# Thread: Analysis: Absolute Continuity & Differentiation

1. ## Analysis: Absolute Continuity & Differentiation

Let A $\subset$[0, 1] be a Borel set such that 0 < m(A $\cap$ I) < m(I) for all interval I $\subset$ [0, 1]. Let
F(x) = m([0, x] $\cap$ 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 $F = \int_E F'$ implies that F is 0 on the set E. But, wouldn't this contradict the fact that F is strictly increasing?

2. Originally Posted by southprkfan1
Let A $\subset$[0, 1] be a Borel set such that 0 < m(A $\cap$ I) < m(I) for all interval I $\subset$ [0, 1]. Let
F(x) = m([0, x] $\cap$ 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 $F = \int_E F'$ 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

$F(x) = \int_{[0,x]} \chi_A(x)dm$ --> $F'(x) = \chi_A(x)$

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