$\sigma$-algebra generated by trace is trace of generated $\sigma$-algebra I started studying measure theory by myself using a book by D. Werner. One thing (apparently easy to prove) that I can't work out is the following (${\mathcal P}(S)$ denotes the power set of S):
${\bf Definition}$: Let $S$ be  a set, $E \subset S$ and ${\mathcal E} \subset {\mathcal P}(S)$. Define
\begin{equation}
{\mathcal E}\cap E:=\{A \in {\mathcal P}(E): \exists F \in {\mathcal E} \text{ with } A = F\cap E  \}
\end{equation}
to be the trace of ${\mathcal E}$ on $E$. 
I could show that ${\mathcal E} \cap E$ is again a $\sigma$ algebra on $E$, if ${\mathcal E}$ is a $\sigma$-algebra on S, but for the following I am lost. The claim is: 
\begin{equation}
\sigma({\mathcal E \cap E}) = \sigma({\mathcal E}) \cap E
\end{equation}
Maybe this is really trivial, but right now I have no clue how to start.
 A: From $\mathcal E\subseteq\sigma(\mathcal E)$ it follows immediately that: $$\mathcal E\cap E\subseteq\sigma(\mathcal E)\cap E\tag1$$
You proved that the RHS is a $\sigma$-algebra on $E$ so $(1)$ allows the conclusion:$$\sigma(\mathcal E\cap E)\subseteq\sigma(\mathcal E)\cap E\tag2$$
Now have a look at the collection: $$\mathcal A:=\{F\in\wp(S)\mid F\cap E\in\sigma(\mathcal E\cap E)\}$$
It can be proved that $\mathcal A$ is a $\sigma$-algebra on $S$ (give this a try yourself, and let me know if you get stuck), and this obviously with $\mathcal E\subseteq\mathcal A$, so that also:$$\sigma(\mathcal E)\subseteq\mathcal A$$or equivalently:$$\sigma(\mathcal E)\cap E\subseteq\sigma(\mathcal E\cap E)\tag3$$

Actually the statement $\sigma(\mathcal E\cap E)=\sigma(\mathcal E)\cap E$ can be written as:$$\sigma(i^{-1}(\mathcal E))=i^{-1}(\sigma(\mathcal E))$$where $i:E\to S$ denotes the inclusion.
This can be recognized as a special case of:$$\sigma(f^{-1}(\mathcal E))=f^{-1}(\sigma(\mathcal E))\tag4$$ where $f:T\to S$ is a function.
For a proof of the more general $(4)$ have a look at this answer.
A: Let $A\in \mathcal{E}\cap E$, then $A= F\cap E$ for some $F\in \mathcal E$. Since $\mathcal{E}\subset \sigma(\mathcal E)$, it follows that $A \in \sigma(\mathcal{E})\cap E$. This proves
$$
\mathcal{E}\cap E \subset \sigma(\mathcal{E})\cap E. 
$$
Hence for the generated $\sigma$-Algebras you have$$
\sigma(\mathcal{E}\cap E) \subset \sigma(\sigma(\mathcal{E})\cap E). 
$$
But  the trace of $\sigma$-Algebras is a $\sigma-$Algebra, i.e. $\sigma(\sigma(\mathcal{E})\cap E) = \sigma(\mathcal{E})\cap E$ and you obtain
$$
\sigma(\mathcal{E}\cap E) \subset \sigma(\mathcal{E})\cap E. 
$$
Can you do the other inclusion yourself?
