An algebra is a $\sigma$-algebra if for all $E_1 \subset E_2 \subset\cdots$, $E_k \in A$, one has $\cup E_k \in A$

Question

Let $X$ be a non-empty set and $A$ an algebra on $X$. I want to show that $A$ is a $\sigma$-algebra on $X$ if it has the property : for $E_k \in A, k\in \mathbb{N}$ and $E_1\subset E_2\subset\cdots $ then
$$\bigcup\limits_{k=1}^{\infty} E_{k}\in A.$$ I know that i have to take random sets $E_n \in A$ and create the relationship $E_1\subset E_2\subset\cdots$ but i have no idea on how to do that. Can someone help me?

Answer 1

Let $(A_{n})_{n\in\mathbb{N}}\subseteq\mathcal{A}$ a sequence of elements of a algebra, then

\begin{eqnarray*} M_{1} & = & A_{1}\\ M_{n} & = & \bigcup_{i=1}^{n}A_{i} \end{eqnarray*}

Then $(M_{n})_{n\in\mathbb{N}}$ is a monotone class (i.e. $M_{1}\subseteq M_{2}\subseteq\dots$). If $ \mathcal{A}$ is closed for increasing unions, then

\begin{equation*} \bigcup_{n\in\mathbb{N}}^{}A_{n} = \bigcup_{n\in\mathbb{N}}^{}M_{n}\in\mathcal{A} \end{equation*}

Since the sequence $(A_{n})_{n\in\mathbb{N}}$ was arbitrary, it is concluded that $\mathcal{A}$ is closed under countable unions, then $\mathcal{A}$ is a $\sigma$-álgebra.

The idea is that for any collection of sets, we can adjust them by the definition for $(M_{n})_{n\in\mathbb{N}}$, and use this property.