Suppose a markov chain is also a martingale taking values in non-negetive integers where 0 is an absorbent state. Suppose the starting value is X which is greater than 0. Now as it is a positive martingale we know that it converges almost surely. But is it true that the martingale converges to 0 almost surely?
I also assumed that 0 is the only absorbent state.
Yes. Call this chain $(X_n)_{n\ge 0}$. Because it's a non-negative martingale, the limit $X_\infty:=\lim_nX_n$ exists a.s. Because $X_n$ takes only integer values, it must be that $X_n=X_\infty$ for all $n>n_0(\omega)$ for an a.s. finite $n_0(\omega)$. This means that $X_\infty$ must be in the (only!) absorbing state $0$.