Timeline for Is "There exists an unbounded non-measurable set but no bounded non-measurable set" consistent with $\mathsf{ZF}$?
Current License: CC BY-SA 4.0
34 events
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| Jun 8, 2021 at 12:13 | history | edited | Clement Yung | CC BY-SA 4.0 |
added 40 characters in body
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| Jun 1, 2021 at 11:39 | vote | accept | Clement Yung | ||
| May 19, 2021 at 15:14 | comment | added | Emil Jeřábek | @ElliotGlazer Oh, I see! Thank you. | |
| May 19, 2021 at 14:08 | comment | added | Elliot Glazer | @EmilJeřábek Suppose $X$ is measurable against bounded intervals. Assume $Y$ has finite outer measure (otherwise trivial). Fix an open cover $C$ of $Y$ of measure less than $\lambda^*(Y)+\epsilon.$ Finitely many intervals $\{I_k\}_1^n$ in $C$ contain all but $\frac{\epsilon}{2}$ of the measure. For each $k \le n,$ choose covers $I_k \cap X \subset C_{1, k}$ and $I_k \setminus X \subset C_{2, k}$ such that $\lambda(C_{1,k})+\lambda(C_{2,k})<\lambda(I_k)+\frac{\epsilon}{2n}.$ The rest is clear. | |
| May 19, 2021 at 8:52 | comment | added | Emil Jeřábek | @ElliotGlazer Really? At the risk of sounding like a broken record: how do you prove that without countable choice? Note that this would answer question 2 negatively. | |
| May 19, 2021 at 5:01 | answer | added | Elliot Glazer | timeline score: 12 | |
| May 18, 2021 at 19:24 | comment | added | Elliot Glazer | @EmilJeřábek Whether you have the test sets be bounded intervals or arbitrary $Y \subset \mathbb{R},$ you get equivalent definitions. | |
| May 18, 2021 at 18:31 | answer | added | Elliot Glazer | timeline score: 2 | |
| May 18, 2021 at 17:36 | comment | added | Gerald Edgar | So .. We have a fine definition (see @EmilJeřábek) for "Measurable set" that makes sense in ZF. The fact that the collection of such sets is a sigma-algebra may require AC: but that is not required to answer the question in the title. | |
| May 18, 2021 at 17:15 | comment | added | Asaf Karagila♦ | @Peter: Not in this case. This case is more akin to saying that we are changing the definition of "compact" to mean "indiscrete" because then we can prove in ZF Tychonoff's theorem. | |
| May 18, 2021 at 17:14 | comment | added | Asaf Karagila♦ | @Emil: Exactly. You don't. Measures in the context of measure theory are defined to be $\sigma$-additive. If they are not, then they are not called "measures" but "finitely additive measures". This can be excused when working with something like coded Borel sets, where we have a semblance of $\sigma$-additivity in that if we have a sequence of codes, then the sets generated is also coded. But this definition is awful to work with, and there's a reason why we make it abundantly clear that this is "coded Borel sets" and not "Borel sets". | |
| May 18, 2021 at 17:10 | comment | added | Peter LeFanu Lumsdaine | @AsafKaragila: “Why that definition?” — that kind of question doesn’t usually have a more definitive answer than “because it’s the/a definition that works well for re-developing the theory in the weaker foundation”; and the definition Emil gave is at least somewhat established in the literature on measurability in choiceless settings, suggesting it’s generally accepted as a good one. If you want to argue for a different definition, the burden of proof is more on you than on Emil. (Of course, best would be if OP could tell us which definition they have in mind.) | |
| May 18, 2021 at 14:59 | comment | added | Emil Jeřábek | What exactly do you mean by the Borel measure, and how do you construct it other than taking something like the definition I gave above, restricted to Borel sets? | |
| May 18, 2021 at 14:50 | comment | added | Asaf Karagila♦ | @Emil: Because why is that that definition, or not the one where you complete the Borel measure? You can define a lot of things in a lot of ways, you need to justify why your choice of definition is better or worse. | |
| May 18, 2021 at 14:32 | comment | added | Emil Jeřábek | The Lebesgue measure may be weird, but why would it not be well defined? I just defined it. | |
| May 18, 2021 at 14:30 | comment | added | Asaf Karagila♦ | @Emil: If $\Bbb R$ is the countable union of countable sets, then the Lebesgue measure is not well-defined since every set is Borel. If you want to talk about coded sets, that's your problem and I will have no part of this discussion, since that's purposefully kneecapping yourself (I'm not saying it's not interesting, it's just too hard to be actually useful). | |
| May 18, 2021 at 14:28 | comment | added | Emil Jeřábek | You need countable choice to show that a measurable set is a difference of a $G_\delta$ set and a null set. And how do you show that $f$ preserves null stes? This is still the same problem. | |
