# Borel subsets of the unit square

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.

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Are you asking whether by taking countable unions, intersections and countable complements successively you can get all Lebesgue measurable subsets of $I^2$ from open sets? That's not even true for $I$. –  t.b. Nov 3 '11 at 1:00
@t.b.: Edited, thanks! –  user17240 Nov 3 '11 at 1:58
@Willie: Thanks! I missed the notification of the edit (before it was a duplicate of what I linked to). I removed the possible duplicate link: why don't you close and re-open? –  t.b. Nov 7 '11 at 13:36
@t.b. good idea. @ All: closing a re-opening to clear the votes. –  Willie Wong Nov 7 '11 at 13:38

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.

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