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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 $\mathbb{R} \times \mathbb{R}^{n}$ ?

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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
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already answered here mathoverflow.net/questions/38795/borel-sets-on-rn –  Valerio Capraro Feb 3 '12 at 12:48
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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
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closed as too localized by Emil Jeřábek, Andreas Blass, Bill Johnson, Mark Meckes, Anthony Quas Feb 3 '12 at 15:20

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1 Answer

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.

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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
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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
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