Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a lebesgue integrable function. I want to show that almost everywhere $ \lim\limits_{n \rightarrow \infty}{f(x+n)} = \lim\limits_{n \rightarrow \infty}{f(x-n)} = 0 $
This looks so unintuitive for me. Is $x$ even important? I'm just starting with measure theory but for all I know with $f(x)=x$ is obviously lebesgue-integrable and $\lim\limits_{n \rightarrow \infty}{f(x+n)} = \infty $.
The only thing I can imagine is that all $n$ are a null set that does not change the integral but why does my intution fail?
As I already mentioned it might be enough to look at constant $x$ or at least that $x$ is from a intervall. And as I have seen in other proofs it makes things easier to just look at positie functions and then transfer it to general functions.
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 $[l,l+1]$.