Express the Markov property using the transition matrix

Question

Let $\{X_t; t \in \mathbb{N}_0\}$ be a discrete Markov chain with finite state space $\Omega$ and transition matrix $P$. Let $\{\mathcal{F}_t; t \in \mathbb{N}_0\}$ be the corresponding filtration.

I saw in a paper (but without justification) that the Markov property can be expressed as

$$\mathbb{E}[f(X_{t+1})|\mathcal{F}_t] = P f(X_t).$$

My question is:

How do I come to this? From which definition of Markov property do I need to start from? And is $f$ any function, or does it need to be bounded and measurable?

Any help would be appreciated.

Answer 1

Presumably, $\ f:\Omega\rightarrow V\ $ where $\ V=\mathbb{R}^n\ $$\big($or maybe $\ \mathbb{C}^n\ \big)$ and $\ Pf:\Omega\rightarrow V\ $ is the function defined by $$ (Pf)(\omega)=\sum_{s\in\Omega}P_{\omega s}f(s)\ . $$ For a discrete finite-state Markov chain, the Markov property reduces to $$ \mathbb{P}\big(X_{t+1}\in A\,|\,\mathcal{F}_t\big)=\mathbb{P}\big(X_{t+1}\in A\,|\,X_t\big)\ , $$ or equivalently, $$ \mathbb{P}\big(X_{t+1}=s\,|\,\mathcal{F}_t\big)=\mathbb{P}\big(X_{t+1}=s\,|\,X_t\big) $$ for $\ s\in\Omega\ $. Also, by the definition of the transition matrix $\ \mathbb{P}\big(X_{t+1}=s\,|\,X_t\big)=P_{X_ts}\ $. So \begin{align} \mathbb{E}\big(f\big(X_{t+1}\big)\,\big|\,\mathcal{F}_t\big)&=\sum_{s\in\Omega}f(s)\,\mathbb{P}\big(X_{t+1}=s\,|\,\mathcal{F}_t\big)\\ &=\sum_{s\in\Omega}f(s)\,\mathbb{P}\big(X_{t+1}=s\,|\,X_t\big)\\ &=\sum_{s\in\Omega}P_{X_ts}f(s)\\ &=\big(Pf\big)\big(X_t\big)\ . \end{align} I can't see how the identity could be made sense of unless $\ f\ $ is a function on $\ \Omega\ $, in which case it's automatically measurable with respect to any $\ \sigma$-algebra on its codomain and the discrete $\ \sigma$-algebra on $\ \Omega\ $.