If $f$ is integrable on $\mathbb R$, there exists a nullset $N$ with $\liminf_{n\to\infty}f(x+n)=0 \ \text{ for all } x\in\mathbb{R}\setminus N$ 
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be Lebesgue integrable. Show that there exists a nullset $N$, such that
  $$\liminf_{n\to\infty}f(x+n)=0 \ \text{ for all } x\in\mathbb{R}\setminus N$$

Things I've considered is Fatou to show that the integral over $f$ is $0$ which is equivalent to $f=0$ almost everywhere. However, I somehow can't put things correctly to get the proof.
I'd appreciate any help.
 A: There are two nice ways to show this (or something very similar)

First note that a simple change of variables yields
$$
\int_0^\infty |f(x+n)| \, dx = \int_n^\infty |f(y)| \, dy = \int_\Bbb{R} |f(y)| \chi_{(n,\infty)}(y) \, dy \to 0.
$$
Here the last convergence is justified by dominated convergence, because the integrand converges to $0$ pointwise and is dominated by $|f|$, which is integrable by assumption.
Using Fatou's Lemma, we conclude
$$
0 \leq \int_0^\infty \liminf_n |f(x+n)| \, dx \leq \liminf_n \int_0^\infty |f(x+n)| \, dx = 0.
$$
But a nonnegative function has integral zero iff it vanishes almost everywhere. Hence, we have shown
$$
\liminf_n |f(x+n)| = 0 \text{ for almost all } x > 0.
$$
A similar argument shows the same convergence for almost all $x < 0$.
The only problem here is that this shows the claim for $|f|$ instead of $f$.

We can fix this problem using the so-called periodization trick. Using monotone convergence and a change of variables, we get
\begin{eqnarray*}
\infty > \int_\Bbb{R} |f(x)| \, dx &=& \sum_{k=-\infty}^\infty \int_{[0,1) + k} |f(x)| \, dx \\ &=& \sum_{k=-\infty}^\infty \int_{[0,1)} |f(x+k)| \, dx \\ &=&  \int_{[0,1)} \sum_{k=-\infty}^\infty |f(x+k)| \, dx .
\end{eqnarray*}
Since a function with finite integral has to be finite-valued almost everywhere, we see
$$
\sum_{k=-\infty}^\infty |f(x+k)| < \infty
$$
for almost all $x \in [0,1)$, i.e. for all $x \in [0,1) \setminus N$, with $N$ of measure zero.
It is now easy to see that $N' := \bigcup_{k \in \Bbb{Z}} N + k$ is also a null-set and that
$$
\sum_{k=-\infty}^\infty |f(x+k)| < \infty
$$
holds for all $x \in \Bbb{R}\setminus N'$, hence for almost all $x$.
In particular, we get $|f(x+k)| \to 0$ as $k\to\infty$ for almost all $x$. This is an even stronger statement then the one you are interested in.
