Theoretical definition Markov chain

Question

I have trouble understanding the theoretical definition of a Markov chain (the intuitive meaning is clear). My trouble arises when reading Durrett, page 233, the book can be found here: https://services.math.duke.edu/~rtd/PTE/PTE4_1.pdf

He defines a measurable space $(S, \mathcal{S})$ and defines a transition probability. He then says that $X_n$ is a Markov chain (w.r.t. to $\mathcal{F}_n$) if P($X_{n+1} \in B$ | $\mathcal{F}_n$) = p($X_n$|B). Are we assuming $X_n: S \rightarrow \mathbb{R}$?

My confusion is mostly due to his theorem 6.1.1. "$X_n$ is a Markov chain (with respect to $\mathcal{F}_n = \sigma(X_0, ... , X_n))$ ) with transition probability $p$. What is he claiming? Given a measurable space $(S, \mathcal{S})$ and some transition probability p, a sequence of r.v. $X_n$ on $S$ is automatically a Markov chain (with respect to $\sigma(X_0, ... , X_n))$? What I also not get is that (for me), when speaking of a random variable $X_n$, we have to have a probability measure - otherwise, what's the point? The transition probabilities do not serve as a replacement for that (?)

I hope my question is clear, or it emphasizes what is not clear to me :)

Answer 1

P($X_{n+1} \in B$ | $\mathcal{F}_n$) = p($X_n$|B) basically means that the probability of going from $X_n$ to $X_{n+1}$ does not depends on the whole path $\mathcal{F}_n = \sigma(X_0, ... , X_n)$, but only on the last r.v $X_n$. The theorem shows a way to construct a Marcov chain when we have those probabilities. And yes, the transition probabilities are driving the probability distributions of the r.v $X_j, \text{ for } j = \{1,2,...\} $

Answer 2

There are two spaces here: the basic probability space $(\Omega,\mathcal F,P)$ and a measurable space $(S,\mathcal S)$. $X_n$'s are random elements with values in $S$ in the sense $X_n^{-1}(B) \in \mathcal F$ for all $B \in \mathcal S$.