# Continuous path of stochastic processes

In my probability book a stochastic process is defined as a measurable map $X: \Omega \rightarrow S^T,$ where $S^T$ is equipped with the sigma algebra of cylinder events. Our professor mentioned that the canonical image space of $X$ with its sigma algebra may be a bad choice in many situations, as plenty of events are not measurable w.r.t. this sigma-algebra.

Now, a stochastic process is continuous, if $$P(\{\omega; t \mapsto X(\omega,t) \text{ is continuous}\})=1.$$

What I don't understand is: Why is this event measurable?

Or differently: If you read such a definition of a stochastic process, what kind of definition of stochastic processes do you have in mind such that all of this makes sense?

You are right; the event

$$\{\omega; t \mapsto X_t(\omega) \, \text{is continuous}\}$$

is, in general, not measurable - this is because the set of continuous functions is not contained in the sigma algebra of cylinder sets, see this question.

There are at least two ways to evade this measurability issue:

1. We say that $X$ has almost surely continuous sample paths if there exists a measurable $P$-null set $N$ such that $$\{\omega; t \mapsto X_t(\omega) \, \text{is not continuous}\} \subseteq N.$$
2. Instead of considering $X$ as a mapping $X: \Omega \to S^T$ we assume that $X: \Omega \to C^T$ where $C^T$ denotes the set of continuous functions $f: T \to S$. Moreover, we endow $C^T$ with the trace $\sigma$-algebra of the cylinder sets.
• I have difficulties understanding the second appoach. If you know that $X$ is a mapping into the continuous functions, then you know that it is a continuous function for any $\omega.$ So I see a problem with this approach: You require $P(t \mapsto X_t)=1$ is continuous, but you already assume this in your definition. So in that case, the requirement is rather redundant, don't you think? – user167575 Oct 25 '15 at 8:44
• @user167575 The only assumption is that $X$ maps $\Omega$ to $C^T$... this implies automatically that all sample paths are continuous. (in particular, if we use this approach, we don't need a probability measure on $\Omega$ to talk about continuity.) – saz Oct 25 '15 at 8:49
• okay, but I feel as if this is an even stronger requirement, as we have continuity not only a.e., but everywhere. So for instance: Is it possible require in the case of Brownian motion that $B: \Omega \rightarrow C^T$? – user167575 Oct 25 '15 at 9:18
• @user167575 Yes, in the case of Brownian motion it works very well. The point is basically: Assume $X$ is a process with almost surely continuous sample paths (e.g. in the sense which I suggested in the first approach). Then we can define a new process $Y$ by setting $$Y(\omega) := \begin{cases} X(\omega), & \text{if X(\omega) is continuous} \\ 0, & \text{otherwise} \end{cases}.$$ Then, we have $Y: \Omega \to C^T$. Since $X$ has almost surely continuous sample paths, the processes $X$ and $Y$ are "almost the same" ... so, roughly, instead of $X$ we can consider $Y$. – saz Oct 25 '15 at 9:27