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A local analysis textbook I have used has the following exercise:

Let $X$ be a finite-dimensional, $Y$ a separable Banach-space, $f\colon X\rightarrowtail Y$ any function. Show that $f'$ is Borel.

I have no idea how to approach this. I looked at the proof of Rademacher, but even that only gives Lebesgue-measurability as far as I can see, and this has no assumptions on the function at all. Any help would be appreciated.

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    Out of curiosity (I'm not sure if this statement is true or not): Have you actual *need* for this statement or is it out of interest that you ask?2011-05-11
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    Can you do the case $X = Y = \mathbb R$ ?2011-05-11
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    I do not need it for an application now, it is more out of interest.2011-05-12
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    Even if $X=Y=\mathbb{R}$ I cannot approach this. I do not know what possible method could apply to functions like $id^2_\mathbb{R}\cdot\chi_A$ for any $A$. Should I bring this mathoverflow?2011-05-12
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    Define $f_n(x):=\frac{f(x + 1/n) - f(x)}{1/n}$... then...2011-05-12
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    Then... what then?2011-05-12
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    @Jonas T: This obviously works under the assumption of Borel or Lebesgue measurability, but we have *no assumptions whatsoever*.2011-05-12
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    I just remembered that I actually [answered this question on MO](http://mathoverflow.net/questions/49418/) in the case of $X=Y=\mathbb{R}$ (it would be more precise to say that I provided some links to references). The extension to $X$ real and $Y$ separable should then be possible with some thinking as @GEdgar pointed out. (meta: include an `@username` somewhere in your comment, so that the user you want to address get notified), as I did in my two comments.2011-05-12
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    @Theo: Oh, right I missed that. I believe there is a proof in Bogachev. I should look it up.2011-05-12
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    @Theo, @scineram, are you sure this is even true for an arbitrary function? I don't think it is without at least boundedness... this seems a little rough for an exercise. Scineram, perhaps you can post some more context? Which book exactly?2011-05-13
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    @Glen: I must admit that I haven't checked this in much detail since I had difficulty in seeing any need for such a statement (even for functions from $\mathbb{R} \to \mathbb{R}$, to which I restrict in the following). It seems to me that where the derivative exists, it should be automatically Borel (even Baire class $1$) since $f$ has to be continuous there, so if one believes the theorem I mention in the answer on MO, one should be able to "drop" the $G_{\delta}$ and the $G_{\delta\sigma}$-null and modify $f'$ accordingly by setting it to be zero there.2011-05-13
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    @Glen: My guess is however that the suggestion by Jonas T is the one intended. I agree that a precise reference to the book would be a very good thing to have.2011-05-13
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    @Theo that's probably true. I agree with you wholeheartedly on the applications question: I can't really think of a good use for such a statement. Somewhat interesting nonetheless.2011-05-13

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