Chapman-Kolmogorov equations

The identity kernel is a two-sided identity: directly from its Dirac [F1, F2] sections, $(IK)(x,A)=K(x,A)$ and $(KI)(x,A)=\int 1_A(y)K(x,dy)=K(x,A)$. Thus $K^{m+0}=K^m=K^m{K}^0$, including $m=0$. If $K^{m+n}=K^m{K}^n$, then [F1]--[F2] give $K^{m+n+1}=K^{m+n}K=(K^m{K}^n)K=K^m(K^{n}K)=K^m{K}^{n+1}\mathrm{.}$ Induction proves the kernel identity for all $m,n\ge 0$.

Fix $m$ and bounded measurable $f$. For $n=0$, $f(X_m)$ is [F1, F3, F4, step 1.1] ${\mathcal{F}}_m$-measurable and hence is its own conditional expectation; it is also $K^{0}f(X_m)$. Suppose the formula holds at $n$. By [F3] at time $m+n$, [F4], and the induction hypothesis applied to the bounded measurable function $Kf$, $\begin{array}{rl}{\displaystyle \mathbb{E}[f(X_{m+n+1})\mid {\mathcal{F}}_m]}& {\displaystyle =\mathbb{E}[\mathbb{E}(f(X_{m+n+1})\mid {\mathcal{F}}_{m+n})\mid {\mathcal{F}}_m]}\\ {\displaystyle }& {\displaystyle =\mathbb{E}[Kf(X_{m+n})\mid {\mathcal{F}}_m]=K^n(Kf)(X_m)=K^{n+1}f(X_m)\mathrm{.}}\end{array}$ The last equality is the definition of kernel composition from step 1.1. Induction proves the conditional-expectation formula. Choice is used only by [F3]--[F4], which operate on conditional-expectation classes.

Taking $f=1_A$ in step 2.1 gives the displayed conditional-probability [F3, step 2.1] formula; conversely that formula for all $A$ gives the bounded-function formula by the preceding lemma. Empty and full $A$ yield respectively zero and one on both sides.