I'm self studying Measure Theory by Bartle's book and I have a doubt in the proof of a lemma. Before I say what is the lemma and what is my doubt, I would like to say that the book is available here if anyone wants to see and I would like to introduce some notation according to the Bartle's book that I believe it's necessary to understand the proof of the lemma.
$\textbf{Notation:}$
(i) $\overline{\mathbb{R}} := \mathbb{R} \cup \{ -\infty, +\infty \} $ is the extended real number system and the function with codomain $\overline{\mathbb{R}}$ are said to be extended real valued functions.
(ii) $\textbf{X}$ in bold is the $\sigma-$algebra of a set $X$.
(iii) The collection of all extended real valued $\textbf{X}-$measurable function on $X$ is denoted by $M(X, \textbf{X})$.
$\textbf{Lemma 2.8}$ An extended real valued function $f$ is measurable if and only if the sets
$$A := \{ x \in X \ ; \ f(x) = +\infty \} \hspace{0.5cm} \text{and} \hspace{0.5cm} B := \{ x \in X \ ; \ f(x) = - \infty \}$$ belong to $\textbf{X}$ and the real-valued function $f_1$ defined by
$$f_1(x) := f(x), x \notin A \cup B$$
$$f_1(x) := 0, x \in A \cup B$$
is measurable.
$\textbf{Proof:}$
If $f \in M(X, \textbf{X})$, it has already been noted that $A$ and $B$ belong to $\textbf{X}$. Let $\alpha \in \mathbb{R}$ and $\alpha \geq 0$, then
$$\{ x \in X \ ; \ f_1(x) > \alpha \} = \{ x \in X \ ; \ f(x) > \alpha \} \ \backslash \ A$$
If $\alpha < 0$, then
$$\{ x \in X \ ; \ f_1(x) > \alpha \} = \{ x \in X \ ; \ f(x) > \alpha \} \cup B$$
The proof continues, but my doubt emerges here. I would like to know why $\{ x \in X \ ; \ f_1(x) > \alpha \} = \{ x \in X \ ; \ f(x) > \alpha \} \cup B$? I don't know why $\{ x \in X \ ; \ f_1(x) > \alpha \}$ have points of $B$ since $\alpha$ is fixed.
Thanks in advance!
This equality is valid for $\alpha <0$, so \begin{align}\{ x \in X \ ; \ f_1(x) > \alpha \}&=\{ x \in X \ ; \ f(x) > \alpha \}\cup\{ x \in X \ ; \ f_1(x) =0 \}\\[1ex] &=\{ x \in X \ ; \ f(x) > \alpha \}\cup (A\cup B)\\[1ex] &=\{ x \in X \ ; \ f(x) > \alpha \}\cup B \end{align} since $\;A\subset \{ x \in X \ ; \ f(x) > \alpha \}$.