JT
Posts:
1,448
Registered:
4/7/12


Re: Half Measure and Correction
Posted:
Mar 25, 2013 5:03 PM


On 25 mar, 14:07, David C. Ullrich <ullr...@math.okstate.edu> wrote: > On Sun, 24 Mar 2013 13:55:12 0700 (PDT), Butch Malahide > > > > > > > > > > <fred.gal...@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.
You are all of course free to add and subtract null quantities to eachother (and even try add them to numbers), even multiply then but when you start to divide them and raise numbers into power of zero it start to raise my suspicion something really went terribly wrong along the way.

