# Borel sigma algebra generated by open intervals proof understanding.

In lecture notes I am working through it says that the fact that the Borel sigma algebra is generated by the open intervals follows from the fact that open and closed sets are Lebesgue measurable (which we have proved). How?

it is clear that $\ \sigma(A) \subset \sigma(O)$ where $\ A$ is the collection of open intervals and $\ O$ is the collection of open sets. But to prove $\ \sigma(O) \subset \sigma(A)$ I would think you would have to prove that any open set is the countable union of open intervals (which I know from online resources outside of the lecture notes to be ture) but how does the fact that open and closed sets are lebesgue measurable imply this?

The definition of the Borel $\sigma$-algebra is the $\sigma$-algebra that is generated by the open sets of a topological space $X$. Call this $\mathscr{B}$. Then $\mathscr{B}$ is the smallest $\sigma$-algebra containing the open sets of $\mathbb{R}$. Since $\sigma(A) \subset \mathscr{B}$, as you said, it suffices to show that every open set is contained in $\sigma(A)$. But this follows since open intervals form a countable basis for the open sets in $\mathbb{R}$, as you said. To my knowledge, this does not have anything to do with Lebesgue measurability; the two are a priori separate ideas.
• Sorry perhaps my question is not clear, the definition you use is the one I am using as well. What I am asking for is help with proof of the result: "The borel sigma algebra can be generated by the open intervals rather than the open sets" so in my symbols I want to prove $\ \sigma(A) = \sigma(O)$. – Hugh Kinnear Apr 15 '18 at 7:03