try to understand proof of equivalence of borel-sigma

Question

When we prove $\sigma((a,b)) = \sigma((a,b])$. All we have to show is that $(a,b]$ can be represented by a countable intersection of $(a,b+1/n)$. And the similar job is done for the other way.

I am not sure why this is enough to show that two sigma algebra is equivalent. I hope someone can explain, and my explanation is clear enough. Thanks,

Answer 1

When you show that $(a,b]$ is the countable intersection of $(a, b + \frac{1}{n})$, you are showing that $\{(a,b]:a, b \in \mathbb{R}\} \subseteq \sigma((a,b))$. Since $\sigma((a,b])$ is the "smallest" $\sigma$-algebra that contains $\{(a,b]:a, b \in \mathbb{R}\}$, we have then that $\sigma((a,b]) \subseteq \sigma((a,b))$.

If you can show that for all $a,b \in \mathbb{R}$, $(a,b) \in \sigma((a,b])$, then you can conclude that $\sigma((a,b)) \subseteq \sigma((a,b])$. Then the double inclusion will imply that $\sigma((a,b)) =\sigma((a,b])$.

Does that make sense?

Answer 2

That $\bigcap_{n}(a,b+n^{-1}) = (a,b]$ shows that $(a,b] \in \sigma (\{ (a,b) \mid a,b \in \mathbb{R}\})$. That $(a,b) = \bigcup_{n}(a,b-n^{-1}]$ shows that $(a,b) \in \sigma ( \{ (a,b] \mid a,b \in \mathbb{R} \})$.