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 $|f_n| \leq M$ on $E$, $\forall n$.
Then, if $\{f_n\} \rightarrow f$ pointwise on $E$, then $\lim_{n \rightarrow \infty} \int_E f_n = \int_E f$.
I was wondering, would this theorem still hold if $m(E) < \infty$ but we drop the assumption that $\{f_n\}$ is uniformly bounded on $E$? I've been thinking about this awhile and feel like I've fallen into tunnel vision, I would appreciate a fresh perspective! Thanks!
Best Answer
No, the classic counterexample is a sequence of triangles of greater and greater height and smaller and smaller width on the interval $[0,1]$ with Lebesgue measure.
The sequence is $$ f_n(x)=\begin{cases} -n^2(x-1/n),&0\leq x\leq 1/n\\ 0,& 1/n<x\leq 1 \end{cases} $$ Then $f_n(x)\to 0=f(x)$ pointwise, $m(E)=\ell([0,1])=1$, but $\int_0^1f_n(x)\mathrm dx=1/2\not\to 0=\int_0^1 f(x)$.
Note the functions are even continuous.