Exercise 1.4.12 (Restriction of a $\sigma$-algebra)

We copy-paste the proof from Exercise 1.4.2.

(i)

We have $\varnothing \in B$, and therefore $\varnothing =\varnothing \cap Y\in B{\text{↾}}_Y$.

(ii)

Let $E^{\prime }\in B{\text{↾}}_Y$, i.e., $E^{\prime }=E\cap Y$ for some $E\in B$. Then $X∕E\in B$ as well, and we have $Y∕E^{\prime }=Y∕(E\cap Y)=Y∕E=(X∕E)\cap Y$.

(iii)

Let ${(E_n^{\prime })}_{n\in \mathbb{N}}$ be a countably infinite sequence of sets in $B{\text{↾}}_Y$. Then $\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:E_n^{\prime }=E_n\cap Y$ for a countable infinite collection ${\left(E\right)}_{n\in \mathbb{N}}$ in $B$. We have ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n\in B$ and therefore $$\left(\bigcup _{n=1}^{\infty }{E}_n\right)\cap Y=\bigcup _{n=1}^{\infty }\left(E_n\cap Y\right)=\bigcup _{n=1}^{\infty }{E}_n^{\prime }$$

is contained in $B{\text{↾}}_Y$.