= Solution
For $s\geq0$ and $\xi\in\mathbb R$, define the <characteristic curve> $X(t;s,\xi)$ by
$$
\dot X(t;s,\xi)=F(t,X(t;s,\xi)),
\qquad
X(s;s,\xi)=\xi.
$$
The bounded derivative $F_x$ makes $F(t,\cdot)$ globally <Lipschitz continuity>[Lipschitz], uniformly in $t$. On each finite time interval, $|F(t,x)|\leq|F(t,0)|+L|x|$, so <Gronwall inequality> prevents finite-time escape. The <Picard-Lindelof theorem> therefore gives a unique trajectory for every finite $t\geq0$. Differentiation in $\xi$ gives
$$
\partial_\xi X(t;s,\xi)
=\exp\left(\int_s^tF_x(r,X(r;s,\xi))\,dr\right)>0,
$$
so the <characteristic flow map> is a $C^1$ increasing diffeomorphism.
Along a characteristic, the <chain rule> changes the equation into
$$
\frac d{dt}u(t,X(t;0,\xi))=g(t,X(t;0,\xi)).
$$
Tracing $(t,x)$ backward by the flow therefore gives
$$
u(t,x)=u_0(X(0;t,x))+\int_0^tg(s,X(s;t,x))\,ds.
$$
The regularity of the flow makes this a $C^1$ <classical solution>. Conversely, every classical solution obeys the same ordinary differential equation along every characteristic, so the formula also proves uniqueness.
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