebra and $\mathcal{E}_1\subset \mathcal{G}$ (if $A\in\mathcal{E}_1$ then $A\cap Y\in \sigma(\mathcal{E}_1\cap G)$ implying that $A\in\mathcal{G}$) we have
 
\[\sigma(\mathcal{E}_1)\subset \sigma(\mathcal{G}) = \mathcal{G}.\]
 
(The last equality holds since $\mathcal{G}$ is a $\sigma$-algebra.)
 
Now if $A\in\sigma(\mathcal{E}_1)$ then $A\in\mathcal{G}$ and hence $A\cap Y\in\sigma(\mathcal{E}_1\cap Y)$. Thus
 
$$F=\sigma(\mathcal{E}_1) \cap Y \subset \sigma(\mathcal{E}_1\cap G)=\sigma(\mathcal{E}_2),$$
that is 
\[F \subset \sigma(\mathcal{E}_2).\]