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

Suppose $f: [0, \infty) \to \mathbb R$ is continuous and 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 continuous a.e.?

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    $\begingroup$ In your definition, "the sequence converges" in what topology? $\endgroup$ – Alexandre Eremenko Mar 3 at 14:26
  • $\begingroup$ Err, the standard topology on the reals. $\endgroup$ – James Baxter Mar 3 at 14:35
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    $\begingroup$ f(x+np) is a value sir. Pointwise if you wanna view them as functions. Also I’m editing the question a little. $\endgroup$ – James Baxter Mar 3 at 14:43
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    $\begingroup$ Can you give an example where $F$ is not constant? $\endgroup$ – Sam Hopkins Mar 3 at 14:44
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    $\begingroup$ Yes, though the construction is a little sketchy, but imagine a series of bump functions that take the value 1 at all points of the form na or n, for n integer. By making the bumps thin out, we can make F converge to 1 at 0 and 0 everywhere else. Hence the edit.. $\endgroup$ – James Baxter Mar 3 at 14:50
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$F$ must be constant. 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 $[0,a),$ so by the Baire category theorem there is an interval $[c,d]\subset C_N$ for some $0<c<d<a$ and some $N.$ Shrinking the interval $[c,d]$ if necessary we can ensure that $f([c,d]+aN)$ lies in an interval $[t-\epsilon/3,t+\epsilon/3].$ This implies that $\lim_{m\to\infty}f(x+am)\in [t-2\epsilon/3,t+2\epsilon/3]$ for $x\in[c,d],$ which gives $f(x+an)\in[t-\epsilon,t+\epsilon]$ for all $x\in[c,d]$ and $n\geq N.$ But for any $x\in[0,1)$ the sequence $x+n$ lies in the set $[c,d]+a\mathbb N$ infinitely often, giving $\lim_{n\to\infty}f(x+n)\in [t-\epsilon, t+\epsilon]$ for all $x\in[0,1).$ So $\sup F-\inf F\leq 2\epsilon$ for all $\epsilon,$ which means $F$ is constant.

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