TITLE: Continuous and bounded variation does not imply absolutely continuous
QUESTION [15 upvotes]: 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.

REPLY [18 votes]: 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

REPLY [9 votes]: 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.