I'm trying to furnish a proof for the following identity, and would like to have my proof checked.

Suppose $(\Omega, \mathcal{F}, \mathbf{P})$ is a probability space on which random variable $X: \Omega \to \mathbf{R}$ is defined. If $X$ is nonnegative, then $$ \mathbf{E} X= \int_0^\infty \mathbf{P}\{X \geq t\}\, dt. $$

Here's the proof: $$ \mathbf{E}X = \int_\Omega X(\omega) d\mathbf{P}(\omega) = \int_\Omega \left(\int_{\mathbf{R}_+} \mathbf{1}_{[0, X(\omega)]}(t) ~dt \right) ~d\mathbf{P}(\omega). $$ The first equality is trivial: $\lambda([0, X(\omega)]) = X(\omega)$, where $\lambda$ denotes Lebesgue measure.

By Fubini's Theorem, we may interchange the measures. $$ \mathbf{E}X = \int_{\mathbf{R}_+} \left(\int_\Omega\mathbf{1}_{[0, X(\omega)]}(t) ~d\mathbf{P}(\omega) \right) ~dt = \int_0^\infty \mathbf{P}\{X \geq t\} \, dt. $$ The last equality arises by obseving that for $t \geq 0$, $0 \leq t \leq X(\omega)$ iff $X(\omega) \geq t$, and the set $X^{-1}([t, \infty)) \in \mathcal{F}$.

The only thing that isn't immediate is the use of Fubini: the map $(t, \omega) \mapsto \begin{cases} 1, & X(\omega) \geq t \\ 0 & \text{else} \end{cases}$ needs to be $\mathcal{F} \otimes \mathcal{B}_\mathbf{R}$ measurable, i.e., the set $$ A = \{(\omega, t) : X(\omega) \geq t\} $$ needs to be measurable. If $\phi_n \uparrow X$ are nonnegative simple functions, then $A_0 = \cap_n A_n$, where $$ A_n = \{(\omega, t) : X(\omega) - t > -1/n\}. $$ Notice that $A_n = \cup_m \{ (\omega, t) : \phi_m(\omega) - t > -1/n\}$. Note that $\phi_m = \sum_i c_i \mathbf{1}_{F_i}$, where $F_i$ are disjoint and $c_i > 0$. Finally, notice that $$ \{ (\omega, t) : \phi_m(\omega) - t > -1/n\} = \cup_{i=1}^n (F_i \times [0, c_i + 1/n)) \cup (\cap_i F_i^c) \times [0, 1/n). $$ This set is hence measurable, and now the result follows since the product sigma algebra is closed under intersection and union.

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    $\begingroup$ Yes. $ $ $ $ $ $ $\endgroup$ – Did Dec 10 '18 at 2:51
  • $\begingroup$ Your question is very specific, and user Did has already given you the confirmation. Nonetheless, I'd still like to make a link to a closely related post. $\endgroup$ – Lee David Chung Lin Dec 10 '18 at 4:33

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