If $f$ is integrable on $\mathbb R$, there exists a nullset $N$ with $\liminf_{n\to\infty}f(x+n)=0 \ \text{ for all } x\in\mathbb{R}\setminus N$

Question

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.

Answer 1

There are two nice ways to show this (or something very similar)

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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 $f$.

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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.