I'm reading Existence of solutions to a higher dimensional mean-field equation on manifolds and get stuck on Lemma6. When $\lambda>\Lambda_1$, with $\Lambda_1=(2 m-1) ! \operatorname{vol}\left(S^{2 m}\right)$. They proved that the following functional is not bounded below so they wanted to find the saddle point $$ I_\lambda(u)=\frac{1}{2} \int_M|\Delta_g^{m/2} u|^2 d \mu_g-\frac{\lambda}{2 m} \log \left(\int_M e^{2 m u} d \mu_g\right) $$ on $$ E:=\left\{u \in H^m(M): \int_M u d \mu_g=0\right\}. $$ And equip $E$ with the norm $\|u\|:=\left(\int_M\left|\Delta_g^{\frac{m}{2}} u\right|^2 d \mu_g\right)^{\frac{1}{2}}$. The critical point of this functional is the solution of the mean-field equation $$\left(-\Delta_g\right)^m u+\lambda=\lambda \frac{e^{2 m u}}{\int_M e^{2 m u} d \mu_g}$$ on a unit volume closed Riemannian manifold $(M, g)$ of dimension $2 m$.

In Lemma5, they have proved that **there exists a bounded sequence $\left(u_n\right)$ in $E$ such that $I_\mu^{\prime}\left(u_n\right) \rightarrow 0$ and $I_\mu\left(u_n\right) \rightarrow c_\mu$.**

Then they assume that $u_n$ converges weakly in $E$ and almost everywhere to a function $u$, and proved that $e^{2 m u_n}$ and $e^{2 m u}$ are uniformly bounded in $L^4$.

My question arises in the next step, they wrote that :

**by dominated convergence one has for $N>0$
$$\tag{1}
\min \left\{e^{2 m u_n}, N\right\} \rightarrow \min \left\{e^{2 m u}, N\right\} \quad \text { in } L^2\left(M, d \mu_g\right)
$$
as $n \rightarrow \infty$ and that
$$
\sup _{n \in \mathbb{N}}\left\|\min \left\{e^{2 m u_n}, N\right\}-e^{2 m u_n}\right\|_{L^2}^2 \leq \frac{1}{N^2} \sup _{n \in \mathbb{N}}\left\|e^{2 m u_n}\right\|_{L^4}^4 \rightarrow 0 \quad \text { as } N \rightarrow \infty,
$$
we infer that $e^{2 m u_n} \rightarrow e^{2 m u}$ in $L^2$.**

I wonder why they need to choose a $N$ and wrote (1) ? Actually I don't even know how the dominated convergence is used here.