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Let $A\subset [0,1]$ be a Borel set such that $0<m(A\cap I)<m(I)$ for any subinterval $I$ of $[0,1]$. Let $F(x)=m([0,x]\cap A)$. Show that $F$ is absolutely continuous and strictly increasing on $[0,1]$, but $F'=0$ on a set of positive measure.

My Work:

Let $\epsilon>0$. Then we can find a finite set of disjoint intervals $(a_1,b_1),(a_2,b_2),...,(a_N,b_N)$ of $[0,1]$ such that $\displaystyle (b_i-a_i)<\frac{\epsilon}{N}$ . Then we have $\displaystyle \sum_{i=1}^N (b_i-a_i)<\epsilon$. Now $F(b_i)=m([0,b_i]\cap A)<m([0,b_i])=b_i$ and $F(a_i)=m([0,a_i]\cap A)<m([0,a_i])=a_i$. If I can show that $F(b_i)-F(a_i)<(b_i-a_i)$ for all $i$ then I can show that $F$ is absolutely continuous. But now I am stuck in proving this. Can anyone please give me a hint?

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    $\begingroup$ You say "we can find ... $(a_1, b_1), \ldots, (a_N, b_N)$" but $N$ isn't mentioned before. This would suggest that you were allowed to choose $N$ yourself. In particular you could choose $N=1$. But that isn't correct according to the definition of absolute continuity...The correct statement would start with something like: "Let $\epsilon > 0$. Then given any finite set of disjoint intervals, whose union has measure less than..." $\endgroup$
    – mathmandan
    Dec 11 '14 at 22:05
  • $\begingroup$ So is it correct if I start like this: Let $\epsilon>0$. Suppose $(a_1,b_1),(a_2,b_2),...,(a_N,b_N)$ is a finite set of disjoint intervals of $[0,1]$ such that $\displaystyle \sum_{i=1}^N (b_i-a_i)<\epsilon$... $\endgroup$
    – Extremal
    Dec 11 '14 at 22:20
  • $\begingroup$ Yes, that would be better (note that by saying this, you'd be saying $\delta = \epsilon$). Definition of absolutely continuous function here: en.wikipedia.org/wiki/Absolute_continuity#Definition $\endgroup$
    – mathmandan
    Dec 11 '14 at 22:25
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Hint: You might want to think about calculating with $F(b_i) - F(a_i)$ directly, rather than using $F(x) < x$.

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