# a counter example to the fact that bounded convergence theorem holds if we drop the assumption that sequence is uniformly bounded.

I am searching for a counter example to the fact that bounded convergence theorem holds if we drop the assumption that sequence is uniformly bounded.

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 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 $\lim\limits_{n \to \infty} \int_E f_n = \int_E f.$

• What examples have you tried? – Clayton Mar 12 '17 at 1:41

Let $E=[0,1]$ and define $f_n(x)=(n+1)x^n$. Then $f=\lim f_n\equiv0$ a.e., while $\int_E f_n =1$.

• If you let $E=[0,1)$ then $f_n$ converges everywhere on $E$ to $0.$ – DanielWainfleet Mar 12 '17 at 5:24

Take $E = [0,2]$. Consider this sequence of "triangle functions":

$$f_n(x) = \begin{cases} n^2x & 0 \leq x < \frac 1 n \\ n(2 - nx) & \frac 1 n \leq x < \frac 2 n \\ 0 & \frac 2 n \leq x \leq 2.\end{cases}$$

Is this sequence uniformly bounded or not?

What is the value of $\int_E f_n$ for each $n \in \mathbb N$?

What does $f_n$ converge to pointwise? And what is the integral of this pointwise limit?

A "standard" example, on the interval $(0,1]$ equipped with the Lebesgue measure on the Borel sets: $f_n(x) = n\cdot1_{(0,1/n]}(x)$. Then $\lim_nf_n(x)=0$ for all $x\in(0,1]$ and $\int_{(0,1]}f_n(x)\,dx=1$ for all $n$.