Converse of mean value theorem almost everywhere? Let $f: \mathbb R \to \mathbb R$ be a $C^1$ function.
We say a point $c \in \mathbb R$ is a mean value point of $f$ if there exists an open interval $(a,b)$ containing $c$ such that $f’(c) = \frac{f(b) - f(a)}{b-a}$.

Question: Is it true that (Lebesgue) almost every point in $\mathbb R$ is a mean value point of $f$?

 A: Let $U$ be an open and dense subset of $\mathbb{R}$ with finite measure. Let $g: \mathbb{R} \to \mathbb{R}^{\ge 0}$ be a continuous function with $\{g = 0\} = U^c$. Then define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = g(0)+\int_0^x g(t)dt$. Then $f \in C^1$, and $f' \equiv g$ means $f$ is strictly increasing (since any interval $(x,y)$ contains an interval lying in $U$ on which $g$ is strictly positive). So we have a strictly increasing $C^1$ function with zero derivative everywhere except for a finite measure set.
A: There was a series of articles in the Mathematical Monthly that considered near variants of this problem. These were

*

*Jingcheng Tong & Peter A. Braza (1997) A Converse of the
Mean Value Theorem, The American Mathematical Monthly, 104:10, 939-942, DOI:
10.1080/00029890.1997.11990743
In this article, Tong and Braza give two forms of converse to the mean value theorem. They say that a differentiable function $f$ on $(a, b)$ satisfies the Weak Form of the converse at a point $c \in (a, b)$ if $f'(c) = \frac{f(\beta) - f(\alpha)}{\beta - \alpha}$ for some interval $(\alpha, \beta) \subset (a, b)$, and $f$ satisfies the Strong Form if in addition $c \in (\alpha, \beta)$.
They show that if $f'(c)$ is not a global extremum of $f$, then the Weak Form is satisfied at $c$. If further $c$ is not an accumulation point of the set $\{ x : f'(x) = f'(c) \}$, then it satisfies the Strong Form.


*J. M. Borwein & Xianfu Wang (1998) The Converse of the Mean Value
Theorem May Fail Generically, The American Mathematical Monthly, 105:9, 847-848, DOI:
10.1080/00029890.1998.12004975
In this article, Borwein and Wang give an example that has a dense, uncountable set of points $c$ that fail the Weak Form (as described above), and thus are not mean value points (or what is sometimes called mean value abscissas). But this set still has measure $0$.


*H. Fejzić & D. Rinne (1999) More on a Mean Value Theorem Converse, The
American Mathematical Monthly, 106:5, 454-455, DOI: 10.1080/00029890.1999.12005069
In this article (which is one of those examples of an article in the AMM that casually includes pretty sophisticated math), Fejzić and Rinne describe a construction of a function $f$ that is strictly increasing but which has zero derivative on a set of positive measure.
They also show that the Lebesgue measure of points $c$ that are mean value abscissas is positive.
In short, it's not true that almost every (in the Lebesgue measure sense) point is a mean value abscissa.
