I want to show that if $$u(x):= \int_{x_0}^x v(t) \; dt$$, that then $u$ is locally absolutely continuous. The proof in Lemma 3.31 in A First Course in Sobolev Spaces suggests that we first want to show that $v$ is equi-integrable, and then that $u$ is locally absolutely continuous. I am however struggling with both steps.
Suppose $v\in L^p(I)$ is equi-integrable. Then for every $\epsilon>0$ there exists $\delta>0$ such that \begin{align*} \int_F |v(x)|\;d\mu(x) \leq \epsilon, \end{align*} for every lebesgue measruable set $F\subset I$ with $\mu(F)\leq \delta$. Let $F$ be any finite disjoint collection of intervals $(a_i, b_i)$ with $a_i,b_i\in I$. Then by the equi-integrability
\begin{align*} \epsilon \geq \int_F |v(x)| \geq \int_F v(x)\; d\mu(x) = \sum_{i=1}^n \int_{a_i}^{b_i} v(x)\; d\mu(x) = \sum_{i=1}^n u(b_i) - u(a_i). \end{align*}
This is close to the definition of absolute continuity, but not quite, since we're missing absolute value bars around the differences in the sum on the right.
To show that $v$ is equi-integrable (a uniformly integrable collection with only one element) I use this mathstackexchange post as suggested by David Mitra