Let $I^2:=[0,1]^2\subseteq \mathbb{R}^2$ be the closed unit square in the plane. Open and closed subsets of $I^2$ are Borel measurable for trivial reasons. Also, every set obtained from open and closed subsets of $I^2$ by taking at most a countable number of unions, intersections and complements is also Borel measurable.
My question is: are there Borel measurable subsets of $I^2$ which are not of the type just described?
Thank you.
You have to be careful when you say "at most a countable number". Your answer is "YES" but you have to go up the countable ordinals... Let $\cal A_0$ be the collection of all closed sets. For any ordinal $\eta$, let $\cal A_{\eta+1}$ be the collection of all countable unions of sets from $\cal A_\eta$, together with their complements. If $\lambda$ is a limit ordinal, let $\cal A_\lambda = \bigcup_{\eta<\lambda} \cal A_\eta$. Then, in fact, every Borel set lies in $\cal A_\eta$ for some countable ordinal $\eta$. But $\cal A_\eta \ne \cal A_\zeta$ if $\eta \ne \zeta$ are countable ordinals.
By definition, a Borel measurable set is a set in the Borel sigma algebra, which on a topological space is defined to be the collection of all sets that can be formed by taking countable union, countable intersection, and relative complementation from the open sets.
So tautologically, there are no other sets that is Borel measurable.
If however, you are looking for completions of the Borel measurable sets (for example, the Lebesgue measurable sets), then the answer could become more interesting. (See the question t.b. linked to.)
