Do we have d/dx F(x) = d/dx ∫ (over [a,x]) f dm = f(x)?

Suppose the real valued f is defined on [a, b] and, for every c in [a,b], f is Lebesgue integrable, with the Lebesgue measure, over [a, c]. Define F(x) = ∫ (over [a,x]) f dm, x in [a, b], m = Lebesgue measure. Then, is it true that F is differentiable and F'(x) = f(x) for every x in [a, b]?





Thank you

Do we have d/dx F(x) = d/dx ∫ (over [a,x]) f dm = f(x)?
No.





As a trivial counter-example, let f(0)=1 and f(x)=0 elsewhere.





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As to the new question, yes. This is a standard theorem; see, e.g., H.L. Royden (1988), REAL ANALYSIS, p. 107.





The proof is actually rather clever, involving a sequence of approximating functions...





If this links right, you should also be able to see a proof here (theorem 5.5 or so):


http://books.google.com/books?id=JypBdru...








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