# Thread: F not absolutely continuous

1. ## F not absolutely continuous

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

It satisfies

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

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 $\epsilon$ exists $\delta$ s.t. whenever a disjoint (finite) sequence $[x_1,y_1],\cdots [x_n,y_n]$ of sets with $\sum_k|x_k-y_k|<\delta$ we have $\sum_{k}|F(y_k)-F(x_k)| < \epsilon$.

I can't think of any reason why this is...

2. ## Re: F not absolutely continuous

Ok sorry, it might be important that F is monotone increasing, and F(0) =0, F(1) = 1. Like i said, it's a distribution function...It looks a lot like this: Cantor Function -- from Wolfram MathWorld

3. ## Re: F not absolutely continuous

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 $\sum_k|x_k-y_k|<\delta$ for any given $\delta$ while $\sum_{k}|F(y_k)-F(x_k)|$ is some constant.

- Hollywood