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Re: Measure and Density
Posted:
Feb 23, 2013 11:26 AM
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On Thu, 21 Feb 2013 21:03:16 -0800 (PST), Butch Malahide
<fred.galvin@gmail.com> wrote:
>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
locally integrable
>measurable
>function (such as your chi_A - chi_B), which integrates to zero over
>each interval, is zero almost everywhere?
Yes.
This is immediate from the Lebesgue Differentiation Theorem:
If f is locally integrable, then for almost every x,
f(x) = lim_{h->0} 1/(2h) int_{x-h}^{x+h} f(t) dt.
Quoting Wade, there may be a more elementary way to see it.
In fact there is almost surely a more elementary way...
>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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