I have a sequences of function $(u_n)_{n\in \mathbb{Z}}: \mathbb{R}^3\times [0,\infty)\rightarrow \mathbb{R}^3$ that satisfies \begin{equation}\lim_{n\to \infty}\int_{B(0,R)} u_n(x,t)dx<\infty\end{equation} where $B(0,R)$ is a bounded cube with rational lenght. And let $a:\mathbb{R}^3\times [0,\infty)\rightarrow \mathbb{R}^3$ be a continuous functions in $L^2(\mathbb{R}^3)$ then I want to show that \begin{equation} \lim_{n\to \infty }\int_{\mathbb{R}^3}u_n(x,t)a(x,t)dx <\infty \end{equation}
How can I do? I want to use a density argument but I don't know whether the cube is dense in $\mathbb{R}^3$. Can someone help me?
Let's note $A_n=\{ x\in \mathbb{R} ^3 | a(x)>u_{n}(x) \}$ so $$ \lim_{n\to \infty }\int_{\mathbb{R}^3}u_n(x,t)a(x,t)dx= \lim_{n\to \infty }\int_{A_{n}}u_n(x,t)a(x,t)dx +\lim_{n\to \infty }\int_{A_{n}^c}u_n(x,t)a(x,t)dx \leq \lim_{n\to \infty }\int_{A_{n}}a^2(x,t)dx +\lim_{n\to \infty }\int_{A_{n}^c}u_{n}^2(x,t)dx \leq \lim_{n\to \infty }\int_{ \mathbb{R}^3 }a^2(x,t)dx +\lim_{n\to \infty }\int_{ \mathbb{R}^3 }u_{n}^2(x,t)dx \leq \lim_{n\to \infty }\int_{ \mathbb{R}^3 }a^2(x,t)dx +\lim_{n\to \infty } \Big( \int_{ \mathbb{R}^3 }u_{n}(x,t) \Big) ^2 = \lim_{n\to \infty }\int_{ \mathbb{R}^3 }a^2(x,t)dx +\Bigg( \lim_{n\to \infty }\int_{ \mathbb{R}^3 }u_{n}(x,t) \Bigg) ^2 $$
The second number is finite because it is given in the question and the second is finite becaue $a \in L^2(\mathbb{R}^3)$