Prove that for all $\lim_{n\to\infty} \frac 1n \sum_{j=0}^{n-1} \chi_B(T^{j}(x_0)= \frac 1k \sum_{j=0}^{k-1} \chi_B(T^{j}(x_0)$

Question

Let $(X,\mathfrak B , \mu)$ be a probability space and $T : X\to X$ be a measure preserving transformation. Let $x_0 \in X$ be a periodic point of least period $k \ge 1$. Prove that for all $B \in \mathfrak B$,

$\lim_{n\to\infty} \frac 1n \sum_{j=0}^{n-1} \chi_B(T^{j}(x_0)= \frac 1k \sum_{j=0}^{k-1} \chi_B(T^{j}(x_0)$, where $\mathfrak B$ is the borel measurable set. I was trying to prove it using some bound on the left and on the right such that both converges to $\frac 1k \sum_{j=0}^{k-1} \chi_B(T^{j}(x_0)$ but in vain. Anyway I got the answer and thanks a lot.

Answer 1

Let $f = χ_B$. Note $\frac 1n \sum_{j=0}^{n-1} \chi_B(T^{j}(x_0))=m \frac{f(x_0) + f(T(x_0)) + · · · + f(T ^{k−1} (x_0))}{mk + r}+ \frac 1{mk + r} \sum_{j=0}^{r} \chi_B(T^{j}(x_0))$

where $n = mk + r, m = [n/k]$ (integer part). Now let $n, m \to ∞$ to get

$\lim_{n\to ∞} \frac 1n \sum_{j=0}^{n-1} \chi_B(T^{j}(x_0))= \frac 1k \sum_{j=0}^{k-1} \chi_B(T^{j}(x_0))$