Probability Theory – Problem 26: Prove that

Suppose $g$ is a function of period 1 that is continuous on $[0,1]$. Let $X$ be a random variable uniformly distributed on $[0,1]$, and let $\alpha$ be an irrational number. Define $Z_{n}=g(X+(n-1)\alpha)$. Prove that $$\operatorname*{lim}_{n\to\infty}{\frac{1}{n}}\sum_{i=1}^{n}Z_{i}=\int_{0}^{1}g(u)\mathrm{d}u\qquad a.s.$$

Step 1. Reformulate the problem in terms of a deterministic initial value $x$

Let $X$ be uniformly distributed on $[0,1]$, let $\alpha$ be irrational, and define $$Z_n = g\bigl(X + (n-1)\alpha\bigr),\quad n\ge1.$$ The function $g$ is continuous on $[0,1]$ and has period $1$, i.e., for every $u\in\mathbb{R}$, $$g(u+1)=g(u).$$

For a real number $y$, denote its fractional part by $$\{y\} := y-\lfloor y\rfloor\in[0,1),$$ so that by periodicity $g(y)=g(\{y\})$.

Fix a deterministic $x\in[0,1]$ and consider the sequence $$Y_n(x)=g\bigl(x+(n-1)\alpha\bigr),\quad n\ge1.$$ The random setting is recovered by replacing $x$ with the random variable $X$. If we can show that for every deterministic $x\in[0,1]$, $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}Y_i(x) = \int_0^1 g(u)\,du,$$ then in the random setting, for almost every sample point $\omega$, $X(\omega)$ takes some value $x\in[0,1]$, and hence $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}Z_i(\omega) = \int_0^1 g(u)\,du.$$

It therefore suffices to prove: for every deterministic $x\in[0,1]$, $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} g\bigl(x+(i-1)\alpha\bigr) = \int_0^1 g(u)\,du.$$ Set $$y_i := \{x+(i-1)\alpha\}\in[0,1),\quad i\ge1.$$ Since $g$ has period $1$, the above is equivalent to $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} g(y_i) = \int_0^1 g(u)\,du.$$

Step 2. Use the Weyl equidistribution criterion to show that $\{x+(n-1)\alpha\}$ is equidistributed on $[0,1]$

Equidistribution modulo $1$: A sequence $\{y_i\}_{i\ge1}\subset[0,1)$ is said to be equidistributed on $[0,1]$ if for every interval $[a,b]\subset[0,1]$, $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{[a,b]}(y_i) = b-a.$$

Weyl's criterion: $\{y_i\}$ is equidistributed on $[0,1]$ if and only if for every nonzero integer $k\in\mathbb{Z}\setminus\{0\}$, $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} e^{2\pi i k y_i} = 0.$$

For the sequence $y_i=\{x+(i-1)\alpha\}$ in this problem, take any $k\ne 0$ and consider $$S_n(k) =\frac{1}{n}\sum_{i=1}^{n} e^{2\pi i k y_i} =\frac{1}{n}\sum_{i=1}^{n} e^{2\pi i k (x+(i-1)\alpha)}.$$ Factoring out the common term gives $$S_n(k) = \frac{e^{2\pi i k x}}{n}\sum_{j=0}^{n-1} e^{2\pi i k \alpha j}.$$ Let $$r := e^{2\pi i k\alpha}.$$ Since $\alpha$ is irrational, $k\alpha$ is not an integer, so $r\ne 1$. Thus $$\sum_{j=0}^{n-1} r^j = \frac{1-r^n}{1-r},$$ and hence $$S_n(k)= e^{2\pi i k x}\,\frac{1-r^n}{n(1-r)}.$$ Therefore $$|S_n(k)| \le \frac{|1-r^n|}{n|1-r|} \le \frac{2}{n|1-r|}\xrightarrow[n\to\infty]{}0.$$

This holds for all $k\ne0$. By Weyl's criterion, for every $x\in\mathbb{R}$, the sequence $$y_i=\{x+(i-1)\alpha\}$$ is equidistributed on $[0,1]$.

