Ergodic flow in tori

Solution 1:

I guess now that I know the answer to this question, I shouldn't leave it unanswered! Basically, we define the probability measure $\mu_Y$ for each $Y > 0$ by $$\mu_Y(B) = \frac{1}{Y} \int\limits_{\{y \in [0,Y] : (e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}) \in B\}}{dy}$$ for each Borel set $B \subset \mathbb{T}^n$. The fact that $$\lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{g(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}) \: dy} = \int_{H}{g(z) \: d\mu_H(z)}$$ for any continuous function $g : \mathbb{T}^n \to \mathbb{C}$, where $\mu_H$ is the normalised Haar measure on $H$, implies that the probability measures $\mu_Y$ are converging weakly to $\mu_H$. By the Portmanteau theorem, this is equivalent to $$\mu_H(B) = \lim_{Y \to \infty} \mu_Y(B) = \lim_{Y \to \infty} \frac{1}{Y} \int\limits_{\{y \in [0,Y] : (e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}) \in B\}}{dy}$$ for every continuity set $B \subset \mathbb{T}^n$; that is, for every Borel set $B$ whose boundary in $\mathbb{T}^n$ has $\mu_H$-measure zero.