Absolute continuity of $f:[a,b] \rightarrow \mathbb{R}$ implies differentiability a.e.

Question

This question is inspired by this question. It addressed the same problem, but the books recommend seemed to use a lot of tools I was not familiar with, so I'm having a lot of trouble following. Obviously, for a bounded $f\in \mathcal{L}^1([a,b])$ the function $$F:[a,b]\rightarrow \mathbb{R}, x \mapsto \int_{a}^x f(t)dt$$

is absolutely continuous. Now I want to show that absolute continuity does imply differentiability a.e. and $F'=f$ a.e.

From reading this I found that you could maybe show this result by showing that $F$ is of bounded variation, which implies it's a difference of monotonically increasing functions, which implies diff. a.e. . Is there a more elementary or direct approach to prove this result for example avoiding bounded variations and lebesgue points?

The result you're proving is essentially the Lebesgue differentiation theorem. So you're basically asking for a “Simple” proof of Lebesgue's Differentiation Theorem for dimension 1?. One unusual though not elementary proof of this is at https://math.stackexchange.com/questions/2356808/how-to-prove-the-lebesgue-density-theorem-using-martingales (stated for indicator functions, but should work more generally) — Dap, Nov 27 '17 at 12:27

score 5 · Accepted Answer · answered Nov 27 '17 at 10:30

5

Well bounded variation is way a bigger class than AC. You can show by rather elementary means that any AC function on the line is the difference of two increasing function, and then prove that increasing functions are differentiable a.e.

The proof is very elementary and done by hands, see this for example: http://www.math.unipd.it/~monti/AR_2014_15/App_GDM_2013_14.pdf

answered Nov 27 '17 at 10:30

Diesirae92

1,157

3

Specifically, the function $F$ in the question can be written as $F(x)=\int_a^x|f(t)|dt - \int_a^x(|f(t)|-f(t))dt$ which is a difference of increasing functions. – Dap Nov 27 '17 at 11:26

Absolute continuity of $f:[a,b] \rightarrow \mathbb{R}$ implies differentiability a.e.

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