In the proof of Theorem 4.14 in Brézis book, we have $(\Omega_n)$ be a sequence of measurable sets in $\Omega$ such that $\Omega=\bigcup_{n=1}^{\infty} \Omega_n$ and $|\Omega_n|< \infty$ for all $n \in \mathbb{N}$ and $\chi_n=\chi_{\Omega_n}$. In the last part of the proof the author consider $$T_n h=\begin{cases}h, & \hbox{if} |h| \leq n,\\ \frac{nh}{|h|}, & \hbox{if} |h|>n.\end{cases}$$
Finally there is a statement: choose $g=T_n h$ (truncation of $h$) in (13) and to observe that $\chi_n T_n h \rightarrow h$ in $L^1(\Omega)$ for all $h \in L^1(\Omega)$.
My question: How to prove that $\chi_n T_n h \rightarrow h$ in $L^1(\Omega)$ for all $h \in L^1(\Omega)$?
I already checked that $\chi_n T_n h \rightarrow h$ pointwisely, I'm stuck to show that this sequence is dominated in order to apply the Lebesgue's Dominated Convergence Theorem.
Notice that $h/|h|=\textrm{sgn}(h)$. Therefore $T_nh = h$ if $|h|<n$ while $T_nh=-n$ if $h\leq -n$ and $T_nh=n$ if $h \geq n$. Therefore $|T_nh|\leq |h|\in L^1$. So $|\chi_nT_nh-h|\leq\chi_n|T_nh|+|h|\leq 2|h|\in L^1$ and you use DCT to prove $L^1$-convergence.