This paper came out of an attempt to solve the following general problem: Suppose $\{Y_n, n \geqq 1\}$ is a stationary process with a finite state-space $J$. Under what conditions can we express it as a function of a finite Markov chain? More precisely, when can we find a stationary Markov chain $\{X_n, n \geqq 1\}$ with a finite state-space $I$ and a function $f$ on $I$ onto $J$ such that the process $\{f(X_n)\}$ has the same distribution as $\{Y_n\}$? We do not solve the general problem here but for mixing processes we obtain a theorem which is the best possible in a certain sense. Suppose $\epsilon$ denotes a state of $J$ and suppose $s, t$ denote finite sequences of states of $J$. If $s = \epsilon_1 \cdots \epsilon_n$, let $p(s) = P\lbrack (Y_1, \cdots, Y_n) = s\rbrack$. For each $\epsilon$, define $n(\epsilon)$ to be the largest integer $n$ such that we can find $s_i, t_i (i = 1, \cdots, n)$ such that the matrix $\|p(s_i\epsilon t_j)\|$ is nonsingular. Gilbert [4] has shown that if $\{Y_n\}$ is a function $f$ of a finite Markov chain $\{X_n\}$ and if $f$ takes $N(\epsilon)$ states of $I$ into the state $\epsilon$ of $J$, then $n(\epsilon) \leqq N(\epsilon)$. If $n(\epsilon) = N(\epsilon)$, then $\{Y_n\}$ is said to be a regular function of a Markov chain. Thus a necessary condition for $\{Y_n\}$ to be a function of a finite Markov chain is that $\sum n(\epsilon)$ is finite. It is proved here that if $\sum n(\epsilon) < \infty$ and if the process $\{Y_n\}$ is mixing, then there exists a positive integer $m^\ast$ such that for every $m \geqq m^\ast$ the process $\{Y_{nm + 1}, n \geqq 0\}$ is a function of a Markov chain with $\sum n(\epsilon)$ states. An example is constructed to show that $m^\ast$ cannot, in general, be brought down to 1. Thus the whole process $\{Y_n, n \geqq 1\}$ may still not be a function of a Markov chain with $\sum n(\epsilon)$ states.