Integral of a Gaussian process

Let $(\Omega,\Sigma,P)$ be a probability space and $X: [0,\infty) \times \Omega \to \mathbb{R}$ be a Gaussian process (i.e. all finite linear combinations $\sum_i a_i X_{t_i}$ are Gaussian random variables). If the process is continuous, it seems to be clear that the process $Y_t (\omega) = \int_0^t X_s(\omega) ds$ is a Gaussian process.

Is it true that $Y_t$ is a Gaussian process even if $X$ is only assumed to be measurable? I am working through the derivation of Kalman Filter in the text by Bernt Oksendal Eq 6.2.10 (fifth edition) and this seems to be a way to show that $M_t$ (in the book) is a Gaussian process.

Question 1: Is $Y_t(\omega)$ well-defined?

Answer: No, in general, $Y_t(\omega)$ is not well-defined; we need some additional assumption on the integrability of $X$ to ensure that $$\int_0^t |X_s(\omega)| \, ds <\infty$$ for $t>0$. This is e.g. satisfied if $X$ has continuous sample paths or $\sup_{t \leq T} \mathbb{E}(|X_t|)<\infty$ for any $T>0$. (To see that it $Y_t$ is in general not well-defined, just consider $X_t := t^{-1} Z$, $t>0$, for $Z \sim N(0,1)$; then $X$ is Gaussian, but the integral $\int_0^t X_t \, ds$ does not exist.)

Question 2: Is $\omega \mapsto Y_t(\omega)$ a random variable for fixed $t \geq 0$?

Answer: If the process $X: (0,\infty) \times \Omega \to \mathbb{R}$ is jointly measurable, then $Y_t$ is a random variable for each $t \geq 0$. Otherwise, measurability of $Y_t$ might fail.

Question 3: Is $(Y_t)_{t \geq 0}$ Gaussian?

Answer: If $t \mapsto X_t(\omega)$ is Riemann integrable, this follows by approximation the integral by Riemann sums; see e.g. this question. (Note that a bounded function $f:[0,T] \to \mathbb{R}$ is Riemann integrable if, and only if, the points in $[0,T]$ where $f$ is discontinuous is a Lebesgue null set.)

Edit: Okay, so somewhat more detailed: For any (Riemann) integrable function $f:[0,t] \to \mathbb{R}$ it is known that the (Riemann) integral

$$\int_0^t f(s) \, ds$$

can be approximated by Riemann sums

$$\sum_{j=0}^{n-1} f(s_j) (t_{j+1}-t_j)$$

where $0=t_0 < \ldots <t_n = t$ is a partition of the interval $[0,t]$ and $s_j \in [t_j,t_{j+1}]$. In particular, if we choose $s_j = t_j := t \frac{j}{n}$, we find

$$\int_0^t f(s) \, ds = \lim_{n \to \infty} \frac{t}{n} \sum_{j=0}^{n-1} f \left( t \frac{j}{n} \right).$$

Applying this in order (stochastic) setting, we get

$$\int_0^t X_s \, ds = \lim_{n \to \infty} \frac{t}{n} \sum_{j=0}^{n-1} X_{t j/n};$$

and $\frac{t}{n} \sum_{j=0}^{n-1} X_{t j/n}$ is Gaussian because $X$ is Gaussian.