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...