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Let $A \subset [0,1]$ be a Borel set such that $0 < m(A\cap I) < m(I)$ for every subinterval $I$ of [0,1].

a. Let $F(x) = m([0,x] \cap A)$. The $F$ is absolutely continuous and strictly increasing on $[0,1]$, but $F' = 0$ on a set of positive measure.

b. Let $G(x) = m([0,x] \cap A) - m([0,x]\backslash A)$. Then $G$ is absolutely continuous on [0,1], but $G$ is not monotone on any subinterval of $[0,1].

I am stuck on this problem and would sincerely appreciate help. Thank you.

user44081
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    Where are you stuck? Can you see $F$ and $G$ are absolutely continuous and $F$ is strictly increasing? – 23rd Dec 06 '12 at 21:56
  • @richard: I actually cannot see this. Could you please explain why this is so? Thank you. – user44081 Dec 06 '12 at 22:45
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    Sorry, it seems that you are not quite familiar with the concepts involved in the question, so I have no idea how to provide a comprehensive answer to you. – 23rd Dec 07 '12 at 03:45

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I'll get you rolling on part (a).

$F$ is actually Lipschitz. To see this, without loss of generality assume $x > y \in [0,1]$. Then we have $$(A \cap [0,y]) \cup (A \cap (y,x]) = A \cap [0,x],$$ and since these are disjoint sets $$m(A \cap [0,y]) + m(A \cap (y,x]) = m(A \cap [0,x]).$$ Hence, $$|F(x) - F(y)| = |m(A \cap [0,x])-m(A \cap [0,y])| = |m(A \cap (y,x])| < |x-y|.$$

Travis
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