
Re: Half Measure and Correction
Posted:
Mar 24, 2013 4:55 PM


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? 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.

