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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.

Questions:

  • 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.?

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    $\begingroup$ 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
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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.

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