Bounded Convergence Theorem Proof
Sebastian Wright 
If $(f_n)$ is a pointwise convergent, uniformly bounded sequence of measurable functions on and interval $I:[0,1]$, then:
$$\begin{align*} \lim_{n\to\infty} \int_I f_n\, d\mu = \int_I f\, d\mu \end{align*}$$
Are we really necessary to use Fatou's Lemma here?
$\endgroup$ 51 Answer
$\begingroup$Here is a proof of the Bounded Convergence Theorem using Egorov's Theorem:
Egorov's Theorem: Let $\forall n: f_n:E\to\mathbb{R}$ be measurable, $m(E)<\infty, f_n\to f$ on $E$. Then $\forall \epsilon>0, \exists F_\epsilon\in\tau^c: F_\epsilon\subseteq E, m(E-F_\epsilon)<\epsilon$ and $f_n\stackrel{u.}{\to} f$ on $F_\epsilon$.
The Bounded Convergence Theorem: Let $\forall n: f_n:E\to\mathbb{R}$ be measurable, $m(E)<\infty, f_n\to f$ on $E$. Then if $\exists M\geq0,\forall n,\forall x\in E: |f_n(x)|\leq M$, then $\int_E f_n\to\int_E f$.
Proof of the Bounded Convergence Theorem:
If $m(E)=0$, then $\int_E f_n=0\to0=\int_E f$, so suppose $m(E)>0$.
Let $\epsilon>0$. Since $\{f_n\}_n$ is uniformly bounded by $M$ and $f_n\to f$ pointwise, $\forall x\in E,\exists N': |f(x)|\leq |f_{N'}(x)|+1 \leq M+1$, so that $f$ is bounded, and consequently $\{|f_n-f|\}_n$ is uniformly bounded by $2M+1$.
By Egorov's Theorem, $\exists F\in\tau^c: F\subseteq E, m(E-F)<\dfrac{\epsilon}{2(2M+1)}$ and $f_n\stackrel{u.}{\to} f$ on $F$. Further, since $f_n\stackrel{u.}{\to} f$ on $F$ we have that $\exists N,\forall n\geq N,\forall x\in F: |f_n(x)-f(x)|<\dfrac{\epsilon}{2m(E)}$. Then for $n\geq N$ we have
\begin{align} 0&\leq\left|\int_E f_n-\int_E f\right| \leq\int_E|f_n-f|\\ &= \int_F |f_n-f| + \int_{E-F} |f_n-f| \\ &< \int_F \dfrac{\epsilon}{2m(E)} + \int_{E-F} (2M+1) \\ &=\dfrac{\epsilon}{2m(E)}m(F)+ (2M+1)m(E-F) \\ &<\dfrac{\epsilon}{2}+(2M+1)\dfrac{\epsilon}{2(2M+1)}=\epsilon.\checkmark \end{align}
$\endgroup$
