countable unions and intersections of analytic sets are analytic

A property of analytic sets (http://planetmath.org/AnalyticSet2) which makes them particularly suited to applications in measure theory is that, in common with $\sigma$-algebras (http://planetmath.org/SigmaAlgebra), they are closed under countable unions and intersections.

Theorem 1.

Let $\mathrm{(}X\mathrm{,}\mathrm{F}\mathrm{)}$ be a paved space and ${\mathrm{(}{A}_n\mathrm{)}}_{n\mathrm{\in }\mathrm{N}}$ be a sequence of $\mathrm{F}$-analytic sets. Then, ${\mathrm{\bigcup }}_n{A}_n$ and ${\mathrm{\bigcap }}_n{A}_n$ are $\mathrm{F}$-analytic.

A consequence of this is that measurable sets are analytic, as follows.

Corollary.

Let $\mathrm{F}$ be a nonempty paving on a set $X$ such that the complement (http://planetmath.org/Complement) of any $S\mathrm{\in }\mathrm{F}$ is a union of countably many sets in $\mathrm{F}$.

Then, every set $A$ in the $\sigma$-algebra generated by $\mathrm{F}$ is $\mathrm{F}$-analytic.

For example, every closed subset of a metric space $X$ is a union of countably many open sets. Therefore, the corollary shows that all Borel sets are analytic with respect to the open subsets of $X$.

That the corollary does indeed follow from Theorem 1 is a simple application of the monotone class theorem.First, as the collection $a(ℱ)$ of $ℱ$-analytic sets is closed under countable unions and finite intersections, it will contain all finite unions of finite intersections of sets in $ℱ$ and their complements, which is an algebra (http://planetmath.org/RingOfSets). Then, Theorem 1 says that $a(ℱ)$ is closed under taking limits of increasing and decreasing sequences of sets. So, by the monotone class theorem, it contains the $\sigma$-algebra generated by $ℱ$.