On Borel sets of the extended reals

[psie]

TL;DR

I am reading Folland's text on real analysis. He defines the Borel $\sigma$-algebra on the extended reals and says this coincides with the usual definition of the Borel $\sigma$-algebra being generated by open sets. I don't see how they coincide.

On page 45 in Folland's text on real analysis, he writes that we define Borel sets in $\overline{\mathbb R}$ by $\mathcal B_{\overline{\mathbb R}}=\{E\subset \overline{\mathbb R}: E\cap\mathbb R\in \mathcal B_{\mathbb R}\}$. Then he remarks that this coincides with the usual definition of the Borel $\sigma$-algebra if we make $\overline{\mathbb R}$ into a metric space with metric $d(x, y) = |\arctan x - \arctan y|$.

I don't see how the two coincide and would be grateful if someone could explain in more detail. I see how $d$ is a metric on $\overline{\mathbb R}$, where e.g. $\arctan(\infty)$ should evaluate to $\pi/2$, but I don't see how $\mathcal B_{\overline{\mathbb R}}=\{E\subset \overline{\mathbb R}: E\cap\mathbb R\in \mathcal B_{\mathbb R}\}$ coincides with the usual definition of $\mathcal B_{\overline{\mathbb R}}$ being generated by the open sets in ${\overline{\mathbb R}}$. In other words, how to show $$\sigma(\{\text{open sets in }\overline{\mathbb R}\})=\{E\subset \overline{\mathbb R}: E\cap\mathbb R\in \mathcal B_{\mathbb R}\}?$$

 

[psie]

Ok, here's an attempt at a solution:

$\subset$: If $E$ is open in $\overline{\mathbb R}$ then by definition of subspace topology, $E\cap\mathbb R$ is an open set in $\mathbb R$, thus belongs to $\mathcal B_\mathbb R$. We've shown $\{\text{open sets in }\overline{\mathbb R}\}\subset \{E\subset \overline{\mathbb R}: E\cap\mathbb R\in \mathcal B_{\mathbb R}\}$, hence also the $\sigma$-algebra generated by $\{\text{open sets in }\overline{\mathbb R}\}$ is a subset of $\{E\subset \overline{\mathbb R}: E\cap\mathbb R\in \mathcal B_{\mathbb R}\}$.

$\supset$: Let $E\in \{E\subset \overline{\mathbb R}: E\cap\mathbb R\in \mathcal B_{\mathbb R}\}$. Then $E = (E\cap \mathbb{R})\cup (E\cap \{\infty, -\infty\})$ and $E\cap \mathbb{R}$ is a Borel set of $\mathbb R$. Now, since $\mathbb R\subset \overline{\mathbb R}$, the Borel $\sigma$-algebra on $\mathbb R$ is the restriction of the Borel $\sigma$-algebra of $\overline{\mathbb R}$ to $\mathbb R$, i.e. $\mathcal B_\mathbb R=\{A\cap \mathbb R:A\in \mathcal B_{\overline{\mathbb R}}\}$. Because $E\cap \mathbb{R}\in \mathcal B_\mathbb R$, there's a Borel set $E_0$ in $\mathcal B_{\overline{\mathbb R}}$ such that $E\cap \mathbb{R}=E_0\cap\mathbb R$, and since $\mathbb R\in \mathcal B_{\overline{\mathbb R}}$ ($\mathbb R$ is open in $\overline{\mathbb R}$, hence Borel in $\overline{\mathbb R}$), $E\cap\mathbb R$ is a Borel set in $\overline{\mathbb R}$. $E\cap \{\infty, -\infty\}$ also belongs to $\mathcal B_{\overline{\mathbb R}}$, since it is finite (it is an intersection of singleton, i.e. closed sets, and hence Borel).

 

[Gavran]

By usual definition $E\cap\mathbb{R}\in\mathcal{B}_\mathbb{R}$ means that $E\cap\mathbb{R}$ is a Borel set formed from open sets in $\mathbb{R}$ through the operations of countable union, countable intersection (countable intersection can be got from countable union and relative complement (De Morgan law) and it can be skipped) and relative complement. Every open set in $\mathbb{R}$ is in $\overline{\mathbb{R}}$ too because of $\mathbb{R}\subset\overline{\mathbb{R}}$. This means that $E\cap\mathbb{R}$ is a set formed from open sets (the same open sets as in the first sentence, so open rays are not included ) in $\overline{\mathbb{R}}$ through the operations of countable union, countable intersection and relative complement. After including open rays into the open sets it can be said that $E$ is a set formed from the open sets in $\overline{\mathbb{R}}$ through the operations of countable union, countable intersection and relative complement.
Every set $E$ can be equal to $E\cap\mathbb{R}$ or can be formed by countable union of $E\cap\mathbb{R}$ and open rays.