Why this doesn't contradict Monotone and Dominated Convergence Theorem?

Question

$$\lim_{n \to \infty} \int_0^\infty f_n(x)dx \ne \int_0^\infty \lim_{n \to \infty} f_n(x)dx$$

where : $f_n=ne^{-nx}$ for all $x \in [0,\infty)$ $n \in \mathbb{N}$

Can somebody help me with explanation?

Answer 1

The sequence $\{f_n\}$ is not increasing, so that discards MCT.

For DCT, I cannot find an easy argument that any function $g$ with $g\geq f_n$ for all $n$ cannot be integrable. But it has to be the case, because otherwise DCT would give equality.

Answer 2

Note that $e^{-nx} \geq 1-nx$. Therefore if $x \in \left [0,\frac{1}{2n} \right ]$, $f_n(x) \geq n/2$, so $\int_0^{\frac{1}{2n}} |f_n(x)| dx \geq 1/4$. Since $\lim_{n \to \infty} \frac{1}{2n} = 0$, we conclude that $f_n$ is not uniformly integrable. Any dominated sequence is uniformly integrable, so your sequence is not dominated either. It's also easy to see that it is not monotone.