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I know that differentiability implies the existence of the derivative as a function. But I'm curious — can this derivative somehow be "bad" in the sense of not being measurable?

More precisely, is it possible to construct a function such that is differentiable everywhere, but its derivative is not Lebesgue measurable?

Or does differentiability inherently guarantee measurability of the derivative?

If not, what's the obstruction — is it something about the Borel structure or the construction of measurable functions that prevents such an example?

dom_cobb
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1 Answers1

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To elaborate on geetha290km's comment, if $f$ is differentiable everywhere then it is continuous everywhere, so $$f_n(x) := {f(x+1/n) - f(x)\over 1/n}$$ is continuous everywhere. The derivative $f'(x) = \lim_{n\to\infty} f_n(x)$ is therefore the pointwise limit of continuous (in particular, measurable) functions and is therefore measurable. The technical term for a pointwise limit of continuous functions is Baire class 1.

The derivative, however, can fail to be Lebesgue integrable. See for example Counterexamples in Probability and Real Analysis by Gary L. Wise and Eric B. Hall.

Timothy Chow
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