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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
up vote 2 down vote accepted

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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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.)

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No need to link Wikipedia. I have described the hierarchy many times, as well others, right here on the site. – Asaf Karagila Nov 7 '11 at 13:43
@Asaf: feel free to edit and improve (with a proper link). – Willie Wong Nov 8 '11 at 9:03
But that would violate my usual approach of being lazy, yet complaining in the comments!! :-) – Asaf Karagila Nov 8 '11 at 12:25

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