Continuous and bounded variation does not imply absolutely continuous

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

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.