Is every Markov Process a Martingale Process? According to the definition (2.3.6) of a Markov Process in Shreve's book titled Stochastic Calculus for Finance II:

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, let $T$ be a fixed positive number, and let $\mathcal F(t)$, $0\leqslant t\leqslant T$, be a filtration of sub-$\sigma$-algebras of $\mathcal F$. Consider an adapted stochastic process $X(t)$, $0\leqslant t\leqslant T$. Assume that for all $0\leqslant s\leqslant t\leqslant T$ and for every nonnegative, Borel-measurable function $f$, there is another Borel-measurable function $g$ such that $$\mathbb E\left[f(X(t))\mid\mathcal F(s)\right] = g(X(s)). $$ Then we say that the $X$ is a Markov process.

it seems obvious to me that every Markov Process is a Martingale Process (Definition 2.3.5):

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, let $T$ be a fixed positive number, and let $\mathcal F(t)$, $0\leqslant t\leqslant T$, be a filtration of sub-$\sigma$-algebras of $\mathcal F$. Consider an adapted stochastic process $M(t)$, $0\leqslant t\leqslant T$. If $$\mathbb E\left[M(t)\mid\mathcal F(s)\right] = M(s)\quad\textrm{for all } 0\leqslant s\leqslant t\leqslant T,$$ we say this process is a martingale.

Can someone please tell me if this is correct?
Thanks!
 A: Just to point out the error in your logic more directly:
Suppose $X(t)$ is a Markov process.  Then if we take $f(x) = x$, it is true that for all $s < t$ there exists a function $g$ such that $E[X(t) \mid \mathcal{F}(s)] = g(X(s))$.  For instance if $X(t)$ is an Ornstein-Uhlenbeck process, that function $g$ is something like $g(x) = e^{-(t-s)} x$.  
However, in order for $X(t)$ to be a martingale, we would need specifically to end up with $g(x) = x$.  In general this need not happen.
The converse isn't true either.  If we know $X(t)$ is a martingale and we try to see whether it is a Markov process, we know that if we take $f(x) = x$ there is a function $g$ that works (namely $g(x)=x$), but if we take $f$ to be some other function, the definition of martingale does not guarantee that we can find a corresponding $g$ at all.
A: For a simple counterexample, let $X_t=t$ and $\mathcal F_t$ be the natural filtration. Then for $s<t$ and nonnegative measurable $f$, $$\mathbb E[f(X_t)\mid\mathcal F_s] = f(X_s+t-s)=:g(X_s) $$ so that $X_t$ is Markov, but $$\mathbb E[X_t\mid \mathcal F_s] = t\ne X_s, $$
so that $X_t$ is not a martingale.
