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I'm trying to understand the concept of $\sigma$-algebras.

Let $S=\{1,2,3\}$. Would a sigma algebra on $S$ be $\Sigma=\{\emptyset,\{1,2,3\},\{1\},\{2,3\},\{2\},\{1,3\}\}$?

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2  
Probably a typo: $\{3,4\} \not \subset S$. –  t.b. May 27 '12 at 22:34
    
Thanks @t.b. Sorry, I was trying out an example myself, not a typo. Why isnt $\{3,4\} \in S$? –  Steven May 27 '12 at 22:43
1  
$S = \{1,2,3\}$, so $4 \notin S$. –  Henry T. Horton May 27 '12 at 22:45
    
Thanks, I edited my original post, is $\Sigma=\{\emptyset,\{1,2,3\},\{1\},\{2,3\},\{2\},\{1,3\}\}$ a sigma algebra on $S$? –  Steven May 27 '12 at 22:46
3  
No, $\{2,3\} \cap \{1,3\} = \{3\} \notin \Sigma$. –  t.b. May 27 '12 at 22:47

3 Answers 3

up vote 3 down vote accepted

To elaborate on the concept as it lies in general: $\sigma$-algebra is basically a collection that is closed under countable set operations and complements. That is, all countable unions and intersections of elements of the $\sigma$-algebra remain elements of the $\sigma$-algebra, and same with complements. If the concept seems vague for the moment, I may try to motivate you by pointing out $\sigma$-algebras have a key role in the foundations of measure (and thus probability-) theory.

The given set $\Sigma$ is in fact not a $\sigma$-algebra. Since $\sigma$-algebras are closed under countable intersections, then $\{1,3\},\{2,3\}\in\Sigma$ should imply $\{3\}=\{1,3\}\cap\{2,3\}\in\Sigma$, which as you can see is not the case. The smallest $\sigma$-algebra that would contain $\Sigma$, denoted by $\sigma(\Sigma)$, is the power-set of $\{1,2,3\}$: this follows from the previous note that (in this case) in particular the singletons $\{1\},\{2\},\{3\}$ must elements of $\sigma(\Sigma)$. Since every subset of $\{1,2,3\}$ is a countable union of these singletons, then every subset of $\{1,2,3\}$ must be in $\sigma(\Sigma)$.

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a $\sigma$-algebra A is just a subset of $2^\Omega$ (the set of all subsets) for any set $\Omega$ with properties: 1. the set $\Omega$ is in A, 2. complements of sets in A are in A and 3. the union of countable subsets of A is in A

so given $\Omega = \{1,2,3\}$, than $A_1 = \{\emptyset,\{1,2,3\}\}$ is the smallest possible $\sigma$-algebra, $A_2 = \{\emptyset,\{1\},\{2\},\{3\},\{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}$, the biggest.

If $\{1\} \in A$, than $\{2,3\} \in A$, and if $\{1\}, \{2\} \in A$, than $A = A_2$ (why?) your example is no $\sigma$-algebra, because its not closed under complements

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Of course notation like $2^{\Omega}$ is probably outside Steven's background. –  GEdgar May 27 '12 at 23:20
    
The given example is actually closed under complements; so this is not where it fails to be a $\sigma$-algebra. –  Thomas E. May 29 '12 at 13:40

If you partition a finite set into a collection of indivisible "atoms", then the collection of all possible unions of the atoms (along with the empty set) forms a sigma algebra.

For example, the atoms $\{1\}$ and $\{2,3\}$ generate the sigma algebra $\{\emptyset, \{1\},\{2,3\},\{1,2,3\}\}$.

It's a fun exercise to show that all sigma algebras on a finite set arise from partitions in this way.

On the other hand, in the infinite case sigma algebras are not necessairily generated by some partition, and this is what makes them interesting. For example the borel sigma algebra on the real numbers contain all singleton points, but the sigma algebra is not the powerset of the reals.

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