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Let $\mathbb{T}^n = { (z_1,\ldots,z_n) \in \mathbb{C}^n : |z_l| = 1, \; 1 \leq l \leq n }$ denote the $n$-torus, and let $t_1, \ldots, t_n$ be arbitrary real numbers. Then it can be shown that the topological closure $H$ in $\mathbb{T}^n$ of the one-parameter subgroup $$H' = {(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}) \in \mathbb{T}^n : y \in \mathbb{R}}$$ is an $r$-dimensional subtorus of $\mathbb{T}^n$, where $0 \leq r \leq n$ is the dimension over $\mathbb{Q}$ of the span of $t_1, \ldots, t_n$, and furthermore 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 $H$.

My question is, for which sets $B \subset \mathbb{T}^n$ do we have that $$\lim_{Y \to \infty} \frac{1}{Y} \int\limits_{{y \in [0,Y] : (e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}) \in B}}{dy} = \mu_H(B). $$ I have a feeling that some sort of approximation by continuous functions should tell you, but I can't seem to make it work for some reason.

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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.

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