Counterexample for Bounded convergence theroem if we relax finite measure condition

Question

I come across Bounded convergence theorem which states that $\left\{f_n\right\}_{n=0}^{\infty}$ sequnce of measurable function on X which is of finite measure such that there $M\in \mathbb R$ such that $|f_n(x)|\leq M, \forall x\in X$ and $f_n(x)\to f$ then $\int f=\lim_{n\to \infty}\int f_n(x)$

I was thinking if we relax finite measure of X then above should not be true.

I thought following example

$f_n(x)=x/n$ and $X=[0,\infty)$.

$\int f_n(x)=\infty$ for every n.

Now but f(x)=0

So done.

Is my example is correct?

Any help will be appreciated

Answer 1

Your example is not counterexample because $x/n$ is not bounded on $S=[0,\infty]$.

One of counterexample for this theorem is $f_n(x)=\frac{1}{n+1}$. It is bounded because $|f(x)|\le1$. And $\lim_{n\to\infty}\int_S f(x)=\infty$, but $\int_S \lim_{n\to\infty} f(x)=\int_S0=0$