Suppose we have a Markov Chain $(X_i)$ taking values in a discrete state space $S$. I want to show that given any event $F$ involving $X_n,X_{n+1},X_{n+2},...$,
$$P(F|X_0=x_0,..,X_n=x_n)=P(F|X_n=x_n).$$
Also how does one show that $P(F|H,X_n=x_n)=P(F|X_n=x_n)$ where $H$ is any event involving $X_0,X_1,..X_{n-1}$?
I tried using the measure theoretic definition of Markov chains i.e $P(X_{n+1}\in A|X_0,..,X_n)=P(X_{n+1}\in A|X_n)$ but I don't seem to be getting anywhere.
Any ideas? It seems obvious but I cant seem to show it rigorously.
By a monotone class argument, it suffices to prove that for every $k$ and any subsets $A_{n},\dots,A_{n+k}$, $$ \mathbb P\left(\{X_n\in A_n,\dots,X_{n+k}\in A_{n+k}\}\mid \{X_0=x_0,\dots,X_n=x_n\}\right)=\mathbb P\left(\{X_n\in A_n,\dots,X_{n+k}\in A_{n+k}\}\mid X_n=x_n\right). $$ First observe that if $x_n\in A_n$, then the sets $\{X_n\in A_n\}$ can be removed and if $x_n\notin A_n$, both terms are equal to $0$. Therefore, we are reduced to prove that $$\tag{*} \mathbb P\left(\{X_{n+1}\in A_{n+1},\dots,X_{n+k}\in A_{n+k}\}\mid \{X_0=x_0,\dots,X_n=x_n\}\right)=\mathbb P\left(\{X_{n+1}\in A_{n+1},\dots,X_{n+k}\in A_{n+k}\}\mid X_n=x_n\right). $$ As $A_i$, $n+1\leqslant i\leqslant n+k$ is a subset of a discrete set, it suffices to prove (*) when each $A_i$ contains one element, denoted $x_i$, that $$\tag{**} \mathbb P\left(\{X_{n+1}=x_{n+1},\dots,X_{n+k}= x_{n+k}\}\mid \{X_0=x_0,\dots,X_n=x_n\}\right)=\mathbb P\left(\{X_{n+1}= x_{n+1},\dots,X_{n+k}=x_{n+k}\}\mid X_n=x_n\right). $$ Then we use $$ \mathbb P(A\cap B\mid\mathcal C)=\frac{\mathbb{P}(A\cap B\cap C)}{\mathbb P(C)}=\frac{\mathbb{P}(A\mid B\cap C)}{\mathbb P(C)} \mathbb{P}( B\cap C) $$ with $A=\{X_{n+k}=x_{n+k}$, $B= \{X_{n+1}=x_{n+1},\dots,X_{n+k-1}= x_{n+k-1}\}$ and $C=x_{n+1},\dots,X_{n+k}= x_{n+k}\}\mid \{X_0=x_0,\dots,X_n=x_n\}$ to get a simpler expression for the left hand side of (**), and a similar trick for the right hand side.