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A measurable space is a set $S$, together with a nonempty collection, $\mathcal{S}$, of subsets of $S$ satisfying the following two conditions:

  1. For any $A$, $B$ in the collection of $\mathcal{S}$, the set $A-B$ is also in $\mathcal{S}$

  2. For any $A_1, A_2, ... \in \mathcal{S}$, their union is in $\mathcal{S}$.

Source

Is $\mathcal{S}$ a sigma algebra? The definition of sigma algebra states it's a collection of subsets of $X$ closed under countable union, complementation, and that it contains the empty set.

Point 1. guarantees it contains the empty set, point 2. states its closed under countable union. But how 1. and 2. imply that it's closed under complement? How to prove that $\mathcal{S}$ contains $S$ as well, as it should (because it contains the empty set), given it really is a sigma-algebra?

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  • $\begingroup$ In your 2nd paragraph you say "complementation" and in your 3rd paragraph you say "countable complement". I don't know what "countable complement" means. Did you intend to write "countable intersection"? $\endgroup$ Commented May 27, 2016 at 18:40
  • $\begingroup$ A $\sigma$-algebra is a collection of subsets of $X$. $\endgroup$ Commented May 27, 2016 at 18:41
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    $\begingroup$ Consider $S=\{1,2\}$ with $\mathcal{S}=\{\{1\},\emptyset\}$. Or, better yet, $S=\{1\}$ and $\mathcal{S}=\{\emptyset\}$. $\endgroup$ Commented May 27, 2016 at 18:44
  • $\begingroup$ Worth noting that all the examples from the cited document place $S$ in $\mathcal{S}$ by fiat as there is no way to do so from (1.) and (2.). $\endgroup$ Commented May 27, 2016 at 18:49
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    $\begingroup$ 1) and 2) do not guarantee that the collection is closed under complementation and (equivalently) do not guarantee that the collection contains $S$. $\endgroup$
    – drhab
    Commented May 27, 2016 at 19:03

1 Answer 1

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$\mathcal S$ is not necessarily a $\sigma$-algebra.

For instance let $S=\{1,2,3,4\}$ and let $\mathcal S=\wp(\{1,2,3\}$).

Then $\mathcal S$ satisfies the conditions, but $\{1,2,3,4\}\notin\mathcal S$.

$\mathcal S$ is a so-called $\sigma$-ring.


edit:

Another example inspired by the comment of Samuel and emphasizing that there is an essential difference:

Let $S$ be an uncountable set and let $\mathcal S$ denote the collection of countable subsets of $S$.

Note that in this case (unlike the one above) the collection cannot be identified as a $\sigma$-algebra on a subset of $S$.

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  • $\begingroup$ However it is a $\sigma$-algebra on a subset of $S$. Need this be the case? $\endgroup$ Commented May 27, 2016 at 18:46
  • $\begingroup$ Does it mean the definition given in this documemt is wrong? $\endgroup$ Commented May 27, 2016 at 18:46
  • $\begingroup$ @user4205580 It can happen that measures are defined on collections like these. So it goes too far to say that the document is wrong. $\endgroup$
    – drhab
    Commented May 27, 2016 at 18:48
  • $\begingroup$ @SamuelCoskey No. Take for instance the collection of countable subsets of an uncountable set. $\endgroup$
    – drhab
    Commented May 27, 2016 at 18:49
  • $\begingroup$ I have added a link. $\endgroup$
    – drhab
    Commented May 27, 2016 at 18:53

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