Continuous and bounded variation does not imply absolutely continuous

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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