TITLE: Measurability of $\xi$ in the mean value theorem
QUESTION [28 upvotes]: Suppose $f\in C^1(\mathbb{R})$, by mean value theorem, for any $x\in (0,\infty)$, there exists $\xi(x)\in (0,x)$ such that $$\frac{f(x)-f(0)}{x}=f'(\xi(x)).$$ My question is:

Question: Can $\xi(x)$ always be chosen to be a measurable function in $x$?

REPLY [5 votes]: How about: Define $\xi(x)$ as the least number $\xi$ such that 
$$
\frac{f(x)-f(0)}{x}=f'(\xi)
$$
From the $C^1$ hypothesis we know $f'$ is continuous.  
That does not quite work, because we may get $\xi(x) = 0$ sometimes.  
OK, define $\xi(x)$ as the least number $\xi \ge x/2$ such that
$$
\frac{f(x)-f(0)}{x}=f'(\xi)
$$
if there is such a $\xi$, otherwise the greatest number $\xi \le x/2$ such that
$$
\frac{f(x)-f(0)}{x}=f'(\xi)
$$