Step in the proof of the closure theorem for SBV I'm working through a section of Ambrosio's Functions of Bounded Variation and Free Discontinuity Problems and have gotten stuck at a step in the proof of the closure theorem for $SBV(\Omega)$ (details of this aren't very important for my question, I think).
We have a sequence $(u_k)_k$ converging in $L^1$ to some $u$, a uniformly integrable sequence $(\nabla u_k)_k$ converging weakly in $L^1$ to some $a$ and a Lipschitz function $\psi \in C^1(\mathbb{R})$. The author claims that we can then use Vitali's dominated convergence theorem to conclude
$$
(\psi'(u_k) - \psi'(u)) \nabla u_k\xrightarrow[] {L^1} 0 \,.
$$
I've been thinking about this for a couple of days now, but I really can't see how to get there. In particular, I don't know how to conclude that this sequence goes to 0 almost everywhere without having bounds on $\Vert \nabla u_k \Vert$. What am I missing?
 A: Using the helpful advice of @GiorgiMetafune in the comments, this is the solution I came up with:

Fix $\epsilon > 0$, and denote the Lipschitz constant of $\psi$ by $K$. By uniform integrability of $(\nabla u_k)_k$, there exists $\delta > 0$ such that
$$
\sup_{k \geq 1} \int_E \Vert \nabla u_k \Vert \, dx \leq \epsilon / 4K 
$$
whenever $\lambda(E) \leq \delta$. Using Egorov's theorem, we obtain a measurable subset $G \subseteq \Omega$ such that $\lambda(\Omega \setminus G) \leq \delta$  and $\psi'(u_k) \xrightarrow[] {L^\infty} \psi'(u)$ on $G$. Now we can split the integral
$$
\int_\Omega \vert \psi'(u_k) - \psi'(u) \vert \Vert \nabla u_k \Vert \, dx
$$
over $G$ and its complement. For the first component, we use the uniform convergence and the boundedness of the $L^1$-norms of $(\nabla u_k)_k$ to obtain
$$
\int_G \vert \psi'(u_k) - \psi'(u) \vert \Vert \nabla u_k \Vert \, dx \leq \vert \psi'(u_k) - \psi'(u) \vert_\infty \sup_{k \geq 1} \int_G \Vert \nabla u_k \Vert \, dx \,,
$$
and thus we have $\int_G \vert \psi'(u_k) - \psi'(u) \vert \Vert \nabla u_k \Vert \, dx \leq \epsilon / 2$ for $k$ sufficiently large.
The second component can be estimated by using the boundedness of $\psi'$ and the fact that $\lambda(\Omega \setminus G) \leq \delta$:
$$
\int_G \vert \psi'(u_k) - \psi'(u) \vert \Vert \nabla u_k \Vert \, dx \leq 2K \sup_{k \geq 1} \int_G \Vert \nabla u_k \Vert \, dx \leq \epsilon / 2
$$

