# An algebra $A$ is a $\sigma$-algebra iff it’s closed under countable unions

An algebra $A$ is a $\sigma$-algebra if and only if $A$ is closed under countable increasing unions.

Proof: Suppose $\{E_j\}_{1}^{\infty}\subset A$ and $E_1\subset E_2\subset \ldots$ Set $$F_k = E_k \setminus \big[\bigcup_{1}^{k-1}E_j\big] = E_k \cap \big[\bigcup_{1}^{k-1}E_j\big]^{c}$$ Then the $F_k$\s belong to $A$ and are disjoint, and $$\bigcup_{1}^{\infty}E_j = \bigcup_{1}^{\infty}F_k$$ Therefore, $\bigcup_{1}^{\infty}E_j\in A$

I am not sure if this is correct, any suggestions is greatly appreciated

Clearly a $\sigma$-algebra, $\mathscr{A}$ on a set $X$ is by definition closed under "all kinds of" countable unions of sets that belongs to $\mathscr{A}$.
Now suppose $\mathscr{A}$ is an algebra on $X$, which is closed under countable increasing unions.
Pick an arbitrary sequence of sets $\left\{A_{n}\right\}_{n=1}^{\infty}\subset \mathscr{A}$ and consider the sequence $\left\{B_{k}\right\}_{k=1}^{\infty}$, defined by $$B_{k}= \bigcup_{n=1}^{k}A_{n}$$
Now indeed $B_{k}\subseteq B_{k+1}$ and $B_{k}\in \mathscr{A}$, for all $k\geq1$, since $\mathscr{A}$ is an algebra. Moreover $$\bigcup_{n=1}^{\infty}A_{n}=\bigcup_{k=1}^{\infty}B_{k}\in \mathscr{A}$$