# Proof that $\sigma$-algebras are closed under countable disjoint unions

From Folland's Analysis, suppose $\{E_j\}_1^\infty \subset \mathcal{A}$. Set

$$F_k = E_k \setminus \bigcup_{j=1}^{k-1}E_j$$

Then the $F_k$'s are disjoint, and $\bigcup_1^\infty E_j = \bigcup_1^\infty F_k$.

My question is, what if you let $x \in \bigcap_1^\infty E_j$? Then wouldn't $x \in \bigcup_1^{k-1}E_j$ for all $k$, which would imply $x \not \in F_k \implies \bigcup_1^\infty E_j \neq \bigcup_1^\infty F_k$?

I'm sure I'm just missing something really simple here, but I can't for the life of me figure out what it is!

Notice that $F_1 = E_1$. So every element $x \in \bigcap_{j=1}^\infty E_j \subseteq E_1$ is contained in $F_1$.