If $\lim\limits_{A \rightarrow \infty} \sup\limits_{n} \frac{\int_{|x|>A}x^2 dF_n(x)}{\int_\mathbb Rx^2 dF_n(x)}=0$ then $\{F_n\}$ is tight 
Suppose $X_n$, $n \geq 1$, are random variables with distribution functions $F_n$ satisfying $EX_n^2 < \infty$ for all $n$ and $$\lim_{A \rightarrow \infty} \sup_{n} \frac{\int_{\{x: |x|>A\}}x^2 dF_n(x)}{\int_{\mathbb R}x^2 dF_n(x)}=0.$$ Show that $\{F_n\}$ is tight.

By the limit in the condition, we know that, when $A$ is large enough, the fraction will be small, and we can obtain an estimate of $\frac{EX_n^2}{A^2}<2$, but I don't know how  to start from here to estimate $$\lim_{A \rightarrow \infty} \sup_{n}P(|X_n|>A).$$
 A: Let us fix $\varepsilon\gt 0$. We want to find some $a$ such that 
$$\sup_n \mathbb P\left\{\left|X_n\right|\gt a\right\}\leqslant 2\varepsilon.$$
We assume that $\varepsilon<1/2$. By assumption, we can find $a$ such that for each $n$, 
$$\mathbb E\left[X_n^2\mathbf 1\left\{\left|X_n\right|\gt a\right\}\right]\lt \varepsilon\mathbb E\left[X_n^2\right]=\varepsilon\mathbb E\left[X_n^2\mathbf 1\left\{\left|X_n\right|\gt a\right\}\right]+\varepsilon\mathbb E\left[X_n^2\mathbf 1\left\{\left|X_n\right|\leqslant a\right\}\right].$$
Bounding the second term of the right hand side by $\varepsilon a^2$, we get $$\left(1-\varepsilon\right)\mathbb E\left[X_n^2\mathbf 1\left\{\left|X_n\right|\gt a\right\}\right]\lt \varepsilon a^2.$$
Using $$\mathbb E\left[X_n^2\mathbf 1\left\{\left|X_n\right|\gt a\right\}\right]\geqslant a^2\cdot\mathbb P\left\{\left|X_n\right|\gt a\right\},
$$ we infer that 
$$\sup_n \mathbb P\left\{\left|X_n\right|\gt a\right\}\leqslant \frac{\varepsilon}{1-\varepsilon}\leqslant 2\varepsilon,$$
which gives tightness.
