Markov chain, define $\mathbb{P}_x$ as "conditioning on $X_0$" Given a markov chain $\{X_n :\Omega \rightarrow S\}$ ($S$ countable) with transition probability $p$, according to Durrett, to define $\mathbb{P}_\mu$ on the path space $S^\mathbb{N}$, the first step is to fix some initial distribution $\mu$ for $X_0$, then apply Kolmogorov Extension theorem to this specific case 
$$ \mathbb{P}(X_0\in B_0, \cdots X_n \in B_n) = \int_{B_0} \mu(dx_0) \int_{B_1} p(x_0, dx_1) \cdots\int_{B_n} p(x_{n-1}, dx_n)$$
to obtain a measure ${\mathbb P}_\mu$ on the path space $(S^{\mathbb{N}}, \mathcal{S}^{\mathbb{N}})$.
For different $x,y \in S$, we actually apply Kolmogorov Extension theorem twice (for $X_0 \sim \delta_x$ and $X_0 \sim \delta_y$) to obtain ${\mathbb{P}}_x$ and ${\mathbb{P}}_y$.
I want to define  $\mathbb{P}_x$ as "conditioning on $X_0$".
For some measurable set $A\subset S^\mathbb{N}$, I want to say
$${\mathbb{P}}_x(A) = {\mathbb{E}_?}[1_A |  X_0 = x]$$


*

*we need a probability measure first on $S^{\mathbb{N}}$ to even define the conditional expectation on the right, how should we choose this probability measure. 

*Isn't this a bad notation because the the right hand side is defined regardless what the distribution of $X_0$ is...Provided we have a probability measure 
$${\mathbb{E}_?}[1_A |  X_0 = x]$$ 
is a $\mu_{X_0}$ a.e defined function of $x$.
One problem I have is that since the chapter on Markov chain is done completely on $S^{\mathbb{N}}$ without referring back to the $\Omega$ probability space at all, even the Markov property on wiki
$$\mathbb{P}[X_n\in B| \sigma(X_0, \cdots,  X_k)] = \mathbb{P}[X_n\in B| \sigma(X_k)]$$
looks strange to me, and I have not done probability at undergraduate level. Is what I am trying to do above correct?
 A: I'm not completely sure what your question here is, but let me try to clarify a few things. This was too long for a comment, so I'm posting it as an answer.
Consider a probability space $(\Omega, \mathcal{F}, \mathbf{P})$, and define an $S$-valued Markov chain $X=\{X_n\}_n$ on $\Omega$, with transition function $p$. Let $Y=\{Y_n\}_n$ be the collection of evaluation functions defined on $S^{\mathbb{N}}$ (i.e. $Y_n(\omega) := \omega(n)$ for $\omega \in S^{\mathbb{N}}$). Then saying that $\mathbb{P}_x$ (as you defined it in your question) is the "law of the chain conditioned on $X_0 = x$" (where $x \in S$) is to say the that the two stochastic processes $X$ and $Y$, defined respectively on the probability spaces $(\Omega, \mathcal{F}, \mathbf{P}(\cdot|X_0=x))$and  $(S^{\mathbb{N}}, \mathcal{S}^{\mathbb{N}}, \mathbb{P}_x)$, are equal in law. By this, one means that for any $n \geq 1$, and $B_1,\ldots,B_n \subseteq S$, $$ \mathbf{P}(X_1 \in B_1,\ldots, X_n \in B_n|X_0=x) = \mathbb{P}_x(Y_1 \in B_1,\ldots, Y_n \in B_n).$$
As for your confusion with the Markov property, note that in the beginning you are declaring your chain $X$ to be Markov. The relation you have given at the end of your question (but now under the measure $\mathbf{P}$ if our present notation is to be followed) is then a sufficient mathematical expression of the same (at least when $X$ is a discrete time stochastic process). On the other hand, you can prove the Markov property for $Y$, i.e. $$ \mathbb{P}_x(Y_n \in B | Y_1,\ldots, Y_{n-1}) = \mathbb{P}_x(Y_n \in B|Y_{n-1})$$ from your definition of $\mathbb{P}_x$ on $(S^{\mathbb{N}} \mathcal{S}^{\mathbb{N}})$.
