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Re: Measure and Density
Posted:
Feb 21, 2013 10:08 AM
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On Wed, 20 Feb 2013 15:40:29 -0800 (PST), Butch Malahide
<fred.galvin@gmail.com> wrote:
>On Feb 20, 4:46 pm, W^3 <82nd...@comcast.net> 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
Very good. Gold star for Butch, or Fred or whoever you are.
This is an illustration of one of Littlewood's three principles:
A measurable set is almost open. I know another one is
a measurable function is almost continuous; I'm not sure
about the third, maybe almost everywhere convergence
is almost uniform.
Of course the "almost" is a little fuzzy, but they're useful
things to keep in mind.
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