Let $(X_n)_{n\in \mathbb{N}}$ be a homogeneous Markov Chain where $X_n : \Omega \to E$ for all $n\in\mathbb{N}$, $(\Omega, \mathcal{F}, \mathbb{P})$ is a probability space and $(E,\mathcal{A})$ is a measure space.
If $K: E \times \mathcal{A} \to [0,1]$ is the Markov Kernel of $(X_n)_{n\in\mathbb{N}}$, then we define the first iteration of $K$ as $K^{(1)} = K$ and the $t+1$-th iteration as $K^{(t+1)}(x,A) = \int_E K^{(t)}(y, A)K(x,dy)$ for all $t \geq 1$.
I want to understand why $\mathbb{P}\left( X_{s+t} \in A | X_s = x \right) = K^{(t)}(x,A)$. It is intuitive, but I want to know the strict justification.
In fact, I can see that it all turns out to be the same problem as: if $p$ is a probability in $E$ and $X_0 \sim p$, then $\mathbb{P}(X_1 \in A) = \int_{E} K(y,A) p(dy)$, but I could not show that.
Obs: $p(dy)$ and $K(x,dy)$ means that we are integrating regarding to the measures $p$ and $K(x,\cdot)$, respectively.