What is the Haar measure on the Bohr compactification $b\mathbb{Z}$ of the integers?

Question

What is the Haar measure on the Bohr compactification $b\mathbb{Z}$ of the integers?

We (my collaborators and I) suspect that it is $$ \int_{b\mathbb{Z}} d\mu_H f(n) = \lim_{N\to\infty}\frac{1}{2N+1} \sum_{n=-N}^{N} f(n) ~. $$ This is finite and translation invariant, as required, but we are not sure how to prove that nothing "weird" happens at infinity.

This suspicion is motivated by the observation that the Haar measure on the Bohr compactification of the real numbers $b\mathbb{R}$ is [1] $$ \int_{b\mathbb{R}} d\mu_H f(x) = \lim_{L\to\infty}\frac{1}{2L} \int_{-L}^{L}dx\, f(x) ~, $$ where $dx$ is the Lebesgue measure.

I'm coming at this from the perspective of a theoretical physicist learning some harmonic analysis in order to better understand what's called "polymer quantum mechanics" [2], a method for quantizing simple physical systems that is not equivalent to the standard quantization. Harmonic analysis is new to me, hence I'm not confident how to prove our conjectured measure. (The paper [3] seems like it might give this measure, or at least something similar, but we're not entirely sure.)

UPDATE [10/12/25]: As I learn more about the Haar measure on locally compact Abelian groups, it seems that the "mean value" or "invariant mean" (suitably defined) in general defines a Haar measure on compact groups; see for example [4-5]. Though lacking an actual proof that the proposed measure is the correct invariant mean, this would suggest the answer to the question is in the affirmative.

[1] An introduction to harmonic analysis, Katznelson (2004).

[2] Polymer quantum mechanics and its continuum limit by Corichi, Vukasinac, and Zapata (Phys. Rev. D76 044016 (2007)) is not a bad place to begin.

[3] Lacunarity and the Bohr topology by Kunen and Rudin (Mathematical Proceedings of the Cambridge Philosophical Society 126 117-137 (1999), link).

[4] The Joys of Haar Measure by Diestel and Spalsbury (American Mathematical Society (2014)), Ch. 5.

[5] Notes on representations of compact groups, Dragan Milicic (link)

Answer 1

This is correct, but needs a bit of interpretation. Here is some of the relevant background:

• The algebra $C(b\mathbb{Z})$ of continuous complex-valued functions on the Bohr compactification $b\mathbb{Z}$ is canonically isomorphic to the algebra $AP(\mathbb{Z}) \subseteq \ell^\infty(\mathbb{Z})$ of Bohr almost periodic functions on $\mathbb{Z}$. This canonical isomorphism is isometric for uniform norms. If you are interested in C*-algebras, a more precise statement is that the Bohr compactification of $\mathbb{Z}$ is the Gelfand spectrum of $AP(\mathbb{Z})$.
• There are many equivalent definitions of $AP(\mathbb{Z})$ but the one most convenient for present purposes is that it is the norm-closure of the algebra of trig polynomials. By a trig polynomial, I mean a finite linear combination of "characters". Each complex number $z$ of norm $1$ defines a corresponding character $\chi_z$ on $\mathbb{Z}$ by $\chi_z(n) = z^n$. A trig polynomial is a finite linear combination $\sum_i a_i \chi_{z_i}$.
• The Haar measure of $b\mathbb{Z}$ is the unique translation-invariant probability measure on $b \mathbb{Z}$. By the appropriate version of the Riesz representation theorem, we may think of it as a translation invariant "state" (meaning positive linear functional of norm $1$) on $C(b\mathbb{Z})$. Making the identification $C(b\mathbb{Z})=AP(\mathbb{Z})$, we may think of it as a translation invariant state $$\tau:AP(\mathbb{Z}) \to \mathbb{C}.$$ This invariant state $\tau$ is uniquely determined by what it does to the characters: $\tau(\chi_z) = 1$ for $z=1$ and $\tau(\chi_z) = 0$ for $z \neq 1$. If you like, this is forced by the translation invariance because the characters are eigenvectors for translation, $\chi_z(n+m) = z^m \chi_z(n)$. Also, $\chi_1 \equiv 1$ and $\tau(1)=1$ holds for any state.

What you are claiming (or should be claiming) is that if we define a sequence of states $\tau_n : AP(\mathbb{Z}) \to \mathbb{C}$ by $$\tau_n(f) = \frac{1}{2n+1} \sum_{k=-n}^n f(k)$$ then \begin{align}\tag{$\star$}\tau_n(f) \to \tau(f) && \forall f \in AP(\mathbb{Z}). \end{align}

Indeed, ($\star$) is easy to prove if $f$ is a character, so by linearity it follows that ($\star$) also holds if $f$ is a trig polynomial. Does ($\star$) continue to hold when we pass to the closure? Yes, because $\tau_N$ and $\tau$ are states; they are bounded linear functionals of norm $1$. Given any $f \in AP(\mathbb{Z})$ and $\epsilon > 0$ we can find a trig polynomial $g$ such that $\|f-g\| < \epsilon$ (uniform norm). A simple calculation shows that $$ | \tau_n(f) - \tau(f) | \leq 2\epsilon + | \tau_n(g)- \tau(g) |.$$ Thus, for all sufficiently large $n$, we have $| \tau_n(f)- \tau(f) | < 3\epsilon$.

As a final remark, note that ($\star$) can definitely be false if we don't insist $f \in AP(\mathbb{Z})$. If we work in the larger algebra $\ell^\infty(\mathbb{Z})$ we can taken functions $f: \mathbb{Z} \to \{0,1\}$ which are $0$ for a long time, then $1$ for a longer time, then $0$ for an even longer time, and so on, in such a way that the limit of $\tau_n(f)$ as $n \to \infty$ does not exist.