Bounded-function form of the Markov property

Assume [F1]. If $s={\sum }_{j=1}^r{a}_j{1}_{A_j}$ is a nonnegative measurable [F1, F2] simple function, then, for $B\in {\mathcal{F}}_n$, finite additivity of the integral and [F1] give $\mathbb{E}[1_{B}s(X_{n+1})]={\sum }_j{a}_j\mathbb{E}[1_B{1}_{\{X_{n+1}\in A_j\}}]=\mathbb{E}[1_{B}Ks(X_n)]\mathrm{.}$ The variable $Ks(X_n)$ is ${\mathcal{F}}_n$-measurable by [F2], so it is a version of $\mathbb{E}[s(X_{n+1})\mid {\mathcal{F}}_n]$.

Let $0\le f\le M$. By [F3], choose simple $s_j\uparrow f$, replacing [F2, F3, F4, F5, step 1.1] $s_j$ by $s_j\wedge M$ if necessary. Then $K{s}_j(x)\to Kf(x)$ for every $x$; this follows from [F4] with the probability measure $K(x,\cdot )$ and majorant $M$. For each $B\in {\mathcal{F}}_n$, [F4] under $\mathbb{P}$ on both sides of the identity from step 1.1 gives $\mathbb{E}[1_{B}f(X_{n+1})]=\mathbb{E}[1_{B}Kf(X_n)]\mathrm{.}$ Thus $Kf(X_n)$ is a conditional-expectation version; [F5] makes the equality an equality of almost-everywhere classes.

For bounded real $f$, apply step 2.1 to $f^{+}$ and $f^{-}$. Subtracting the [F2, F5, step 2.1] two defining event-integral identities shows that $Kf(X_n)=K{f}^{+}(X_n)-K{f}^{-}(X_n)$ is a version of the desired conditional expectation. This proves the bounded-function form.

Conversely, put $f=1_A$. Then $Kf(X_n)=K(X_n,A)$, so the asserted [F1, step 3.1] bounded-function identity is exactly [F1] for $A$. Together with the forward direction through step 3.1, this proves the equivalence.