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I know that a continuous function which is a BV may not be absolutely continuous. Is there an example of such a function? I was looking for a BV whose derivative is not Lebesgue integrable but I couldn't find one.

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The Devil's staircase function does the trick.

Its derivative is almost surely zero with respect to Lebesgue measure, so the function is not absolutely continuous.

See http://mathworld.wolfram.com/DevilsStaircase.html

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    $\begingroup$ This is indeed the most standard example of a function which has BV but is not AC. It might be helpful to the OP to know that such a function is (more?) commonly referred to as the Cantor function: en.wikipedia.org/wiki/Cantor_function. $\endgroup$ – Pete L. Clark Sep 14 '10 at 23:00
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Byron already answered your main question, but your last sentence is another matter. You want a BV function whose derivative is not integrable, but such things don't exist. In particular, if $f$ is monotone on $[a,b]$, then $f'$ exists a.e., is Lebesgue integrable, and $\int_a^b f' \leq f(b)-f(a)$. Thus half of the fundamental theorem of calculus holds, so to speak. General BV functions are differences of monotone functions, so their derivatives are also Lebesgue integrable.

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  • $\begingroup$ Thanks for pointing out my error. $\endgroup$ – Digital Gal Sep 14 '10 at 22:14

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