Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
Yes, that's right. –  Rasmus Feb 17 '12 at 8:33
1  
Clearly, any closed set is in $\mathcal B(\mathbb R)$. –  user23211 Feb 17 '12 at 10:22
add comment

1 Answer

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}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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