Theorem (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 is a number $M\geq 0$ for which $|f_n| \leq M$ for all $n$. If $\{f_n\} \to f$ pointwise on $E$, then $\lim\limits_{n \to \infty} \int_E f_n = \int_E f.$
Why is it important in this theorem for it to be uniformly pointwise bounded as opposed to just pointwise bounded? Is this because if it is not uniformly bounded than it is certainly not uniformly convergent.
Best Answer
If you avoid the requirement of uniform boundedness then there is a counterexample $$ f_n=n^2 1_{[0,n^{-1}]} $$
But there are examples when the theorem holds even if the sequence of functions is not uniformly pointwise bounded. For example $$ f_n=n^{1/2}1_{[1,n^{-1}]} $$
The most general requirement on boundedness of $f_n$ when theorem still holds is $$ \forall n\in\mathbb{N}\quad\forall x\in E\quad |f_n(x)|\leq F(x) $$ for some integrable $F:E\to\mathbb{R}_+$. You can also weaken the condition of pointwise convergence just to convergence in measure $$ \forall\varepsilon>0\quad\lim\limits_{n\to\infty}\mu(\{x\in E:|f_n(x)-f(x)|>\varepsilon\})=0 $$