Let $I \subseteq \mathbb R$ be an interval, let $X \subseteq \mathbb R^n$ and $U \subseteq \mathbb R^m$ be open sets, and let $f:X\times U\to \mathbb R^n$ be continuous. Assume that $f$ is locally Lipschitz in the state variable uniformly on compact subsets of $U$: for every compact set $K_X\subset X$ and every compact set $K_U\subset U$, there exists $L\ge 0$ such that $|f(x,u)-f(y,u)|\le L|x-y|$ for all $x,y\in K_X$ and all $u\in K_U$. Let $u:I\to U$ be Lebesgue measurable and assume that for every compact interval $K\subseteq I$ there exists a compact set $C_K\subset U$ such that $u(t)\in C_K$ for a.e. $t\in K$. If $t_0\in I$ and $x_0\in X$, then there exists an interval $J\subseteq I$, containing $t_0$ as a relative interval in $I$, and a locally absolutely continuous map $x:J\to X$ such that $x(t_0)=x_0$ and $\dot{x}(t)=f(x(t),u(t))$ for a.e. $t\in J$. Moreover, this solution is unique on its interval of definition: if $y:L\to X$ is another locally absolutely continuous solution on an interval $L\subseteq I$ containing $t_0$ with $y(t_0)=x_0$, then $x(t)=y(t)$ for all $t\in J\cap L$. There is a unique maximal such solution $x_{\max}:J_{\max}\to X$, where $J_{\max}\subseteq I$ is an interval containing $t_0$, in the sense that every other solution $y:L\to X$ with initial condition $(t_0,x_0)$ satisfies $L\subseteq J_{\max}$ and $y=x_{\max}|_L$.