Let $f \in C^1(\mathbb{R})$, let $g \in L^\infty(\mathbb{R})$, and let $u \in L^\infty(\mathbb{R} \times (0,\infty))$ be a weak solution of the scalar conservation law: \begin{align*} u_t + f(u)_x &= 0, \quad (x,t) \in \mathbb{R} \times (0,\infty), \\ u(\cdot, 0) &= g, \end{align*} in the sense that for every $\phi \in C_c^\infty(\mathbb{R} \times [0,\infty))$: \begin{align*} \int_0^\infty \int_{-\infty}^\infty \bigl(u\,\phi_t + f(u)\,\phi_x\bigr)\,d\mathcal{L}^1(x)\,d\mathcal{L}^1(t) + \int_{-\infty}^\infty g(x)\,\phi(x,0)\,d\mathcal{L}^1(x) = 0. \end{align*} Let $\xi \in C^1((0,\infty))$ be a smooth curve and define the open half-spaces $\Omega^- := \{(x,t) : x < \xi(t),\, t > 0\}$ and $\Omega^+ := \{(x,t) : x > \xi(t),\, t > 0\}$. Suppose $u|_{\overline{\Omega^-}}$ and $u|_{\overline{\Omega^+}}$ are each of class $C^1$, and that $u$ satisfies $u_t + f(u)_x = 0$ pointwise on $\Omega^-$ and $\Omega^+$.

Define the one-sided limits $u^\pm(t) := \lim_{x \to \xi(t)^\pm} u(x,t)$ and the jump notation $[v]_\xi := v^+ - v^-$ for any quantity $v$. Then at every $t \in (0,\infty)$: \begin{align*} \dot{\xi}(t)\,[u]_\xi = [f(u)]_\xi. \end{align*} Equivalently, $\dot{\xi}(t)(u^+(t) - u^-(t)) = f(u^+(t)) - f(u^-(t))$.