Characterization of Borel $\sigma$-Algbebra of extended real line $\overline{\mathbb{R}}$ Consider the extended real line $\overline{\mathbb{R}}$ and define a topology on it as follows:

Definition: $U$ is open in $\overline{\mathbb{R}}$ if $U \cap \mathbb{R}$
is open in $\mathbb{R}$ and in case $\infty \in U$, then $(a,\infty]
\subset U$ for some $a \in \mathbb{R}$ and similarly for $-\infty$.

It is easy to show that this is indeed a topology (and it is actually the order topology). Hence, we can define the Borel $\sigma$-Algebra of the extended real line $\overline{\mathbb{R}}$ as the $\sigma$-Algebra generated by the open sets of $\overline{\mathbb{R}}$ as usual.
Now my lecture notes state the following and proof:

Theorem: $\mathcal{B}(\overline{\mathbb{R}})=\{B \cup A : B \in
\mathcal{B}(\mathbb{R}), A \subset \{-\infty,\infty\}\}$ and the set
$\{(a,\infty]: a \in \mathbb{R}\}$ generates
$\mathcal{B}(\overline{\mathbb{R}})$.
Proof: Since open sets in $\mathbb{R}$ are open in
$\overline{\mathbb{R}}$ as well, it follows that
$\mathcal{B}(\mathbb{R}) \subset \mathcal{B}(\overline{\mathbb{R}})$.
Since open sets in $\overline{\mathbb{R}}$ have open intersection with
$\mathbb{R}$, we know that for a representation $B \cup A \in
\mathcal{B}(\overline{\mathbb{R}})$ with $B \in \mathbb{R}$, we must
have $B \cap \mathbb{R} \in \mathcal{B}(\mathbb{R})$. Hence, it only
remains to show that $\{-\infty\},\{\infty\} \in
\mathcal{B}(\overline{\mathbb{R}})$. But these sets are generated by
open sets of $\overline{\mathbb{R}}$, e.g. $\{\infty\}= \bigcap_{k \in
\mathbb{N}} (k,\infty]$.
...

I will omit the second half of the proof about the generators.
1.) Now I don't really understand what the strategy of the proof of the first part is. Usually, I would just try to prove both inclusions. Is this what the lecturer does in his proof?
One inclusion is clear from the proof:
For $\mathcal{B}(\overline{\mathbb{R}}) \supset \{B \cup A : B \in \mathcal{B}(\mathbb{R}), A \subset \{-\infty,\infty\}\}$ we only have to show that $\mathcal{B}(\mathbb{R}) \subset \mathcal{B}(\overline{\mathbb{R}})$ and $A \subset \mathcal{B}( \overline{\mathbb{R}})$ since $\mathcal{B}(\mathbb{R})$ is a $\sigma$-Algebras.This follows from the first and the last two sentences of the proof. For the other inclusion we could try to prove that $\{B \cup A : B \in
\mathcal{B}(\mathbb{R}), A \subset \{-\infty,\infty\}\}$ is a $\sigma$-Algebra, but it seems the proof proceeds differently.
2.) I don't fully understand how we can deduce the bold part (second sentence of the proof) and also what we are trying to achieve with this step. Could someone please explain?
I can see that any element $\overline{B} \in \mathcal{B}( \overline{\mathbb{R}})$ can be written as $B \cup A$ with $B \in \mathbb{R}$ by simply letting $B=\overline{B} \cap \mathbb{R}$. This would also imply that $A \subset \{-\infty,\infty\}\}$. If $\overline{B}$ isopen, then this follows by definition of the topology on $\overline{\mathbb{R}}$, but not all sets in $\mathcal{B}(\overline{\mathbb{R}})$ are open in $\overline{\mathbb{R}}$.
Could anybody help me out, please?
Thanks a lot!
 A: It is clear that the Borel sets of $\mathbf{R}$ are contained in those of the extended real line $\bar{\mathbf{R}}.$ Also, since $\{\infty\} = \bigcap\limits_{k \in \mathbf{N}} (k, \infty],$ it follows that $\{\infty\}$ is a Borel set in the extended real line, and mutatis mutandis $\{-\infty\}$ is also a Borel set there. Therefore, $\mathscr{X} = \{B \cup A \mid B \in \mathscr{B}_\mathbf{R}, A \subset \{-\infty, \infty\}\} \subset \mathscr{B}_{\bar{\mathbf{R}}}.$ Reciprocally, suppose $O$ is open set in the extended real line and let $B = O \cap \mathbf{R},$ which is therefore open in $\mathbf{R}.$ Clearly $A = O \triangle B \subset \{-\infty, \infty\}$ and therefore $O = B \cup A$ has the desired form: $O \in \mathscr{X}.$ In other words, every open set in the extended real line belongs to $\mathscr{X},$ so to conclude suffices to show that $\mathscr{X}$ is a sigma-field. This is very easy to prove albeit perhaps somewhat tedious. Let me show only that $\mathscr{X}$ is closed under complements. Divide into three cases:

*

*$A$ is empty. The complement of $B$ in $\bar{\mathbf{R}}$ is the union of the complement of $B$ in $\mathbf{R}$ and of $\{-\infty, \infty\},$ which clearly belongs to $\mathscr{X}.$

*$A$ has one point. Assume $A = \{-\infty\},$ the other case being handled similarly. Then the complement of $B \cup A$ in $\bar{\mathbf{R}}$ is equal to the complement of $B$ in $\mathbf{R}$ union $\{\infty\},$ which belongs to $\mathscr{X}.$

*$A = \{-\infty, \infty\}.$ Then the complement in $\bar{\mathbf{R}}$ of $B \cup A$ is the complement in $\mathbf{R}$ of $B$ and this clearly belongs to $\mathscr{X}.$ QED
