Finite expectation of a random variable

Question

$X \geq 0$ be a random variable defined on $(\Omega,\mathcal{F},P)$. Show that $\mathbb{E}[X]<\infty \iff \Sigma_{n=1}^\infty P(X>n) < \infty $.

I got the reverse direction but I am struggling with the $"\implies"$ direction. So far, I have the following worked out:

$\mathbb{E}[X]<\infty$

$\implies \int_0^\infty (1-F(x)) dx < \infty$ (where $F$ is the distribution function of the random variable X)

$\implies \int_0^\infty (1-P(X\leq x)) dx < \infty$

$\implies \int_0^\infty P(X>x) dx < \infty$

Consider $\int_0^\infty P(X>x) dx$

$= \Sigma_{n=1}^\infty \int_{n-1}^n P(X>x) dx$

This is the point I am stuck at. Any help will be deeply appreciated!

Answer 1

Hint: $x \ge \sum_{n=1}^\infty I_{\{x \ge n\}} \ge x-1$ for $x \ge 0$, where $I_A$ is the indicator of $A$ ($1$ when $A$ is true, $0$ when $A$ is false).