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Topic: Half Measure
Replies: 14   Last Post: Mar 25, 2013 5:03 PM

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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: Half Measure and Correction
Posted: Mar 24, 2013 4:55 PM
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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))/(b-a) is continuous on the connected domain {(a,b): a < b},
there is an interval I = (a,b) such that f(a,b) = 1/2.



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