Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is the set { $ \cup_{i \in \mathbb{N}} C_{i} \times D_{i} : C_{i} \in \mathcal{L} \ , D_{i} \in \mathcal{B}^{n} \ $ } a sigma algebra on $\mathbb{R} \times \mathbb{R}^{n}$ ?

share|improve this question

closed as too localized by Emil Jeřábek, Andreas Blass, Bill Johnson, Mark Meckes, Anthony Quas Feb 3 '12 at 15:20

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Let $ X,Y \subset \mathbb{R}^{n}$ Borel sets. Is the set $ X \times Y \subset \mathbb{R}^{2n} $ a Borel set ? –  Santos Feb 3 '12 at 12:29
already answered here mathoverflow.net/questions/38795/borel-sets-on-rn –  Valerio Capraro Feb 3 '12 at 12:48
Valerio, are you saying the answer to the OP's question above is at your link? It seems to be a different question there, although the theme is related. –  Joel David Hamkins Feb 4 '12 at 1:07
add comment

1 Answer 1

Your collection is not closed under complement. To see this, observe that the diagonal $\Delta=\{(x,x)\mid x\in\mathbb{R}\}$ is not in your collection, since the only rectangles it contains are singletons, but there are uncountably many. But the complement of $\Delta$ is the union of countably many open rectangles, so the complement of $\Delta$ is in your collection.

share|improve this answer
This is a good idea ... "diagonal" does not quite make sense for $\mathbb R \times \mathbb R^n$, though. –  Gerald Edgar Feb 3 '12 at 15:15
Yes, Gerald, I had in mind just the case $\mathbb{R}\times\mathbb{R}$. But for the general case, one can use the same idea with $\Delta\times\mathbb{R}^{n-1}$, which is a closed subset of $\mathbb{R}\times\mathbb{R}^n$, whose complement is therefore in the OP's collection, but the set itself is not for similar reasons to what I say in my answer. –  Joel David Hamkins Feb 4 '12 at 0:52
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.