Weyl's equidistribution theorem states that the orbit of a point on the circle under rotation by $\alpha$ becomes asymptotically equidistributed with respect to Lebesgue (Haar) measure whenever $\alpha$ is an irrational multiple of $\pi$. (From the dynamical point of view, this is the statement that irrational rotations are uniquely ergodic.)

What happens if we choose two angles $\alpha$ and $\beta$ and flip a coin at each step to determine which we rotate by? Does the orbit still equidistribute?

Here is a more precise statement of the question, since the above is terse and perhaps open to misinterpretation.

Let $\Omega =\{0,1\}^\mathbb{N}$ be the space of infinite sequences of 0s and 1s, endowed with the $(\frac 12, \frac 12)$-Bernoulli measure $\mu$.

Consider the circle $S^1 = \mathbb{R}/\mathbb{Z}$ and a starting point $x_1\in S^1$. Fix $\alpha_0,\alpha_1\in \mathbb{R} / \mathbb{Z}$. Given $\omega\in \Omega$, define a sequence $x_n(\omega)\in S^1$ iteratively by $x_{n+1} = x_n + \alpha_{\omega(n)}$.

Let $m_n(\omega) = \frac 1n \sum_{k=0}^{n-1} \delta_{x_k(\omega)}$ be the average of the $\delta$-measures at the points $x_1(\omega),\dots, x_n(\omega)$. Let $E\subset \Omega$ be the set of $\omega$ such that $m_n(\omega) \to $ Lebesgue. Note that $\omega\in E$ if and only if $\sigma\omega\in E$, where $\sigma$ is the shift map on $\Omega$, and thus by the ergodic theorem we have $\mu(E)=0$ or $\mu(E)=1$.


  1. What properties of $\alpha_0,\alpha_1$ determine whether we get $\mu(E)=0$ or $\mu(E)=1$? If $\alpha_0=\alpha_1$ is irrational then we have $\mu(E)=1$ as above. On the other hand, if both are rational then $\mu(E)=0$ since $x_n$ takes only finitely many values. If $\alpha_0$ and $\alpha_1$ are linearly dependent over the rationals (but not rational themselves) then I believe we have $\mu(E)=1$, but probably this takes some argument. I don't know what happens if they are linearly independent over the rationals.

  2. Do any of the above answers change if we consider instead of $E$ the set $F$ of $\omega$ with the property that the limit points of the sequence $m_n$ do not depend on the initial point $x_1$? This would allow more general limiting measures. Or we could impose other conditions on the limit points of the sequence $m_n$ and ask the same question.

  3. What happens if we replace $\mu$ with a different $\sigma$-invariant measure on $\Omega$? Do we get a different answer for Bernoulli measures? Markov measures? Gibbs measures? General shift-invariant measures?

I know that's rather broad and open-ended. I'd be quite happy (at first, at least) with an answer to the first question. It seems worth asking all of them since this feels like something that has been probably been studied and may be well understood, although I didn't find anything when I looked. (I found some related results, but nothing that addressed these particular questions.)

  • $\begingroup$ Isn't $m_{n}(\omega)$ the convolution of the Lebesgue measures (given by averaging over the corresponding orbits?, well up to the measure zero set of $x_{n}$'s for which $\alpha$ or $\beta$ appear only finitely many times/ $\endgroup$
    – Asaf
    Dec 6, 2013 at 7:47
  • $\begingroup$ I'm not sure I understand your question: $m_n(\omega)$ is an atomic measure, just the convex combination of $n$ point masses. I don't see how to describe it or its limit points in terms of a convolution. $\endgroup$ Dec 6, 2013 at 13:20
  • $\begingroup$ Dear @VaughnClimenhaga, I never received a notification for the answer I give below. Is there something not clear, or do I miss something about the problem? $\endgroup$ Aug 4, 2016 at 17:59

2 Answers 2


If at least one of $\alpha$ or $\beta$ is an irrational multiple of $\pi$ then you do get equidistribution (almost surely). I'm sure this is a folklore type result; but one place where I know it can be found is a paper of Lagarias and Soundararajan (http://arxiv.org/pdf/math/0509175.pdf) which appeared in JLMS. See Theorem 4.1 there. The context for that paper is that the $(3x+1)$ map where one multiplies either by a number close to $3/2$ or multiplies by $1/2$ at least initially follows such a pattern, and therefore exhibits Benford's law.


