Thread: Sigma Algebra is closed under countable increasing unions

Prove that an algebra $\displaystyle \mathbb {A} $ is a $\displaystyle \sigma $-algebra iff $\displaystyle \mathbb {A} $ is closed under countable increasing unions.

Proof so far:

Suppose that $\displaystyle \mathbb {A} $ is a $\displaystyle \sigma $-algebra, then if $\displaystyle E_1,E_2,... \subset \mathbb {A} $, then I have $\displaystyle \bigcup _{n=1}^{ \infty } E_n \subset \mathbb {A} $

Now, if $\displaystyle E_1 \subset E_2 \subset E_3 ... \subset \mathbb {A} $, then $\displaystyle \bigcup _{n=1}^{ \infty } E_n \subset \mathbb {A} $ since $\displaystyle E_n \in \mathbb {A} \ \ \ \forall n$. This direction is simple.

On the other hand, if A is closed under countable increasing unions, that means that for:

$\displaystyle E_1 \subset E_2 \subset E_3 \subset ... \subset \mathbb {A} $, then I have $\displaystyle \bigcup _{n=1}^{ \infty } E_n \subset \mathbb {A} $ And I need to show that $\displaystyle \mathbb {A} $ is a $\displaystyle \sigma $-algebra.

Suppose that $\displaystyle E_1, E_2, ... \subset \mathbb {A} $, I need to show that:

1. $\displaystyle E^c \in \mathbb {A}$

2. $\displaystyle \bigcup _{n=1}^{ \infty } E_n \subset \mathbb {A} $

How should I proceed from here? Thanks.

Well being closed under complements is a property of an algebra which $\displaystyle \mathbb{A}$ is, thus the first part is a moot point.
As to countable unions, consider $\displaystyle F_n=\displaystyle\bigcup_{j=1}^n E_j$ which is a countable increasing sequence of sets.