Measurability of $\xi$ in the mean value theorem

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 $x$?

Solution 1:

It turns out that you can get away with all three that I mentioned in the comments (Borel of course being the most restrictive). However the argument I have in mind for finding a Borel measurable $\xi$ uses somewhat deep descriptive set theory: Arsenin, Kunugui uniformization for Borel relations with $K_\sigma$ sections (See Kechris' Classical descriptive set theory, Theorem 35.46). This is probably overkill, though, and I would very much like to see a more elementary solution.

To invoke this theorem, one must check that the set $$ A = \{(x,y) : 0 < x < \infty \ \ \mathrm{ and } \ \ 0 < y < x \ \ \mathrm{ and } \ \ (f(x) - f(0))/x = f'(y)\} $$ is a Borel subset of $\mathbb{R} \times \mathbb{R}$ , which follows from continuity of $f$ and $f'$. Also, one must check that for each $x$, the set $\{y : (x,y) \in A\}$ is a nonempty countable union of compact sets. This again exploits continuity of $f'$, since the set $\{y : 1/n \leq y \leq x - 1/n \ \ \mathrm{ and } \ \ f'(y) = c\}$ is compact for each $c$, and the mean value theorem ensures that $\{y : 1/n \leq y \leq x - 1/n \ \ \mathrm{ and } \ \ f'(y) = (f(x) - f(0))/x\}$ is nonempty for some $n$. One then obtains a Borel measurable uniformizing function $\xi$ such that $(x, \xi(x)) \in A$ for all $x \in (0, \infty)$.

Note that if you want only Lebesgue or Baire measurability, you can get away with a somewhat more elementary uniformization theorem (Jankov, von Neumann: Kechris 18.1). Again, this feels like using a sledgehammer, so I hope somebody sees a better argument!

Solution 2:

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 $$ \frac{f(x)-f(0)}{x}=f'(\xi) $$