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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?

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I assume that $\bar{\mathcal{B}}=\mathcal{B}(\bar{\mathbb{R}})$. Let $(\Omega,\mathcal{F})$ be a measurable space. Let $u,v:\Omega\rightarrow\mathbb{\bar{\mathbb{R}}}$ be $\mathcal{F}/\bar{\mathcal{B}}$-measurable functions. Let $A=\{\omega\in\Omega\mid u(\omega)<v(\omega)\}$. We go to show that $A\in\mathcal{F}$. Observe that $$ A=\cup_{r\in\mathbb{Q}}\{\omega\mid u(\omega)<r\}\cap\{\omega\mid r<v(\omega)\}. $$ For each $r\in\mathbb{Q}$, $\{\omega\mid u(\omega)<r\}$, $\{\omega\mid r<v(\omega)\}$ are elements in $\mathcal{F}$. It follows that $A\in\mathcal{F}$ because the union $\cup_{r\in\mathbb{Q}}$ is a countable union.

Similarly, $[u>v]\in\mathcal{F}$. Finally, $[u=v] = \Omega\setminus([u<v]\cup[u>v])\in\mathcal{F}$.

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  • $\begingroup$ In this way, we can avoid explicitly considering arithmetic "subtraction". $\endgroup$ – Danny Pak-Keung Chan May 15 '19 at 22:06

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