Timeline for Is every probability space a factor space of the Haar Measure on some group?
Current License: CC BY-SA 2.5
16 events
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| Sep 16, 2010 at 16:25 | vote | accept | John Wiltshire-Gordon | ||
| Sep 16, 2010 at 16:25 | history | bounty ended | John Wiltshire-Gordon | ||
| Sep 11, 2010 at 11:52 | comment | added | George Lowther | or, what I had in mind, (2) Take the space $2^{\mathbb{R}}$ of zero-one valued functions on the reals, with the 0-1 valued measure which simply assigns probability 1 to any set containing the function with constant value 1. Let P be the restriction of this to the (highly nonmeasurable but full outer measure) subset of $2^{\mathbb{R}}$ consisting of those functions which are zero everywhere outside of a countable set. This space consists of functions which are zero almost everywhere, but equal to 1 almost surely at each point. | |
| Sep 11, 2010 at 11:50 | comment | added | George Lowther | For examples where it might not be possible to find a measure preserving map G->P you could consider the following nasty cases. (1) As suggested by Gerald Elgar, take P to be a highly nonmeasurable subset of [0,1] with outer measure 1 under the Lebesgue measure. Maybe a Vitali set or, supposing the CH fails, a set with cardinality strictly between $\aleph_0$ and $2^\aleph_0$. | |
| Sep 11, 2010 at 11:37 | comment | added | George Lowther | S has full outer measure, so the restriction to S gives an isomorphism $L^2(G)\cong L^2(S)$. In fact, if you are only interested in $L^n(P)$ then this is determined by the measure algebra, and Gerald Elgar's suggestion of using Maharam's structure theorem also works. | |
| Sep 11, 2010 at 7:33 | comment | added | John Wiltshire-Gordon | Good point. I bet a little point-set topology will fix this up, at least in the special case constructed by George Lowther below. | |
| Sep 11, 2010 at 6:16 | comment | added | LSpice | Why does any continuous function on $S$ extend (continuously) at all? Or does one know enough about $S$ to know that the space of continuously extendable functions is still dense in $L^2(S)$? | |
| Sep 11, 2010 at 5:49 | history | edited | John Wiltshire-Gordon | CC BY-SA 2.5 |
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| Sep 11, 2010 at 0:48 | answer | added | Gerald Edgar | timeline score: 3 | |
| Sep 10, 2010 at 23:42 | answer | added | George Lowther | timeline score: 9 | |
| Sep 10, 2010 at 22:08 | answer | added | Ori Gurel-Gurevich | timeline score: 2 | |
| Sep 10, 2010 at 17:43 | history | bounty started | John Wiltshire-Gordon | ||
| Sep 8, 2010 at 23:38 | history | edited | John Wiltshire-Gordon |
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| Sep 7, 2010 at 23:45 | history | edited | John Wiltshire-Gordon | CC BY-SA 2.5 |
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| Sep 7, 2010 at 19:46 | answer | added | user6096 | timeline score: 5 | |
| Sep 7, 2010 at 18:49 | history | asked | John Wiltshire-Gordon | CC BY-SA 2.5 |