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 defined for all $y \in (a+h,b-h)$ as the lower bound of the $z$ such that $\int_{y-h}^{y+h} G(t) dt = 2 h G(z)$ and such that $y \mapsto c_y$ is not Lebesgue measureable?**