[proofplan] We first express each value $Z(\omega)$ as the [Lebesgue measure](/page/Lebesgue%20Measure) of the interval of levels below it. This converts $Z$ into the integral of the indicator function $\mathbb 1_{\{Z>u\}}$, and Tonelli's theorem then permits the order of integration to be exchanged because the integrand is nonnegative. The tail bound follows by splitting the resulting tail integral at $a$ and evaluating the remaining Gaussian half-line integral. [/proofplan]

[step:Represent $Z$ by the measure of its superlevel intervals] For $u\in[0,\infty)$, define the [measurable function](/page/Measurable%20Function) \begin{align*} H:\Omega\times[0,\infty)&\to[0,\infty) \end{align*} \begin{align*} (\omega,u)&\mapsto \mathbb 1_{\{Z>u\}}(\omega). \end{align*} The map $H$ is $\mathcal F\otimes\mathcal B([0,\infty))$-measurable because \begin{align*} \{(\omega,u)\in\Omega\times[0,\infty):H(\omega,u)=1\} = \{(\omega,u)\in\Omega\times[0,\infty):u<Z(\omega)\} \end{align*} is measurable as the preimage of the [open set](/page/Open%20Set) $(0,\infty)$ under the measurable map $(\omega,u)\mapsto Z(\omega)-u$. For each $\omega\in\Omega$, the section $u\mapsto H(\omega,u)$ is the indicator of the interval $[0,Z(\omega))$. Hence \begin{align*} \int_0^\infty H(\omega,u)\,d\mathcal L^1(u)=Z(\omega). \end{align*} [/step]

[step:Use Tonelli's theorem to identify the expectation with the tail integral]Since $H$ is nonnegative and measurable on the product [measure space](/page/Measure%20Space) \begin{align*} (\Omega\times[0,\infty),\mathcal F\otimes\mathcal B([0,\infty)),\mathbb P\otimes\mathcal L^1), \end{align*} Tonelli's theorem for nonnegative [measurable functions](/page/Measurable%20Functions) applies. Therefore \begin{align*} \mathbb E[Z] = \int_\Omega Z(\omega)\,d\mathbb P(\omega). \end{align*} Using the pointwise identity from the previous step gives \begin{align*} \mathbb E[Z] = \int_\Omega\left(\int_0^\infty H(\omega,u)\,d\mathcal L^1(u)\right)d\mathbb P(\omega). \end{align*} Tonelli's theorem then yields \begin{align*} \mathbb E[Z] = \int_0^\infty\left(\int_\Omega H(\omega,u)\,d\mathbb P(\omega)\right)d\mathcal L^1(u). \end{align*} For each $u\in[0,\infty)$, \begin{align*} \int_\Omega H(\omega,u)\,d\mathbb P(\omega)=\mathbb P(Z>u). \end{align*} Thus \begin{align*} \mathbb E[Z]=\int_0^\infty \mathbb P(Z>u)\,d\mathcal L^1(u), \end{align*} with equality allowed to take the value $+\infty$.[/step]

