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?


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.