Call a function $f: [0, \infty) \to \mathbb R$ nearly eventually almost periodic with period $p > 0$ if for a.e. $x \in [0, p)$, the sequence ${f(x + np)}_{n \in \mathbb N}$ converges.

Suppose $f: [0, \infty) \to \mathbb R$ is continuous and nearly eventually almost periodic of periods $1$ and $a$, where $a$ is irrational and $0 < a < 1$. Define $F: [0, 1) \to \mathbb R$ by $F(x) := \lim_{n \to \infty} f(x + n)$. Is $F$ necessarily constant a.e.? That is, is $F$ equal a.e. to a constant function?

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    $\begingroup$ You already posted a very similar question. The very least you should do is link to it and point out the difference, and perhaps comment thereon (I imagine you're changing the question because you're unsatisfied with the answer you got, but you should state this). $\endgroup$ – Gro-Tsen Mar 9 at 11:55
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    $\begingroup$ Perhaps MO has some nearly eventually almost-periodic behavior itself. $\endgroup$ – Matt Cuffaro Apr 30 at 14:07
  • $\begingroup$ Possible duplicate of Eventually almost periodic functions $\endgroup$ – Alex M. Apr 30 at 14:34

I tried to prove a positive answer by copying and modifying a bit Dap’s answer to your very similar question, so a main contribution to this answer belongs to @Dap.

Let $$Z_a=\{x\in[0,a)| \mbox{ the sequence }\{f(x + np)\}_{n \in \mathbb N}\mbox{ does not converge}\}.$$ Then $Z_a$ has measure $0$. Consider an $\epsilon>0.$ The sets $$C_N=\{x\in[0,a)\mid |f(x+an)-f(x+am)|\leq \epsilon/3\text{ for all }n,m\geq N\}$$ are closed and cover $[0,a)\setminus Z_a$, so a measure of $C_N$ is positive for some $N$. For each $t\in (2\epsilon/3)\Bbb Z$ put $$D_t=\{ x\in[0,a)\mid f(x+aN) \in [t-\epsilon/3,t+\epsilon/3]\}.$$ Since the function $f$ is continuous, the sets $D_t$ are closed and cover $[0,a)$, so a measure of $G=C_N\cap D_t$ is positive for some $t$.

This implies that $$\lim_{m\to\infty}f(x+am)\in [t-2\epsilon/3,t+2\epsilon/3]\text{ for all } x\in G,$$ which gives $$\phantom{abcdefghi}f(x+an)\in[t-\epsilon,t+\epsilon]\text{ for all }x\in G\text{ and }n\geq N.$$

Let $$Z_1=\{x\in[0,1)|\mbox{ the sequence }\{f(x + np)\}_{n \in \mathbb N}\mbox{ does not converge}\}.$$ Then $Z_1$ has measure $0$. If Lemma below holds then for almost any $x\in[0,1)\setminus (Z_a\cup Z_1)$ the sequence $\{x+n\}$ lies in the set $G+a\mathbb N$ infinitely often, giving $$\lim_{n\to\infty}f(x+n)\in [t-\epsilon, t+\epsilon]\text{ for all }x\in[0,1) \setminus Z_a.\phantom{abc}$$ So $\sup F-\inf F\leq 2\epsilon$ for all $\epsilon,$ which means $F$ is constant.

Lemma. For almost any $x\in[0,1)$ the sequence $\{x+n\}$ lies in the set $G+a\mathbb N$ infinitely often.

Put $G’=G+a\mathbb N$ and for each $n\in N$ put $G’_n=\bigcup_{k \in\Bbb N,\, k\ge n} G’-k$ and $G’’=\bigcap_{n\in\Bbb N} G’_n$. It suffices to show that a set $G’’\cap [0,1]$ has measure $1$. For this we have to show that for each $n\in\Bbb N$ a set $G’_n\cap [0,1]$ has measure $1$. This should be a known result, but I don’t found an exact reference. I guess that a map $$T=T_{1-a}: [0,1]\to [0,1],\, x\mapsto x-a-\lfloor x-a\rfloor$$ is an irrational rotation, which is ergodic. Put $E=G’_n\cap [0,1]$. It is easy to check that $T^{-1}(E)\subset E$ and, since $a$ is irrational, $E$ has positive measure. Then, since $T$ is ergodic, $E$ has measure $1$, see third property here.


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