The authors prove the existence of minimal and maximal absolutely continuous solutions \(x:[0,1]\to \mathbb{R}\) to the initial value problem \(x'(t)=f(t,x(t))\), \(x(0)=0\). Here, \(f:[0,1]\times \mathbb{R}\to \overline{\mathbb{R}}\) satisfies standard “measurability” assumptions and a nonstandard “continuity” assumption: for every \(x\in \mathbb{R}\), the function \(t\to f(t,x)\) is Lebesgue measurable; there exists a Lebesgue integrable function \(M:[0,1]\to \overline{\mathbb{R}}\) such that \(|f(t,x)|\leq M(t)\) for almost all \(t\in[0,1]\) and for all \(x\in \mathbb{R}\); \(\limsup_{y\uparrow x}f(t,y)\leq f(t,x)\leq \liminf_{y\downarrow x}f(t,y)\) for almost all \(t\in[0,1]\) and for all \(x\in \mathbb{R}\). Applications concern the scalar equation \(x'(t)=q(x(t))f(t,x(t))\) where \(q\) is Lebesgue measurable as well as the vector equation \(x'(t)=f(t,x(t))\) where \(x_k\to f_i(t,\cdots,x_{k-1},x_k,x_{k+1},\cdots)\) is nondecreasing for each \(i\neq k\) [cf. J. Szarski, Differential inequalities, Warszawa: PWN-Polish Scientific Publishers (1967; Zbl 0171.01502)].