Define the two rotations $T_i(x) = x+\alpha_i$ and define the $[T_1,T_2]$-endomorphism of the space $(0,1) \times \{0, 1\}^\mathbb{N}$ by $$[T_1,T_2]\bigl(x, (\epsilon_0, \epsilon_1, \ldots)\bigr)=\bigl(T_{\epsilon_0}(x), (\epsilon_1, \epsilon_2, \ldots)\bigr).$$ Your first question is highly close to the ergodicity of $[T_1,T_2]$.

Let me translate this problem in the language of probability theory. When $E$ is an endomorphism of a Lebesgue space $(Z,\nu)$, take a random variable $Y_0 \sim \nu$ and define the stationary process $(Y_n)$ by $Y_n = E^n(Y_0)$ for $n \geq 0$. The ergodicity of $E$ means that the averages $n^{-1}\sum_{i=0}^{n-1}f(Y_i)$ go to $E[f(Y_0)]$ for all $f \in L^1$.

Now, still in the general context of the endomorphism $E$, define the decreasing sequence of $\sigma$-fields $(\mathcal{F}_n)$ by ${\cal F}_n=\sigma(X_n, X_{n+1},\ldots)$. Ergodicty of $T$ is implied by the degeneracy of ${\cal F}_\infty$, and this property is called the exactness of $E$.

Now let us go back to your context. Ergodicity of $[T_1,T_2]$ implies that your averages of $\delta$-measures go to the Lebesgue measure for almost all $x_0$ in $(0,1)$ and almost all $(\epsilon_0, \epsilon_1, \ldots)$.

I claim that $[T_1,T_2]$ is exact when $\alpha_1-\alpha_0$ is irrational. Here the random variables $Y_n$ are $Y_n=\bigl(X_n, (\epsilon_n, \epsilon_{n+1}, \ldots)\bigr)$. The reversed filtration $(\mathcal{F}_n)$ is also generated by the process $\bigl(X_n-n\alpha_0, (\epsilon_n, \epsilon_{n+1}, \ldots)\bigr)$. The point is the following one: one would get this process (I mean a process with the same distribution) from the $[\textrm{Id},T]$-endomorphism, where $\textrm{Id}$ is the identity of $(0,1)$ and $T(x) = x + (\alpha_1-\alpha_0)$. In Meilijison's paper Mixing properties of a class of skew-products, it is proved that $[\textrm{Id},T]$ is exact whenever $T^2$ is ergodic.

  • $\begingroup$ A generalization of Meilijison's theorem: hal.archives-ouvertes.fr/hal-01987993/document (section 1). $\endgroup$ Feb 28, 2019 at 19:11
  • $\begingroup$ I still don't understand why a random shift process as $T_i (x) = x + a_i$ on circle is not Ergodic, where $a_i$ are independent identically distributed random variables with bounded second moment. Based on Theorem 1 and Corollary 1 of Lacey et al. (1994)(numdam.org/article/AIHPB_1994__30_3_353_0.pdf), a random rotation (shift) is representative sequence of a flow. Now any function $f \in L^2$ can be represented as a flow. Take the Lebesgue space $([0,1], \mathcal{B}[0,1], \lambda)$, then $\frac{1}{n}\sum_{i=1}^n f(T_i(x_0))$ should be $\int_0^1 f(x)dx$ or I am missing something? $\endgroup$
    – Mohamad
    Oct 16, 2019 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.