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 $f’(c) = \frac{f(b) - f(a)}{b-a}$.



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


