# F not absolutely continuous

#### Dinkydoe

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...

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#### hollywood

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