2
$\begingroup$

In the text of Probability Essentials by J.Jacod & P.Protter, a theorem:

The Borel $ \sigma $- algebra of $R $ is generated by intervals of the form $(-\infty,a ]$, where $a \in Q$.

As far as I've known the Borel sigma algebra is generated by all open subsets of $R$, which surely contains sets like (x, y), here x is irrational. So my question is how can 'a', in theorem, a rational generate such set (x, y).

$\endgroup$
4
$\begingroup$

$(x,y) = \displaystyle\bigcup_{\substack{a,b\in\mathbb{Q} \\ x<a<b<y}} (a,b)$.

$\endgroup$
  • $\begingroup$ could you please explain it a bit in detail? $\endgroup$ – Dylan Zhu Oct 13 '13 at 18:34
  • $\begingroup$ @DylanZhu Filling in more details: $(a,b'] = (-\infty,b'] \cap (-\infty,a']^c$, and $(a,b) = \cup_{b'\in\mathbb{Q}, b'<b} (a,b']$. Let me know if you have any specific questions. $\endgroup$ – Slade Oct 13 '13 at 19:04
  • $\begingroup$ yeah, i think i still dont know how can you make sure that b is close enough to y, that there wont be other irrationals between them... $\endgroup$ – Dylan Zhu Oct 13 '13 at 19:43
  • 1
    $\begingroup$ The rationals are dense. For any $(x,x+\epsilon)$, pick $n > 1/\epsilon$, then some rational $t/n$ must land in there. $\endgroup$ – Slade Oct 13 '13 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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