Given a sequence of reals $(a_n)_{n > 0}$, let $f: [0, 1] \to R$ be the generalised raindrop function defined:

$f(x) = a_q$ if $x$ is rational, with denominator $q$ in lowest form; $0$ otherwise.


  • What are necessary and sufficient conditions on $a_n$for $f$ to be differentiable a.e.?

  • If $f$ is discontinuous on a set of positive Lebesgue measure, is $f$ discontinuous a.e.?

  • 2
    $\begingroup$ I think with the result that a.e. irrationals have irrationality measure 2, we can get a pretty decently weak condition. $\endgroup$ – James Baxter Mar 4 at 3:46
  • $\begingroup$ I think so, though I haven’t worked out the details. There may be some subtleties that trip me up haha. $\endgroup$ – James Baxter Mar 4 at 3:59
  • $\begingroup$ So this question sounds like a homework question to me... $\endgroup$ – Anthony Quas Mar 5 at 1:38
  • $\begingroup$ The homework problem that hasn’t been answered $\endgroup$ – James Baxter Mar 5 at 2:05

For question 2: If $a_n \to 0$ as $n \to \infty$, then $f$ is continuous at all irrationals, and thus a.e., as $\lim_{t \to x} f(t) = 0$ for every $x$.

If $\limsup_{n \to \infty} a_n > \varepsilon > 0$, then $\{x: f(x) > \varepsilon\}$ is dense, and $f$ is discontinuous everywhere.

For question $2$: if $a_q \ge c q^{-2}$ for some $c > 0$, $f$ is nondifferentiable everywhere, while if $a_q = O(q^{-\eta})$ for some $\eta > 2$, $f$ is differentiable a.e. But I don't think you can say it is differentiable a.e. if $a_q = o(q^{-2})$.

EDIT: In fact (see Khinchin's book, "Continued Fractions") for almost every $x$ there are infinitely many $p/q$ with $|x-p/q| < 1/(q^2 \ln(q))$, but for $\varepsilon > 0$, only finitely many with $|x -p/q| < 1/(q^2 \ln(q)^{1+\varepsilon})$. In particular, for $a_q = q^{-2}/\ln(q)$ which is $o(q^{-2})$, $f$ is nondifferentiable almost everywhere.


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.