Lebesgue integral of Gaussian process is Gaussian. Let $X:\Omega\times\mathcal{T}\rightarrow\mathbb{R}$ be a measurable stochastic process, Lebesgue integrable on $\mathcal{T}$ for a.e. $\omega\in\Omega$: $$\int_{\mathcal{T}} |X(\omega,t)|\,dt<\infty.$$ Suppose that $X$ is Gaussian, that is, for every $t_1,\ldots,t_m\in\mathcal{T}$ and $m\in\mathbb{N}$ the random vector $(X(t_1),\ldots,X(t_m))$ follows a multivariate normal law. I want to prove that the random variable defined by 
$$\omega\mapsto \int_{\mathcal{T}} X(\omega,t)\,dt $$ is normal.
If the integral were interpreted as a Riemann integral, then we could express $\int_{\mathcal{T}} X(\omega,t)\,dt$ as a limit of Riemann sums, which are clearly normal (see the accepted answer here). But I do not know how to prove that $\int_{\mathcal{T}} X(\omega,t)\,dt$ is normal when the integral is in the Lebesgue sense. We know that the Lebesgue integral is a limit of step functions, but those step functions may not be normal.
 A: Let me assume (for simplicity) that $E[X(t)]=0$ for each $t$. If $Y:=\int_{\mathcal T} X(t)\,dt$ is to be Gaussian, then it must have finite moments. So, let's also assume that 
$$
\int_{\mathcal T} \sqrt{E[X(t)^2]}\,dt<\infty.
$$
This is more than enough (by Cauchy-Schwarz) to ensure that 
$E[Y^2]<\infty$, and will be needed below to justify the use of Fubini's theorem.  Let $L^2$ denote that Hilbert space of square-integrable random variables on the probability space $(\Omega,\mathcal F, P)$ of the $X(t)$, and let $\Gamma$ be the closure in $L^2$ of the linear span of $\{X(t): t\in\mathcal T\}$. Each random variable in $\Gamma$ is normally distributed (with mean $0$). To show that $Y$ is normally distributed it suffices to show that $Y\in\Gamma$; or, what is the same, that $Y\in(\Gamma^\perp)^\perp$. (Here $\perp$ indicates orthogonal complement.) That is, we must show that if $Z\in\Gamma^\perp$ (namely, if $Z\in L^2$ and $E[ZG]=0$ for every $G\in\Gamma$) then $E[ZY]=0$ as well. So fix $Z\in\Gamma^\perp$. Then for each $t\in\mathcal T$, $X(t)\in\Gamma$, and so $E[ZX(t)]=0$.
Consequently, by Fubini's theorem,
$$
E[ZY]=\left[Z\int_{\mathcal T}X(t)\,dt\right]=\int_{\mathcal T} E[ZX(t)]\,dt =0.
$$
