Timeline for Are measurable automorphism of a locally compact group topological automorphisms?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| May 8, 2011 at 19:29 | comment | added | Theo Buehler | Actually, there is a further major mistake: the reason for the existence of the paper seems to be a confusion two completely different meanings of "Baire sets" (the one Banach uses is the $\sigma$-algebra generated by the Borel sets and the meager sets) and the other one is the $\sigma$-algebra by the compact $G_{\delta}$'s. I give a reference to Banach's work in my answer here mathoverflow.net/questions/57616/… and François's answer to the same question points to Pettis' standard results Julien quotes. | |
| May 8, 2011 at 19:14 | history | edited | Matthew Daws | CC BY-SA 3.0 |
Clarification; mention measurability
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| May 8, 2011 at 19:06 | comment | added | Matthew Daws | @Julien: Thanks for the heads-up about the erratum! Yeah, I coming to the same conclusion (given the much better papers listed above). | |
| May 8, 2011 at 15:14 | comment | added | Julien Melleray | This paper by Neeb is actually not so nice - first, it seems to ignore all the literature on the subject (there are simpler proofs of that theorem of Banach, without the unnecessary assumptions on arcwise connectedness, and they were available before publication of his paper; actually, the result appears in textbooks, like Kechris' book); second, there is a major error in the part about representation theory (claiming that the operator norm topology and strong operator topologies have the same Borel sets, which is dead wrong). There seems to be an erratum of sorts for that paper (MR1747686 ). | |
| May 6, 2011 at 13:09 | comment | added | Gerald Edgar | The basic thing in the proof: if $A$ is a measurable set with postive measure, then $A A^{-1}$ contains a neighborhood of the identity. | |
| May 6, 2011 at 12:43 | comment | added | Matthew Daws | I should say that this refers to "Borel set" not "Completed Borel for Haar measure". The latter seems like a somewhat stronger condition, a priori. | |
| May 6, 2011 at 12:39 | history | answered | Matthew Daws | CC BY-SA 3.0 |