Convergence of function in $L^1$ space Let $\Omega \subset \mathbb{R}^{n}$ be bounded and open. Assume that $a: \Omega \times \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is a Caratheodory function which means $a(x,.,.)$ is continuous on $\mathbb{R} \times \mathbb{R}^{n}$ for almost all $x \in \Omega$ and $a(.,s,\xi)$ is measurable in $\Omega$ for every $(s,\xi) \in \mathbb{R} \times \mathbb{R}^{n}$. We  further assume that $$|a(x,s,\xi)| \leq k(x) + \beta(|s|^{p-1}+|\xi|^{p-1})  $$ for almost every $x \in \Omega$, for every $(s,\xi) \in \mathbb{R} \times \mathbb{R}^{n}$ and for some $k \in L^{p'}(\Omega), \beta \geq 0, p > 1$ where $p' = \frac{p}{p-1}$.
If we consider $u_{m} \rightarrow u$ in $L^{\infty}(\Omega)$ and $u_{m} \rightarrow u$ uniformly, would it then follow that $$a(x,u_{m}, \xi) \rightarrow  a(x,u,\xi) \text{   } \text{  in  } L^{1}(\Omega)$$
Proposed idea:
Does this simply follow by noting that if $$|a(x,u_{m}(x),\xi) - a(x,u(x),\xi)| \rightarrow 0$$ then $$\int_{\Omega}|a(x,u_{m}(x),\xi) - a(x,u(x),\xi)|dx \rightarrow 0$$
 A: Yes, even if the convergence is pointwise almost everywhere. Use the Lebesgue dominated convergence theorem with the fact Omega and a are bounded.
The implication for uniform convergence could also follow from the embedding of bounded functions in absolutely integrable functions.
Edit: Hi, as promised I have returned. 
You can use the bound 
$0 \geq \int_{\Omega} |a(x,u_m(x),\xi) - a(x,u(x),\xi)| dx \leq |\Omega| \sup_{x \in \Omega} |a(x,u_m(x),\xi) - a(x,u(x),\xi)| =: |\Omega| \|u_m - u \|_{L^{\infty}}$
Therefore convergence in $L^{\infty}$ implies convergence in $L^1$ provided $|\Omega| < \infty$ ($| \cdot |$ is the Lebesgue measure in this setting, think area for 'nice' sets). 
Arguably you can also use the fact that convergence in $L^{\infty}$ implies convergence pointwise almost everywhere, and use my first comment. Be sure that your assertion is true, your assumptions are very strong for this result. In fact you can strengthen your convergence beyond simply $L^1$ to $L^p$ by using Holders inequality.
