Primitive of an $L^1$ function is continuous 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.  
 A: The fundamental result you need is that if $f \in L^1[a,b]$, then the measure 
$\mu A = \int_A |f(x)| dx$ is absolutely continuous with respect to the
Lebesgue measure. This means that for all $\epsilon>0$ there is some $\delta >0$ such that if $m A < \delta$ then $\mu A < \epsilon$.
A: Fact. If $\,f\in L^1[a,b]$, then for every $\varepsilon>0$, there exists a $\delta>0$, such that $I\subset [a,b]$ and $\mu(I)<\delta$ implies that $\int_I\lvert\, f\rvert \,dx<\varepsilon$.
Proof. If we set $E_M=\{x\in[a,b]: \lvert\, f(x)\rvert\ge M\}$, then $\lim_{M\to\infty}\int_{E_M}\lvert\, f\rvert\,dx=0$. To see this, let
$$
f_M(x)=\left\{\begin{array}{ll} f(x) & \text{if} & \lvert\,f(x)\rvert\le M, \\ 0 & \text{otherwise.}\end{array}\right.
$$
Then, the Monotone Convergence Theorem implies that $\lim_{M\to\infty} \int_{[a,b]}\lvert\, f_M\rvert \,dx=\int_{[a,b]}\lvert\, f\rvert \,dx$. But,
$$
\int_{E_M}\lvert\, f\rvert\,dx= \int_{[a,b]}\lvert\, f\rvert \,dx-\int_{[a,b]}\lvert\, f_M\rvert \,dx.
$$
Now let $\varepsilon>0$. First we obtain $M>0$, such that $\int_{E_M}\lvert\, f\rvert\,dx<\varepsilon/2$, and then we set
$$
\delta=\frac{\varepsilon}{2M}.
$$
Then, if $\mu(I)<\delta$, 
$$
\int_I\lvert\,f\rvert\,dx=\int_{I\cap E_M}\lvert\,f\rvert\,dx+
\int_{I\smallsetminus E_M}\lvert\,f\rvert\,dx\le
\int_{E_M}\lvert\,f\rvert\,dx+
\int_{I}M\,dx < \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.
$$
