Limit of lebesgue-integrable functions
Solution 1:
Let $g(x)=\displaystyle\sum_{k=-\infty}^{\infty}|f(x+k)|$, then $\displaystyle\int_{0}^{1}g(x)dx=\sum_{k=-\infty}^{\infty}\int_{k}^{k+1}|f(x)|dx=\int_{\bf{R}}|f(x)|dx<\infty$, so for a.e. $x\in[0,1]$ we have $\displaystyle\sum_{k=-\infty}^{\infty}|f(x+k)|<\infty$, so $f(x+n)\rightarrow 0$ as $|n|\rightarrow\infty$. Similar argument works for other interval $[l,l+1]$.