How is this function a member of $L^{1}(0, \frac{1}{b})$?

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The function in question is:

$$ F_{n}(x) = \sum_{k \in \mathbb{Z}} f\left(x-\frac{k}{b}\right) g^{\ast}\left(x - na - \frac{k}{b}\right) $$

Where $\ast$ denotes complex conjugation, $f, g \in L^{2}$, $n \in \mathbb{N}$ and $a, b \in \mathbb{R}$ are fixed but arbitrary.

I suppose I don't understand how this function is $\frac{1}{b}$ periodic when $k$ ranges over all of $\mathbb{Z}$ and $b$ is given in $\mathbb{R}$.