Let $\mathcal{A} \subseteq \mathcal{P}(\Omega)$, $\Omega$ a set. Then isn't the set of call countable unions of countable intersections of elements or complements of elements of $\mathcal{A}$ equal to $\sigma(\mathcal{A})$? More precisely, $\sigma(\mathcal{A}) = \{ \cup_{i=1}^{\infty} \cap_{j=1}^{\infty} A_j^i : A = A_j^i$ or $A = A_j^i$ for some $A \in \mathcal{A}$, for all $i,j\}$. If that's not true, then I'm pretty sure that if you also contain the countable $\cap$ of countable $\cup$ 's, then it is true.
I found it difficult trying to prove that $\cap_{i=1}^{\infty} \cup_{j=1}^{\infty} A_j^i = \cap_{j=1}^{\infty} \cup_{i=1}^{\infty} A_j^{k(i,j)}$, where $k :\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ is some onto function. So that's why I've also said it might have to include the $\cap\cup$'s.
What do you think, is it true?
Best Answer
The question automatically came back (since it has no answer).
As Andrés noted, the answer is "no" in general. To generate a sigma-algebra "from within" is much more complex than this.
Let $\Omega$ be a set, let $\mathcal{A} \subseteq \mathcal{P}(\Omega)$. Recursively define sets $\mathcal A_\alpha$ as $\alpha$ ranges over all ordinals:
\begin{align} \mathcal A_0 &:= \mathcal A\\ \mathcal A_{\alpha+1}&:= \text{countable unions of sets and complements of sets from } \mathcal A_{\alpha}\\ \mathcal A_\lambda &:= \bigcup_{\alpha < \lambda} \mathcal A_\alpha \text{ for limit ordinal }\lambda \end{align} Then: $$ \sigma(\mathcal A) = \mathcal A_{\omega_1} $$ where $\omega_1$ is the least uncountable ordinal. And, in general, all the $\mathcal A_\alpha$ for $\alpha < \omega_1$ are different, and (therefore) are not sigma-algebras.
In particular, if $\Omega = \mathbb R$ and $$\mathcal A = \{(a,b) \subseteq \mathbb R : -\infty < a < b < +\infty\},$$ then all $\mathcal A_\alpha$ for $\alpha < \omega_1$ are different.