Suppose $(X,\mathbb{X})$ is a measure space and $f: X \to [0,\infty]$ is measurable. Is it true that $f$ when considered as a function $f: X \to \bar{\mathbb{R}} := [-\infty,\infty]$ is also measurable?
My reasoning is as follows. Suppose $f: X \to [0,\infty]$ is measurable and denote by $\mathbb{B}_{\bar{\mathbb{R}}}$ the Borel $\sigma$-algebra on the extended real line. Then, the $\sigma$-algebra on $[0,\infty]$ is the $\sigma$-algebra $\{A \in \mathbb{B}_{\bar{\mathbb{R}}}: A\subseteq [0,\infty]\}$. If $A \in \mathbb{B}_{\bar{\mathbb{R}}}$, we would have (when $f$ is considered an $\bar{\mathbb{R}}$-valued funcion): $$f^{-1}(A) = \{x \in X: f(x) \in A\} = \{x \in X: f(x) \in A \cap [0,\infty]\} = f^{-1}(A\cap [0,\infty])$$ which belongs to the $\sigma$-algebra $\mathbb{X}$.
Is my reasoning correct? As a consequence, if $f: X \to \bar{\mathbb{R}}$ is such that $f^{+} := \operatorname{max}\{f,0\}$ and $f^{-} := \operatorname{max}\{-f,0\}$ are both measurable functions, then $f$ and $|f|$ are both measurable because they are sums of measurable functions $f = f^{+}-f^{-}$ and $|f| = f^{+}+f^{-}$.
Your reasoning is correct, as long as you justify why $B_{[0, \infty]} = \{A \in B_{[-\infty, \infty]} : A \subseteq [0, \infty]\}$.
If $X$ is a topological space and $S$ is a subspace of $X$, it is not difficult to prove that $B_S = \{A \cap S : A \in B_X\}$. In fact, we have $\{A \cap S : A \in B_X\} \subset B_S$ since the inclusion map $i \colon S \to X$ is continuous, and we have $B_S \subset \{A \cap S : A \in B_X\}$ since any open set in $S$ is of the form $O \cap S$ where $O$ is open in $X$.
If we also know that $S \in B_X$, then $\{A \cap S : A \in B_X\} = \{A \in B_X : A \subseteq S\}$. Thus in this case $B_S = \{A \in B_X : A \subseteq S\}$.
Putting $X = [-\infty, \infty]$, $S = [0, \infty]$ gives what you want.