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I understand that an algebra $F \subset 2^\Omega$ is called a $\sigma$-algebra if it additionaly satisfies:

$(A_i)_{i \in \mathbb{N}}$ with $A_i \in F$ pairwise disjoint, then also $\cup_{i \in \mathbb{N}} A_i \in F$

, i.e., each finite union of pairwise disjoint subsets must again be in $F$.

However, I have difficulties imagining an example for an algebra which is not $\sigma$-algebra.

Can there be an algebra which is no $\sigma$-algebra e.g. on $\Omega = \{1,2,3,4,5,6\}$ ?

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    $\begingroup$ your definition is not correct. the sets do not need to be pairwise disjoint. $\endgroup$ – Loreno Heer Nov 7 '14 at 16:21
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    $\begingroup$ Not a finite set like you gave. The only algebras that aren't sigma algebras exist on infinite sets only. $\endgroup$ – Alan Nov 7 '14 at 16:21
  • $\begingroup$ It looks like you wrote "finite" where you meant "infinite." Of course, a regular algebra has finite unions, and the $\sigma$ part means that this is strengthened to be countable unions. The ordinary definition I'm familiar with does not say anything about being disjoint. $\endgroup$ – rschwieb Nov 7 '14 at 16:22
  • $\begingroup$ Any algebra on a finite set would also be a $\sigma$ algebra: since there are only finitely many possible sets, any countable union would be, in fact, rerwriteable as a finite union. $\endgroup$ – rschwieb Nov 7 '14 at 16:25
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Consider:

$$\mathcal{C}=\{A \subset \mathbb{Z} : A \text{ is finite or } \mathbb{Z}\backslash A \text{ is finite} \}$$

Then it's algebra: it's closed under finite sum and complement $\emptyset, \mathbb{Z} \in \mathcal{C}$, but not $\sigma$-algebra, for example $\{2n\}\in\mathcal{C}$ for all $n \in \mathbb{Z}$, but:

$$\cup_{n \in \mathbb{Z}}\{2n\} \not\in \mathcal{C}$$

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