# Borel sigma algebra are generated by open sets

The Borel sigma algebra $$B(\mathbb{R}^d)$$ is generated by all (left) half-open intervals. I need to show that it is also generated by all open intervals:

So $$(a,b)=\bigcup(a,b-1/n]$$ so $$(a,b)$$ is in the Borel set generated by half open intervals.

Similarly $$(a,b]=\bigcup(a,b+1/n)$$ hence $$(a,b]$$ is in the Borel set generated by all open intervals.

• I think you are looking at $\mathcal{B}(\mathbb{R})$ (1 dimensional). Commented Oct 18, 2022 at 23:00

I believe you mean $$(a,b] = \bigcap (a, b+1/n)$$ (rather than $$\bigcup$$) which is indeed in the $$\sigma$$-algebra generated by the open intervals. Besides this typo, you're entirely correct!
• Thank you just one more question: How can we say in the first part that since $(a,b)$ can be written as a countable union of half-open sets this implies that the Borel algebra of open sets is also generated by half-open sets? Commented Oct 18, 2022 at 23:20
• You're going the other direction. We know that every borel set can be written in terms of the open sets $(a,b)$. So since we can write $(a,b)$ in terms of the half open sets $(p,q]$, this means we can also write every borel set in terms of the half opens. Indeed, just write it in terms of opens $(a,b)$, then write each $(a,b)$ as a union of half opens. You can think of this as showing a certain set $\mathcal{C}$ of vectors is spanning by writing every basis vector in terms of vectors in $\mathcal{C}$. Commented Oct 19, 2022 at 2:03