Step 3. Limiting average for step functions

First verify the conclusion for simple step functions. Let $$s(u) = \sum_{j=1}^{m} c_j\,1_{(a_j,b_j]}(u),$$ where $0\le a_j<b_j\le1$ and $c_j\in\mathbb{R}$. Then $$\int_0^1 s(u)\,du = \sum_{j=1}^{m} c_j (b_j-a_j).$$

By the equidistribution of $y_i$, for each interval $(a_j,b_j]$, $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{(a_j,b_j]}(y_i) = b_j-a_j.$$ Hence $$\begin{aligned} \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} s(y_i) &= \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} \sum_{j=1}^{m} c_j\,1_{(a_j,b_j]}(y_i) \\ &= \sum_{j=1}^{m} c_j \left( \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} 1_{(a_j,b_j]}(y_i) \right) \\ &= \sum_{j=1}^{m} c_j (b_j-a_j) = \int_0^1 s(u)\,du. \end{aligned}$$

Thus for every step function $s$, $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} s(y_i) = \int_0^1 s(u)\,du.$$

Step 4. Uniform approximation of the continuous function $g$ by step functions

Since $g$ is continuous on the compact interval $[0,1]$, it is uniformly continuous and bounded. For any $\varepsilon>0$, there exists a step function $s_\varepsilon$ such that $$\sup_{u\in[0,1]} |g(u)-s_\varepsilon(u)| \le \varepsilon.$$

For a fixed $x$ and the corresponding sequence $y_i$, set $$A_n := \frac{1}{n}\sum_{i=1}^{n} g(y_i),\quad B_n := \frac{1}{n}\sum_{i=1}^{n} s_\varepsilon(y_i).$$ Then $$|A_n-B_n| \le \frac{1}{n}\sum_{i=1}^{n} |g(y_i)-s_\varepsilon(y_i)| \le \sup_{u\in[0,1]} |g(u)-s_\varepsilon(u)| \le \varepsilon.$$

By the result of the previous step for step functions, $$\lim_{n\to\infty} B_n = \int_0^1 s_\varepsilon(u)\,du.$$ Therefore $$\begin{aligned} \left|A_n - \int_0^1 g(u)\,du\right| &\le |A_n-B_n| + \left|B_n-\int_0^1 s_\varepsilon(u)\,du\right| + \left|\int_0^1 s_\varepsilon(u)\,du - \int_0^1 g(u)\,du\right| \\ &\le \varepsilon + \left|B_n-\int_0^1 s_\varepsilon(u)\,du\right| + \int_0^1 |s_\varepsilon(u)-g(u)|\,du \\ &\le \varepsilon + \left|B_n-\int_0^1 s_\varepsilon(u)\,du\right| + \varepsilon. \end{aligned}$$ Letting $n\to\infty$, the middle term tends to $0$, yielding $$\limsup_{n\to\infty}\left|A_n - \int_0^1 g(u)\,du\right| \le 2\varepsilon.$$ Since $\varepsilon>0$ is arbitrary, $$\lim_{n\to\infty}A_n =\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} g(y_i) =\int_0^1 g(u)\,du.$$

This holds for every $x\in[0,1]$.

Step 5. Return to the random variable $X$ and $Z_n$ to obtain almost sure convergence

For each sample point $\omega$, set $x=X(\omega)\in[0,1]$. Then $$Z_i(\omega) = g\bigl(X(\omega)+(i-1)\alpha\bigr) = g\bigl(x+(i-1)\alpha\bigr).$$ By the conclusion of Step 4, for this $x$, $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} Z_i(\omega) = \int_0^1 g(u)\,du.$$ Since this conclusion holds for all $x\in[0,1]$, it holds for almost every $\omega$ as well, i.e., $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} Z_i = \int_0^1 g(u)\,du \quad \text{a.s.}$$

QED.