probability - How does $E(|X|)=int_0^{infty}P[|X|ge x]dx$?

As the title states, how does $E(|X|)=\int_0^{\infty}P[|X|\ge x]dx$ ? The only assumption being that $E(|X|)\le \infty$

Mybe I can use the identity function in some way, since $E[1_{X\ge x}]=P[X\ge x]$?

Thanks in advance!

Answer

This is, at its heart, a consequence of Tonelli's Theorem, which is a lot like Fubini's Theorem. For any non-negative random variable $Y$ with finite expectation, you can write
$$
\begin{align*}
\int_0^{\infty}P(Y\geq y)\,d\mu(y)&=\int_0^{\infty}\int_{\Omega}1_{\{Y(\omega)\geq y\}}\,dP(\omega)\,d\mu(y)\\
&=\int_{\Omega}\int_0^{\infty}1_{\{Y(\omega)\geq y\}}\,d\mu(y)\,dP(\omega)\\
&=\int_{\Omega}Y(\omega)\,dP(\omega)\\
&=\mathbb{E}[Y],
\end{align*}
$$

where $(\Omega,\mathcal{F},P)$ is our probability space and $\mu$ is Lebesgue measure on $\mathbb{R}$.

(Note that we can definitely apply Tonelli's Theorem here, as $P$ and $\mu$ are both $\sigma$-finite and $1_{\{Y(\omega)\geq y\}}$ is a non-negative function.)