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

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W^3

Posts: 29
Registered: 4/19/11
Re: Measure and Density
Posted: Feb 21, 2013 11:43 PM
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In article <pnrdi8hdtq1jf8ug3fhnmr4n9a5jnk44cc@4ax.com>,
quasi <quasi@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
>
> quasi


True, it follows from: If E is a measurable subset of [0,1], then for
a.e. x in [0,1], 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. There may be a more
elementary way to see it.



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