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By Darboux's theorem if $f:D\to\mathbb R$ is differentiable then $f'$ satisfies the intermediate value property $-$ even if it is discontinuous. In particular I am interested in the following:

Assume $f'(a)<f'(b)$ for some $a<b$. We know that then $f'$ assumes every value in the interval $I=[f'(a),f'(b)]$ within $[a,b]$. Does this imply that $(f')^{-1}\big(I\big)$ has postive measure?

Intuitively it seems like it must be true; but how to prove it? This question is related to a comment I made on this thread.

Note that if $g:D\to\mathbb R$ has the intermediate value property, but is not the derivative of a differentiable function, then it can be quite volatile $-$ see for instance Conways base 13 function. Here, the pre-image of $I$ could only be described as a mess.

Some related stuff:

Hyperplane
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  • Related: https://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be/112133 – Hans Lundmark Dec 18 '17 at 07:37
  • We know that derivatives are certainly measurable. Hence it suffices to show that the pre-image of an interval under any measurable map has positive measure. Not sure if it's true though. – MathematicsStudent1122 Dec 19 '17 at 00:11

1 Answers1

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With those assumptions, $(f')^{-1}(I)$ has positive measure. There is a proof here: https://mathoverflow.net/questions/266377/how-quickly-can-the-derivative-of-an-everywhere-differentiable-function-change-s. Here is an alternate argument for anyone interested.

Assume not, so $(f')^{-1}(I)$ has zero measure. By scaling and adding a linear function we may assume $I=[0,1],$ specifically $f'(a)=0$ and $f'(b)=1.$ We will construct a nested sequence of closed intervals $[a_n,b_n]$ such that the divided difference $d_n=\frac{f(b_n)-f(a_n)}{b_n-a_n}$ satisfies $d\in(0,1/3)$ for odd $n$ and $d\in(2/3,1)$ for even $n.$

To get started, if $f(b)-f(a)$ is positive, by continuity there is some $x\in (a,b]$ such that $(f(x)-f(a))/(x-a) \in (0,1/3)$; in this case start at $n=1$ with $a_1=a$ and $b_1=x.$ If $f(b)-f(a)$ is negative, there is some $x\in [a,b)$ with $(f(b)-f(x))/(b-x)\in (2/3,1)$; in this case start at $n=0$ instead of $n=1,$ and take $a_0=x$ and $b_0=b.$

Assume we have constructed $[a_n,b_n]$ with $n$ even, so $d_n\in(2/3,1).$ This means the (Henstock-Kurzweil) average value of $f'$ on $[a_n,b_n]$ is less than $1.$ Since $(f')^{-1}([0,1])$ has zero measure, there is some $x\in [a_n,b_n]$ with $f'(x)<0.$ Either $(f(b_n)-f(x))/(b_n-x)$ or $(f(x)-f(a_n))/(x-a_n)$ is positive, because $d_n$ is a convex combination of these ratios. In the first case we take $a_{n+1}=x$ and take $b_{n+1}\in(x,b_{n+1}]$ such that $d_{n+1}\in(0,1/3),$ which exists by continuity. In the second case is take $b_{n+1}=x$ and $a_{n+1}\in[a_n,x)$ such that $d_{n+1}\in(0,1/3).$ Similarly, given $[a_n,b_n]$ with $n$ odd, the same argument applied to $x-f(x)$ gives a $[a_{n+1},b_{n+1}]$ with $d\in(2/3,1).$

This completes the construction of the sequence of intervals $[a_n,b_n].$ If $\lim (b_n-a_n)>0$ then $d_n$ would converge to $(f(\lim b_n)-f(\lim a_n))/\lim(b_n-a_n).$ If $\lim (b_n-a_n)=0$ then $a_n$ and $b_n$ converge to some common point $x,$ but then $d_n$ would have to converge to $f'(x).$ In either case $d_n$ has to converge but, by construction, it doesn't.

Dap
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    For what it's worth, this is called the Denjoy-Clarkson property in the literature. For three successively stronger versions of this property (by Zygmunt Zahorski in 1950, by Clifford E. Weil in 1973, by Peter S. Bullen and D. N. Sarkhel in 1996), see the paragraph in my 21 May 2008 sci.math post that begins with "Derivatives also satisfy stronger versions of the intermediate value property". – Dave L. Renfro Feb 10 '18 at 10:34