Hello @DanielFischer!!!
Suppose that we have $x=(\overline{x_k}), y=(\overline{y_k}) \in \mathbb{Z}_p \setminus{\{0\}}$, there are $n,m \in \mathbb{N}$ such that $x_n \not\equiv 0 \mod{p^{n+1}}$, $y_m \not\equiv 0 \mod{p^{m+1}}$. This means that $x_n=p^{n'}u_0, u_0 \in \mathbb{Z}_p^{\star}, n' \leq n$, $y_m=p^{m'}v_0, v_0 \in \mathbb{Z}_p^{\star}, m' \leq m$. Then we have that $x \cdot y=p^{n'+m'}u_0v_0$. We set $l=n+m+1$. We have that $x_l \not\equiv 0 \mod{p^{n+1}}$ and $y_l \not\equiv 0 \mod{p^{m+1}}$. How do we deduce from this that that the greatest power of p that divides $x_l$ is $p^…

@DanielFischer So can we justify as follows that $x \cdot y \neq 0$?

$x=(\overline{x_k}), y=(\overline{y_k}) \in \mathbb{Z}_p \setminus{\{0\}}$, so there are $n,m \in \mathbb{N}$ such that $x_n \not\equiv 0 \mod{p^{n+1}}$, $y_m \not\equiv 0 \mod{p^{m+1}}$. This means that $x_n=p^{n'}u_0, u_0 \in \mathbb{Z}_p^{\star}, n' \leq n$, $y_m=p^{m'}v_0, v_0 \in \mathbb{Z}_p^{\star}, m' \leq m$. Then we have that $x \cdot y=p^{n'+m'}u_0v_0$.

Since $n' + m'\in \mathbb{N}$, which follows from $n',m'\in\mathbb{N}$ and the product of units is a unit again we conclude that $x \cdot y \neq 0$.

@DanielFischer So is it right like that? :/
$x=(\overline{x_k}), y=(\overline{y_k}) \in \mathbb{Z}_p \setminus{\{0\}}$, so there are $n,m \in \mathbb{N}$ such that $x_n \not\equiv 0 \mod{p^{n+1}}$, $y_m \not\equiv 0 \mod{p^{m+1}}$.

This means that $x=p^{n'}u_0, u_0 \in \mathbb{Z}_p^{\star}, n' \leq n$.
If we had $n'>n$ then $x_n=p^{n'}u_0 \mod{p^{n+1}}=p^{n+1+(n'-n-1)}u_0 \mod{p^{n+1}}$ and $n'-n-1 \geq 0$, so $x_n \equiv 0 \mod{p^{n+1}}$, which is a contradiction.

In the same way, $y=p^{m'}v_0, v_0 \in \mathbb{Z}_p^{\star}, m' \leq m$. Then we have that $x \cdot y=p^{n'+m'}u_0v_0$.

@DanielFischer Let $\Omega$ be open bounded in $\mathbb{R}^{N}$. Consider Caratheodory function $c: \Omega \times \mathbb{R} \rightarrow \mathbb{R}$. Also, assume $u_{n}(x) \rightarrow u(x)~~a.e.~x \in \Omega$, then by the continuity of the Caratheodory function $c$, it follows that $c(x,u_{n}) \rightarrow c(x,u)~~a.e. x \in \Omega$.

Also assume that $|c(x,u_{n})| \leq k$, for some $k > 0$.

Can you see how it follows by the Dominated Convergence Theorem that results in
$$\int_{\Omega}c(x,u_{n})dx \rightarrow \int_{\Omega}c(x,u)dx$$