Additive subgroups of the reals and irrational rotation

We present a useful characterization of the additive subgroups of $\mathbb{R}$.

Theorem. Let $G$ be a subgroup of $(\mathbb{R}, +)$. Then precisely one of the following holds:

(i) $G = d\mathbb{Z}$ for some $d \in \mathbb{R}$;

(ii) $G$ is dense in $\mathbb{R}$.

Proof: If $G = \{0\}$, then $G = 0\mathbb{Z}$. Assume $G$ is nontrivial. Let \[ d = \inf\{g: 0 < g \in G\}.\] Note that $d \geq 0$ since for every nonzero $y \in G$, $ky \in G$ and $ky > 0$ for some $k \in \mathbb{Z}$.
Case 1: $d > 0$. We claim that $d \in G$. If not, then for every $\epsilon > 0$ there exists $g \in G$ with $g \in (d, d + \epsilon)$. It follows that for every $\epsilon > 0$ there are $g, h \in G$ such that $0 < g - h < \epsilon$, thus there exists $x \in G$ with $0 < x < \epsilon$ (let $x = g - h$). This contradicts to $d > 0$. We also have $d = \min\{|g|: 0 \neq g \in G\}$. Now fix any $0< g \in G$ and consider $S = \{g - nd: n = 0, 1, 2, \ldots\}$. The set $S$ is discrete, we can find $n \geq 0$ such that $g - nd \geq 0$ and $g - (n + 1)d < 0$. Since $g - nd, g - (n +1)d \in G$, $g - (n + 1)d \leq -d$, so $g - nd \leq 0$ and $g = nd$. Therefore $G = d\mathbb{Z}$.
Case 2: $d = 0$. Arguing as in the beginning of case 1, we find that 0 is a cluster point of $G \cap \mathbb{R}_+$. Let $(a, b)$ be any nonempty bounded open interval of $\mathbb{R}$. Take $\epsilon = b - a$ and pick some $g \in G \cap (0, \epsilon)$. For sufficiently large $n$, $ng \geq a$, and we take the smallest such $n$. Then $ng - g < a$, i.e. $ng \in (a - g, a]$, thus $ng + g \in (a, a + g] \subseteq (a, b)$. Therefore $G \cap (a, b) \neq \emptyset$. It follows that $G$ is dense in $\mathbb{R}$.

The above result can be used to prove the density of the orbit of an irrational rotation in the unit circle.

Corollary. (i) For an irrational $t$, $\mathbb{Z} + t\mathbb{Z}$ is dense in $\mathbb{R}$.
(ii) For an irrational $t$, $\{e^{2\pi int}: n \in \mathbb{Z}\}$ is dense in the unit circle.

Proof: (i) Note that $\mathbb{Z} + t\mathbb{Z}$ is a subgroup of $(\mathbb{R}, +)$. If $\mathbb{Z} + t\mathbb{Z} = a\mathbb{Z}$ for some $a \in \mathbb{R}$, then $1 = an$ for some $n \in \mathbb{Z}$, so $a \in \mathbb{Q}$. But $t = am$ for some $m \in \mathbb{Z}$, so $t \in \mathbb{Q}$, contradiction. By Theorem, $\mathbb{Z} + t\mathbb{Z}$ must be dense in $\mathbb{R}$.
(ii)  Let $e^{2\pi i x}$ be a point on the unit circle. By (i), there exist sequences $m_k, n_k$ of integers such that $|m_k + n_k t - x| \rightarrow 0$. Then $|e^{2 \pi i n_k t} - e^{2 \pi i x}| = |e^{2\pi i x}||e^{2 \pi i (n_k t - x)} - 1| = |e^{2\pi i (m_k + n_k t - x)} - 1| = O(|m_k + n_k t - x|) \rightarrow 0$. Hence $\{e^{2\pi int}\}_{n \in \mathbb{Z}}$ is dense in the unit circle.

Some results in measure theory

Proposition. Let $\mu$ be a positive Borel measure on $\mathbb{R}$ and $\epsilon > 0$. Then for almost every $x$, we have \[ \int_\mathbb{R} \frac{d\mu(t)}{|x - t|^{1 + \epsilon}} = +\infty. \]
Proof: For each positive integer $N$, define $E_N = \{x \in \mathbb{R}: \int_\mathbb{R} d\mu(t) / |x - t|^{1 + \epsilon} \leq N\}$. Let $I$ be any bounded closed interval with $E_N \cap I$ nonempty. Pick $x \in E_N \cap I$ and write $I = [x - \delta_1, x + \delta_2]$. We have \[ N \geq \int_\mathbb{R} \frac{d\mu(t)}{|x - t|^{1 + \epsilon}} \geq \int_I \frac{d\mu(t)}{|x - t|^{1 + \epsilon}} \geq \frac{\mu(I)}{\max\{\delta_1, \delta_2\}^{1 + \epsilon}} \geq \frac{\mu(E_N \cap I)}{(\delta_1 + \delta_2)^{1 + \epsilon}}, \] i.e. $\mu(E_N \cap I) \leq Nm(I)^{1 + \epsilon}$ where $m(I)$ is the length of the interval $I$. Now let $[a, b] \subseteq \mathbb{R}$ and $n \in \mathbb{N}$. Decompose $[a, b]$ into a union of $n$ closed subintervals of length $(b - a)/n$, then \[ \mu(E_N \cap [a, b]) \leq nN\left(\frac{b - a}{n}\right)^{1 + \epsilon} \rightarrow 0 \] as $n \rightarrow \infty$. Hence $\mu(E_N \cap [a, b]) = 0$. It follows that $\mu(E_N) = 0$ for all $N \in \mathbb{N}$ and our result follows.

