Proving that $Z_t=g(X_t)$ is a martingale if and only if $\mathbb{E}(Z_t|Z_s)=Z_s \ \forall t>s \geq 0$. ($X_t$ Markov)

Question

I want to prove the next property:

Let $X_t$ be a Markov process (so $\mathbb{E}(X_t|\mathcal{F}_s)=\mathbb{E}(X_t|X_s)$ where $\mathcal{F_t}=\sigma(X_s, s\leq t)$).

Suppose that $Z_t=g(X_t)$ where $g: \mathbb{R} \longrightarrow\mathbb{R}$ is an injective function. Then, $Z_t$ is a martingale adapted to $ \mathcal{F}_t $ if and only if $\mathbb{E}(Z_t|Z_s)=Z_s \ \forall t>s \geq 0$.

So by starting with $\mathbb{E}(X_t|\mathcal{F}_s)=\mathbb{E}(X_t|X_s)$ I have to prove that $\mathbb{E}(Z_t|\mathcal{F}_s)=\mathbb{E}(Z_t|Z_s)$.

I do not know how to start, so any hint or reference will be very usefull.

Answer 1

Since $(X_t)$ is a Markov process, $\mathbb{E}(g(X_t)\mid {\cal F}_s)=P_{s,t}g(X_s)$ where $P_{s,t}$ is the transition kernel. Therefore $\mathbb{E}(g(X_t)\mid {\cal F}_s)$ is measurable with respect to $\sigma(X_s)=\sigma(Z_s).$