While proving the bounded convergence theorem my book makes use of Egoroff's theorem and then proceeds to take limsup and liminf's to reach it's conclusion. I'm confused as to why taking the limit is not justified here.
Bounded Convergence Theorem
Given a measure space $(D, \Sigma, \mu)$. Let $(f_n,\; n\in \mathbb{N})$ be a uniformly bounded sequence of real valued $\Sigma$-measurable functions on a set $D \in \Sigma$ with $\mu(D)<\infty$. Let $f$ be a bounded real-valued $\Sigma$-measurable function on $D$. If $(f_n\; n\in \mathbb{N})$ converges to $f$ a.e. on $D$ then \begin{equation} \lim_{n\rightarrow \infty} \int_D |f_n-f|\;d\mu = 0 \end{equation}
Sketch Proof
By Egoroff's theorem, for every $\eta > 0$ there exits a set $E$ such that $\mu(E)<\eta$ such that $f_n$ converges to $f$ uniformly on $D\setminus E$. Now \begin{align} \int_D |f_n-f|\;d\mu &= \int_{D\setminus E}|f_n-f|\;d\mu + \int_E |f_n-f|\;d\mu \\ &= \int_{D\setminus E}\epsilon\;d\mu + \int_{E}2M\;d\mu \\ &\leq \epsilon \mu(D\setminus E) + 2M \mu(E) \\ &< \epsilon \mu(D)+2M\eta \end{align} At this point, why can I not take limits to conclude that \begin{equation} \lim_{n\rightarrow \infty} \int_D |f_n-f|\;d\mu \leq 0 \end{equation}
The proof in my book takes the limsup and liminf and uses the squeeze theorem to conclude. I'm happy that if $f_n \rightarrow f$ pointwise I would need to follow that approach as the limit may not exist. As convergence is uniform I was under the impression that the limit would exist.
Many thanks for all the help.