this is a question on ergodic theory. Suppose I have an integer $N \geq2$ and a probability space $(\sum^{+} , B, \mu_{p})$, where $\mu_{p}$ is the Bernouilli measure with respect to probability vector $p = (p_{1},...p_{N}$, $B$ is Borel sigma algebra, and $\sum^{+}$ is the shift space. I'd like to show that for $\mu_{p}$ almost all $x \in \sum^{+}$ we have
$$\lim_{n\rightarrow \infty} \frac{1}{n} \log \mu (C_{n}(x_{0},...,x_{n+1})) = \sum_{i=0}^{N-1} p_{i} \log p_{i}.$$
I know that the right hand side is equal to $\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=0}^{N-1} \log p_{i}$, and then I should somehow apply Birkhoff's ergodic theorem to the induced $\sigma^{+}$ one sided shift map (which is known to be mixing and hence ergodic). But I just cannot formalise this argument and I am not sure if it is even corret. I would be grateful if you could point me in the right direction?
Let $f(x) = \log (p_{x_0})$, where $x = (x_0,x_1,x_2,\ldots)$. We have $$ \log \mu (C_n (x_0,x_1,\ldots,x_{n-1}) = \sum_{k=0}^{n-1} f(\sigma^k (x)) $$ where $x$ is any element of $C_n (x_0,\ldots,x_{n-1})$. By Birkhoff's theorem $$ \lim_{n\to\infty} \frac{1}{n} \log \mu (C_n (x_0,x_1,\ldots,x_{n-1}) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} f(\sigma^k x) = \int f \, d\mu = \sum_{i=0}^{N-1} p_i \log p_i $$ holds for a.e. $x$.