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Topic: Measure and Density
Replies: 14   Last Post: Feb 23, 2013 11:26 AM

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quasi

Posts: 10,451
Registered: 7/15/05
Re: Measure and Density
Posted: Feb 22, 2013 12:53 AM
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W^3 wrote:
>quasi wrote:
>> Butch Malahide wrote:
>> >W^3 wrote:
>> >>
>> >> Is it possible that there exist 0 < c < d < 1 such that
>> >> cm(I) < m(S /\ I) < dm(I) for all nonempty open intervals
>> >> I contained in (0,1)?

>> >
>> >No. If S is a (Lebesgue) measurable subset of the real line
>> >with m(S) > 0, and if d < 1, then there is a nonempty interval
>> >I such that m(S /\ I) > dm(I). Sometime in the previous
>> >millennium I took a class in measure theory, using the
>> >textbook by Halmos, and I recall that this was proved in an
>> >early chapter.
>> >
>> >More is true:
>> >
>> >http://en.wikipedia.org/wiki/Lebesgue's_density_theorem

>>
>> A possibly related question ...
>>
>> Prove or disprove:
>>
>> If A,B are measurable subsets of [0,1] such that
>> m(A /\ I) = m(B /\ I) for all open intervals I contained in
>> [0,1], then m(A\B) = 0.

>
>True, it follows from: If E is a measurable subset of [0,1],
>then for a.e. x in [0,1],


I think you meant to say "for a.e x in E".

>m(E /\ (x-d,x+d))/(2d) -> 1 as d -> 0. Apply this to A\B to
>obtain a contradiction if m(A\B) > 0.


Yes, I see it now. Thanks.

>There may be a more elementary way to see it.

quasi



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