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Does the following example show that the union of $\sigma$-algebras may fail to be a $\sigma$-algebra?

Let $X=\{a,b,c,d\}$ and let $\mathscr{A}=\{\emptyset,X,\{a,b\},\{c,d\}\}$ and $\mathscr{B}=\{\emptyset,X,\{a,c\},\{b,d\}\}$. Then $\mathscr{A}$ and $\mathscr{B}$ are $\sigma$-algebras while $\mathscr{A}\cup \mathscr{B}=\{\emptyset,X,\{a,b\},\{c,d\},\{a,c\},\{b,d\}\}$ is not a $\sigma$-algebra since it is not closed under intersection.

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  • $\begingroup$ The example you gave is perfectly fine. It is a very well known result that every $\sigma$-algebra is closed under (at most countable) intersection. So, since $\mathscr{A}\cup \mathscr{B}$ is not closed under (finite) intersection, $\mathscr{A}\cup \mathscr{B}$ cannot be a $\sigma$-algebra. $\endgroup$ – Ramiro Oct 12 '15 at 16:09
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The example you give is fine, but recall that a $\sigma$-algebra is not defined as being closed under intersections. This follows from its other properties, such as being closed under countable unions. The example you gave is not closed under unions because $\{a,b\}\cup\{a,c\}=\{a,b,c\}$ is not in your algebra.

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    $\begingroup$ you are right, but some books define $sigma-algebras$ using intersections ;-) $\endgroup$ – Milad Oct 12 '15 at 14:41
  • $\begingroup$ If so, then I see no issues here. $\endgroup$ – Alex S Oct 12 '15 at 14:42

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