🚀 go-pugleaf

In article <ggrzgca@rpi.edu> sassoj@rs6412.ecs.rpi.edu (John J. Sasso Jr.) writes:
>   Now, the Cantor set has Lebesgue measure 0 for the Lebesgue measure defined
>in spaces R^1, R^2, etc...  However, the Cantor does have a fractal dimension
>(or Hausdorff dimension) between 0 and 1 (I cannnot recall the value), and
log(2)/log(3)

>I'll call this dimension Df.  Is it possible to define a Lebesgue measure in
>a space of dimension Df (R^Df) such that the measure of the Cantor set is a 
>value other than 0?
It is possible to define a measure such that the Cantor set has positive 
measure, and any set contained in its complement has zero measure. It is
possible to define infinitely many such measures.

But requesting that it be a Lebesgue measure is not something you can do,
as you will easily realize on some reflection, because there is no 
Euclidean space of noninteger dimensionality. Lebesgue measure is defined
through use of very special properties of R^n: it is the only measure 
to be invariant under rototranslations and to coincide with regular volume
on rectangles. 

There is no canonical space of dimension Df, there is no such space that
will satisfy the axioms of linear algebra etc etc, be a Lie group, etc. etc.
These are the properties of R^n you use to make one measure (Lebesgue)
stand out of the crowd of infinitely measures. 

Sure enough, you can, however, make a measure whose support will coincide 
with the cantor set. But there is no canonical such measure. 

Best Regards,
Marcelo Magnasco
marcelo@mjf.rockefeller.edu	magnasco@research.nj.nec.com