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Definition of Borel on $\mathbb R$ ($\mathcal B(\mathbb R)$) is that it's the $\sigma$-algebra generated by all open sets in $\mathbb R$.

OK, if I take some open set like $C = (0,1)$, by definition of $\sigma$-algebra the complement of $C$ must also be in $\mathcal B(\mathbb R)$, so something like $(-\infty, 0] \cup [1, \infty)$ must be in $\mathcal B(\mathbb R)$. Is that right?

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Yes, that's right. – Rasmus Feb 17 '12 at 8:33
Clearly, any closed set is in $\mathcal B(\mathbb R)$. – user23211 Feb 17 '12 at 10:22

That is right you also have all closed sets and the countable union s of closed s sets what we call $F_{\sigma}$ and their complements which are call ed $ G_{\delta}$. However you can find Borel sets that aren't neither $ F_{\sigma}$ or $ G_{\delta}$.

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