# What does it mean to say a function is differentiable with respect to lebesgue measure?

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

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The part of whether one needs to know about the Radon-Nikodym derivative first is only answered implicitly in that Radon-Nikodym doesn't show up there. But I think Radon-Nikodym is rather easy compared to Vitali's covering lemma. – t.b. Feb 19 '11 at 11:09
To clarify things, can you provide a definition for what it means to say "$f$ is differentiable with respect to the lebesgue measure almost everywhere" I could not infer this from the wiki page. Your help without the sarcastic & patronising remarks would be appreciated. – aukie Feb 20 '11 at 15:22
@Theo: You make a good point and I think I agree with it. – Did Feb 20 '11 at 17:42

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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@aukie: Thanks for your appreciation. The last sentence of your post indicates that you did not understand the wiki page I referred to, so let me try to rephrase it. First, in dimension $\ge 2$, Lebesgue theorem deals with limits along very general subsets (not only balls or products). Second, even in dimension $1$, a striking feature of Lebesgue theorem is that, for any integrable function $f$, the set of points where a primitive of $f$ has no derivative cannot be large, irrespective of the regularity of $f$. At the time, mathematicians were stunned by this result, but you know better... – Did Feb 20 '11 at 9:42
@Didier Piau ... I find comments such as "but you know better" and "I wonder what part of your question is not covered" condesending... I didn't ask the question to be mocked. I get it you are intelligent! thats brilliant. Regarding the wiki link, I dont think it clearly explains the meaning of the statement. Ofcourse the Lebesgue Differentiation Theorem is impressive. As far as I understand it in the one dimensional case, it seems to be generalising the fundamental theorem of calculus. But I didnt ask when the primitive of a function is differentiable a.e. – aukie Feb 20 '11 at 14:56
@aukie Wow! Somebody might want to cool down here... In case you are stil interested in mathematical facts, let me recapitulate: the statement a function is differentiable with respect to the [L]ebesgue measure almost everywhere means nothing unless translated as Lebesgue differentiation theorem, whose statement is: Every integrable function is such that any of its primitives is differentiable everywhere except on a set whose Lebesgue measure is zero. .../... – Did Feb 20 '11 at 17:35
.../... Thus my answer to your post, indicating that Lebesgue differentiation theorem was the way to put some sense in an otherwise ill-formed statement. In this context, what do you think of the first sentence of your aswer to your own post? Note that two different objects are involved here, an integrable function, often denoted by $f$, and its integrals on various subsets, and that nowhere in your post nor in your answer is mentioned the difference between $f$, about which some hypothesis is made (integrability), and its integrals, about which a conclusion is reached .../... – Did Feb 20 '11 at 17:35
.../... (almost everywhere differentiability). Note also that in higher dimensions, the definition of differentiability itself is a part of the statement of the theorem and that you mention nowhere in your question that you are interested in the 1D setting only (if this is indeed the case). So, if we leave aside your inflammatory (and unfounded) accusations, where does all this leave us? With a badly formulated question, to which a constructive answer was nevertheless proposed, answer which you met less than charitably, answering yourself your post (!). .../... – Did Feb 20 '11 at 17:36