
Re: Measure and Density
Posted:
Feb 21, 2013 10:08 AM


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.

