A process $X=(X_t)_{t\geq0}$ defined on some probability space $(\Omega, \Sigma, \mathbb P)$ and taking values in some measurable space $(Y,\mathcal Y)$ is called a Markov process if
$$ \mathbb E[f(X_t)|\mathcal F_s] = E[f(X_t)|\mathcal X_s], \quad \forall f \text{bounded measurable}, $$
where $\mathcal F_t : = \sigma(X_u, 0\leq u\leq t)$. With this definition, I would like to ask if $X$ always admits a family of transition kernel $P = (P_{s,t})_{0\leq s\leq t}$ such that:
i) For all $0\leq s \leq t, Y \in \mathcal Y$, $x\mapsto P_{s,t}(x,A)$ is measurable
ii) For all $x \in Y$, $\mathcal Y \ni A \mapsto P_{s,t}(x,A)$ is a probability measure on $Y$.
iii) $$ P_{s,t}(x,A) = \int_Y P_{s,u}(x,dy)P_{u,t}(y,A), \quad \forall u \in (s,t)$$
and is this family unique?
I got this question because I have read a few books in which their approaches to Markov processes are different. One starts with transition kernels, others start with the general definition above but later "freely" take $(s,t,x,A)\mapsto \mathbb P(X_t \in A|X_s =x)$ for transtion kernels. However, it's not clear what $\mathbb P(X_t \in A|X_s =x)$ stands for!
What follows is my attempt in understanding this. First, take $f = \mathbb 1_A, A\in \mathcal Y$ then we have the Markov property written as
$$ \mathbb P(X_t\in A|\mathcal F_s) = \mathbb P(X_t \in A|X_s). $$
Since $\mathbb P(X_t \in A|X_s)$ is $\sigma(X_s)$-measurable, there is a measurable function, call $x\mapsto P'_{s,t}(x,A)$ such that
$$ P(X_t \in A|X_s)(\omega) = P'_{s,t}(X_s(\omega),A), \forall \omega \in \Omega.$$
And then we mean $\mathbb P(X_t \in A|X_s =x) = P'_{s,t}(x,A)$
The first problem arises is how do we know for a fixed $x\in Y$ $P'_{s,t}(x,\cdot)$ is even a measure? The uniqueness of $P'_{s,t}(\cdot,A)$ is only in the sense that if $P''_{s,t}(\cdot,A)$ is another measurable function then $\mathbb P(P''_{s,t}(X_s,A) = P'_{s,t}(X_s,A)) = 1$. If we can somehow entangle this, we could have $P'$ have the properties i), ii), iii) above and I'm stuck here...