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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? $\endgroup$ Commented Jan 30, 2014 at 12:46
  • $\begingroup$ The post is tagged with proof-verification. $\endgroup$
    – mrk
    Commented Jan 30, 2014 at 12:48
  • $\begingroup$ x @saadtaame That is a fact, not a question. $\endgroup$ Commented Jan 30, 2014 at 12:49
  • $\begingroup$ Ok sir, now it's a question. $\endgroup$
    – mrk
    Commented Jan 30, 2014 at 12:50
  • $\begingroup$ what is $2^\omega$ ? $\endgroup$
    – Thomas
    Commented Jan 30, 2014 at 12:58

2 Answers 2

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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? $\endgroup$
    – mrk
    Commented Jan 30, 2014 at 14:04
  • $\begingroup$ That is not stated in the definition. But it is true, so should be mentioned in the solution. $\endgroup$
    – GEdgar
    Commented Jan 30, 2014 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. $\endgroup$
    – mrk
    Commented Jan 30, 2014 at 14:14
  • $\begingroup$ Exactly right. So include that in your solution. $\endgroup$
    – GEdgar
    Commented Jan 30, 2014 at 14:20
  • $\begingroup$ Alright. Thank you for the time. $\endgroup$
    – mrk
    Commented Jan 30, 2014 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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  • $\begingroup$ On your last statement, I'd have said that since the family of $\sigma$ - algebras can be an arbitrary one and for an element in the intersection of this family holds true that it has to be in all of the $\sigma$-algebras, point 2 of their statement should even hold true for arbitrary Js and therefore hold true for countable index sets. $\endgroup$
    – Leoncino
    Commented Apr 2 at 14:19

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