Thread: Analysis: Absolute Continuity & Differentiation

Let A [0, 1] be a Borel set such that 0 < m(A I) < m(I) for all interval I [0, 1]. Let
F(x) = m([0, x] A), where m is lebesgue measure. Show that:
1. F(x) is absolutely continuous and strictly increasing on [0, 1]
2. F'(x) = 0 on a set of positive measure.

I've shown 1, but can't prove 2. In fact, 2 would seem to be false because if F is Abs Cont and F' is zero on a set E, then implies that F is 0 on the set E. But, wouldn't this contradict the fact that F is strictly increasing?

Originally Posted by southprkfan1

Let A [0, 1] be a Borel set such that 0 < m(A I) < m(I) for all interval I [0, 1]. Let
F(x) = m([0, x] A), where m is lebesgue measure. Show that:
1. F(x) is absolutely continuous and strictly increasing on [0, 1]
2. F'(x) = 0 on a set of positive measure.

I've shown 1, but can't prove 2. In fact, 2 would seem to be false because if F is Abs Cont and F' is zero on a set E, then implies that F is 0 on the set E. But, wouldn't this contradict the fact that F is strictly increasing?

Got the answer from another source

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So, F'(x) is 0 on [[0,1]\A], which has positive measure.