Let's say we want to model the fivefold throw of a fair coin. Then we define the corresponding process via the outcome of the throws, i.e. we set

$$X_t := \begin{cases} 1 & \text{t-th throw is head} \\ 0 & \text{otherwise} \end{cases}$$

for $t \in \{1,\ldots,5\}$. Now, since we have a fair coin, the probability that $X_t$ equals $1$ is 0.5 for each $t$. In probability theory, this is translated in the following abstract way: For a probability space $(\Omega,\mathcal{A},\mathbb{P})$, the random variables $X_t$ have to satisfy $\mathbb{P}(X_t=1)=\tfrac{1}{2}$. This means in particular that we do not care how the probability space looks like; only the *distribution* of the random variables is of importance.

Moreover, for any $\omega \in \Omega$ the mapping $t \mapsto X_t(\omega)$ is a realization of our process. If we throw the coin five times and observe e.g. $$0 \, \, 1 \, \, 0 \, \, 0 \, \, 1,$$ then there exists $\omega \in \Omega$ which "symbolizes" this outcome, i.e.

$$(X_1(\omega),X_2(\omega),X_3(\omega),X_4(\omega),X_5(\omega))=(0,1,0,0,1).$$

Usually, we are interested in questions like "What is the probability that we throw head 3 out of 5 times?"; this probability equals $\mathbb{P}(\sum_{t=1}^5 X_t = 3)$. This question can be answered if we know the (finite dimensional) distributions of the stochastic process $(X_t)_t$.

So, basically a stochastic process (on a given probability space) is an abstract way to model actions or events we observe in the *real world*; for each $\omega \in \Omega$ the mapping $t \mapsto X_t(\omega)$ is a realization we might observe. The likeliness of the realization is characterized by the (finite dimensional) distributions of the process.

## Best Answer

Take $\mathcal{F}=\{\emptyset,\Omega\}$. Then any non-deterministic process is non measurable. For a more interesting example, consider a a family of i.i.d. random variables $(X_t:t\in[0,1])$ s.t. $\mathsf{E}X_0=0$ and $\mathsf{E}X_0^2=1$. It is easy to see that if $\{X_t\}$ is measurable, then for any $0\le a<b\le 1$, $$ \mathsf{E}\left[\int_{[a,b]}X_t\, dt\right]^2=\int_{[a,b]^2}\mathsf{E}[X_sX_t]\, dsdt=0 $$ (by Fubini), i.e., $X_t=0$ for almost all $\omega\in\Omega$ and $t\in[0,1]$. Thus, $$ \mathsf{E}\left[\int_0^1X_t^2\,dt\right]=0\ne 1=\int_0^1 \mathsf{E}X_t^2\, dt. $$