4
$\begingroup$

Suppose I have a set $\Omega$, and I partition $\Omega$ into $4$ sets $A_{1}, A_{2}, A_{3}, A_{4}$.

Clearly, the $A_{i}$'s are pairwise disjoint and their union equals $\Omega$. Now, it's also clear to me that the $\sigma$-algebra generated by these four sets is just the set of all possible unions of these, as well as $\emptyset$.

I'm wondering if the only sub-$\sigma$-algebras of this $\sigma$-algebra are only the $\sigma$-algebra $\{\Omega, \emptyset \}$ and $\sigma(A_{1})$, $\sigma(A_{2})$, $\sigma(A_{3})$, and $\sigma(A_{4})$. For each $i$, $\sigma(A_{i}) = \{ \emptyset, \Omega, A_{i}, A_{i}^{c} \}$, which shows that the intersection of any of four the $\sigma$-algebras I listed above (which has to be a sub-$\sigma$-algebra itself) is just $\{\Omega, \emptyset \}$.

Is it true that these are the only sub-$\sigma$-algebras? What if we can partition $\Omega$ into countably many sets? Does the same statement hold?

$\endgroup$
  • $\begingroup$ I think it's not hard to show that any $\sigma$-algebra $\mathcal{F}$ on a finite or countable set $\Omega$ consists of a partition of $\Omega$ into finitely or countably many sets, together with all unions of those sets. (Consider the atoms of $\mathcal{F}$.) It follows that any sub-$\sigma$-algebra of $\mathcal{F}$ is similarly generated by a coarsening of this partition. Incidentally, this explains why "interesting" probability theory can only be done on uncountable probability spaces. $\endgroup$ – Nate Eldredge Mar 11 '15 at 17:10
  • $\begingroup$ @NateEldredge My question didn't specify that $\Omega$ is countable. So you think it's only true if $\Omega$ is countable? $\endgroup$ – layman Mar 11 '15 at 21:07
3
$\begingroup$

I think the following are also sub-$\sigma$-algebras:

$$\sigma(A_1\cup A_2),\quad\sigma(A_1\cup A_3),\quad\sigma(A_1\cup A_4).$$

More generally, let $\{A_1,\ldots,A_n\}$ be a partition of $\Omega$ and let $S=\hat{\sigma}(A_2,A_3,\ldots,A_n)$ be a $\sigma$-algebra of $\Omega\setminus A_1$.

Then $\sigma(A_1\cup B)$ is a sub-$\sigma$-algebra of $\sigma(A_1,\ldots,A_n)$ for all $B\in S$.

To illustrate, let $n=4$ so that

\begin{eqnarray*} S &=& \hat{\sigma}(A_2,A_3,A_4) \\ &=& \{\emptyset,\;\;A_2,\;\;A_3,\;\;A_4,\;\;A_2\cup A_3,\;\;A_2\cup A_4,\;\;A_3\cup A_4,\;\;A_2\cup A_3\cup A_4 \}. \end{eqnarray*}

The $\sigma$-sub-algebras of $\sigma(A_1,\ldots,A_4)$ then are:

$$\qquad\sigma(A_1),\quad\sigma(A_1\cup A_2),\quad\sigma(A_1\cup A_3),\quad\sigma(A_1\cup A_4),\quad\sigma(A_1\cup A_2\cup A_3),\quad\sigma(A_1\cup A_2\cup A_4),\quad\sigma(A_1\cup A_3\cup A_4),\quad\sigma(\Omega).$$

I think this would apply also to countable partitions.

$\endgroup$
  • $\begingroup$ Ok, now the question is are each of these sub-$\sigma$-algebras non-trivial? For example, I had a feeling that $\sigma(A_{1} \cup A_{3}) = \sigma(\Omega)$. That's what spurred my question in the first place. Can you show that they are not equal? $\endgroup$ – layman Mar 16 '15 at 12:33
  • $\begingroup$ Also, I think the elements of your $S$ are incorrect. You didn't add $A_{2} \cup A_{3} \cup A_{4}$, and consequently you didn't add the complement of this set, which is $A_{1}$. Thus, $S$ is actually $\sigma(\Omega)$. $\endgroup$ – layman Mar 16 '15 at 12:37
  • $\begingroup$ @user46944 $\sigma(\Omega)$ is just $\{\emptyset,\;\Omega\}$. $\sigma(A_1\cup A_3)=\{\emptyset,\;A_1\cup A_3,\;A_2\cup A_4,\;\Omega\}$. $\endgroup$ – Mick A Mar 16 '15 at 12:37
  • $\begingroup$ @user46944 Note that $S$ is a $\sigma$-algebra of $\Omega\setminus A_1$ so $A_1$ cannot appear in it. $\endgroup$ – Mick A Mar 16 '15 at 12:39
  • $\begingroup$ Why would you construct $S$ in that way? We want $S$ to be a sub-$\sigma$-algebra of $\sigma(\Omega)$, which is a $\sigma$-algebra on $\Omega$. $\endgroup$ – layman Mar 16 '15 at 12:41

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.