show that $g$ is not Lebesgue integrable

Question

For all $n\in \mathbb N$ let $f_n : (0,1)\rightarrow \mathbb R, x\mapsto \frac{2nx}{(1+nx^2)^2}$. Furthermore let $g: (0,1) \rightarrow \mathbb R$ be a function so that for all $x\in (0,1)$ and all $n\in \mathbb N$ $|f_n(x)|\leq g(x)$. Show that $g$ is not Lebesgue integrable.
I think I have to show that $\int_{(0,1)}|g| d\lambda=\infty$, right? How can I show this, maybe with convergence theorems?

Answer 1

If possible let $g$ be integrable i.e. $\displaystyle\int_{(0,1)}g\,d\lambda <\infty$

Then we have by Dominated Convergence theorem that

$$\int_{(0,1)}\lim_{n\to\infty}f_{n}\,d\lambda=\lim_{n\to\infty}\int_{(0,1)}f_{n}\,d\lambda$$

$\displaystyle f_{n}=\begin{cases} 0\,,x=0\\\dfrac{\frac{2x}{n}}{\frac{1}{n^{2}}+x^{4}+\frac{2x^{2}}{n}}\,,x\neq 0\end{cases}$

So $f_{n}\to 0$ pointwise

Hence we have $$\int_{(0,1)}\lim_{n\to\infty}f_{n}\,d\lambda = \int_{(0,1)}\,0\,d\lambda =0 $$ .

But $\int_{(0,1)}f_{n}=\int_{0}^{1}\frac{2nx}{(1+nx^{2})^{2}}\,dx$ .

Substitute $1+nx^{2}=t$ to get $2nx\,dx=dt$ and hence

$$\int_{0}^{1}f_{n}(x)dx = \int_{1}^{n+1}\frac{1}{t^{2}}\,dt = 1-\frac{1}{n+1}$$

This means that $$\lim_{n\to\infty}\int_{(0,1)}f_{n}\,d\lambda = \lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)=1$$

Which implies that $0=1$ a contradiction.

Hence $g$ cannot be integrable.