probability theory - A more advanced proof for showing $mathbb{E}[X]=int_{0}^{infty}mathbb{P}(Xgeq t)dt$

I want to show that for a non-negative random variable $X: \Omega \to [0,+\infty)$ in a probability space $(\Omega, \mathcal{F}, \mathbb{P})$:

$\mathbb{E}[X]=\int_{0}^{\infty}\mathbb{P}(X\geq t)dt$

I know that I can replace $\mathbb{P}(X\geq t)$ with $\mathbb{P}(X > t)$ and then I'd have:

$\int_{0}^{\infty} (1-F_{X}(x))dx = \int_{0}^{\infty}\int_{x}^{\infty} f_{X}(y)dydx = \int_{0}^{\infty}\int_{0}^{y}dx\: f_{X}(y)dy = \int_{0}^{\infty}yf_{X}(y)dy = \mathbb{E}[X]$

And I know that both sides of the equation can be $\infty$. I established the relation for X being an indicator function, and now I want to establish it for X as a simple non-negative function and then for non-negative measurable functions.

I'd appreciate any help.