1
$\begingroup$

In a proof I am reading, the writer takes the sigma-algebra generated from an algebra. Since an algebra is closed under finite unions/intersections and complements I assumed that you just "threw in" all the countable unions of sets in the algebra into the algebra to get the sigma-algebra. Therefore, since the countable union of countable unions is countable, every set in the sigma-algebra should be able to be written as the countable union of sets in the algebra (that is my naive impression at least).

However, the proof is implying that some sets in the sigma-algebra cannot be written as the countable union of sets in the algebra. Could someone explain why or give me an example (maybe the algebra of open sets on the reals has one).

Thank!

$\endgroup$
  • $\begingroup$ If you do that, then there's no guarantee what you get is closed under complementation. $\endgroup$ – Lord Shark the Unknown Sep 26 '18 at 16:31
  • $\begingroup$ Is there an example? $\endgroup$ – user56628 Sep 26 '18 at 16:38
1
$\begingroup$

Firstly, I must be pedantic and say that the collection of all open sets in $\mathbb R$ is not an algebra (it is not closed under complementation).

Now that the pedantry is over with, let's get to business. Consider the algebra $\mathcal A$ of all finite unions of half-open intervals in $\mathbb R$ of the form $[a,b)$ for some $a<b$ (along with the associated rays $(-\infty, b)$ and $[a,\infty)$). I claim the $\sigma$-algebra this generates is the Borel $\sigma$-algebra on $\mathbb R$ (this isn't too difficult to see). Now let $\mathcal A_\sigma$ denote the collection of all countable unions of elements of $\mathcal A$. This contains all open sets, as each open set is a countable union of open intervals, and open intervals are countable unions of half-open intervals. But I claim that the closed set $$K=\{0\}\cup\{\frac{1}{n}:n\in\mathbb N\}$$ is not in $\mathcal A_\sigma$. To see this, suppose $K=\cup_{n\in\mathbb N}E_n$ where $E_n\in\mathcal A$. Then $0\in E_k$ for some $k$, hence there is some $\varepsilon>0$ such that $[0,\varepsilon)\subset E_k\subset\cup_{n\in\mathbb N}E_n=K$, which is clearly false.

To obtain a $\sigma$-algebra, once you add all countable unions you have to add the complements of all countable unions. Then do this again. And again. And again. And once you've done this countably many times, take the union over everything. Then start over. And once you've done this uncountably many times, you finally have a $\sigma$-algebra. For a better account of this, see the notes at the end of chapter 2 of Folland's Real Analysis.

$\endgroup$
  • $\begingroup$ recently I posted a very similar question math.stackexchange.com/questions/2888908/…. Is your specific example of the set $K$ which is not generated by countable unions remain if we added the complements of this unions? $\endgroup$ – Eduardo Sep 26 '18 at 18:04
  • $\begingroup$ You say "To obtain a σ-algebra, once you add all countable unions you have to add the complements of all countable unions." But since the algebra is closed under complements, isn't the complement of a countable union of sets in the algebra itself a countable union of sets in the algebra? $\endgroup$ – user56628 Sep 26 '18 at 18:09
  • $\begingroup$ @Eduardo Yes, $K^c\in\mathcal A_\sigma$. $\endgroup$ – Aweygan Sep 26 '18 at 18:40
  • $\begingroup$ @user56628 No, it would be a countable intersection of elements of the algebra, which isn't necessarily a countable union of elements of the algebra (as my counterexample shows). $\endgroup$ – Aweygan Sep 26 '18 at 18:41
  • $\begingroup$ Yes I believe the example does show that my statement is incorrect $\endgroup$ – user56628 Sep 26 '18 at 18:45

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.