Problem:
Let $f$ be a Lebesgue integrable function on $\mathbb{R}.$ Prove that the series $$\sum\limits_{n=-\infty}^{+\infty}f(x+n)$$ converges absolutely for a.e. $x \in \mathbb{R}.$
What I have done:
$\sum\limits_{n=-\infty}^{+\infty}f(x+n)$ converges absolutely for a.e. $x \in \mathbb{R}.$ This is true iff $\sum\limits_{n=-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}f(x+y)d\mu$ is finite where $\mu$ is lebesgue measure. This is also true iff, $\int\limits_{R}\int\limits_{-\infty}^{+\infty}f(x+y)d\mu d\nu$ is finite where $\nu$ is the counting measure. Iff by the Fubini, if $f$ is $-\mu\times \nu$ measurable. But I dont know here can I say that since $f$ is measurable and integrable then $F(x,y)=f(x+y)$ is $-\mu\times \nu$ meadurable.
Comment:
If this approach is not OK please let know. For the alternative way, please give me a hint.
We may assume that $f$ is non-negative. Then by the translation invariance of the Lebesgue integral, the sum is always infinite unless $f$ is identically zero:
$$ \int_{\mathbb{R}} \sum_{n=-\infty}^{\infty} f(x+n) \, dx \stackrel{\text{(Tonelli)}}{=} \sum_{n=-\infty}^{\infty} \int_{\mathbb{R}} f(x+n) \, dx = \infty \cdot \int_{\mathbb{R}} f(x) \, dx. $$
That being said, your first claim on 'if and only if' condition for absolute convergence is not true as long as we prove that $\sum_{n=-\infty}^{\infty} f(x+n)$ converges even if the above integral diverges.
Indeed, you can try the following variant:
\begin{align*} \int_{[0,1)} \sum_{n=-\infty}^{\infty} f(x+n) \, dx &= \sum_{n=-\infty}^{\infty} \int_{[0,1)} f(x+n) \, dx \\ &= \sum_{n=-\infty}^{\infty} \int_{[-n,1-n)} f(x) \, dx \\ &= \int_{\mathbb{R}} f(x) \, dx < \infty, \end{align*}
from which we find that $\sum_{n=-\infty}^{\infty} f(x+n)$ converges absolutely for a.e. $x \in [0, 1)$. Since the sum is $1$-periodic, the proof is done.