4
$\begingroup$

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 $\Omega$?

$\endgroup$
1
  • 5
    $\begingroup$ Yes, because by the first and third properties $\Omega$ is in the sigma field. $\endgroup$
    – mark999
    Mar 22 '16 at 5:56
3
$\begingroup$

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: $A_1\cap A_2=(A_1^c\cup A_2^c)^c.$

$\endgroup$
3
  • 3
    $\begingroup$ $\cap_{i=1}^{\infty} A_i = (\cup_{i=1}^{\infty} A^\mathrm{c}_i)^\mathrm{c}$ $\endgroup$ Mar 22 '16 at 13:22
  • $\begingroup$ @JuhoKokkala D'oh. That's a good point. Edited. $\endgroup$
    – Sycorax
    Mar 22 '16 at 13:22
  • 2
    $\begingroup$ Saying that "$\mathscr F$ is a subset of $\Omega$" is inaccurate: F is a set of subsets of Omega. $\endgroup$
    – amoeba
    Mar 22 '16 at 13:49
1
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ The second inclusion is immediate and requires no elaboration, because by definition the union of subsets of $\Omega$ is a subset of $\Omega$, whence $O\subseteq\Omega$. $\endgroup$
    – whuber
    Mar 22 '16 at 15:11
  • $\begingroup$ 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. $\endgroup$
    – ekvall
    Mar 22 '16 at 16:24
  • $\begingroup$ @whuber Added something to satisfy readers of your opinion too. $\endgroup$
    – ekvall
    Mar 22 '16 at 16:48
  • 1
    $\begingroup$ The Axiom of Union 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 $O\subseteq\Omega$. $\endgroup$
    – whuber
    Mar 22 '16 at 17:12
  • $\begingroup$ 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. $\endgroup$
    – ekvall
    Mar 22 '16 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.