
Re: Measure and Density
Posted:
Feb 22, 2013 12:03 AM


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.

