For all $n \in \mathbb{N}$ let $f_n:(0,1) \rightarrow \mathbb{R}, f_n= \frac{2nx}{(1+nx^2)^2}$. Futher, let $g:(0,1) \rightarrow \mathbb{R}$ be a function and for all $x \in (0,1)$ and for all $n \in \mathbb{N}$: $|f_n| \leq g$. Show that $g$ is not Lebesgue integrable.
I tried to find an $n$ such that the integral for $f_n$ diverges to then use the dominated convergence theorem, but I couldnt quite find one.
If $g$ was Lebesgue integrable, then by the Dominated Convergence Theorem we would have that
$$\lim_{n\to\infty}\int_{(0,1)} f_n~\mathrm{d}\lambda=\int_{(0,1)} f~\mathrm{d}\lambda,$$
where, for $x\in(0,1)$,
$$f(x)=\lim_{n\to\infty}f_n(x).$$
Let's compute these limits. If $x\in(0,1)$, then
$$f(x)=\lim_{n\to\infty}\frac{2nx}{(1+nx^2)^2}$$
is an indeterminate form on the form $\frac{\infty}{\infty}$, and so, using L'Hôpital's rule, we have that
$$f(x)=\lim_{n\to\infty}\frac{2x}{2(1+nx^2)x^2}=\lim_{n\to\infty}\frac{1}{x(1+nx^2)}=0.$$
But now we also have, using the fact that the Lebesuge integral will equal the Riemann integral, and with the substitution $\phi=1+nx^2$, $\mathrm{d}\phi=2nx~\mathrm{d}x$, that
$$\int_{(0,1)} f_n~\mathrm{d}\lambda=\int_0^1\frac{2nx}{(1+nx^2)^2}~\mathrm{d}x=\int_1^{1+n}\frac{\mathrm{d}\phi}{\phi^2}=\biggl[-\frac{1}{\phi}\biggr]_1^{1+n}=1-\frac{1}{n+1}.$$
But this means that
$$\lim_{n\to\infty}\int_{(0,1)} f_n~\mathrm{d}\lambda=1$$
while
$$\int_{(0,1)} f~\mathrm{d}\lambda=0,$$
and so, as $1\neq 0$, such a $g$ cannot be Lebesgue integrable.
