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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.

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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.

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  • $\begingroup$ How do you change the change $\int_{[0,1)} \sum_{n=-\infty}^{\infty} f(x+n) \, dx\sum_{n=-\infty}^{\infty} \int_{[0,1)} f(x+n) \, dx$ ??? $\endgroup$ – Hamed Baghal Ghaffari Dec 23 '17 at 18:14
  • $\begingroup$ Also why the summation is 1-periodic? may I say $\sum_{n=-\infty}^{\infty} f(x+n)=\sum_{n=-\infty}^{\infty} f(x+1+n)$ $\endgroup$ – Hamed Baghal Ghaffari Dec 23 '17 at 18:15
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    $\begingroup$ We have a mighty weapon called Tonelli’s theorem, which allows to interchange the order of integration as long as the integrand is non-negative (and jointly measurable, of course). For your second question, again the answer comes either from absolute convergence or from non-negativity. $\endgroup$ – Sangchul Lee Dec 23 '17 at 18:18
  • $\begingroup$ thank you. If I use counting measure and integral instead of summation? $\endgroup$ – Hamed Baghal Ghaffari Dec 23 '17 at 18:24
  • $\begingroup$ That’s right. That is actually an important detail, since conditionally convergent series cannot be turned into an integral w.r.t. the counting measure. $\endgroup$ – Sangchul Lee Dec 23 '17 at 18:37
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The problem statement is to show that for almost every $x$ the series of function values $\sum_n f(x+n)$ converges, not that the series of integrals of the function converges.

Your idea to use Fubini is a good one, but you seem to be muddled in the execution. I cannot understand the notation in your second displayed paragraph: what is $y$, what does it have to do with $n$. More importantly, I cannot understand your "This is true iff...". Why?

You should maybe start over, applying Fubini to the product space $[0,1)\times\mathbb Z$ (with Lebesgue measure cross counting measure) and integrand $g(u,n)=f(u+n)$.

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