
Re: Measure and Density
Posted:
Feb 23, 2013 11:26 AM


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_{xh}^{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.

