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

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David C. Ullrich

Posts: 21,553
Registered: 12/6/04
Re: Half Measure and Correction
Posted: Mar 25, 2013 9:07 AM
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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))/(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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