A time-inhomogeneous chain may require enlarged state

If the displayed process were a homogeneous $K$-chain, its transition from [F1] $X_0=0$ to $X_1=0$ would force $K(0,\{1\})=\mathbb{P}(X_1=1\mid {\mathcal{F}}_0)=0.$ But its next transition from the same state $X_1=0$ to $X_2=1$ would force $K(0,\{1\})=\mathbb{P}(X_2=1\mid {\mathcal{F}}_1)=1,$ a contradiction. The witness uses the same state twice, so changing only the row at state $1$ cannot repair it.

Put $\tilde{S}={\mathbb{N}}_0\times \{0,1\}$ with its power set and let [F1] $b_0=b_1=0$, $b_n=1$ for $n\ge 2$. Define $T(n,x)=(n+1,b_{n+1}),\tilde{K}((n,x),A)=1_A(T(n,x))\mathrm{.}$ Every section is a Dirac probability and every evaluation is measurable. For $Y_n=(n,X_n)$ one has $Y_{n+1}=T(Y_n)$, so its conditional transition is the same kernel $\tilde{K}$ at every time. Hence [F1] now holds on the enlarged state space.

The failed conclusion is therefore specifically the existence of one [step 1.1, step 1.2] homogeneous kernel on the unaugmented state space, not the Markov nature of the time-augmented evolution. The probabilities zero and one, times zero/one/two, and both state endpoints are explicit. Choice is used only to phrase the conditional-probability identities in [F1].