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According to first mean value theorem for integration, if $G \ : \ [a,b] \to \mathbb{R}$ is a continuous function, there exists $x \in (a,b)$ such that $$\int_a^b G(t) dt = G(x)(b-a)$$

Assume $G$ is a continuous function defined on $[a,b]$. For $0 < h < \frac{b-a}{2}$ $$\overline{G} \ : \ y \mapsto \int_{y-h}^{y+h} G(t) dt$$ is defined for $y \in [a+h,b-h]$. Applying the first mean value theorem for integration, for all $y \in [a+h,b-h]$, there exists $c_y \in (y-h,y+h)$ with $$\overline{G}(y)=\int_{y-h}^{y+h} G(t) dt = 2 h G(c_y)$$

Taking for $G$ a constant function, $c_y$ can by any point in $(y-h,y+h)$. Hence we can pick up it in a way for which $y \mapsto c_y$ is not a Lebesgue measureable function.

Question: can one find a continuous function $G$ for which $c_y$ is uniquely defined for all $y \in (a+h,b-h)$ and such that $y \mapsto c_y$ is not Lebesgue measureable?

Jean-Pierre (http://www.mathcounterexamples.net)

mathcounterexamples.net
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  • In fact I have the feeling that my question is not so relevant... Is the condition on the unicity of $c_y$ not implying that $G$ is monotonic. Hence $y \mapsto c_y$ is also monotonic and measurable? – mathcounterexamples.net Jun 06 '15 at 09:39
  • Are you also assuming that $c_y$ does not depend on $h$ ? – charmd Jun 24 '18 at 07:57
  • I can think of a result which might be of some interest to you (as it seems close to the core of your problem): if $f$ is differentiable and for all $x<y$, the set of $z\in (x,y)$ such that $f'(z)=\frac{f(y)-f(x)}{y-x}$ is an interval, then $f$ or $-f$ is convex (thus $f'$ monotonic) – charmd Jun 24 '18 at 08:00
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    I suppose that $h$ is defined and fixed in the question. – mathcounterexamples.net Jun 24 '18 at 08:42

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