
Re: Half Measure and Correction
Posted:
Mar 25, 2013 9:07 AM


On Sun, 24 Mar 2013 13:55:12 0700 (PDT), Butch Malahide <fred.galvin@gmail.com> wrote:
>On Mar 24, 1:49 pm, David C. Ullrich <ullr...@math.okstate.edu> wrote: >> On Sun, 24 Mar 2013 10:37:46 0700 (PDT), Butch Malahide >> > <fred.gal...@gmail.com> wrote: >> >On Mar 24, 10:53 am, David C. Ullrich <ullr...@math.okstate.edu> >> >wrote: >> >> [. . .] >> >> So the more interesting version of the question, >> >> in any case less trivial, amounts to this: Is there >> >> a measurable set D such that >> >> >> 0 < m(D intersect I) < m(I) >> >> >> for every open interval I, >> >> >Didn't we just have that thread? >> >> Yes. >> >> >http://groups.google.com/group/sci.math/msg/0cfe35786f2279f0?hl=en >> >> >> and such that m(D intersect [0,1]) = 1/2 ? >> >> >OK, that's different. >> >> Precisely! heh. > >But not so very different, is it?
I was joking. Of course once we know that there exists D with 0 < m(D intersect I) < m(I) for every interval I it follows that there exists such a D with the second condition.
By any of at least three arguments:
1. The one I had in mind.
2. The one you give below.
3. By saying "Fine, now how in the world could it be that the value 1/2 is somehow excluded?"
>Let D be a measurable set such that >0 < m(D intersect I) < m(I) for every interval I. It will suffice to >find an interval I such that m(D intersect I)/m(I) = 1/2. Since m(D) > >0, there is an interval J such that m(D intersect J)/m(J) > 1/2; >likewise, since m{R\D) > 0, there is an interval H such that m(D >intersect H)/m(H) < 1/2. Since the function f(a,b) = m(D intersect >(a,b))/(ba) is continuous on the connected domain {(a,b): a < b}, >there is an interval I = (a,b) such that f(a,b) = 1/2.

