Timeline for Subset of the reals with zero inner measure and "full" outer measure in $\mathsf{ZF}+\mathsf{DC}$
Current License: CC BY-SA 4.0
5 events
when toggle format | what | by | license | comment | |
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Mar 24 at 17:42 | vote | accept | David Fernandez-Breton | ||
Feb 29 at 17:25 | comment | added | Andreas Blass | I hope that the modification in my previous comment isn't another bazooka. It might become simpler if one first observes that $A$ (resp. $B$) can be taken to be $F_\sigma$ (resp. $G_\delta$), so $B-A$ is $G_\delta$. Also, it might help to observe that the measure on $B-A$ (extended by $0$ to the rest of $[0,1]$) is absolutely continuous wwith respect to Lebesgue measure. | |
Feb 29 at 17:19 | comment | added | Andreas Blass | @DavidFernandez-Breton I agree that the isomorphism to thhe standard $[0,1]$ is like a bazooka and should rather be used to destroy a whole ant hill. The particular ant here should allow a simpler attack. I'd suggest defining a function $f:[0,1]\to[0,1]$ by setting $f(x)=\mu([0,x]\cap B-A)$, where $\mu$ is the scaled mesure as in my answer. The idea is that the restriction of $f$ to a slight modification of $B-A$ should give the desired isomorphism. The modification referfs to worries about the map being literally a bijection; $f$ itself should suffice for a.e. bisection. | |
Feb 29 at 17:06 | comment | added | David Fernandez-Breton | Perhaps I should have asked explicitly whether there's a more "elementary" proof of this fact. Using the fact that every standard Borel space has a measure-preserving Borel isomorphism to $[0,1]$ seems akin to "using a bazooka to kill an ant". I had suspected that, this being in the one-dimensional space $[0,1]$ where the definition of Lebesgue measure has a clear intuitive underpinning in geometry, there should be some clever but elementary argument somewhere. Anyway, if after some time it seems no such argument is available, I'll accept this answer as the best one... | |
Feb 29 at 16:50 | history | answered | Andreas Blass | CC BY-SA 4.0 |