Consider the sequence of functions $f_n = n \chi_{[0, \frac{1}{n}]}$ defined on $\mathbb R$, which is equipped with Lebesgue measure, and take the sigma-algebra of subsets of $\mathbb R$ to be the Borel sigma-algebra.
In a previous question (found here), I asked for help in showing that $f_n \to 0$ almost everywhere.
I then continued to work out the following: \begin{align*} \lim \int f_n d\lambda &= \lim \int n\chi_{[0, \frac{1}{n}]} d\lambda \\ &= \lim n \int \chi_{[0, \frac{1}{n}]} d\lambda \\ &= \lim n \lambda([0, \frac{1}{n}]) \\ &= \lim n\frac{1}{n} \\ &= \lim 1 \\ &= 1, \end{align*} i.e. $\int f_n d\lambda$ does not converge to zero.
Can anyobody perhaps explain to me why? I understand that $f_n$ is unbounded for $x \in [0, \frac{1}{n}]$ and so we cannot apply the Lebesgue Dominated Convergence Theorem, but I see that we do have that $f$ is monotonically increasing and $f \in M^+$, so should the Monotone Convergence Theorem not apply?
Can somebody perhaps explain to me why it does not converge to zero?
The sequence $f_n = n\chi_{]0,1/n]}$ is not monotone: when $x\in (1/2,1]$, $f_1(x)=1$ while $f_n(x) = 0$ for all $n\ge 2$. The only convergence result you can use is the Fatou's lemma, $$\int_0^1 f \le \liminf_{n\to \infty} \int_0^1 f_n ,$$ but it says nothing interesting.