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Show that the Borel sigma-algebra on $\mathbb{R}$, denoted $B_R$ is generated the open intervals in $\mathbb{R}$.

My attempt: Let $I$ be the collection of all open intervals, let $\sigma I$ be the sigma- algebra generated by $I$, and let $\tau$ be the standard topology on $\mathbb{R}$. Since every open set in $\mathbb{R}$ is the union of countably many open intervals, and since sigma-algebras are closed under countable unions, we have that every open set is in $\sigma I$, and so $B_R=\sigma\tau\subseteq\sigma I$. But of course, every open interval is an open is also an open set in $\mathbb{R}$, so $I\subseteq\tau$ which implies $\sigma I\subseteq\sigma\tau=B_R$.

Does this look ok?

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Yes, this is correct. Another way to approach the problem would be to show that $\sigma I$ is a $\sigma$-algebra that contains the Borel $\sigma$-algebra.

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