Measure theory exercise From measure theory volume 1 by Fremlin, exercise 111Xf:
Let $X$ be a set, $\mathcal{A}$ a family of subsets of $X$, and $\Sigma$ the $\sigma$-algebra of subsets of $X$ generated by $\mathcal{A}$. Suppose that
$Y\subset X$. Show that  ${\{E \cap Y : E \in \Sigma}\}$  is the σ-algebra of subsets of $Y$ generated by ${\{A \cap Y : A \in \mathcal{A}}\}$.
I'm able to show that it's a $\sigma$-algebra, and clearly it contains the $\sigma$-algebra generated by ${\{A \cap Y : A \in \mathcal{A}}\}$, but I don't know how to show ${\{E \cap Y : E \in \Sigma}\}$ $\subseteq\sigma (\{{A \cap Y : A \in \mathcal{A}}\})$.
 A: First note that what you want to show is not true in general. You forgot to take the $\sigma$ algebra generated by $\{A\cap Y\mid \dots\}$ instead of just $\{A\cap Y\mid \dots\}$.
To prove the corrected statement, use the good set principle, i.e. set
$$
G = \{E\in \Sigma \mid E\cap Y\in \sigma(\{A\cap Y\mid A\in \mathcal{A}\})\}.
$$
Show that $G$ is a $\sigma$ algebra and that $\mathcal{A}\subset G$.
Then think about why that implies your claim.
A: Let $\mathcal A' = \left\{E\cap Y : E\in\mathcal A\right\}$ and $\Sigma' = \left\{E\cap Y : E\in\Sigma\right\}$. Let
$$\mathcal F = \left\{E\in\Sigma : E\cap Y\in\sigma(\mathcal A')\right\}$$
Then $\mathcal A\subseteq\mathcal F\subseteq\Sigma=\sigma(\mathcal A)$ and $\mathcal F$ is an $\sigma$-algebra (show this). Thus $\mathcal F =\Sigma$.
Now, for any $E\in\Sigma'$, we have $E = F\cap Y$ for some $F\in\Sigma$. But by previous result, $F\cap Y\in\sigma(\mathcal A')$ for all $F\in\Sigma$, so $E\in\sigma(\mathcal A')$. This proves $\Sigma'\subseteq\sigma(\mathcal A')$.
