Let $f,g\in C^1([0,1],\mathbb{R}^n)$, and then we define $h$ as a peicewise $C^1$ mapping as follows $$h(t)=\begin{cases} f(t) & t\in[0,\frac{1}{2}[\\ g(t) &t\in]\frac{1}{2},1] \end{cases} $$ I know that if I define the following function : $x(t)=\int_0^t f(s)ds$ for all $t\in [0,\frac{1}{2}[$ then $x$ is an absolutely continuous function because $f \in L^1([0,1],\mathbb{R}^n)$.
So my question is that can I say that $$y(t)= \int_0^t h(s)ds,$$ is an absolutely continuous function on $[0,1]$ ?
$\int_0^{1} |h(s)|ds=\int_0^{1/2} |f(s)|ds+\int_{1/2}^{1} |g(s)|ds <\infty$. The indefinite integral of any integrable function is absoluetly continuous. So the answer is YES,