The product sigma-algebra on $S^T$ is generated by the projections
$$\Pi_t: S^T \to S, f \mapsto f(t) \qquad (t \in T)$$
i.e. it's the smallest sigma-algebra such that projections $\Pi_t$ ($t \in T$) are measurable. This means that a mapping $X: \Omega \to S^T$ is measurable (with respect to the product sigma-algebra on $S^T$) iff the mappings
$$\Pi_t \circ X: \Omega \to S, \omega \mapsto X(\omega)(t) = X_t(\omega)$$
are measurable for all $t \in T$.
Let $Y$ a set, $(Y_i,\mathcal{A}_i)$ measure spaces ($i \in I$) and $f_i: Y \to Y_i$ arbritary mappings ($i \in I$). Denote by $\sigma(f_i,i \in I)$ the $\sigma$-algebra generated by the mappings $f_i$. Moreover, let $(\Omega,\mathcal{A})$ a measure space and $g: \Omega \to Y$ a mapping. Then the following statements are equivalent.
- $g$ is $\mathcal{A}/\sigma(f_i,i \in I)$-measurable
- $\forall i \in I: f_i \circ g$ is $\mathcal{A}/\mathcal{A}_i$-measurable
(Here: $Y:=S^T$, $I:=T$, $Y_i := S$, $f_i := \Pi_i$ for $i \in T$.)
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
Yes, the measurable space $(X, \mathcal{M})$ ($\mathcal{M} \subset 2^X$ being a $\sigma$-algebra) should be the same for all $t$ in the index set. If you want the process to occupy different sets $X_t$ at different times, then the right things to do is to take $X$ to be the disjoint union of all the $X_t$. There is no requirement that $F_t$ be surjective, so there is no harm if some of the state space is not used at some times.
For a random walk on a graph, the state space $X$ is just the vertex set of the graph, typically with the discrete $\sigma$-algebra. Then for each $t \in T$, $F_t$ is some random element of $X$, i.e. a random vertex of the graph. Formally, if your underlying probability space is $(\Omega, \mathcal{F}, P)$, then $F$ is a map from $\Omega \times T$ into $X$, with the property that for each $t \in T$, the map $\Omega \ni \omega \mapsto F(\omega, t) \in X$, denoted $F_t$, is measurable.