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?