Timeline for Models with fixed cardinality of non-Lebesgue measurable sets
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 3, 2022 at 3:24 | comment | added | Noah Schweber | You can make this much simpler with symmetric differences: if $X$ is a subset of the Cantor set (hence null) then $X\triangle A$ is measurable iff $A$ is. | |
| May 18, 2021 at 7:16 | vote | accept | Clement Yung | ||
| May 15, 2021 at 9:18 | comment | added | Emil Jeřábek | Thank you. Note that you don’t actually need $C$ measurable: $A\cup C$ is non-measurable for every subset $C\subseteq(-\infty,0)$. | |
| May 15, 2021 at 8:31 | history | edited | Clement Yung | CC BY-SA 4.0 |
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| May 15, 2021 at 8:18 | comment | added | Clement Yung | Thank you all for the comments. With that in mind, is it consistent with $\mathsf{ZF}$ that there exists a non-measurable set but all bounded sets are measurable? | |
| May 15, 2021 at 8:17 | history | edited | Clement Yung | CC BY-SA 4.0 |
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| May 15, 2021 at 8:11 | comment | added | Emil Jeřábek | Wojowu’s objection is warranted, however, countable union is not needed for the argument. ZF certainly proves that the union of two measurable sets is measurable, thus if $X$ is non-measurable, then $X\cap[0,+\infty)$ or $X\cap(-\infty,0]$ is non-measurable, and this gives plenty of room to make the argument work. | |
| May 15, 2021 at 6:57 | comment | added | Wojowu | I'm not sure that ZF proves that countable union of (Lebesgue-)measurable sets is measurable. | |
| May 15, 2021 at 6:46 | history | answered | Clement Yung | CC BY-SA 4.0 |