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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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    @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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It seems my professor whose book I originally met this problem in published the proof himself, generalizing the previous result of Federer for continuous functions. I was really surprised how much simpler it was than I expected.

In fact, I noticed the proof only requires that $L(X;Y)$ is separable, hence so are $X'$ and $Y$, since they embed into it. Given these, either of them being finite-dimensional is sufficient. It also raises the question whether $L(X;Y)$ can be separable when both spaces are infinite-dimensional. I read somewhere that this is an open problem when $X=Y$.