F not absolutely continuous

Dec 2009
I have the following function \(\displaystyle F:[0,1]\to[0,1]\) (it is a distr. function. Not so important)

It satisfies

\(\displaystyle F(x) = 1/2\) for \(\displaystyle x\in [1/4,3/4]\)
\(\displaystyle F(1-x) = 1-F(x)\) for all \(\displaystyle x\)
\(\displaystyle F(x)= 2F(x/4) \) for \(\displaystyle x\in [0,1/4]\)

I want to show F is not absolutely continuous w.r.t. Lebesgue measure. Should be straightforward but i dont understand...

It is continuous, but not absolutely continuous.

Apparantly, according to the definition it means the following does not hold: For all \(\displaystyle \epsilon\) exists \(\displaystyle \delta\) s.t. whenever a disjoint (finite) sequence \(\displaystyle [x_1,y_1],\cdots [x_n,y_n]\) of sets with \(\displaystyle \sum_k|x_k-y_k|<\delta\) we have \(\displaystyle \sum_{k}|F(y_k)-F(x_k)| < \epsilon\).

I can't think of any reason why this is...
Last edited:
Mar 2010
I'm not sure I understand how F is constructed. Is it the limit of the process you describe?

Also, I think the definition of absolute continuity allows the sequence of intervals to be countably infinite. Given that, you can probably construct a sequence of intervals with \(\displaystyle \sum_k|x_k-y_k|<\delta\) for any given \(\displaystyle \delta\) while \(\displaystyle \sum_{k}|F(y_k)-F(x_k)|\) is some constant.

- Hollywood