Let $a,b\in\mathbb R$ with $a<b$, and let $f:[a,b]\to\mathbb C$ be absolutely continuous. Then $f$ is differentiable at $\mathcal L^1$-almost every point of $(a,b)$. Moreover, there exists a function $g\in L^1((a,b),\mathcal B((a,b)),\mathcal L^1)$ such that $g(x)=f'(x)$ for $\mathcal L^1$-almost every point $x\in(a,b)$ at which $f$ is differentiable, and for every $x\in[a,b]$,

\begin{align*} f(x)=f(a)+\int_{(a,x)} g(t)\,d\mathcal L^1(t). \end{align*}

Equivalently, after choosing this $L^1$ representative of the almost-everywhere derivative and denoting it by $f'$, one has, for every $x\in[a,b]$,

\begin{align*} f(x)=f(a)+\int_{(a,x)} f'(t)\,d\mathcal L^1(t). \end{align*}