I would like to show that if $(X_n)$ is an irreducible recurrent Markov chain (on a countable space), and $f \geq 0$ is harmonic, then it is constant.
I do the following: for any $x$, $f(X_n)$ is a positive martingale for the probability measure that sets $X_0=x$. Therefore for any $x$, $f(X_n)$ converges almost surely to $f(x)$.
Now fix $x,y$. We know that almost surely starting from $x$, $f(X_n)$ will converge to $f(x)$ and we also know (by recurrence and irreducibility) that almot surely the sequence $Y_0,Y_1,\ldots$ starting at $y$ will pass through $x$ at some time $\tau$. Then the law of the sequence $Y_\tau\ldots$is the same as the law of the sequence $X_n$ so in fact $f(x)=f(y)$.
Is this correct?