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).


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

  • $\begingroup$ 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

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