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.


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.

  • $\begingroup$ How do you change the change $\int_{[0,1)} \sum_{n=-\infty}^{\infty} f(x+n) \, dx\sum_{n=-\infty}^{\infty} \int_{[0,1)} f(x+n) \, dx$ ??? $\endgroup$ – Hamit Dec 23 '17 at 18:14
  • $\begingroup$ Also why the summation is 1-periodic? may I say $\sum_{n=-\infty}^{\infty} f(x+n)=\sum_{n=-\infty}^{\infty} f(x+1+n)$ $\endgroup$ – Hamit Dec 23 '17 at 18:15
  • 1
    $\begingroup$ 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? $\endgroup$ – Hamit 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

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 $g(u,n)=f(u+n)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.