[guided]The point of introducing $H$ is to turn the [random variable](/page/Random%20Variable) $Z$ into a two-variable indicator. Define \begin{align*} H:\Omega\times[0,\infty)&\to[0,\infty) \end{align*} \begin{align*} (\omega,u)&\mapsto \mathbb 1_{\{Z>u\}}(\omega). \end{align*} Before applying Tonelli's theorem, we verify its measurability hypothesis. The set on which $H$ equals $1$ is \begin{align*} \{(\omega,u)\in\Omega\times[0,\infty):H(\omega,u)=1\} = \{(\omega,u)\in\Omega\times[0,\infty):u<Z(\omega)\}. \end{align*} The coordinate projection \begin{align*} \pi_2:\Omega\times[0,\infty)&\to[0,\infty) \end{align*} \begin{align*} (\omega,u)&\mapsto u \end{align*} is $\mathcal F\otimes\mathcal B([0,\infty))$-measurable, and the composition \begin{align*} Z\circ\pi_1:\Omega\times[0,\infty)&\to[0,\infty) \end{align*} with the first coordinate projection $\pi_1:\Omega\times[0,\infty)\to\Omega$ is also measurable. Hence the map \begin{align*} G:\Omega\times[0,\infty)&\to\mathbb R \end{align*} \begin{align*} (\omega,u)&\mapsto Z(\omega)-u \end{align*} is measurable, and the preceding superlevel set is $G^{-1}((0,\infty))$. Since $(0,\infty)$ is Borel in $\mathbb R$, this proves that $H$ is $\mathcal F\otimes\mathcal B([0,\infty))$-measurable. For fixed $\omega\in\Omega$, the function $u\mapsto H(\omega,u)$ equals $1$ exactly for those levels $u$ below $Z(\omega)$ and equals $0$ above that level. Therefore \begin{align*} \int_0^\infty H(\omega,u)\,d\mathcal L^1(u)=\mathcal L^1([0,Z(\omega)))=Z(\omega). \end{align*} We now want to integrate this identity over $\Omega$. The integrand $H$ is nonnegative, so Tonelli's theorem applies without any prior integrability assumption. This is the reason the identity is valid even when $\mathbb E[Z]=+\infty$. We obtain \begin{align*} \mathbb E[Z] = \int_\Omega\left(\int_0^\infty H(\omega,u)\,d\mathcal L^1(u)\right)d\mathbb P(\omega). \end{align*} Tonelli's theorem permits the order of integration to be exchanged: \begin{align*} \mathbb E[Z] = \int_0^\infty\left(\int_\Omega H(\omega,u)\,d\mathbb P(\omega)\right)d\mathcal L^1(u). \end{align*} For each fixed $u\in[0,\infty)$, the inner integral is the expectation of the indicator of the event $\{Z>u\}$: \begin{align*} \int_\Omega H(\omega,u)\,d\mathbb P(\omega) = \int_\Omega \mathbb 1_{\{Z>u\}}(\omega)\,d\mathbb P(\omega) = \mathbb P(Z>u). \end{align*} Substituting this into the previous display gives \begin{align*} \mathbb E[Z]=\int_0^\infty \mathbb P(Z>u)\,d\mathcal L^1(u). \end{align*} The equality is an equality of extended nonnegative quantities, so no finite-expectation assumption is being hidden.[/guided]

[step:Split the tail integral at the threshold $a$] Assume now that $a,b>0$ and that \begin{align*} \mathbb P(Z>a+u)\le e^{-u^2/b^2} \end{align*} for every $u>0$. By the tail identity, \begin{align*} \mathbb E[Z] = \int_0^a \mathbb P(Z>u)\,d\mathcal L^1(u) + \int_a^\infty \mathbb P(Z>u)\,d\mathcal L^1(u). \end{align*} Since probabilities are bounded above by $1$, \begin{align*} \int_0^a \mathbb P(Z>u)\,d\mathcal L^1(u)\le a. \end{align*} For the second integral, use the change of variables $v=u-a$, which maps $[a,\infty)$ onto $[0,\infty)$ and preserves one-dimensional Lebesgue measure. Then \begin{align*} \int_a^\infty \mathbb P(Z>u)\,d\mathcal L^1(u) = \int_0^\infty \mathbb P(Z>a+v)\,d\mathcal L^1(v). \end{align*} The assumed tail estimate holds for every $v>0$, and the endpoint $v=0$ does not affect the [Lebesgue integral](/page/Lebesgue%20Integral). Hence \begin{align*} \int_0^\infty \mathbb P(Z>a+v)\,d\mathcal L^1(v) \le \int_0^\infty e^{-v^2/b^2}\,d\mathcal L^1(v). \end{align*} [/step]

[step:Evaluate the Gaussian half-line integral] Define \begin{align*} \phi:[0,\infty)&\to[0,\infty) \end{align*} \begin{align*} v&\mapsto e^{-v^2/b^2}. \end{align*} Using the substitution $w=v/b$, so that $v=bw$ and $d\mathcal L^1(v)=b\,d\mathcal L^1(w)$, we get \begin{align*} \int_0^\infty e^{-v^2/b^2}\,d\mathcal L^1(v) = b\int_0^\infty e^{-w^2}\,d\mathcal L^1(w). \end{align*} By the classical Gaussian half-line integral formula, \begin{align*} \int_0^\infty e^{-w^2}\,d\mathcal L^1(w)=\frac{\sqrt{\pi}}{2}. \end{align*} Therefore \begin{align*} \int_0^\infty e^{-v^2/b^2}\,d\mathcal L^1(v)=\frac{\sqrt{\pi}}{2}b. \end{align*} Combining this estimate with the split tail bound gives \begin{align*} \mathbb E[Z]\le a+\frac{\sqrt{\pi}}{2}b. \end{align*} In particular the right-hand side is finite, so $Z$ is integrable under the stated tail assumption. [/step]