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Let $A\subset \mathbb{R}$ and $F: A\rightarrow\mathbb{R}$ be a function that is monotonic.

Must $\dfrac{dF}{dx}$ exist?

I'm wondering about this in the context where $F$ is a cumulative distribution function and want to come up with a pathological CDF that is continuous but not differentiable. This is weird to me, since a CDF is an integral, so the CDF must be differentiable. However, we really define the density as a Radon-Nikodym derivative of the measure that induces the CDF, not the other way around.

So must $F$ have a (usual, not RN) derivative?

Dave
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  • It was a long time ago, so I don't remember the details, but I recall seeing a construction of such a function. It was similar to the Cantor middle third function. – herb steinberg May 06 '20 at 02:40
  • I was thinking of the Cantor distribution, but its CDF is differentiable (and equal to zero) almost everywhere. – Dave May 06 '20 at 02:42
  • You are right. I don't think I can be of further help. – herb steinberg May 06 '20 at 02:47
  • I believe this post will guide you to finding a proof that such a function cannot exist. Unless maybe your $A$ is particularly disconnected. – K.Power May 06 '20 at 03:01
  • Does a super disconnected $A$ help me for the ultimate goal of having this apply to a CDF? – Dave May 06 '20 at 03:09
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    No. First, CDF-s are defined on entire $\mathbb{R}$, and second, if $A$ is "super disconnected", e.g. discrete, derivative does not even make sense since the difference quotients are not defined at nearby points. – Conifold May 06 '20 at 03:16
  • As Conifold pointed out, your ultimate goal is impossible, because CFDs are by definition defined on all of $\mathbb R$. For your title question, the only counterexample would necessarily be defined on some set with no interior, for example $\mathbb Q$, and then you would have to redefine your definition of derivative to make sense in the subspace topology, and proceed from there. – K.Power May 06 '20 at 03:39
  • https://en.wikipedia.org/wiki/Weierstrass_function May be of some help. However monotone requirement could be a problem. – herb steinberg May 06 '20 at 21:00
  • @Dave http://mathonline.wikidot.com/lebesgue-s-theorem-for-the-differentiability-of-monotone-fun A cdf is differentiable almost everywhere. – herb steinberg May 09 '20 at 22:02

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