(To be continued)

Some results in real analysis

Proposition. Let $f: [a, b] \rightarrow \mathbb{R}$ be a function such that left-hand limit $f(x-)$ and right-hand limit $f(x+)$ of $f$ exist at every $x \in [a, b]$. Then the number of discontinuities of $f$ is at most countable.

Proof: Define a real-valued function $g$ on $[a, b]$ by $g(x) = \max\{|f(x) - f(x+)|, |f(x) - f(x-)|\}$. Then $f$ is continuous at $x$ iff $g(x) = 0$. Let $D_n = \{x \in [a, b]: g(x) \geq 1/n\}$. The points of discontinuity of $f$ is exactly $\bigcup_{n = 1}^\infty D_n$. We prove that each $D_n$ is finite.

Suppose on the contrary that $D_n$ is infinite. Then there is a sequence $\{x_n\}$ of distinct points in $D_n$ which converges to $x \in [a, b]$. Since $f(x+), f(x-)$ exist, we can choose some $\delta > 0$ such that $|f(y) - f(x+)| < 1/4n$ whenever $0 < y - x < \delta$ and $|f(y) - f(x-)| < 1/4n$ whenever $0 < x - y < \delta$. Therefore $|f(y) - f(z)| < 1/2n$ whenever $y, z \in (x - \delta, x)$ or $y, z \in (x, x + \delta)$. We can find some $k$ such that $0 < |x_k - x| < \delta$. If $x_k > x$, then $|f(y) - f(x_k)| < 1/2n$ whenever $y \in (x, x + \delta)$, let $y \rightarrow x_k$ from both left and right we get $|f(x_k+) - f(x_k)| \leq 1/2n$ and $|f(x_k-) - f(x_k)| \leq 1/2n$, so $g(x) \leq 1/2n < 1/n \leq g(x)$, contradiction. If $x_k < x$ then similarly we also obtain a contradiction. Hence $D_n$ must be finite. QED.

Proposition. Let $f_n: U \rightarrow \mathbb{R}$ be a sequence of continuous functions on $U \subseteq \mathbb{R}^d$ such that $f_1 \leq f_2 \leq \cdots$ and $\{f_n\}$ converges pointwise to a continuous function $f$ on $U$. Then $f_n \rightarrow f$ uniformly on compact subsets.

Proof: Note that $f_1 \leq \cdots \leq f_n \leq \cdots \leq f$. Let $K \subseteq U$ be compact and $\epsilon > 0$. For any $x \in K$, since $f_n(x) \rightarrow f(x)$, there is $N_x \in \mathbb{N}$ such that $f(x) - f_n(x) < \epsilon/3$ whenever $n \geq N_x$. For any $n \in \mathbb{N}$, by uniform continuity of $f$ and $f_n$ on $K$, there is $\delta_n > 0$ such that for any $x, y \in K$, $|f(x) - f(y)| < \epsilon/3$ and $|f_n(x) - f_n(y)| < \epsilon/3$ whenever $|x - y| < \delta_n$. For every $x \in K$, if $y \in B(x, \delta_{N_x})$, then
$\begin{align*}
&f(y) - f_n(y) \leq f(y) - f_{N_x}(y) \\ &= [f(y) - f(x)] + [f(x) - f_{N_x}(x)] + [f_{N_x}(x) - f_{N_x}(y)] \\ &< \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon
\end{align*}$
whenever $n \geq N_x$. Since $\{B(x, \delta_{N_x}): x \in K\}$ is an open cover of $K$, then there are $x_1, \ldots, x_k \in K$ such that $K = \bigcup_{j = 1}^k B(x_j, \delta_{N_{x_j}})$. Take $N = \max\{N_{x_1}, \ldots, N_{x_k}\}$. For $n \geq N$, if $x \in K$, then $x \in B(x_j, \delta_{N_{x_j}})$ for some $1 \leq j \leq k$, so $f(x) - f_n(x) < \epsilon$. Hence $f_n \rightarrow f$ uniformly on $K$. QED.


(To be continued)