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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 $x$?

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    @Theo: It is not an obstacle in itself, but I misread the OP's question and thought that he asked if $\xi$ was measurable or not. So, thanks for pointing it out.2011-09-14

2 Answers 2

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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)

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    Thanks for clarifying! I guess I got distracted by your comment and forgot that this problem was dealing with continuous functions. The minimization trick works in general to get Lebesgue measurable functions, but I'm still not sure about Borel (but again I could be missing something).2011-09-14
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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!

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    Tha$n$ks! It seems there is a really big machinery to deal with measurability problem.2011-10-08