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Definition of sigma-algebra

  1. Oct 4, 2005 #1
    I had a quick question concerning the definition of a [itex]\sigma[/itex]-algebra [itex]\mathcal{F}[/itex] over a set [itex]\Omega[/itex]. Most sources I've seen (e.g. http://en.wikipedia.org/wiki/Sigma-algebra ) require that [itex]\Omega[/itex] or the empty set be an element of [itex]\mathcal{F}[/itex].

    Is this necessary? I ask because I am looking at "Probability: Theory and Examples" by Durrett, and he does not state that as a requirement. He only requires that an element's complement be in [itex]\mathcal{F}[/itex] and that countable (possibly infinite) unions of elements (in the set) remain in the set. Additionally, he says that [itex]\mathcal{F} \neq \emptyset[/itex], but this does not necessarily imply that the empty set is in [itex]\mathcal{F}[/itex].

    So, has Durrett just forgotten to include this? Do his later results assume this requirement? Or is it the case this is an unnecessary requirement?
     
  2. jcsd
  3. Oct 4, 2005 #2
    Specifically, he states:

    if [itex]A_i \in \mathcal{F}[/itex] is a countable sequence of sets then [itex]\cup_i A_i \in \mathcal{F}[/itex]

    I think this is my answer. Let the sequence consist of only the set [itex]\mathcal{F}[/itex]. Then [itex]\mathcal{F}[/itex] (and hence the empty set as well) is in [itex]\mathcal{F}[/itex].

    Correct?
     
    Last edited: Oct 4, 2005
  4. Oct 4, 2005 #3
    let [itex]\mathcal{F}[/itex] be a sigma algebra over a set [itex]\Omega[/itex]
    since [itex]\mathcal{F}[/itex] in noempty there exists an [itex]A\in\mathcal{F}[/itex] since [itex]\mathcal{F}[/itex] is a sigma algebra [itex]A^c\in\mathcal{F}[/itex] and [itex]A\bigcup A^c=\Omega\in\mathcal{F}[/itex]
     
  5. Oct 4, 2005 #4
    a sigma algebra R on a set X is a nonempty collection of sets satisfying the following:
    i) R closed under complements
    ii) R closed under countable unions
    & that's all

    we can derive the fact that the set X on which the algebra is defined, is in R and also the empty set. the empty set is in every sigma algebra because if E is in R, then E\E (=empty set) is in R since R is closed under complementation. also E union E' = X is also in R. so no, the definition doesn't need to include anything about the empty set or the underlying set X is in the algebra.
     
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