Fall 2018 - Problem 4

Solution.

1. Given a Borel set $E$, we have

   $$m\p{R_\alpha\p{E}} = \int_0^1 \chi_{R_\alpha\p{E}}\p{e^{2\pi ix}} \,\diff{x} = \int_0^1 \chi_E\p{e^{2\pi i\p{x-\alpha}}} \,\diff{x} = m\p{E},$$

   since the Lebesgue measure is translation invariant and $e^{2\pi ix}$ is $1$-periodic.

2. If $\alpha$ is not a root of unity, then $\alpha$ must be irrational. Indeed, if this were not the case, then $\alpha = \frac{a}{b}$ for some integers $a, b$, so

   $$\p{e^{2\pi i\alpha}}^b = \p{e^{2\pi ia/b}}^b = e^{2\pi ia} = 1,$$

   i.e., $\alpha$ is a root of unity. Since $\alpha$ is irrational, it follows that $\set{\alpha^n \mid n \in \Z}$ are all distinct. Otherwise, if $\alpha^n = \alpha^m$, then

   $$e^{2\pi in\alpha} = e^{2\pi im\alpha} \implies e^{2\pi i\p{n-m}\alpha} = 1 \implies \p{e^{2\pi i\alpha}}^{n-m} = 1,$$

   i.e., $\alpha$ would be a root of unity. Hence, the powers of $\alpha$ are all distinct, so they form an infinite subset of the compact space $\T$. Thus, they must accumulate, so for any $\epsilon > 0$, there exist $n, m \in \Z$ so that

   $$\abs{e^{2\pi i\p{n-m}\alpha} - 1} = \abs{e^{2\pi in\alpha} - e^{2\pi im\alpha}} \leq \epsilon.$$

   Thus,

   $$\T = \bigcup_{k=1}^\infty B\p{\p{e^{2\pi i\p{n-m}\alpha}}^k, \epsilon},$$

   so the powers of $\alpha$ are dense.

3. Let $\mu$ be a Borel probability measure which is $\alpha$-invariant on $\T$. We will show that $\mu$ is rotation invariant: let $\beta \in \T$ and $\epsilon > 0$. By density, there exists $\set{\alpha^{n_k}}_k$ so that $\alpha^{n_k} \to \beta$ in $\T$. Then

   $$\mu\p{E} = \mu\p{R_{\alpha^{n_k}}\p{E}} = \int_0^1 \chi_{R_{\alpha^{n_k}}\p{E}}\p{e^{2\pi i\theta}} \,\diff\mu\p{\theta} = \int_0^1 \chi_E\p{e^{2\pi i\p{\theta-n_k\alpha}}} \,\diff\mu\p{\theta}.$$

   Since $\mu$ is a probability measure, the integrand is bounded by $1$, which is $L^1\p{\mu}$, so by dominated convergence,

   $$\mu\p{E} = \lim_{n\to\infty} \int_0^1 \chi_E\p{e^{2\pi i\p{\theta-n_k\alpha}}} \,\diff\mu\p{\theta} = \int_0^1 \chi_E\p{e^{2\pi i\p{\theta-\beta}}} \,\diff\mu\p{\theta} = \mu\p{R_\beta\p{E}},$$

   so $\mu$ is rotation-invariant. Now, observe that

   $$\begin{gathered} 1 = \mu\p{\T} = \sum_{k=0}^{n-1} \mu\p{\br{\frac{k}{n}, \frac{k+1}{n}}} = n\mu\p{\br{\frac{k}{n}, \frac{k+1}{n}}} \\ \implies \mu\p{\br{\frac{k}{n}, \frac{k+1}{n}}} = \frac{1}{n} = m\p{\br{\frac{k}{n}, \frac{k+1}{n}}} \end{gathered}$$

   for any $k = 0, \ldots, n-1$, so $\mu$ agrees with $m$ on rational intervals, hence on any interval, by dominated convergence. Thus, because intervals generate the Borel $\sigma$-algebra, it follows that $\mu = m$, which was what we wanted to show.