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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$\begingroup$ Let $ X,Y \subset \mathbb{R}^{n}$ Borel sets. Is the set $ X \times Y \subset \mathbb{R}^{2n} $ a Borel set ? $\endgroup$– SantosCommented Feb 3, 2012 at 12:29
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1$\begingroup$ already answered here mathoverflow.net/questions/38795/borel-sets-on-rn $\endgroup$– Valerio CapraroCommented Feb 3, 2012 at 12:48
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1$\begingroup$ 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. $\endgroup$– Joel David HamkinsCommented Feb 4, 2012 at 1:07
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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.
answered Feb 3, 2012 at 13:36
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$\begingroup$ This is a good idea ... "diagonal" does not quite make sense for $\mathbb R \times \mathbb R^n$, though. $\endgroup$ Commented Feb 3, 2012 at 15:15
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1$\begingroup$ 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. $\endgroup$ Commented Feb 4, 2012 at 0:52