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?
Best Answer
This is not quite an answer, but I'm going to post it anyway because I think it may lead to a real answer.
If we modify your question so that instead of $c_y$ we consider $$d_y = d(y) = \inf\left\{z \in [y-h, y+h] : \int_{y-h}^{y+h} G(t)\,dt = 2 h G(z) \right\}$$ (note the closed interval instead of open), then we can show that the function $d$ is lower semicontinuous, and in particular is Lebesgue measurable. Fix a sequence $y_n \to y_0$ and set $d_0 = \liminf_{n \to \infty} d(y_n)$. We have to show $d(y_0) \le d_0$. Passing to a subsequence, we may assume $d_0 = \lim_{n \to \infty} d(y_n)$. Note that $y_n -h \le d(y_n) \le y_n + h$ by definition, so $y_0 - h \le d_0 \le y_0 + h$; in particular $d_0$ is finite.
Now for each $n$, by definition of $d$ there exists $z_n$ such that $d(y_n) \le z_n \le d(y_n) + \frac{1}{n}$ and $$\int_{y_n-h}^{y_n+h} G(t)\,dt = 2 h G(z_n). \tag{*}$$ By the squeezing lemma, $z_n \to d_0$ and so by continuity of $G$ we have $2hG(z_n) \to 2hG(d_0)$. On the other hand, by the dominated convergence theorem we have $$\int_{y_n-h}^{y_n+h} G(t)\,dt \to \int_{y_0-h}^{y_0 + h} G(t)\,dt.$$ (To see this, set $F_n = 1_{[y_n - h, y_n + h]}$ and note that $F_n(t) \to F_0(t)$ for all $t \ne y_0 \pm h$; in particular $F_n \to F_0$ almost everywhere. So $F_n G \to F_0 G$ almost everywhere, and is dominated by $G$ on some bounded interval large enough to contain all $[y_n - h, y_n + h]$.)
So when we pass to the limit in (*), we obtain $$\int_{y_0 - h}^{y_0 + h} G(t)\,dt = 2 h G(d_0).$$ Therefore, $d(y_0) \le d_0$.
If we work with $c(y)$ instead, most of the argument above still works. But the very last line could fail because we might have $c_0 = y_0 - h$ in which case we cannot conclude $c(y_0) \le c_0$. I think there should be a way to fix this, but it's too late at night here to keep thinking about it...