$f_1, f_2,\dots f_m$ are continuous maps on $[0,1]\to [0,1]$ and they form a Markov chain as follows $X_{n+1}= f_{\omega_n}(X_n)$ where $\omega_0, \dots, \omega_n$ are i.i.d discrete random variables taking values in $\{1,2,\dots,m\}$. We are given that there is an unique invariant measure for the above system and also $|X_n(x, \omega)-X_n(y, \omega)|\to 0$ almost surely as $n\to \infty$, where $X_n(x,\omega)$ is a realization of the chain starting from a point $x$. Could anyone tell me how to show $\lim_{n\to \infty} \frac{1}{n+1}\sum\limits_{k=0}^{n}f(X_k(x,\omega))\to \int f d\mu$ for all continuous function $f$ for all $x$ and for almost all $\omega$.
I started with $x=X_0\sim \mu$ then $X_n=(f_{\omega_{n-1}}\circ f_{\omega_{n-2}}\circ\cdots\circ f_{\omega_0})(X_0)= X_n(x, \omega)$ right?