Closure of the set of weak solutions of conservation laws Consider the conservation law $$u_t + q(u)_x = 0 \quad \tag{CL}$$
A function $u$ is a weak solution of $(CL)$ if $u \in L^\infty_\text{loc}((0, \infty)\times \mathbb{R})$ and $$\int_0^\infty \int_{-\infty}^{+\infty} uv_t + q(u)v_x \ \  dt \, dx = 0,$$ for every test function $v \in C_c^\infty((0,\infty)\times \mathbb{R}).$

Let $\{u_\epsilon\}$ be a sequence of weak solutions of $(\text{CL})$ and $u \in L^\infty((0,\infty) \times \mathbb{R})$. 
How does one prove that if, for every $\epsilon$, $\Vert u_\epsilon
 \Vert_{L^\infty((0,\infty)\times \mathbb{R})} < B$ for some $B > 0$
   and $u_\epsilon \to u$ in $L^1_\text{loc}((0,\infty) \times \mathbb{R})$,
   then $u$ is a weak solution of $(\text{CL})$?

Also, is it possible to prove a stronger similar theorem?
 A: Let $K\subset (0,\infty)\times \mathbb{R}$ be a compact set containing the support of $\nu$. Then $u_\epsilon \to u$ in $L^1(K)$. By passing to a subsequence $u_k:=u_{\epsilon_k}$, we may assume $u_k \to u$ almost everywhere in $K$. Therefore $q(u_k) \to q(u)$ almost everywhere on $K$, and thus
$$u_k\nu_t + q(u_k)\nu_x \longrightarrow u \nu_t + q(u)\nu_x$$
almost everywhere on $K$ as $k\to \infty$. Since $\nu_x$ and $\nu_t$ are continuous functions on the compact set $K$, they are bounded. Provided $q$ is a locally bounded function, the sequence $q(u_k)$ is also uniformly bounded, hence
$$C:=\sup_k\| u_k\nu_t + q(u_k)\nu_x\|_{L^\infty(K)} < \infty.$$
So the sequence $u_k\nu_t + q(u_k)\nu_x$ is uniformly bounded by the $L^1(K)$ function $C$. So by the dominated convergence theorem
$$0 = \lim_{k\to \infty} \iint_K u_k\nu_t + q(u_k)\nu_x dt dx = \iint_K u \nu_t + q(u)\nu_x dtdx =\int_{-\infty}^\infty\int_0^\infty u \nu_t + q(u)\nu_x dtdx .$$
Since this holds for arbitrary $\nu$ smooth with compact support, $u$ is a weak solution.
