Prove that stochastic prcess trajectories are continous. Consider gaussian process $\{X_{t}, t \in [0,1]\}$ with zero mean and covariance function $R(X_{s}, X_{t}) = \min (s,t) -st$. 

We want to know does this process has continous trajectories, i.e. $\lim_{t \to t_0} X_t = X_{t_0}$ a.e.

My attempt : 

First of all let's prove that this process continous in $L_{2}$. Consider $\mathbb{E}(X_{t+h} - X_{t})^{2} = h \to 0$ (the same result may be proved by proving continuity of covariance function).

Next we have that it's continous in measure (because of convergence in average). But we know that probability of $\mathbb{P}(\{ \lim X_t \to X_{t_0} \})$ is continious function, so we may use Riesz theorem which give us subsequence converging almsot sure. Am I right? Hard problem for me actually, I don't know how it can be proven only by definitions.
Assume that such a process $X_t$ exists and is continuous (i.e. $X_t$ is a Brownian bridge).

Take $U$ an independent uniform variable over $[0,1]$ and let $Y_t=X_t+1_{U.t \in \mathbb Q}$.

Then for all $t$, $\mathbb P(X_t=Y_t)=\mathbb P(U.t \not \in \mathbb Q)=1$ so $Y_t$ is a gaussian process with the same mean and covariance as $X_t$, however almost surely it is continuous nowhere.

This shows in general that mean and variance of a stochastic process never guarantee its continuity.
Assume that such a process $X_t$ exists and is continuous (i.e. $X_t$ is a Brownian bridge).

Take $U$ an independent uniform variable over $[0,1]$ and let $Y_t=X_t+1_{U.t \in \mathbb Q}$.

Then for all $t$, $\mathbb P(X_t=Y_t)=\mathbb P(U.t \not \in \mathbb Q)=1$ so $Y_t$ is a gaussian process with the same mean and covariance as $X_t$, however almost surely it is continuous nowhere.

This shows in general that mean and variance of a stochastic process never guarantee its continuity.