I'm actually following a Brownian motion course, unfortunately I'm stuck with the following. Consider a probability space $(\Omega, \mathcal{F},\mathbb{P})$ and stochastic process $(X_t)_{t \in \mathbb{R}^+}$ on this probability space.
We are interested in the property $(*)$ which is,
Property ($*$): It is possible to find some probability space $(\tilde{\Omega}, \mathcal{L},\mathbb{Q})$, and some random process $(Y_t)_{t \in \mathbb{R}^+}$ on this probability space with $(Y)_{t\in \mathbb R^+}\stackrel{\text{fdd}}=(X)_{t\in \mathbb R^+}$ and such that there exists a measurable set with probability 1, such that for all $\omega$ in this set, $t\mapsto Y_t(\omega)$ is continuous on $\mathbb{R}^+$.
Here, $(Y)_{t\in \mathbb R^+}\stackrel{\text{fdd}}=(X)_{t\in \mathbb R^+}$ means that the two processes have the same finite dimensional distributions.
Now suppose that our $(X_t)_{t \in \mathbb{R}^+}$ have the $(*)$ property. The statement is the following: suppose $J$ is a countable dense subset of $\mathbb{R}^+$ (for instance $\mathbb{Q}^+$) and one has a collection of random variables $(Z_q)_{q\in J}$ with the same fdds as $(X_q)_{q \in J}$ (this is really a property about the law in particular we do not suppose anything about the probability space of $(Z_q)_{q\in J}$), then $(Z_q)_{q\in J}$ almost surely has a continuous extension. In addition this extension $(\tilde{Z_t})_{t \in \mathbb{R}^+}$ is such that $(\tilde{Z}_t)_{t\in \mathbb R^+}\stackrel{\text{fdd}}=(X_t)_{t\in \mathbb R^+}$.
I've tried to prove the above without success for a long time thus I will greatly appreciate if any of you can help me with that :)
Have a great day.