W^3
Posts:
29
Registered:
4/19/11


Re: Measure and Density
Posted:
Feb 21, 2013 11:43 PM


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 /\ (xd,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.

