As a homework assignment, I have been asked to provide a concrete example for why the countable union of sigma algebras is not always a sigma algebra.
The example that I had thought of was to take the sigma algebras of singletons over the natural numbers. i.e. $\mathscr A_i = \bigl\{\{i\},\{i\}^c, \emptyset,\Bbb N \bigr\}$. Then for example, $\mathscr A_1 \bigcup \mathscr A_2$ would contain {1,2} but not its complement.
However, another student told me that this would not work because not only do you take the unions of the elements {i}, you also take the unions with $\{ \emptyset\}$ and $\{\Bbb N\}$, so the countably infinite union would contain the union $\{ \emptyset\}\bigcup\{ \emptyset\} \bigcup \{3\} \bigcup \{4\}\bigcup $… which would be the complement of the union {1,2}. Is this the correct way to take the unions of algebras?
Most of the examples I've seen online would just take the unions of the elements {i} and wouldn't consider the null in the unions. The logic the other student gave me seems correct, I just haven't seen this idea anywhere when reading about this topic. Can someone confirm or refute this?
Best Answer
Here is a simpler example:
Consider the space $\Omega = \{1,2,3\}$, and let $\Sigma_1 = \{ \emptyset, \{1\}, \{2,3\}, \Omega \}$ and $\Sigma_2 = \{ \emptyset, \{1,2\}, \{3\}, \Omega \}$.
Both are $\sigma$-algebras, but $\Sigma_1 \cup \Sigma_2$ is not as it would have to contain the set $\{1,3\}$.