Measurable function and the Mean Value Theorem Let $\,f:[a,b]\to \mathbb{R}\,$ be continuous on $[a,b]$ and derivable on $(a,b)$. By the mean value property, for all $\,x\in (a,b)\,$ there exists $\,\xi_x\in (a,x)\,$ such that $\,f(x)-f(a)=f'\left(\xi_x\right)(x-a)$. Let $\,h:(a,b)\to\mathbb{R}\,$ be defined as $\,h(x)=\xi_x$. Is the function $h$ measurable? Is $\,f'\!\circ h\,$ measurable?
The motivation for this question is the following: in an exercise in class, I estimated $\,\int_{\mathbb{R}} \left\lvert F(x)-F(-x)\right\rvert dx\,$ for a certain "good" function $\,F$ using $\,\leq \int_{\mathbb{R}} \left\lvert F'\left(\xi_x\right)\right\rvert 2\left\lvert x\right\rvert dx,\,$ but the professor told me to be careful, as he wasn't sure of the measurability of $\,x\mapsto F'\left(\xi_x\right)$.
Edit: Thank you for your answers. Assuming that $x\mapsto \xi_x$ is well-defined (taking $\xi_x$ as the smallest possible), is it measurable?
 A: As already noted by others, $h$ is not well-defined. But you can define $h$ to be measurable: $h$ is actually a well-defined correspondence
$$h(x)=\{ξ\in (a,b);\ f(x)−f(a)=f′(ξ)(x−a)\}$$
It is possible to show $h$ is measurable as a correspondence (see for example corollary 18.8 here) hence by the measurable selection theorem (see theorem 18.13 in the previous link) $h$ admits a measurable selection.
A: The function $x\to\xi_x$ is not well defined, as you stated it. However, if you make it well defined, for instance saying that $\xi_x$ is the smallest number that satisfy the MVT, then $f'\circ h$ is measurable. This is because
$$
(f'\circ h)(x) = \frac{f(x)-f(a)}{x-a}
$$
Since the right hand side is a measurable function, $f'\circ h$ is measurable.
Edit: the only situation where we may not be able to pick the smallest $\xi_x$ is if we have infinitely many. Note: I am assuming that differentiable is intended as "the derivative exists and it is continuous", which is slightly more than the OP specified. If these infinitely many candidate $\xi_x$ are continuous segment, than $f$ is locally a line, and since $f'$ is continuous, I can take the minimum of that segment. The other case is if we have an infinite number of candidates for $\xi_x$ that are disjoint. Clearly, there has to be an infimum. But since $f'$ is continuous, then the infimum must have the same derivative, and therefore be a suitable candidate for $\xi_x$.
