Consider the following result, where integrals are, say, Riemann integrals:

Let $f:\mathbb{R}\to\mathbb{C}$ be a periodic function of period $L$. Then for all $a\in\mathbb{R}$ $$ \int_0^Lf(x)\,dx=\int_a^{a+L}f(x)\,dx. $$

Question: Is a periodic function $f:\mathbb{R}\to\mathbb{C}$ necessarily Riemann-integrable? If not, then clearly the Riemann-integrability of $f$ is implicitly supposed in the proposition, so it is natural to ask whether it is a mistake not to add this hypothesis explicitly?

Thoughts: [Looks like something is wrong with my argument.] I do not think that a periodic function $f:\mathbb{R}\to\mathbb{C}$ is necessarily Riemann-integrable. I think that a periodic extension of, say, $f:[-\sqrt{2},\sqrt{2}]\to\mathbb{C}$ defined by $$ \begin{cases} 1,&x\in\mathbb{Q}\cap[-\sqrt{2},\sqrt{2}],\\ 0,&\text{otherwise} \end{cases} $$ is a counter-example.

Clearly the periodic extension of $f$ is $g:\mathbb{R}\to\mathbb{C}$ where $$ \begin{cases} 1,&x\in\mathbb{Q},\\ 0,&\text{otherwise}. \end{cases} $$ By extending $f$ to obtain $g$, we see that $g$ is $2\sqrt{2}$-periodic. However $2\sqrt{2}$ is irrational, therefore if $a$ is rational then $a+2\sqrt{2}$ is also irrational. Hence $f(a)\neq f(a+2\sqrt{2})$...

What's wrong? Is $g$ really periodic? I think $g$ is periodic for any non-zero rational number.

  • 1
    $\begingroup$ Relatedly math.stackexchange.com/questions/1004426/… The example there is also not Riemann integrable. One way out of the problem you cite would be to impose the strong condition of continuity on $f$. $\endgroup$ – Simon S Nov 10 '14 at 18:33

Another really old post that was never fully answered. Just in case someone else stumbles across it:

It is correct that periodicity does not all imply integrability, no matter what type of integral you use. Just choose any non-integrable function on $[0,1)$ and extend periodically with period $1$ to get a counter-example. How a function behaves outside of $[0,1)$ does not change its integrability in $[0, 1)$.

The problem with the example is that $g$ is not the periodic extension of $f$. The periodic extension $\bar f$ of $f$ with period $a = 2\sqrt 2$ would have value $\bar f(a) = \bar f(0) = f(0) = 1$, but $g(a) = 0$.

$g$ is periodic with any rational period. But $a$ is irrational.


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.