Problem
We have an interval $ I \subseteq [0, 1) $ and the indicator function defined as $ \chi_I(x) = \begin{cases} 1 \text{ if } x \in I\\ 0 \text{ if } x \not \in I \end{cases} $
What we want to show is if the following is true $$ \limsup_{n \rightarrow \infty} \frac{1}{n + 1} \sum_{k = 0}^n \chi_I(k \alpha \mod 1) = |I| $$
where $\alpha$ is an irrational number such that $0 < \alpha < 1$ and $ |I| $ is the length of the interval.
Solution
So my first approach was that I tried to convince myself this numerically using Python. I did the following for some different lengths of the interval:
Suppose that $I = [a, b] \subseteq [0, 1)$ then the length of the interval is calculated as $|I| = b - a$. The sum $\sum_{k = 0}^n \chi_I(k \alpha \mod 1)$ can be written as $$ d_{k + 1} = d_k + \chi_I(k \alpha \mod 1) $$ where $k = 0, 1, ..., n$. Now, for some arbitrary $\varepsilon > 0$ we can pick an index $N > 0$ and we get that
$$|\frac{1}{n + 1}d_{n + 1} - |I| | < \varepsilon, \forall n \ge N $$
This means that the sequence $ \frac{1}{n + 1}d_{n + 1} \rightarrow |I| $ as $n \rightarrow \infty$.
But I'm still stuck because I don't know how to prove this analytically. For the interval $(0, 1)$ we get that $$ \limsup_{n \rightarrow \infty} \frac{1}{n + 1}\sum_{k = 0}^n \chi_{(0, 1)}(k \alpha \mod 1) = \limsup_{n \rightarrow \infty} \frac{1}{n + 1}(\underbrace{0 + 1 + ... + 1}_{= n}) = \limsup_{n \rightarrow \infty} \underbrace{\frac{n}{n + 1}}_{< 1} = 1 $$
Since the length of the interval $(0, 1)$ is just $1$. But how do I show this for any $I = [a, b] \in [0, 1)$ such that $$|I| = b - a = \limsup_{n \rightarrow \infty}\frac{1}{n + 1} \sum_{k = 0}^n \chi_{[a, b]}(k \alpha \mod 1)?$$