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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    $\begingroup$ Can you clarify what you mean by "measurable?" Borel? Lebesgue? Baire? $\endgroup$ – user83827 Sep 14 '11 at 3:57
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    $\begingroup$ @ccc: By saying "measurable" I was expecting at least Lebesgue measurability. But measurability results in other sense are also welcome. $\endgroup$ – Syang Chen Sep 14 '11 at 4:21
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    $\begingroup$ Such a $\xi$ is not unique! $\endgroup$ – Henri Sep 14 '11 at 7:58
  • $\begingroup$ Uhm... what happens if $f$ is a constant function? Is $\xi$ well defined as a map of x? $\endgroup$ – uforoboa Sep 14 '11 at 10:03
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    $\begingroup$ @Henri: Why is non-uniqueness of $\xi$ an obstacle? It seems to be a perfectly reasonable question as asked: is there a measurable function $\xi$ such that the desired equality holds for all $x$? $\endgroup$ – t.b. Sep 14 '11 at 10:58

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

  • $\begingroup$ I guess the point being: measurable selection in the real line where you choose from compact sets doesn't require big machinery. $\endgroup$ – GEdgar Sep 14 '11 at 13:54
  • $\begingroup$ What exactly are you claiming about the measurability of this particular function? I agree that it is Lebesgue measurable (since it's $\sigma(\Sigma^1_1)$-measurable), but I don't see for example why it should be Borel. You can certainly perform a Borel selection from compact sets in general (given that the assignment is Borel, etc.), I just don't see why this choice of function happens to be Borel. $\endgroup$ – user83827 Sep 14 '11 at 15:14
  • $\begingroup$ First prove this: Suppose $f$ and $g$ are continuous. Let the function $\xi$ be defined by: $\xi(x)$ is the least $u$ with $f(x)=g(u)$. Then $\xi$ is lower semicontinuous. So, in particular, $\xi$ is Borel. $\endgroup$ – GEdgar Sep 14 '11 at 18:13
  • $\begingroup$ 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). $\endgroup$ – user83827 Sep 14 '11 at 18:38

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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    $\begingroup$ I took the liberty of adding in links for the convenience of the readers. An alternative reference to Kechris would be Srivastava's A course on Borel sets, there it would be Theorem 5.12.1 on page 219. $\endgroup$ – t.b. Sep 14 '11 at 13:13
  • $\begingroup$ This argument, in retrospect, is really only using compact uniformization, which is much easier to prove. The point is that the sections are not only $K_\sigma$, but $A$ can be written "uniformly" as union of sets with compact sections (which in fact is always possible, but it's the hard part of the $K_\sigma$ uniformization proof). But of course GEdgar's direct argument is simpler still. $\endgroup$ – user83827 Sep 16 '11 at 12:06
  • $\begingroup$ Thanks! It seems there is a really big machinery to deal with measurability problem. $\endgroup$ – Syang Chen Oct 8 '11 at 15:23

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