On Thu, 25 Apr 2013 00:18:49 -0700, William Elliot <email@example.com> wrote:
>On Sun, 21 Apr 2013, firstname.lastname@example.org wrote: > >> More specifically, use the representation of x in C Cantor set in base 3 >> with only 0's and 2's in the expansion of 3, and map >> >> f: x=0.a1a2..... ---> 0.b1b2....... >> >> Wheref(bi)= 0 , if ai=0 , f(bi)=1 , if ai=2 . >> >> This f is continuous, but not absolutely continuous ( which >> preserves sets of measure zero. ) > >Why is f continuous?
Let eps > 0. There exists N such that if x, y are in [0,1] and they have binary expansions that agree in the first N "digits" then |x-y| < eps.
(*) Now there exists delta > 0 such that if s, t are in the Cantor set and |s-t| < delta then the ternary expansions of s and t agree in the first N "digits".
So s, t in C, |s - t| < delta implies |f(s) - f(t)| < eps.
(To see why (*) is true: Go back to the construction of the Cantor set: C is the intersection of the C_n, where C_n is the union of 2^n intervals each of length 1/3^n. Choose delta > 0 so the smallest gap between a pair of these 2^n intervals is larger than delta. Then s, t in C, |s-t| < delta implies that s and t lie in the same component of C_n, hence |s-t| <= 1/3^n.)