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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.

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What do you mean by "just bounded"? There are two ways to relax the condition of uniformly bounded: (1) require that each $f_n$ be a bounded function, or (2) require that $\{f_n\}$ is pointwise-bounded, meaning that $|f_n(x)| \leq \phi(x)$ for some $\phi$. – Jesse Madnick Nov 12 '12 at 7:35
Also, uniform boundedness does not imply uniform convergence. – Jesse Madnick Nov 12 '12 at 7:36
@Jesse I mean pointwise bounded. Uniform convergence implies uniform boundedness. So not uniformly bounded implies not uniformly convergence. – emka Nov 12 '12 at 7:46
Uniform convergence implies uniform boundedness assuming each $f_n$ is a bounded function. For example, consider the sequence $f_n\colon (0,1) \to \mathbb{R}$ given by $f_n(x) = 1/x + 1/n$, which is uniformly convergent but not uniformly bounded. – Jesse Madnick Nov 12 '12 at 18:29
up vote 0 down vote accepted

If you avoid requirement of uniform boundedness then there is a counterexample $$ f_n=n^2 1_{[0,n^{-1}]} $$

But there examples when theorem holds even if sequence of fucntions 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 condition of pointwise convergence just to convergence in measure $$ \forall\varepsilon>0\quad\lim\limits_{n\to\infty}\mu(\{x\in E:|f_n(x)-f(x)|>\varepsilon\})=0 $$

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Consider the example $$f_n(x)=\cases{ nx & if $0\leq x \leq {1\over n}$\cr -n\Bigl(x-{2\over n}\Bigr) & if ${1\over n}\leq x \leq {2\over n}$\cr 0& if ${2\over n}\leq x\leq 1$}$$ Each $f_n$ is piecewise linear and continuous on the interval $[0,1]$, and $f_n(x)\to 0$ as $n\to\infty$ for each $x\in[0,1]$. However, $\int_0^1 f_n(x){\rm d}x=1$ for each $n$, so the conclusion of the Bounded Convergence Theorem does not hold here.

I think this example shows quite clearly what the problem is. You do not want to assume that the convergence is uniform, so there might be a substantial difference between $f(x)$ and each $f_n(x)$ on some set, and you need to make sure that the integral of $|f_n-f|$ over this set will be small as $n\to\infty$.

(Usually this theorem is stated in a slightly stronger form, called the Dominated Convergence Theorem, where the uniform bound $|f_n|\leq M$ is replaced by $|f_n(x)|\leq g(x)$ for each $x$ and $n$, where $g$ is an integrable function.)

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