# smallest sigma algebra can be obtained by countable operations?

Let $\mathcal{A}\subset \mathcal{P}(X)$. Then $\sigma(\mathcal{A})$, the smallest $\sigma$-algebra containing $\mathcal{A}$ is given by the intersection of all $\sigma$-algebras containing $\mathcal{A}$.

Is it true that $\sigma(\mathcal{A})$ exactly consists of the sets which can be obtained by countable union & intersection, and complement of the sets in $\mathcal{A}$?

For example: Is it true that every Borel set can be obtained by countable union & intersection, and complement of open sets in $\mathbb{R}$?

• When you say "obtained", do you mean that you are allowing countable unions of countable intersections of countable unions, etc.? – Zev Chonoles Mar 10 '13 at 8:53
• Every Borel set can be obtained by a tree of countable unions, countable intersections and complements whose branches terminate at open sets, but the tree might be quite deep (any depth below $\omega_1$.) – David Moews Mar 10 '13 at 8:59
• What @DavidMoews says extends to arbitrary generated $\sigma$-algebras. This follows from the fact that if $B\in\sigma(\mathcal{F})$, then there exists countable $\mathcal{C}\subseteq\mathcal{F}$ such that $B\in\sigma(\mathcal{C})$ and the fact that every countably generated $\sigma$-algebra is generated by a real-valued function with the codomain being endowed with the Borel $\sigma$-algebra. – Michael Greinecker Mar 10 '13 at 9:04
• In general, you have to go on and let ${\cal A}(\omega)=\cup_n {\cal A}(n)$, let ${\cal A}(\omega+1)$ be all sets obtained by countable unions, intersections and complements of sets in ${\cal A}(\omega)$, and so on. The sequence goes on through the countable ordinals. Taking the union of ${\cal A}(\alpha)$ for all $\alpha<\omega_1$ will always give you $\sigma({\cal A})$. In some cases the process might terminate earlier. – David Moews Mar 10 '13 at 9:26
• The Joy of Sets by Devlin contains a very readable exposition of the topic. – Michael Greinecker Mar 10 '13 at 11:06