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Fix a set $\Omega$. A $\sigma$-algebra on $\Omega$ is a non-empty collection of subsets of $\Omega$ closed under taking complements and countable unions.

I'd like to prove that (1) for finite $\Omega$, $2^\Omega$ is a $\sigma$-algebra and that (2) the intersection of a family of $\sigma$-algebras is a $\sigma$-algebra. Are these proofs correct?


The def. says "non-empty collection" so the collection contains something and the complement of that something hence the whole things $\Omega$. Since it contains the whole things, it contains its complement, hence the empty set $\emptyset$. This guarantees that $2^\Omega$ and $\bigcap \mathcal{A_i}$ of $\sigma$-algebras $\mathcal{A_i}$ are non-empty.

(Stmt 1) Pf. $2^\Omega$ contains all subsets of $\Omega$. So it's closed under taking complements and countable unions.

(Stmt 2) Pf. Let $\{\mathcal{A_i}\}$ be a family of $\sigma$-algebras.

  1. $A\in \bigcap \mathcal{A_i}\implies A\in \mathcal{A_i} \forall i\implies A^c \in\mathcal{A_i}\forall i\implies A^c\in \bigcap \mathcal{A_i}$
  2. Let $A_j\in \bigcap \mathcal{A_i}$ for $j\in J$. Then $A_j\in \mathcal{A_i} \forall i \forall j$. Therefore $\bigcup A_j\in \mathcal{A_i} \forall i$. Hence $\bigcup A_j\in\bigcap\mathcal{A_i}$
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  • $\begingroup$ Um, what is your question? $\endgroup$ – Henning Makholm Jan 30 '14 at 12:46
  • $\begingroup$ The post is tagged with proof-verification. $\endgroup$ – saadtaame Jan 30 '14 at 12:48
  • $\begingroup$ x @saadtaame That is a fact, not a question. $\endgroup$ – Henning Makholm Jan 30 '14 at 12:49
  • $\begingroup$ Ok sir, now it's a question. $\endgroup$ – saadtaame Jan 30 '14 at 12:50
  • $\begingroup$ what is $2^\omega$ ? $\endgroup$ – Thomas Jan 30 '14 at 12:58
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Your proofs are OK. Well, maybe in (1) you should also state that $2^\Omega$ is nonempty. Similarly in (2) show that $\bigcap \mathcal A_i$ is nonempty.

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  • $\begingroup$ They are empty by def. They all contain $\Omega$ and $\emptyset$. Right? Question: by OK you mean that there are simpler proofs? $\endgroup$ – saadtaame Jan 30 '14 at 14:04
  • $\begingroup$ That is not stated in the definition. But it is true, so should be mentioned in the solution. $\endgroup$ – GEdgar Jan 30 '14 at 14:08
  • $\begingroup$ The def. says "non-empty collection" so the collection contains something and the complement of that something hence the whole things. Since it contains the whole things, it contains its complement, hence the empty set. $\endgroup$ – saadtaame Jan 30 '14 at 14:14
  • $\begingroup$ Exactly right. So include that in your solution. $\endgroup$ – GEdgar Jan 30 '14 at 14:20
  • $\begingroup$ Alright. Thank you for the time. $\endgroup$ – saadtaame Jan 30 '14 at 14:22
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Your proof of statement 1 is nearly correct, you should also proof, as GEdgar mentioned, that the powerset is never empty. This is quite clear from the definition if Omega is nonempty. If Omega is the empty set, then you need to proof a bit more (look e.g. here: https://proofwiki.org/wiki/Power_Set_of_Empty_Set)

Point 1 of your proof of statement 2 is fine.

In point 2 of statement 2 you need to state that J is a countable index set and not an arbitrary, since sigma-algebras are closed only under countable unions.

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protected by Zev Chonoles Aug 15 '16 at 17:20

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