Show X_n converge in probability if and only if $\lim\limits_{n \to \infty} \int_{-\infty}^{\infty}{\frac{x^2}{1+x^2} dF_n(x)} = 0.$ Suppose that $X_1$, $X_n,\ldots$ is a sequence of random variables with distribution functions $F_1(.), F_2(.), \ldots$ respectively. Show that $X_n \xrightarrow{p} 0$ if and only if
$\lim\limits_{n \to \infty} \int_{-\infty}^{\infty}{\frac{x^2}{1+x^2} dF_n(x)} = 0.$
I know how to prove the "only if" part since we can exchange the limit and integral sign and use the fact $\lim\limits_{n \to \infty} F_n(x) = 0$. Can anyone show me how to prove the other part? Thanks.
 A: The condition
$$
\lim\limits_{n \to \infty} \int_{-\infty}^{\infty}{\frac{x^2}{1+x^2} dF_n(x)} = 0$$
can be rephrased as
$$
\lim\limits_{n \to \infty} \mathbb E\left[\frac{X_n^2}{1+X_n^2}\right]= 0.$$
Note that for each positive $\varepsilon$,
$$
\mathbb E\left[\frac{X_n^2}{1+X_n^2}\right]\geqslant \mathbb E\left[\frac{X_n^2}{1+X_n^2}\mathbf{1}_{\{X_n^2>\varepsilon\}}\right]\geqslant \frac{\varepsilon}{1+\varepsilon}\mathbb P\left(X_n^2>\varepsilon\right)
$$
since the map $t\mapsto t/(1+t)$ is increasing.
Moreover,
$$
\mathbb E\left[\frac{X_n^2}{1+X_n^2}\right]
=\mathbb E\left[\underbrace{\frac{X_n^2}{1+X_n^2}}_{\leqslant 1}\mathbf{1}_{\{X_n^2>\varepsilon\}}\right]+\mathbb E\left[\frac{X_n^2}{1+X_n^2}\mathbf{1}_{\{X_n^2\leqslant \varepsilon\}}\right]
\leqslant \mathbb P\left(X_n^2>\varepsilon\right)+\varepsilon.
$$
We thus got
$$
\frac{\varepsilon}{1+\varepsilon}\mathbb P\left(X_n^2>\varepsilon\right)\leqslant \mathbb E\left[\frac{X_n^2}{1+X_n^2}\right]\leqslant \mathbb P\left(X_n^2>\varepsilon\right)+\varepsilon
$$.
