1
$\begingroup$

I am attending a PhD course in Probability, and I have stumbled upon an exercise in which I need to prove that the Borel sigma-field on the real line is countably generated. I think I have managed to do it using sequences of interval whose ends are rationals, like in this question (Showing that the family of Borel sets is countably generated), but I have been told that it is possible to do that using Dynkin $\pi - \lambda$ theorem.

The Borel sigma-field is a $\lambda$- system and it includes $\mathcal{Q}$, the collection of all intervals with rational ends, so that $\sigma(\mathcal{Q}) \subseteq \mathcal{B}(\mathbb{R})$, but I am not able to prove the opposite inclusion (which would lead me to the equality I am looking for).

Thank you in advance.

$\endgroup$
2
$\begingroup$

In general, the Borel $\sigma$-field ${\cal B}(X)$ of a topological space $(X,\tau)$ is just $\sigma(\tau)$, the $\sigma$-field generated by $\tau$.

Therefore, all you need to do is showing that $\sigma({\cal Q})$ contains every open interval (and therefore every open subset -- since every open subset of $\mathbb R$ can be written as a countable union of disjoint open intervals) of $\mathbb{R}$.

For every open interval $I\subseteq\mathbb R$, say $I=(a,b)$, there exists $(a_n)_n$, $(b_n)_n$ sequences in $I\cap\mathbb Q$ such that $a_n\to a$ and $b_n\to b$ (here we're using density of $\mathbb Q$). In that way, $I = (a,b) = \bigcup_{n} (a_n,b_n)$, where $(a_n,b_n)\in{\cal Q}$. Then $I\in\sigma({\cal Q})$, as desired.

$\endgroup$

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.