$f:\mathbb R \to \mathbb R$ is Lebesgue integrable. We need to show that $$\lim_{n \to \infty} f(x-n)=\lim_{n \to \infty} f(x+n)=0$$ for almost all $x \in \mathbb R$.
I have managed to show that
$$A:=\left\{x \in \mathbb R : f(x+n) \not \to 0\}=\bigcup_{m \in \mathbb N} \bigcap_{N \in \mathbb N} \bigcup_{n \ge N}\{x \in \mathbb R: \left|f(x+n)\right| > \frac{1}{m} \right\}$$
Which means that $A$ is measurable. Now I need to show that the mass of $A$ is zero and I guess it should follow from the fact that $f$ is integrable. I have tried pulling it towards Fatou's lemma but without any success. And here I am stuck. Would appreciate a hint.
Integrability of $f$ entails the convergence of the series $$\sum_{n\in\mathbb Z}\int_{[n,n +1]} \left|f\left(x\right)\right|\mathrm d\lambda(x) =\sum_{n\in\mathbb Z}\int_{[0,1]} \left|f\left(x+n\right)\right|\mathrm d\lambda(x)= \int_{[0,1]} \sum_{n\in\mathbb Z} \left|f\left(x+n\right)\right|\mathrm d\lambda(x).$$ This shows that the sequence $\left(f\left(x+n\right)\right)_n$ goes to $0$ as $\left|n\right|\to +\infty$ for almost every $x\in[0,1]$. The same holds for the other $x$.