For $f :\mathbb{R} \rightarrow \mathbb{R}\displaystyle$ we have ${\lim_{n \to +\infty} f(x+n) =0}$ almost everywhere

I want to show that for $f: \mathbb{R} \rightarrow\mathbb{R}$ Lebesgue integrable $\displaystyle{\lim_{n \to +\infty} f(x+n) =0}$ almost everywhere.

It is clear to me that this statement is true and I have tried to prove it by contradiction, by assuming that either $\displaystyle{\lim_{n \to +\infty} f(x+n) =\infty}$ or$\displaystyle{\lim_{n \to +\infty} f(x+n) =K}$ with $K\in \mathbb{R}$ .

In the first case we can follow that $|f|=\infty$ so that $\int_\mathbb{R}f=\infty$ and we have a contradiction.

In the other case we know that there must be an $a\in \mathbb{R}$ so that $\forall x>a$ we have $f_+(x)>K/2$ and by that:

$\int_{{\mathbb{R}}+}|f|>\int_{\mathbb{R}_+}f_+=\int_{[0,a]}f_++ \int_{(a,\infty)}f_+>\int_{[0,a]}f_++\int_{(a,\infty)}K/2=\infty$

Would this be enough to show the statement? Can I even make the assumption that the limit of $f(x+n)$ "exists"? Any hints would be great.

Consider the function $g : [0,1] \to [0, \infty]$ defined by $$g(x) = \sum_{n=0}^{\infty}|f(x+n)|$$ By the monotone convergence theorem, we have $$\begin{aligned} \int_0^1 g(x)\ dx &= \int_0^1 \sum_{n=0}^{\infty}|f(x+n)|\ dx \\ &= \sum_{n=0}^{\infty}\int_0^1 |f(x+n)|\ dx \\ &= \sum_{n=0}^{\infty}\int_n^{n+1} |f(x)|\ dx \\ &= \int_0^{\infty}|f(x)|\ dx \\ & < \infty \end{aligned}$$ which means that $g(x)$ is finite almost everywhere on $[0,1]$. Therefore, the series $$\sum_{n=0}^{\infty}|f(x+n)|$$ converges almost everywhere on $[0,1]$. Since the terms of a convergent series must tend to zero as $n \to \infty$, it follows that $|f(x+n)| \to 0$ for a.e. $x \in [0,1]$.

Note that this implies that $|f(x+n)| \to 0$ a.e. in any interval $[N, N+1]$ because we can define $y = x - N$ (so $y \in [0,1]$), and observe that $|f(x+n)| = |f(y+(n+N))|$. As $n+N \to \infty$ iff $n \to \infty$, it follows from what we proved above that $|f(y+(n+N))| \to 0$ for a.e. $y \in [0,1]$, and hence that $|f(x+n)| \to 0$ for a.e. $x \in [N, N+1]$.

Finally, since $\mathbb R$ is the countable union of the intervals $[N, N+1]$, the result holds for a.e. $x \in \mathbb R$.