The primitive of a continuous function on a compact interval is continuous via the Fundamental Theorem of Calculus. Let $I \subset \mathbb{R}$ be open and let $u': \overline{I} \mapsto \mathbb{R}$ be continuous. Then for any $y \in I$, by the $L^1-L^\infty$ bound from Holder's inequality, we have:
\begin{equation} |u(x)-u(y)|=\left|\int^x_y u'(z) \, dz \right| \leq \int^x_y \left| u'(z) \right| \, dz \leq \| u'\|_{L^{\infty}(\overline{I})} |x-y| \rightarrow 0 \quad \mbox{as } x \rightarrow y, \end{equation} so $u$ is continuous on $I$.
The same argument can be used if $u' \in L^p(I)$ if $p \in (1,\infty]$. However, I am unsure if it is still true if $u' \in L^1(I)$. Can I still conclude that $u$ is continuous?
Edit: fixing some errors in the question.
Edit 2: We use a different argument. The idea is to truncate $u'$ on smaller and smaller supports so that the resulting sequence $u_n'$ will tend to $0$ in the limit. The proof is concluded by the dominated convergence theorem (DCT) because $u_n' \leq u'$.
Fix $y \in I$. For each large positive integer $n$ (large enough of course so that $A_n \subset I$), let \begin{equation} A_n:=\{t \in I: y-1/n<t<y+1/n \}, \end{equation} and the indicator function \begin{equation} {\bf 1}_{A_n}(z):= \left\{ \begin{aligned} 1 \quad \mbox{if } z \in A_n,\\ 0 \quad \mbox{if } z \notin A_n.\\ \end{aligned} \right. \end{equation} Then $u'_n:={\bf 1}_{A_n}u'$ with $u'_n \rightarrow 0$ pointwise (a.e.) as $n \rightarrow \infty$ and $|u_n'| \leq u' \in L^1(I)$ by assumption. Hence, the DCT implies that \begin{equation} u(y+1/n)-u(y-1/n)\leq \int^{y+1/n}_{y-1/n} |u'(z)| \, dz = \int_I |u_n'(z)| \, dz \rightarrow \int_I 0 \, dz=0. \end{equation} That is, for each $\varepsilon > 0$, there exists an $n_0 > 0$ such that $n\geq n_0$ implies $\int^{y+1/n}_{y-1/n} |u'(z)| \, dz < \varepsilon$. Hence, for every $\varepsilon > 0$, there exists $\delta:=1/n_0$ such that \begin{equation} |x-y| < \delta=1/n_0 \quad \mbox{implies} \quad |u(x)-u(y)|\leq \int^{y+1/n}_{y-1/n} |u'(z)| \, dz<\varepsilon. \end{equation}
As Copperhat notes, this actually shows that $v << \mu$ where $v$ is a measure with density $u' \in L^1$ and $\mu$ is the Lebesgue measure.