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

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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: Measure and Density
Posted: Feb 22, 2013 12:03 AM
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On Feb 21, 10:09 pm, quasi <qu...@null.set> 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


You can look up the proof. It's almost the same question: S = A\B, d =
1/2. If m(A\B) > 0, then there is an interval I such that m((A\B) /\
I) > m(I)/2, whence m(A /\ I) > m(I)/2 > m(B /\ I). Or you can follow
W^3's suggestion and use the Lebesgue density theorem, which is more
powerful and takes more work to prove.

For that matter, isn't there a theorem to the effect that a measurable
function (such as your chi_A - chi_B), which integrates to zero over
each interval, is zero almost everywhere? That sounds like it ought to
be true, but it has been over 50 years since I took that measure
theory class, and I don't remember very much.



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