When does continuity with probability one imply mean-square continuity Wikipedia's page on continuous stochastic process defines continuity with probability one if $\lim_{s \to t} X_s = X_t$ almost everywhere and mean-square continuity if the process has finite second moments and $\lim_{s\to t} \mathrm{E}|X_s - X_t|^2$. It then goes on stating that neither implies the other.
My question is:

Under what circumstanes/additional assumptions does one imply the other?

Originally, I thought continuity with probability one implies mean-square (for a second-order process) from the continuity of the norm, i.e. $X_t \to X \Rightarrow \|X_t\| \to \|X\|$, where $\|X\|=E|X|^2$, however perhaps this does not hold because we only have an a.e. convergence.

Is this true and would it then hold if $\lim_{s \to t} X_s = X_t$ $\forall \omega \in \Omega$ ?

Thank you.
 A: I think it is pretty hard to give sufficient conditions in this generality.  The typical statement to get mean-square continuity from pointwise convergence is of course the dominated convergence theorem, but usually we cannot expect to find a nice (uniform) integrable bound for the process.
A result which is sometimes useful to prove mean-square continuity of processes is the following:

Let $(f_n)_{n \in \mathbb{N}}$, $f$ be measurable and $f_n \to f$ almost surely. If $f \in L^2(\mathbb{P})$, then $$f_n \stackrel{L^2}{\to} f \iff \mathbb{E}(|f_n|^2) \to \mathbb{E}(|f|^2).$$

Since $|X_s-X_t| \stackrel{s \to t}{\to} 0$ (for fixed $t$), i.e. the limit $f$ equals $0$, we find by applying the above lemma
$$\mathbb{E}(|X_s-X_t|^2) \to 0 \iff \mathbb{E}(|X_s|^2) \to \mathbb{E}(|X_t|^2).$$
This means that if suffices to check the the $L^2$-norms are convergent.
Example: Let $(B_t)_{t \geq 0}$ be a 1-dimensional Brownian motion. Then $\mathbb{E}(B_t^2) = t$ and therefore the conditon $\mathbb{E}(|B_s|^2) \to \mathbb{E}(|B_t|^2)$ is obviously satisfied. Consequently, $(B_t)_{t \geq 0}$ is mean-square continuous. (Yeah, for Brownian motion we can also prove it directly using the scaling property. However, if we want to check that e.g. Lévy processes with second moments are mean-square continuous, then scaling doesn't work, but the argument presented in this answer works very well.)
Finally, let me give you an example for a square-integrable process which converges with probability $1$, but is not mean-square continuous.
Example: Consider the measurable space $((0,1),\mathcal{B}(0,1))$ endowed with the Lebesgue measure. Define
$$X_0(\omega) := 0 \qquad X_t(\omega) := \frac{1}{t^2} 1_{(0,t)}(\omega).$$
Then $X_s(\omega) \to X_0(\omega)$ as $s \to 0$ for all $\omega \in (0,1)$, but
$$\mathbb{E}(|X_s|^2) = \frac{s}{s^2} = \frac{1}{s}$$
does not converge to $\mathbb{E}(|X_0|^2)=0$. This means in particular that $X_s$ cannot converge to $X_0$ in $L^2$ as $s \to 0$.
