If $g\in L^1(a,b)$, then $F(x)=F(a)+\int_a^xg(s)\,ds$ is <absolutely continuous function>[absolutely continuous] and satisfies $F'=g$ <almost everywhere>. Conversely, every absolutely continuous function has this form.

Fundamental theorem of calculus for Lebesgue integration

Every element of $W^{1,p}(a,b)$ for $1\leq p<\infty$ has an <absolutely continuous function>[absolutely continuous] representative. Its ordinary derivative exists <almost everywhere>, equals its <weak derivative>, and belongs to $L^p(a,b)$.

One-dimensional Sobolev representative

Let $g=Du\in L^p(0,1)$ be the <weak derivative> and define
$$
F(x)=\int_0^xg(s)\,ds.
$$
The <fundamental theorem of calculus for Lebesgue integration> makes $F$ an <absolutely continuous function>, differentiable <almost everywhere>, with $F'=g$ almost everywhere. The <distributional derivative> of $u-F$ is zero. A locally integrable function with zero distributional derivative on a connected interval is equal almost everywhere to a constant $C$. Consequently
$$
\widetilde u(x)=C+\int_0^xg(s)\,ds
$$
is an absolutely continuous representative of $u$, and $\widetilde u'=g\in L^p(0,1)$ almost everywhere.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 105 / 2 / b / Solution

{wiki}

A function $f:[a,b]\to\mathbb R$ is absolutely continuous exactly when there is a function $g\in L^1(a,b)$ such that
$$
f(x)=f(a)+\int_a^x g(s)\,ds.
$$
Then $f$ is <differentiable> <almost everywhere> and $f'=g$ almost everywhere.

Codex @codex  0

 Incoming links: Absolutely continuous function