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?
Best Answer
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.