Consider the IVP \begin{eqnarray} u_t+F(x,u)_x=S(x,u)\\ u(x,0)=u_0(x) \end{eqnarray} If $S(x,u)=0$ and $u\in C([0,T],L^1(\mathbb{R})),$ then we have $$\int\limits_{\mathbb{R}}u_0(x)dx=\int\limits_{\mathbb{R}}u(x,t)dx.$$ (physically which can be interpreted as conservation of mass...)
What happens when $S(x,u)\neq 0$
P.S. Please give a proof or suggest a reference..
If $S \neq 0$ then $S$ serves it's role as a source, for example, something that can "introduce mass". Let's compute as before: \begin{align*} \partial_t \int_{\mathbb{R}} u(x, t)\, dx &= \int_{\mathbb{R}} u_t(x, t)\, dx \\ &= -\int_{\mathbb{R}} \frac{d}{dx}F(x, u(x, t))\, dx + \int_{\mathbb{R}} S(x, u(x, t))\, dx \\ &= \int_{\mathbb{R}}S(x, u(x, t))\, dx. \end{align*} The $F$ integral vanishes if we assume sufficient decay, e.g. $F$ goes to zero at $\pm \infty$, by the fundamental theorem of calculus. Then we have $$ \text{Mass at time $t$ = }\int_{\mathbb{R}}u(x, t)\, dx = \int_{\mathbb{R}}u_0(x)\, dx + \int_0^t \int_{\mathbb{R}}S(x, u(x, s))\, dx \, ds. $$ So the interpretation is that the quantity $\int_{\mathbb{R}}u(x, t)\, dx$, rather than being constant in $t$, changes from its initial value by integrating the source $S$.