Timeline for "Lebesgue-measurable" cardinals and real-closed fields
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 13, 2015 at 2:14 | comment | added | Mike Battaglia | Joel - I pretty much reworded the entire thing. Hope this version makes more sense as to what I was getting at. | |
| Sep 12, 2015 at 23:50 | comment | added | Joel David Hamkins | I would suggest that you ignore the close votes and edit your question, since it has now become a little clearer what you want. (But I would add that the 2-valued measure IS a probability measure, so saying that "it also admits a real-valued probability measure" is not adding much.) | |
| Sep 12, 2015 at 23:47 | comment | added | Mike Battaglia | I think I should probably reword the question more precisely. Since everyone's voting to close, should I just close it at this point, or edit it? | |
| Sep 12, 2015 at 23:43 | comment | added | Mike Battaglia | The point of my question is to note that every measurable cardinal is also real-measurable. That is, in addition to the usual 2-valued measure, it also admits a real-valued probability measure. I am curious if there exists any such choice of real-valued measure and bijection onto the hyperreal unit interval which is translation invariant. Obviously the {0,1}-valued measure and the one I linked to above won't do the trick, but I'm asking if there exists some other measure which does. | |
| Sep 12, 2015 at 23:39 | comment | added | Joel David Hamkins | Such a measure has the property, however, that there are sets of measure $\frac12$, such that all subsets of it have measure either $0$ or $\frac12$, and this will not be true of any translation invariant measure of the kind you seek. | |
| Sep 12, 2015 at 23:37 | comment | added | Joel David Hamkins | Yes, that is correct, using a 2-valued measure you can construct such measures. For example, give yourself $\omega$ many sets of size $\kappa$, and decree that the $n^{th}$ one has value $2^{-n}$, using the original 2-valued measure to measure subsets; now measure subsets of the disjoint union by adding up. | |
| Sep 12, 2015 at 23:33 | comment | added | Mike Battaglia | It says in this reference that "If κ is measurable, then κ has a measure that takes on every value in [0, 1]." Is this not correct, or am I misunderstanding you? | |
| Sep 12, 2015 at 23:27 | comment | added | Joel David Hamkins | Yes, every measurable cardinal is real-valued measurable, but in general the measure has only two values: $0$ and $1$. | |
| Sep 12, 2015 at 23:22 | comment | added | Mike Battaglia | My understanding is that every measurable cardinal is also real-valued measurable, which is what I was driving at. Is that wrong? | |
| Sep 12, 2015 at 23:21 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |