what's the relationship between a.s. continuous and m.s. continuous? suppose that X(t) is a s.p. on T with $EX(t)^2<+\infty$. we give two kinds of continuity of X(t). 


*

*X(t) is continuous a.s.

*X(t) is m.s. continuous, i.e. $\lim\limits_{\triangle t \rightarrow 0}E(X(t+\triangle t)-X(t))^2=0$.


Then, what's the relationship between these two kinds of continuity.
 A: I don't know if there is a clear relation between both concepts. 
For example if you take the Brownian Motion it satisfies 1 and 2 but if you take a Poisson process then it only satisfies 2 (although it satisfies a weaker form of condition 1 which is continuity in probability).  
The question is what do you want to do with those processes ?
Regards
A: For an example which is a.s. but not m.s. continuous, take your time interval to be $[0, \infty]$, and let $X_t$ be a standard one-dimensional Brownian motion started at 0 and stopped the first time it reaches 1.  (That is, if $B_t$ is a standard Brownian motion, take $T = \inf\{t > 0 : B_t = 1\}$ and $X_t = B_{t \wedge T}$.)  Since Brownian motion is recurrent, we have $X_t \to 1$ a.s. as $t \to \infty$, and so by setting $X_{\infty} = 1$ we get an a.s. continuous stochastic process on $[0,\infty]$.  
However, $X_t$ is not m.s. continuous.  If it were, then by Cauchy-Schwarz we would have $E[X_t] \to E[X_\infty] = 1$ as $t \to \infty$.  But $X_t$ is a martingale and so $E[X_t] = 0$ for all $t \in [0,\infty)$.
If you don't like using $[0,\infty]$ as your time interval, then apply a time change: let
$$Y_t = \begin{cases} X_{t/(1-t)}, & t < 1 \\ 1, & t \ge 1.\end{cases}$$
Now $Y_t$ is a.s. continuous but $E[Y_t] = 0$ for $t < 1$, $E[Y_t] = 1$ for $t \ge 1$.  Note that $Y_t$ is a standard example of a local martingale which is not a martingale.
