Proof that absolute continuity implies differentiability a.e.

Question

Can somebody recommend a book/resource that provides a proof that absolute continuity of a function implies its almost-everywhere differentiability?

Answer 1

Rudin's "Real and Complex Analysis" gives this a thorough examination, but this is a brief overview.

Let $\mu$ be an absolutely continuous, complex Borel measure on $\mathbb{R}^n$, and let $f$ be the Radon-Nikodym (RN) derivative of $\mu$ with respect to the Lebesgue measure $\sigma$. Then the symmetric derivative of $\mu$ (written $\mu '=f~a.e.$) with respect to $\sigma$ holds, and we can write $\mu$ as the integral of the symmetric derivative over Borel sets in $\mathbb{R}^n$, ie, \begin{equation*} \mu (B)=\int_B \mu'd\sigma . \end{equation*}

Let $\{ B_i(x)\}$ be a sequence of Borel sets in $\mathbb{R}^n$ converging to $x\in\mathbb{R}^n$ with $f$ integrable over $\mathbb{R}^n$. Then at every Lebesgue point (or by definition, $a.e.$ with respect to $\sigma$), $~f$ can be written in the form \begin{equation*} f(x)=\lim_{i\to\infty}\frac{1}{\sigma(B_i(x))}\int_{B_i(x)}fd\sigma . \end{equation*} Note that a Lebesgue point can be thought of as a point where $f$ does not oscillate much, so the shrinking of $\{B_i(x)\}$ will be "nice" or "well-behaved". Combining this together implies that if $f$ is integrable on $\mathbb{R}$ and \begin{equation*} F(x)=\int^{x}_{-\infty}fd\sigma,~x\in\mathbb{R}, \end{equation*} then $F'(x)=f(x)$ at every Lebesgue point of $f$, and consequently $a.e.$ with respect to $\sigma$.