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Note: All integrals are taken with respect to Lebesgue measure. The symbol $\def\avint{\mathop{\rlap{\raise.15em{\scriptstyle -}}\kern-.2em\int}\nolimits} \avint$ denotes the average integral.

We say a subset $E$ of $[0, \infty)$ is well distributed if for every periodic, locally integrable function $f: [0, \infty) \to \mathbb R$, we have

$$\lim_{x \to \infty} \def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{E \cap [0, x]} f = \avint_{P} f,$$

where the integral on the right hand side is taken over any period $P$ of $f$.

Question: Do there exist well distributed sets $E$ of Banach density zero? That is, such that

$$\lim_{N - M \to \infty} \frac{\mu(E \cap [M, N])}{N - M} = 0.$$

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  • $\begingroup$ What about $E=\cup [n^2+n\sqrt{2},n^2+n\sqrt{2}+1]$? Seems to be equidistributed modulo every $T>0$ by Weyl theorem. $\endgroup$ Commented May 12 at 6:41
  • $\begingroup$ @FedorPetrov Sorry, which Weyl theorem is this? $\endgroup$
    – Nate River
    Commented May 12 at 7:45
  • $\begingroup$ Equidistribution of values of polynomials modulo 1 $\endgroup$ Commented May 12 at 9:00

1 Answer 1

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Let $E:=\cup_{n=1}^\infty [n^2+n\sqrt{2},n^2+n\sqrt{2}+1]$. Clearly it has Banach density 0.

Let $f$ be a $T$-periodic function, $T>0$, $f\in L^1[0,T]$. It is convenient to suppose that $T>1$ (that may always be achieved by replacing $T$ to a multiple of $T$). Subtracting average, we may assume that $\int_0^T f=0$. Then $|\int_a^bf|\leqslant \int_0^T |f|:=C<\infty$ for all real $a<b$. For large $x$ the set $E\cap [0,x]$ has measure $\sqrt{x}+o(\sqrt{x})$, so we want to prove that $\int_{E\cap [0,x]}f(x)dx=o(\sqrt{x})$. Let $[0,x]$ contain $N$ full intervals $[n^2+n\sqrt{2},n^2+n\sqrt{2}+1]$, and possibly a part of $(N+1)$-st such interval. An integral of $f$ over the part of this $(N+1)$-st interval is bounded in absolute value by a constant $C$, by the above observation. So, we need to prove that $$ \frac1N\sum_{n=1}^N \int_{n^2+n\sqrt{2}}^{n^2+n\sqrt{2}+1}f(x)dx=o(1).\tag{1} $$ Left hand side of (1) may be rewritten as $\int_0^T h(x)f(x)$, where $$ h(x)=\frac1N\sum_{n=1}^N\sum_{k=-\infty}^\infty \mathbf{1}(x+kT\in [n^2+n\sqrt{2},n^2+n\sqrt{2}+1]). $$ Since $0\leqslant h(x)\leqslant 1$, by Lebesgue dominated convergence theorem it suffices to prove that pointwise we have $h(x)\to \frac1{T}$. For this goal, we rewrite (for fixed $x$) $$ h(x)=\frac1N \sum_{n=1}^N\mathbf{1}(n^2+n\sqrt{2}\in [x-1,x] \pmod T), $$ where $[x-1,x]$ is the arc of length 1 on the circle $\mathbb{R}/T\mathbb{Z}$ of length $T$. Since the polynomial $(n^2+n\sqrt{2})/T$ has at least one non-free irrational coefficient, this follows from Weyl equidistribution theorem (the original paper is Weyl, H. Uber die Gleichverteilung von Zahlen mod. Eins. Math. Ann. 77(3):313–352, 1916, now it is widely reproduced in literature).

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    $\begingroup$ Ah yes, the application of Weyl’s equidistribution theorem makes sense. Thank you for the answer. $\endgroup$
    – Nate River
    Commented May 12 at 14:06

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