Timeline for Questions about Maharam's classification theorem
Current License: CC BY-SA 4.0
8 events
when toggle format | what | by | license | comment | |
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Oct 14, 2021 at 0:34 | comment | added | Dmitri Pavlov | @TaQ: Yes, exactly. | |
Oct 13, 2021 at 21:32 | comment | added | TaQ | Thank you for the clarification. So the idea is that we have between some subsets of full measure a bijection that preserves the measures of sets in some sets generating the $\sigma$-algebras whence it follows that the function preserves measures of all measurable subsets. | |
Oct 13, 2021 at 0:09 | comment | added | Dmitri Pavlov | @TaQ: Once you remove 1 from [0,1] and sequences $P$ ending in infinitely many 1's from $\{0,1\}^N$, sending a real number r∈[0,1] to its binary expansion establishes a measure-preserving bijection from [0,1) to $\{0,1\}^N∖P$, so in particular, set of measure 0 are the same. To see that the bijection is measure preserving, it suffices to observe that the Borel algebra is generated by sets of sequences with a fixed binary digit in some position $k$. In [0,1], this set corresponds to the union of $2^{k-1}$ half-open intervals of length $2^{-k}$, so the total length is $1/2$ in both cases. | |
Oct 12, 2021 at 18:25 | comment | added | TaQ | "the description of sets of measure $0$" — that is? Can you be more specific? | |
Oct 12, 2021 at 17:05 | comment | added | Dmitri Pavlov | @TaQ: An arbitrary continuous map does not preserve or reflect sets of measure 0. Rather, the specific map that is constructed in the answer does preserve and reflect sets of measure 0. This follows from the description of sets of measure 0 in [0,1] and in $\{0,1\}^N$, for example. | |
Oct 12, 2021 at 15:46 | comment | added | TaQ | "spaces [...] are all isomorphic as measurable spaces, including their sets of measure $0$, [...] isomorphism $[0,1]^2\to[0,1]$ is given by [...]" Do you have a simple proof of this, or is this just a consequence of something given in you article? For example, the map $[0,1]\to[0,1]$ given by $t\mapsto\frac12(ct+t)$ where $c$ is a "Cantor staircase" is even a homeomorphism but it does not map null sets to null sets. | |
Oct 10, 2021 at 15:35 | vote | accept | Ken.Wong | ||
Oct 10, 2021 at 15:25 | history | answered | Dmitri Pavlov | CC BY-SA 4.0 |