Jtdylktuy

say we're in a probability space $(Omega, Sigma, mathbb{P})$ and $X : (Omega, Sigma) to (mathbb{R}, mathscr{B}_{mathbb{R}})$ is random variable

I resulted in the right result but I feel like the math is bad, please check it :

$$begin{align} & int_{0}^{infty}mathbb{P}(X geq t) dt = int_0^{infty}int_{X geq t}dmathbb{P},dt ,,,,\
= & int_{X geq 0}int_{0}^{X}dt,dmathbb{P} ,,(star) = int_{X geq 0}X,dmathbb{P} \
= & int_{Omega}X,dmathbb{P} ,,,, text{due to non-negativity} = mathbb{E}[X]
end{align}$$

where I'm not sure I'm doing legal stuff is $(star)$, I used Fubini-Tonelli theorem to change order of integration, but when I changed bounds of integration I reasoned the following way :

we have : $tgeq 0$ and $Xgeq t$ meaning on the one hand we have $X geq 0$ and $ 0 leq t leq X $

am I doing this right ?

The integral before $(star)$ can be written as $int_Omegaint_{mathbb R}f(t,omega)mathrm dtmathrm dP(omega)$, where $f(t,omega)=1$ if $X(omega)geqslant t$ and $0$ otherwise. Then Fubini's theorem can be used in this setting.