I want to show that $$\lim_{n\rightarrow\infty}f(x+n)=0$$ for almost all $x$ if $f:[0,\infty)\rightarrow[0,\infty)$ is Lebesgue-integrable.
I attempted a proof by contrapositive, testing for the case that $\lim_{n\rightarrow\infty}f(x+n)=a>0$ for many $x$. Then I noticed that this is not sufficient, as it does not account for cases where the limit doesn't exist.
Now I don't see any alternative proof routes, as I'm mentally stuck on the last one. Also, I'm struggling with using the assumption $\lim f(x+n)$ as it takes the limit of discrete steps.
How could one attempt a proof?
By Fubini's theorem $$\int_0^\infty |f(x)|\; dx = \sum_{n=0}^\infty \int_0^1 |f(n+x)| \; dx = \int_0^1 \sum_{n=0}^\infty |f(n+x)| \; dx $$ Thus for almost every $x \in [0,1]$, the series $\sum_{n=0}^\infty |f(n+x)|$ must converge, which implies $f(n+x) \to 0$ as $n \to \infty$.