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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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