Assume that $f:\mathbb{R}^{n}\times \mathbb{R} \to \mathbb{R}$ satisfies
(i) there exists a constant $M>0$ such that $|f(x,t)-f(y,t)|\leq M|x-y|$ for each $x,y\in \mathbb{R}^n$ and each $t\in [0,T]$.
(ii) $f(0,t) \in L^1([0,T]).$
Given the I.V.P
$\begin{cases} x'(t)=f(x,t)\\x(0)=x_0 \end{cases} $
(a), Show it has a unique solution $x=x(t)$ over $t\in[0,T]$.
(b), Show the solution must be absolutely continuous.
Remark: I know the proof of the case where $f$ is assumed to be continuous, in that case one just needs to define a map $K:C \to C$ by $K(x)(t): = x_0+\int_{0}^t f(s,x(s))\mathrm{d}s$ and show it's a contracting map over some $C$ and then apply Banach fixed point theorem to finish the argument. However in this case it's not given explicily that $f$ is continuous, or even integrable over $[0,T]$ for an arbitrary $x\in \mathbb{R}^n$.