I have the following 2 definitions of a Markov chain but I can't see how they are equivalent.
i) Let $X_i$ be a sequence of random variables taking values in a measurable state space $(S,A)$. This process is a discrete Markov chain iff $P(X_{n+1}\in B|\sigma(X_0,X_1,..,X_n))=P(X_{n+1}\in B|\sigma(X_n))$. I'm assuming that $$P(X_{n+1}\in B|\sigma(X_0,X_1,..,X_n))=E[1_{\{X_{n+1}\in B\}}|\sigma(X_0,X_1,..,X_n)]$$
i.e. the conditional expectation of $1_{\{X_{n+1}\in B\}}$ given $\sigma(X_0,X_1,..,X_n)$
ii) The other definition states that $P(X_{n+1}= s|X_0=x_0,X_1=x_1,..,X_n=x_n)=P(X_{n+1}= s|X_n=x_n)$ for all $s,x_0,x_1,x_2,..\in S$. Here, I suppose $$P(X_{n+1}= s|X_0=x_0,X_1=x_1,..,X_n=x_n)=\frac{P(X_{n+1}= s,X_0=x_0,X_1=x_1,..,X_n=x_n)}{P(X_0=x_0,X_1=x_1,..,X_n=x_n)}$$
How are these 2 definitions equivalent?
Edit:
From the following link: https://en.wikipedia.org/wiki/Markov_property it seems to be the case that they are not equivalent. However, if $S$ is a discrete set with the discrete $\sigma$-algebra then they are equivalent.
In this case the following link provides an explanation: Why are these two definitions of Markov property equivalent?
The definitions will be equivalent IF all these random variables are discreet (i.e. their ranges are countable). The $\sigma$-algebra $\sigma(X_1,\ldots, X_n)$ is then generated by the events $\{X_i=x_i\}$.
If $X_{n+1}$ is not discreet, then $P(X_{n+1}=x_{n+1} \mid \ldots)$ could be identically zero, so it's not clear what the second definition means in that case. The first one, however, is universal.
Conditioning on the event $\{X_1=x_1,\ldots, X_n=x_n\}$ is rather problematic when the random variables involved are continuous.