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

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

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.

2. 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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### the union of two sigma algebras is a sigma algebra

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