[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 $+\infty$.
[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]
[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]
