Sigma Algebra is closed under countable increasing unions

**tttcomrader** · September 19th 2009, 04:03 PM

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

Proof so far:

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

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

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

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

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

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

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

How should I proceed from here? Thanks.

**siclar** · September 19th 2009, 07:27 PM

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

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