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From the textbook I'm reading,

A collection of subsets $\mathscr{F}$ of $\Omega$ is called a $\sigma$-field if it satisfies:

  1. empty set in $\mathscr{F}$

  2. if $A_1, A_2, ... \in \mathscr{F}$, then union of $A$'s exist in $\mathscr{F}$

  3. if $A \in \mathscr{F}$, then $A^c\in\mathscr{F}$

and the smallest $\sigma$-field for $\Omega$ is the collection $\mathscr{F} = \{ \emptyset, \Omega \}$.

My question: is the union of all the elements in the $\sigma$-field $\mathscr{F}$ equivalent to $\Omega$?

amoeba
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pigate
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2 Answers2

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Trivially, yes, because if $\emptyset\in\mathscr{F},$ then $\emptyset^C=\Omega\in\mathscr{F},$ by the first and third properties you list.

As an aside, the definition that I'm familiar with is

  1. $\mathscr{F}$ is closed under countable unions
  2. $\mathscr{F}$ is closed under countable intersections
  3. $\mathscr{F}$ is closed under complements

and $\mathscr{F}$ is a set of subsets of $\Omega.$ As Juho points out, this difference doesn't matter, though, because we can make intersections into unions using complements: $A_1\cap A_2=(A_1^c\cup A_2^c)^c.$

Sycorax
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To show that two sets $A$ and $B$ contain the same elements, i.e. that $A=B$, it is often convenient to show the equivalent statement $A\subseteq B$ and $B \subseteq A$, so let us do that.

Let $O$ be the union of all the sets in $\mathscr F$. By properties 1 and 3, $\Omega \in \mathscr F$ and, thus, $\Omega \subseteq O$, which proves the first inclusion.

For the second inclusion, it suffices to show that if $A_i\in \Omega, \forall i \in \mathcal I$, for some index set $\mathcal I$, then $\cup_i A_i \subseteq \Omega$. This is so because $O$ is defined as the union of elements of $\mathscr F$ which are all subsets of $\Omega$ by definition.

I find it illuminating to think about what it would mean if this was not true. Namely, there would exist a point in the union that is not in $\Omega$. But if such a point is in the union, it must, by definition, also be in one of the sets in the union, and all of these are subsets of $\Omega$ so this cannot be. We conclude $O\subseteq \Omega$.

Some notes on this proof: In the comments @whuber expressed the view that the second inclusion is immediate from the definition of a union of subsets. I don't disagree with this view, but have often found the above thought exercise useful and it does, as far as I can tell, constitute a valid proof of the assertion so I leave it in.

ekvall
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    The second inclusion is immediate and requires no elaboration, because *by definition* the union of subsets of $\Omega$ is a subset of $\Omega$, whence $O\subseteq\Omega$. – whuber Mar 22 '16 at 15:11
  • I don't understand the point of that comment @whuber. Both directions are "immediate" by definitions. That doesn't mean you don't have to show them. To me, the contradiction argument illuminates exactly what you wrote. – ekvall Mar 22 '16 at 16:24
  • @whuber Added something to satisfy readers of your opinion too. – ekvall Mar 22 '16 at 16:48
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    The [Axiom of Union](http://mathworld.wolfram.com/Zermelo-FraenkelAxioms.html) in ZFC asserts (among other things) that $x\in\cup\mathscr{F}$ implies there exists $F\in\mathscr{F}$ for which $x\in F$. But since $F\subseteq \Omega$, *a fortiori* $x\in\Omega$. Therefore $O\subseteq\Omega$. – whuber Mar 22 '16 at 17:12
  • I must be expressing myself unclearly, @whuber, because I feel that's exactly what I wrote, albeit in a less elegant way (without the reference to ZFC). I've edited it now, and will leave it for a while. – ekvall Mar 22 '16 at 17:15