Let $C(\mathbb{R})$ be the space of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ with the compact-open topology, and the associated Borel $\sigma$-algebra. Consider the function $p$ from $C(\mathbb{R})$ to $\mathbb{R}_{\geq 0} \cup \{\infty\}$ that maps a continuous function to its period, with the convention that non-periodic functions get mapped to $\infty$. Is the function $p$ a measurable function on $C(\mathbb{R})$? The only way I know of constructing measurable functions is to realize them as iterated lim, limsup, or liminf of a sequence of continuous functions. It's not clear here what continuous functions approximate the period in any reasonable manner.


  1. Is there a way to approximate the period of $f \in C(\mathbb{R})$ using a continuous map from $C(\mathbb{R})$ to $\mathbb{R}$, which on taking appropriate limits, converges to the described function $p$?

  2. Is there some other way of showing that the map $p$ is measurable?

  • 2
    $\begingroup$ One way of extracting the period of a function is taking the Fourier transform, see for example this signal processing post. You get a Dirac comb, the width of which is inversely proportional to the period. This operation is continuous, because the compact-open convergence implies convergence as a tempered distribution. However, I don't know if this is useful for you. $\endgroup$ Commented Oct 29, 2019 at 13:40
  • 1
    $\begingroup$ Is it clear that $p$ is well-defined? To me it is not completely obvious that an arbitrary continuous function is either non-periodic or has a unique period. $\endgroup$
    – B K
    Commented Oct 30, 2019 at 0:57
  • 1
    $\begingroup$ @BK: $\inf \{ t \in \mathbb{R}^+ | (\forall x \in \mathbb{R}) f(x+t) = f(x) \}$ $\endgroup$ Commented Oct 30, 2019 at 1:57

1 Answer 1


Isn't the set $p^{-1}([0, T_0])$ closed for every finite $T_0$? Suppose that $f_n$ has period $T_n \leqslant T_0$ and it converges locally uniformly to $f$. By passing to a subsequence, we may assume that $T_n$ has a limit $T$. Uniform convergence of $f_n$ on $[x, x + T_0]$ implies that $$f(x + T) = \lim f_n(x + T_n) = \lim f_n(x) = f(x),$$ and hence $f$ has period at most $T$.

  • $\begingroup$ Thanks! I guess I was just looking at the problem the wrong way. And from the fact that $p$ is upper semicontinuous, it shouldn't be too hard to find explicitly find continuous functions that approximate the period as well. $\endgroup$ Commented Oct 29, 2019 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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