Consider a mean-zero ($\mu = 0$), unit-variance ($\sigma^2$) Gaussian random process $X(t)$. This process is strictly stationary (the joint-probability distribution does not vary with $t$). The covariance function is isotropic (covariance structure does not depend on $t$) and is given by:
$$C(\tau) = \exp\left(-\frac{\tau^2}{\theta}\right)$$
Where $\tau = t_1 - t_2$.
I am interested in the case where $\theta \rightarrow 0$. In my previous post it was explained that the covariance function $C(\tau)$ becomes the Kronecker delta function:
$$C(\tau) = \delta_\tau = \begin{cases} 0 &\text{if } \tau \neq 0, \\ 1 &\text{if } \tau = 0. \end{cases} $$
However according to Stochastic Filtering Theory by Kallianpur (1980, p. 10) a process with this covariance function is not measurable. AFAIK this means that the integral of such a process is not defined. Consider the following stochastic integral:
$$I = \int_0^LX(t)\, dt$$
Is it correct to say that the integral $I$ is only defined when $\theta > 0$ since $X(t)$ is only measurable if $\theta > 0$?
This post suggests that it may be better to take the correlation function as the Dirac delta function since Dirac delta correlated white noise is measurable and consequently the integral $I$ is defined. However a Gaussian process that is mean-zero with the Dirac delta function is known as Gaussian white noise which is said to have infinite variance which would imply:
$$\text{Var}[X(t)] = \sigma^2 = \begin{cases} 1 &\text{if } \theta \neq 0, \\ \infty &\text{if } \theta = 0. \end{cases} $$
This does not seem correct since the variance of $X(t)$ should be independent of $\theta$. How is the process $X(t)$ defined as $\theta \rightarrow 0$?
Note: sorry if this counts as a duplicate question but I have really been struggling with this topic and could use some guidance specific to my exact situation rather than the general case.