$\sup_{x \ge 0} |x|^{1 + \varepsilon} \mathbb P[X \gt x] \lt \infty$ for arbitrary $\varepsilon > 0$ implies $X$ integrable

Question

Let $X$ be a real-valued random variable with $\mathbb P[X \ge 0] = 1$. I want to show that $\sup_{x \ge 0} |x|^{1 + \varepsilon} \mathbb P[X \gt x] \lt \infty$ for arbitrary $\varepsilon > 0$ implies that $X$ is integrable.

My attempt is to use that $\mathbb E[X] = \int_{[0,\infty)} \mathbb P[X>x] \lambda(dx)$ but I am not sure how to continue from that.

Answer 1

To elaborate on the comment of Dzoooks (thanks to Kavi Rama Murthy for the correction):

\begin{align} E[|X|] = E[X] = \int_0^\infty P[X > x] \, dx &\le 1 + \int_1^\infty (|x|^{1+\epsilon} P[X > x]) |x|^{-(1+\epsilon)} \, dx\\ &= 1 + c \int_1^\infty |x|^{-(1+\epsilon)} \, dx < \infty \end{align}

The first inequality is due to $\int_0^1 P[X>x] \, dx \le \int_0^1 \, dx$.