Is this a special case of the bounded convergence theorem? The proposition in question: 

Let $\{f_n\}$ be a sequence of bounded measurable functions on a set of finite measure $E$.  If $\{f_n\} \rightarrow f$ uniformly on $E$, then $\lim_{n \rightarrow \infty} \int_E f_n = \int_E f$. 

Now the bounded convergence theorem: 

Let $\{f_n\}$ be a sequence of measurable functions on a set of finite measure $E$.  Suppose $\{f_n\}$ is uniformly pointwise bounded on $E$ (that is, there exists $M \geq 0$ for which $\left|f_n\right| \leq M$ on $E$ for all $M$).  If $\{f_n\} \rightarrow f$ pointwise on $E$, then $\lim_{n \rightarrow \infty} \int_E f_n = \int_E f$.

I know it is, I'm just having some issues showing that it is.  
 A: It does not seem so clear that the proposition is a consequence of the bounded convergence theorem. Indeed, what we understand from "a sequence of bounded measurable functions" is that is element of the sequence is bounded, but the bound may a priori depend on $n$. However, with the definition of uniform convergence with $\varepsilon=1$, we know that there exists a $n_0$ such that for all $n\geqslant n_0$, $\sup_{x\in E}\left\lvert f_n(x)-f(x)\right\rvert\leqslant 1$. In particular, with $n=n_0$, we get that 
$$
\sup_{x\in E}\left\lvert  f(x)\right\rvert\leqslant \sup_{x\in E}\left\lvert f_{n_0}(x)-f(x)\right\rvert+\sup_{x\in E}\left\lvert f_{n_0}(x)\right\rvert\leqslant 1+\sup_{x\in E}\left\lvert f_{n_0}(x)\right\rvert.
$$
Therefore, for $n\geqslant n_0$, 
$$
\sup_{x\in E}\left\lvert  f_n(x)\right\rvert\leqslant 2+\sup_{x\in E}\left\lvert f_{n_0}(x)\right\rvert.
$$
hence letting 
$$
M:=2+\max_{1\leqslant k\leqslant n_0}\sup_{x\in E}\left\lvert f_{k}(x)\right\rvert
$$
works for the use of the bounded convergence theorem.
