I am not very familiar with the properties of Riemann-Stieltjes Integral and I would like your opinion if I have correctly solved the following exercise.
Let $X$ a random variable such that $F(u)=\mathbb{P}(|X| >u)\leq e^{-u}$, for every $u>0$. Show that there exists $c>0$ such that, for any $n \in \mathbb{N}$,
\begin{equation} \mathbb{E}\left[X^2\boldsymbol{1}_{A}\right] \leq cn^{2/3}e^{-n^{1/3}}, \end{equation}
where and $A= \{\mid X\mid > n^{1/3}\}$.
$\textbf{ My solution}$:
Let us start by writing $G(u):=1-F(u)$. Note that G is the distribution function of $X$ and $dG(u)=d(1-F(u))=-dF(u)$. (Is this right?). Then,
\begin{equation} \mathbb{E}\left[ X^{2} \boldsymbol{1}_{A}\right] = -\int_{-\infty}^{-n^{1/3}}x^{2}dF(x) -\int_{n^{1/3}}^{\infty}x^{2}dF(x) \leq - \int_{n^{1/3}}^{\infty}x^{2}dF(x). \end{equation} (Why?)
But, \begin{equation} \begin{split} - \int_{n^{1/3}}^{\infty}x^{2}dF(x) & = - \lim_{M \to \infty}\left[M^2F(M) - n^{2/3}F(n^{1/3})-2 \int_{n^{1/3}}^{M}xF(x)dx \right]\\ & = n^{2/3}e^{-n^{1/3}} +2 \int_{n^{1/3}}^{\infty}xe^{-x}dx. \end{split} \end{equation}
In the above equality I used Integration by parts for Riemann–Stieltjes Integral and that $\lim_{ M \to \infty} M^2F(M) =0$. (Is this also right?)
Since $\int_{n^{1/3}}^{\infty}xe^{-x}dx= (1+ n^{1/3})e^{-n^{1/3}}$(simple calculation), we conclude that \begin{equation} \mathbb{E}\left[ X^{2} \boldsymbol{1}_{A}\right] \leq n^{2/3}e^{-n^{1/3}} + 2(1+ n^{1/3})e^{-n^{1/3}}\leq cn^{2/3}e^{-n^{1/3}}, \end{equation} for some constant $c>0$.
I will be very grateful for the possible corrections. Thank you!