Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
My apologies; I only saw Byron's when I posted that comment. –  Akhil Mathew Sep 15 '10 at 2:56

3 Answers 3

up vote 9 down vote accepted

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

share|improve this answer
4  
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. –  Pete L. Clark Sep 14 '10 at 23:00

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.

share|improve this answer
    
Thanks for pointing out my error. –  Digital Gal Sep 14 '10 at 22:14

f(x)=[x] is Of bounded variation on [0,1] but noncontinuous and not abs. cont.

share|improve this answer
2  
The question asks for a continuous function. –  rschwieb Dec 7 '12 at 17:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.