[proofplan] We write each value $X(\omega)$ as the one-dimensional Lebesgue measure of the interval of thresholds $t$ lying below it. This converts $X$ into an integral of the indicator of the subgraph set $\{(\omega,t):t<X(\omega)\}$. Since that indicator is nonnegative and measurable on the product probability-Lebesgue space, Tonelli's theorem permits exchanging the order of integration. The inner integral over $\Omega$ is exactly the tail probability $\mathbb P(X>t)$. [/proofplan]

[step:Declare the product space and the measurable subgraph indicator] Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space in the statement. Let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$, let $\mathcal B([0,\infty)):=\mathcal B(\mathbb R)|_{[0,\infty)}$ denote the Borel $\sigma$-algebra on $[0,\infty)$, and let \begin{align*} X:(\Omega,\mathcal F)&\to([0,\infty),\mathcal B([0,\infty))) \end{align*} be the given measurable map. Let $\mathcal F\otimes\mathcal B([0,\infty))$ denote the product $\sigma$-algebra on $\Omega\times[0,\infty)$, and let $\mathbb P\otimes\mathcal L^1|_{[0,\infty)}$ denote the product measure. Define the open subset \begin{align*} H:=\{(x,t)\in[0,\infty)\times[0,\infty):t<x\}. \end{align*} Define the map \begin{align*} G:(\Omega\times[0,\infty),\mathcal F\otimes\mathcal B([0,\infty)))&\to([0,\infty)\times[0,\infty),\mathcal B([0,\infty))\otimes\mathcal B([0,\infty)))\\ (\omega,t)&\mapsto (X(\omega),t). \end{align*} The first coordinate of $G$ is $X$ composed with the projection $\Omega\times[0,\infty)\to\Omega$, and the second coordinate is the projection $\Omega\times[0,\infty)\to[0,\infty)$; both are measurable, so $G$ is measurable. Hence the set \begin{align*} A:=G^{-1}(H)=\{(\omega,t)\in\Omega\times[0,\infty):t<X(\omega)\} \end{align*} belongs to $\mathcal F\otimes\mathcal B([0,\infty))$. Define its indicator function \begin{align*} \mathbb 1_A:\Omega\times[0,\infty)&\to\{0,1\}\\ (\omega,t)&\mapsto \begin{cases} 1,&(\omega,t)\in A,\\ 0,&(\omega,t)\notin A. \end{cases} \end{align*} Then $\mathbb 1_A$ is a nonnegative measurable function on the product measure space. [/step]

[step:Represent $X$ as the measure of its threshold interval] For each $\omega\in\Omega$, the section \begin{align*} A_\omega:=\{t\in[0,\infty):(\omega,t)\in A\} \end{align*} equals $[0,X(\omega))$. Therefore, by the definition of one-dimensional Lebesgue measure on intervals, \begin{align*} \int_0^\infty \mathbb 1_A(\omega,t)\,d\mathcal L^1(t) = \mathcal L^1([0,X(\omega))) = X(\omega). \end{align*} Integrating this identity with respect to $\mathbb P$ gives \begin{align*} \mathbb E[X] = \int_\Omega X(\omega)\,d\mathbb P(\omega) = \int_\Omega\int_0^\infty \mathbb 1_A(\omega,t)\,d\mathcal L^1(t)\,d\mathbb P(\omega), \end{align*} where the equality is an equality of extended nonnegative integrals. [/step]

[step:Apply Tonelli to exchange the threshold and probability integrals]The function $\mathbb 1_A$ is nonnegative and measurable on $(\Omega\times[0,\infty),\mathcal F\otimes\mathcal B([0,\infty)),\mathbb P\otimes\mathcal L^1|_{[0,\infty)})$. Thus [Tonelli's Theorem](/theorems/???) applies and yields \begin{align*} \int_\Omega\int_0^\infty \mathbb 1_A(\omega,t)\,d\mathcal L^1(t)\,d\mathbb P(\omega) = \int_0^\infty\int_\Omega \mathbb 1_A(\omega,t)\,d\mathbb P(\omega)\,d\mathcal L^1(t). \end{align*}[/step]

[guided]The reason for introducing $\mathbb 1_A$ is that it is nonnegative, so Tonelli's theorem can be used without any prior integrability assumption. We verify the hypotheses carefully. The underlying product measure space is \begin{align*} (\Omega\times[0,\infty),\mathcal F\otimes\mathcal B([0,\infty)),\mathbb P\otimes\mathcal L^1|_{[0,\infty)}). \end{align*} In the first step, the set \begin{align*} A=\{(\omega,t)\in\Omega\times[0,\infty):t<X(\omega)\} \end{align*} was proved to belong to $\mathcal F\otimes\mathcal B([0,\infty))$. Therefore its indicator function \begin{align*} \mathbb 1_A:\Omega\times[0,\infty)&\to\{0,1\} \end{align*} is measurable. It is also nonnegative because its only possible values are $0$ and $1$. Tonelli's theorem applies precisely to nonnegative measurable functions on product measure spaces. Applying it to $\mathbb 1_A$ gives \begin{align*} \int_\Omega\int_0^\infty \mathbb 1_A(\omega,t)\,d\mathcal L^1(t)\,d\mathbb P(\omega) = \int_0^\infty\int_\Omega \mathbb 1_A(\omega,t)\,d\mathbb P(\omega)\,d\mathcal L^1(t). \end{align*} This is the key step: it changes the viewpoint from fixing an outcome $\omega$ and measuring all thresholds below $X(\omega)$, to fixing a threshold $t$ and measuring the set of outcomes for which $X(\omega)>t$.[/guided]

[step:Identify the inner integral with the tail probability] For each $t\in[0,\infty)$, define the event \begin{align*} E_t:=\{\omega\in\Omega:X(\omega)>t\}. \end{align*} Since $X$ is measurable and $(t,\infty)\cap[0,\infty)\in\mathcal B([0,\infty))$, we have $E_t=X^{-1}((t,\infty)\cap[0,\infty))\in\mathcal F$. For every $\omega\in\Omega$, \begin{align*} \mathbb 1_A(\omega,t)=\mathbb 1_{E_t}(\omega). \end{align*} Hence \begin{align*} \int_\Omega \mathbb 1_A(\omega,t)\,d\mathbb P(\omega) = \int_\Omega \mathbb 1_{E_t}(\omega)\,d\mathbb P(\omega) = \mathbb P(E_t) = \mathbb P(X>t). \end{align*} Combining this identity with the preceding two steps gives \begin{align*} \mathbb E[X] = \int_0^\infty \mathbb P(X>t)\,d\mathcal L^1(t). \end{align*} Both sides are extended nonnegative integrals, so the identity remains valid when their common value is $+\infty$. This proves the tail integral formula. [/step]