= Solution
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.
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