Might be an obvious question but is the following correct: if $Z(.)$ is a modification of $X'(.)$, then $Z(.)$ is also a mean square derivative of $X(.)$, because for each $t$, $$\mathbb{E}\left[\left(\frac{X(t+h)-X(t)}{h}- Z(t)\right)^2\right] \leq \mathbb{E}\left[\left(\frac{X(t+h)-X(t)}{h}-X'(t)\right)^2\right]+ \mathbb{E}\left[\left(X'(t)-Z(t)\right)^2\right]$$ the first term tends to 0 as $h \to 0$ by definition, and the second term is 0 because $X'(t)$ and $Z(t)$ are equal almost surely by the fact that $Z(.)$ is a modification of $X'(.)$?

Thank you. Do you know of any citable examples of this being explicitly said? That you can somehow 'pick' the continuous version/modification of a process that's only defined in terms of its mean square properties (such as a Gaussian Process defined in terms of its mean and covariance) without losing generality of whatever it is you're trying to do.

Thank you, that was very helpful. I'm guessing $X$ is cadlag or caglad is an easy way to ensure $\lim_{s\to t} X_s(\omega)$ exists?

Thanks. Could you elaborate a bit more on why modifications are preserved in the limit? In particular why $Y'(.)$ is a modification of $X'(.)$ because they are the limits of two sequences that are modifications of each other.

Is there a condition which ensures the existence of a sample continuous modification is equivalent to the process itself being sample continuous? I suspect separability might work but I'm not sure and I cannot find a source which explicitly states this.