For $f \in L^1(\mathbb{R})$, we can proof, that $\sum_{n\in \mathbb{Z}} |f(t+na)|<\infty$ ALMOST EVERYWHERE for $t\in [0,a]$
($\star$ proof below)
My Question: If we assume $f\in L^1(\mathbb{R})\cap C(\mathbb{R})$, can we conclude $$\sum_{n\in \mathbb{Z}} |f(t+na)|<\infty \text{ for all }t\in [0,a]\ \ \ ?$$
I would especially need to show: $\sum_{n\in \mathbb{Z}} |f(na)|<\infty$
$\star$ proof:
Define $f_n=\chi_{(na,(n+1)a]}\cdot f \Rightarrow\ \sum_{n\in \mathbb{Z}}|f_n|=|f|\ \ \forall x\in \mathbb{R}$
If we use monotone convergence and shifting invariance of the Lebesgue integral we get: $$\infty>\int_\mathbb{R} |f(t)|dt=\int_\mathbb{R}\sum_{n\in \mathbb{Z}}|f_n(t)|dt = \sum_{n\in \mathbb{Z}}\int_\mathbb{R}|f_n(t)|dt=\sum_{n\in \mathbb{Z}}\int_{(na,(n+1)a]}|f(t)|dt=$$ $$=\sum_{n\in \mathbb{Z}}\int_0^a|f(t+na)|dt=\int_0^a\sum_{n\in \mathbb{Z}}|f(t+na)|dt$$ So we can conclude: $\sum_{n\in \mathbb{Z}}|f(t+na)|<\infty$ a.e. for $t\in[0,a]$
No this is not true, it would still be for allmost all $a$. As a counter example consider $a=1$ and $f$ given such that $f(n)=1$ but $\int_{n-1/2}^{n+1/2}f(x)dx = \frac{1}{n^2}$.
If you need me to elaborate a bit more leave a comment.