Here's something that's pretty neat: find a measurable subset AA of [0,1][0,1] such that for any subinterval II of [0,1][0,1], the Lebesgue measure mu(AcapI)mu(AcapI) has 0<mu(AcapI)<mu(I)0<mu(AcapI)<mu(I). There's an explicit construction of such a set in Rudin, who describes such sets as "well-distributed". Balint Virag (and maybe others) found a very slick probabilistic construction.
Let X1,X2,ldotsX1,X2,ldots be i.i.d. coin flips, i.e. X1X1 is 11 with probability 1/21/2 and −1−1 with probability 1/21/2. Consider the (random) series
S:=sumin=1nftyXn/n.,,,S:=sumin=1nftyXn/n.,,,
By the Kolmogorov three-series theorem, it converges almost surely. However, it's a simple exercise to see that for any aa, the event S>aS>a has non-trivial measure: for a>0a>0, there's a positive chance of the first eaea terms of the series being positive, so the eaea-th partial sum is positive, and the tail is independent and positive or negative with equal probability, due to symmetry. For aleq0aleq0, it's trivial, again because of symmetry.
A common way of realizing i.i.d. coin flips on the unit interval is as Rademacher functions: for xin[0,1]xin[0,1], let bnbn be its binary expansion, and Xn(x)=(−1)bnXn(x)=(−1)bn. Realized this way, the random sum SS becomes an almost everywhere finite measurable function from [0,1][0,1] to mathbbRmathbbR. It only takes a bit more work to see that the set S>aS>a is exactly a well-distributed set.
Alex Bloemendal has written this up in a short note, but I'm not sure if he's published it anywhere.
No comments:
Post a Comment