Exercise 8.2.5

Proof. First, let $T=\left\{\mid \subseteq \text{ and }\text{ is a }\sigma \text{-algebra of subsets of }S\right\}$ so that $=\bigcap <!--\; FUNCTION\; APPLICATION\; -->T$. Then we must show that $$ meets the definition of a $\sigma$-algebra:

Regarding (a), consider any $\in T$. Since $$ is then a $\sigma$-algebra it follows that both $\varnothing \in$ and $S\in$ by (a). Then, since $\in T$ was arbitrary, it follows that both $\text{∅}$ and $S$ are in $\bigcap <!--\; FUNCTION\; APPLICATION\; -->T=$.

For (b) suppose that $X\in =\bigcap <!--\; FUNCTION\; APPLICATION\; -->T$ so that $X\in$ for all $\in T$. So consider any such $\in T$ so that clearly $X\in$. Then, since $$ is then a $\sigma$-algebra, it follows that $S-X\in$ by (b). Since $\in T$ was arbitrary, we have that $S-X\in \bigcap <!--\; FUNCTION\; APPLICATION\; -->T=$.

Lastly, for part (c) of the definition, suppose that $X_n\in =\bigcap <!--\; FUNCTION\; APPLICATION\; -->T$ for all $n\in 𝑵$. Let $$ be any element of $T$ so that $X_n\in$ for all natural $n$. Since $$ is a $\sigma$-algebra, it then follows from (c) that both ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n$ and ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n$ are in $$. Since $\in T$ was arbitrary we have that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n$ and ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=0}^{\infty }{X}_n$ are in $\bigcap <!--\; FUNCTION\; APPLICATION\; -->T=$.

Hence we have shown all three parts of the definition so that $$ is indeed a $\sigma$-algebra. □