Let $f:[0,\infty)\to\mathbb R$ be locally Lebesgue integrable and $$g(t):=\int_0^tf(s)\:{\rm d}s\;\;\;\text{for }t\ge0.$$
Using the dominated convergence theorem, it is easy to show that $g$ is continuous, but how can we show that $g$ is absolutely continuous?
If you are talking about absolute continuity on the whole of $[0,\infty)$ (in the sense for every $\epsilon >0$ there exists $\delta >0$ such that $\sum |f(b_i)-f(a_i)| <\epsilon $ whenever $(a_i,b_i)$ are disjoint intervals with $\sum (b_i-a_i) <\delta$) then the result is false. Absolute continuity implies uniform continuity. If $f(s)=s$ then your hypothesis is satisfied but $g$ is not uniformly continuous.
What is true is $g$ is absolutely continuous on $[0,T]$ for any $T <\infty$.
[For a proof note that $f$ is integrable on $[0,T]$ and this implies there exists $\delta >0$ such that $\int_A |f|<\epsilon$ whenever $A$ is measurable with $\lambda (A) <\delta$. ($\lambda$=Lebesgue measure). Ref: Rudin's RCA].