Locally bounded distributional weak solutions of scalar laws

Let $T>0$, let $f\text{\hspace{0.17em}⁣}:\mathbb{R}\to \mathbb{R}$ be continuous, and let $u_0\in L^{\mathrm{\infty }}(\mathbb{R})\cap L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathbb{R})$. A measurable function $u\text{\hspace{0.17em}⁣}:\mathbb{R}\times (0,T)\to \mathbb{R}$ is a locally bounded distributional weak solution of $u_t+f(u{)}_x=0$ with initial datum $u_0$ if for every bounded interval $I\subset \mathbb{R}$ and every $T^{\mathrm{\prime }}<T$ it is essentially bounded on $I\times (0,T^{\mathrm{\prime }})$, and, for every $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R}\times (-\mathrm{\infty },T))$, ${\int }_0^T{\int }_{\mathbb{R}}(u{\phi }_t+f(u){\phi }_x)dxdt+{\int }_{\mathbb{R}}{u}_0(x)\phi (x,0)dx=0.$ The integrals are local: if a test is supported in $I\times [-S,T^{\mathrm{\prime }}]$ with $I$ bounded and $T^{\mathrm{\prime }}<T$, the local essential bound makes $u$ bounded on $I\times (0,T^{\mathrm{\prime }}]$, and continuity of $f$ makes $f(u)$ bounded on that bounded range. The initial integral is finite because $u_0$ is locally integrable and $\phi (\cdot ,0)$ is compactly supported. This is the local integrability version of the weak identity in Distributional weak solutions of the Cauchy problem, whose published formulation assumes a global $L^{\mathrm{\infty }}$ bound. This definition adds no entropy inequality or strong local $L^1$ trace requirement.