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What does it mean to say a function is differentiable with respect to the lebesgue measure almost everywhere. A definition would be helpfull.

Do I need to learn about the Radon Nikodym derivative to understand this.

Thanks

2 Answers 2

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After some thought and since I didn't really get any satisfactory answers. This is what I make of the statement.

The definition of differentiation in the statement is taken in the usual sense (the standard defintion from calculus, for some reason I thought the statement was referring to some new definition of differentiation!). So the statemtnt simply says, The set of points at which the function is not differentiable is a null set w.r.t. the lebesgue measure.

So the statement would have been clearer I think if the author had written

$f$ is differentiable almost everywhere with respect to the lebesgue measure ... I think thats right!!

So no new definition of differentiation, phew! just some confusing terminology.

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    @Didier Piau ... If you dont mind I would like to start again with you. Can I ask you when a statement like "a function $F$ is differentiable wrt the lebesgue measure", is made does this mean, 1) as I mentioned in my "attempted" and not accepted answer. or 2) does it imply that a density $f$ exists such that $F(x)=\int _a^xf(t)dt$, and that the derivative $F'(x):=f(x)$ exists almost everywhere, (where note the derivative of $F$ is defined as $F'(x):=f(x)$ and not by the limit definition of calculus). I've read a statment in the literature and I want to know what it means.2011-02-20
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I wonder what part of your question is not covered by this page.

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    @aukie See my comments appended to your reply to your own post.2011-02-20