| May 18, 2021 at 14:25 | comment | added | Asaf Karagila♦ | I think the point is that with $f(x)=1/x$, null sets are mapped to null sets. And $G_\delta$ sets are mapped to $G_\delta$ set. If you are Lebesgue measurable iff you are almost the same as a $G_\delta$ set, that should just go through. | |
| May 18, 2021 at 14:20 | comment | added | Gerald Edgar | Taking the @EmilJeřábek definition. Suppose $X$ is an unbounded nonmeasurable set. Then there is a bounded interval $Y$ such that $\lambda^*(Y) \ne \lambda^*(Y\cap X)+\lambda^*(Y\smallsetminus X)$. Therefore, the bounded set $Y\cap X)$ is also nonmeasurable. In fact, I believe Lebesgue took this definition. | |
| May 18, 2021 at 14:20 | comment | added | Emil Jeřábek | Hmm. Though if I take only bounded intervals as test sets as I just wrote, it should actually work. I was originally thinking about arbitrary test sets $Y$. So perhaps this part of the definition is a crucial point that needs to be clarified. | |
| May 18, 2021 at 14:12 | comment | added | Emil Jeřábek | @GeraldEdgar I believe the standard definition of Lebesgue measure in this context is as follows. (1) Define the measure of open sets: any such is a countable disjoint union of open intervals (possibly infinite), take the sum of the lengths of the intervals. (2) Define the outer measure $\lambda^*(X)=\inf\{\lambda(U):U\supseteq X\text{ open}\}$. (3) Define $X$ to be measurable if $\lambda^*(Y)=\lambda^*(Y\cap X)+\lambda^*(Y\smallsetminus X)$ for all bounded intervals $Y$ (or possibly a larger collection of $Y$, I’m not sure what is most common in this point). | |
| May 18, 2021 at 14:07 | comment | added | Asaf Karagila♦ | @GeraldEdgar: I'm claiming exactly the opposite. I think that Emil and Wojowu question this theorem's validity, though. | |
| May 18, 2021 at 14:06 | comment | added | Emil Jeřábek | @Asaf This is the point: even with AC, if the function is not Lipschitz, it does not preserve the measure of even open sets in a predictable way, as it may map a bounded set to something of infinite outer measure. | |
| May 18, 2021 at 14:04 | comment | added | Gerald Edgar | There is a theorem (or, in some developments, a definition) ... A set $A$ is measurable if and only if $A \cap [a,b]$ is measurable for every $a<b$. Are you claiming this is cannot be proved in ZF? Perhaps improve the question by including the definition of "measurable". | |
| May 18, 2021 at 14:02 | comment | added | Asaf Karagila♦ | @Emil: Well, it's enough to prove that the function "preserves" the measure of open sets and compact sets, or rather that it changes those measures in a "predictable way". For open sets that's easy. I think it should be the same for compact sets as well. | |
| May 18, 2021 at 14:01 | comment | added | Emil Jeřábek | @AsafKaragila The issue is what I wrote in my comment: how do you prove that if $Y\subseteq(0,1)$ is measurable, then $\{1/x:x\in Y\}$ is measurable, other than by splitting $Y$ to infinitely many pieces bounded away from $0$? | |
| May 18, 2021 at 13:57 | comment | added | Asaf Karagila♦ | Emil, Wojowu, what's the issue with using $1/x$ as a function from $(1,\infty)$ to $(0,1)$ for this? It's enough to consider non-measurable subsets of $(1,\infty)$, after all. | |
| May 18, 2021 at 13:51 | comment | added | Emil Jeřábek | As for question 1, it is consistent with ZF that $\mathbb R$ (and therefore every subset thereof) is a countable union of countable (and therefore null) sets. This certainly implies that Lebesgue measure is not $\sigma$-additive; I’m not sure though whether it also implies the existence of a non-measurable set (i.e., $\neg M_\omega$). | |
| May 18, 2021 at 13:47 | comment | added | Emil Jeřábek | @Wojowu That’s in fact a comment I originally posted on the linked answer, but on further reflection, I don’t think it works without AC. If you cover $f(X)$ by an open set $U$ whose measure is close to the measure of $f(X)$, the preimage of $U$ may still be much larger than $X$ (unless $f$ is bi-Lipschitz, in which case it can’t have bounded image). Basically, you still need to split the set to countably many pieces and approximate each piece by open sets separately, and it requires countable choice to collect the pieces together. | |
| May 18, 2021 at 13:05 | comment | added | Wojowu | For 2, let $f:\mathbb R\to(0,1)$ be a diffeomorphism (= smooth with smooth inverse). I suspect image of any non-measurable set by $f$ is non-measurable as well. This is true under AC but I'm not sure if the proof works in ZF. This would show existence of non-measurable set implies existence of bounded such. | |
| May 18, 2021 at 13:03 | comment | added | Clement Yung | @Wojowu corrected, thank you. | |
| May 18, 2021 at 13:02 | history | edited | Clement Yung | CC BY-SA 4.0 |
added 8 characters in body; edited title
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| May 18, 2021 at 13:02 | comment | added | Wojowu | Did you mean "non-measurable" in your title? | |
| May 18, 2021 at 12:46 | history | asked | Clement Yung | CC BY-SA 4.0 |