[viista]

Let $\mathcal{E}$ be an arbitary collection of subsets of a set $X$, let $A$ be a nonempty subset of $X$, and let
$$
\mathcal{E}\cap A:=\{E\cap A:E\in \mathcal{E}\}.
$$
Show that the $\sigma$-algebra $\sigma_A(\mathcal{E}\cap A)$ on $A$ generated by $\mathcal{E}\cap A$ is equal to $\sigma_X(\mathcal{E})\cap A$, where $\sigma_X(\mathcal{E})$ is the $\sigma$-algebra on $X$ generated by $\mathcal{E}$.