Is a countable union/intersection of sigma-algebra also a sigma-algebra?

**Last_Singularity** · February 4th 2011, 07:31 AM

Suppose that I have a set $\Omega$ over which I define a collection of $\sigma$ -algebra: $\{\beta_i\}$ ; that is, each $\beta_1,\beta_2,...$ is a $\sigma$ -algebra of $\Omega$ . Is the countable union of those also a $\sigma$ -algebra? That is, is $\bigcup_{i \in N} \beta_i = \beta^*$ a $\sigma$ -algebra over $\Omega$ ? How about countable intersection?

1 - It seems pretty clear that $\Omega$ and $\emptyset$ are contained in $\beta^*$ .

2 - It is also clear that if $A \in \beta^*$ , then $A^C \in \beta^*$

So the first two conditions of a $\sigma$ -algebra have been fulfilled: $\beta^*$ contains the full sample space / empty set and it is closed under complements.

3 - But if $A_1,A_2,... \in \beta^*$ , is it true that $A^* = \bigcup_{i \in N} A_i \in \beta^*$ ? I am unable to prove this fact. Yet I cannot find a counterexample.

A counterexample or hint of proof would be very appreciated - thanks.

**roninpro** · February 4th 2011, 08:39 AM

For a counterexample, take $\Omega=\{1,2,3,4,5,6\}$ , $\beta_1=\{\emptyset, \{1\}, \{2,3,4,5,6\}, \Omega\}$ , and $\beta_2=\{\emptyset, \{2\}, \{1,3,4,5,6\},\Omega\}$ .

The set $\beta_1\bigcup \beta_2$ is not a $\sigma$ -algebra because $\{1\}\bigcup \{2\}=\{1,2\}$ is not in $\beta_1\bigcup \beta_2$ .

I think that you will have better luck with intersections.

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