# 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.

• There is an $n_0$ such that $\lvert f_n(x) - f(x)\rvert \leqslant 1$ for all $x\in E$ and all $n \geqslant n_0$. – Daniel Fischer Nov 4 '16 at 20:35

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.