# Error in convservation laws integral equality argument

I'm attempting to solve the following problem from Evans on conservation laws, using the following definition of weak solution

I came up with the following test function construction. Let $$0 < \hat{t} < T$$, $$K$$ be compact such that $$\text{supp}(u) \subset \subset K$$ $$\newcommand{\ep}{\varepsilon}$$ $$v_t = \frac{\chi_{B(\hat{t}, \ep)}}{|B(\hat{t}, \ep)|}\chi_K \qquad v(x,t) = \int_0^t v_t(s) ds$$

then $$v_x = 0$$ on the support of $$u$$ and thus $$F(u)$$ ($$v$$ is constant there), so by the definition of weak solution $$\int_0^T \int_{-\infty}^\infty u(x,t) v_t(t) dxdt = \int_{-\infty}^\infty g(x) v(x, 0) dx = 0$$ and $$\int_0^T \int_{-\infty}^\infty u(x,t) v_t(t) dxdt = \int_{-\infty}^\infty \int_0^T u(x,t) v_t(t)dtdx = \int_{-\infty}^\infty \frac{1}{|B(\hat{t},\ep)|} \int_{B(\hat{t},\ep)} u(x,t) dtdx$$ so letting $$\ep \rightarrow 0$$ we have $$\lim_{\ep \rightarrow 0} \int_{-\infty}^\infty \frac{1}{|B(\hat{t},\ep)|} \int_{B(\hat{t},\ep)} u(x,t) dtdx = \int_{-\infty}^{\infty} u(x,\hat{t}) dx$$ (by the Lebesgue differentiation theorem and bounded convergence) however, this implies that the integral of g has to be zero, which apparently is not required. While this construction isn't itself smooth, it is fairly trivial to regularlize it by mollification. Where is the error here?

The error comes from assuming that v has compact support. While the space dimension doesn't really matter (a smooth cutoff function can be used to smooth the characteristic of the support of u), in time, the test function $$v$$ needs to have have compact support (w/r/t time) in $$[0, \infty)$$ (as opposed to just $$[0,T]$$). Alternatively, the notion of a weak solution can be reformulated to include the boundary terms, and so would become $$\newcommand{\R}{\mathbb{R}}$$ $$\int_0^T \int_{\R} u(x,t) v_t dt + \int_0^T \int_{\R} F(u(x,t))v_x + \int_{\R} u(x,0) v(x,0) - \int_{\R} u(x,T) v(x,T)$$