I am trying to understand this remark but have some difficulties.
More precisely, denote by $\mathcal{M}=\{B\in \mathcal{B}(\mathbb{R}^n): B\cap E\in \mathcal{E}\}$, where $\mathcal{E}=\sigma(S)$ and $S=\{E\cap U: U\ \text{is open in}\ \mathbb{R}^n\}$.
I have shown that all properties of $\sigma$-algebra for $\mathcal{M}$ are true, except complement property.
Suppose $A\in \mathcal{M}$ then $A\in \mathcal{B}(\mathbb{R}^n)$ with $A\cap E\in \mathcal{E}$. Then we have to show that $\mathbb{R}^n-A\in \mathcal{M}$. Easy to see that $\mathbb{R}^n-A\in \mathcal{B}(\mathbb{R}^n)$. But I don't know how to show that $(\mathbb{R}^n-A)\cap E\in \mathcal{E}$.
I know that $\mathcal{E}$ is $\sigma$-algebra. But I don't know on what set it is $\sigma$-algebra.
Would be very grateful if anyone can show detailed answer to my question.
$\mathcal{E}$ is a $\sigma$-algebra on $E$ meaning, for the record, that it is a family of subsets of $E$ that is closed under taking complements (with respect to $E$) and countable unions and it contains the empty set.
As for your other question, $(\mathbb{R}^n - A) \cap E = (\mathbb{R}^n \cap E) - (A \cap E) = E - (A \cap E)$. Do you see why this has to be in $\mathcal{E}$?