Let $f, g: X \to [-\infty, \infty]$ be measurable functions. Is it true that $f - g$ (where it's defined) is measurable? Let $f, g: X \to [-\infty, \infty]$ be measurable functions. Let $X'$ denote the set of all $x$ such that $f(x), g(x) \notin \{-\infty, \infty\}$. Then $X'$ is a measurable set.
Is it true that $h: X' \to \Bbb{R}$ defined by $h(x) = f(x) - g(x)$ is measurable? I think this should be obvious, but I'm having a hard time seeing it. I know if the codomain were $\Bbb{R}$ and $\Bbb{C}$, the sum and difference are measurable (but in that case, we don't need to restrict $X$).
Any help appreciated.

Yes, it is true.
$f-g$ is not defined in the set $B=\Big(\{f=\infty\}\cap\{g=\infty\}\Big)\cup\Big(\{f=-\infty\}\cap\{g=-\infty\}\Big)$. This is measurable set, for example
$$
\{f=\infty\}=\bigcup_{n\in\mathbb{N}}\{f>n\}
$$
Thus, for $a\in\mathbb{R}$,
$$
\begin{align}
\{f-g<a\}=(X\setminus B) \cap\bigcup_{q\in\mathbb{Q}}\{f<q\}\cap\{q-a<g\}\tag{1}\label{one}
\end{align}
$$
which is measurable subset each set $\{f-q\}$, $\{q-a<g\}$ is measurable ($f$ and $g$ are real-extended measurable functions), and the union in $\eqref{one}$ is over a countable set.
