In conjuction with this previous post of mine, I'm starting to think Rene Schilling dropped the ball in chapter 8 of his book on measure theory.
Corollary 8.12 reads:
If $u, v$ are $\mathcal{A}/\bar{\mathcal{B}}$ measurable functions, then $$\{u<v\}, \;\; \{u\leqslant v\}, \;\; \{u = v\}, \;\; \{u \neq v\} \in \mathcal{A}$$
This is presumably a corollary of the preceding results in the chapter (Schilling offers no proof). I would almost agree that this is a corollary, because if the function $u-v$ is well defined, then it is also $\mathcal{A}/\bar{\mathcal{B}}$ measurable (by Corollary 8.10), and thus the sets above are equivalent to the sets
$$\{u-v<0\}, \;\; \{u-v\leqslant 0\}, \;\; \{u -v = 0\}, \;\; \{u -v \neq 0\},$$
which are $\mathcal{A}/\bar{\mathcal{B}}$ measurable by Lemma 8.1 (actually, by an extension of Lemma 8.1 to extended real–valued, $\mathcal{A}/\mathcal{\bar{B}}$-measureable functions).
But the if in the preceding paragraph seems important. If $u$ and $v$ both take values of $+\infty$ or $-\infty$ for the same $x$, then $u-v$ is not well defined, and the above reasoning doesn't work. Yet for these $x$, the expressions of the form $\{u<v\},...$ are still well defined, so we can't just exclude these cases.
How can we resolve this apparently incomplete gap in the above would-be proof? Or is there some other way to prove the corollary that gets around this issue entirely?