I know that every absolutely continuous functions are of bounded(finite) variations but converse need not be true. and the cantor function is wellknown example of function of bounded variation which is not absolutely continuous. I want to know some other examples. Please help me! Thanks in advance!
closed as off topic by George Lowther, Bill Johnson, Emil Jeřábek, Willie Wong, S. Carnahan♦ Jun 22 '12 at 15:48
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$\begingroup$ math.stackexchange.com/questions/4683/… $\endgroup$ – Michael Greinecker Jun 22 '12 at 13:13

4$\begingroup$ And if you don't require continuity, just take a simple jump: $f(x)=0$ for $x<0$ and $f(x)=1$ for $x \ge 1$. $\endgroup$ – Gerald Edgar Jun 22 '12 at 13:24

3$\begingroup$ Other examples: $f(x)=\mu([0,x))$, where $\mu$ is a measure supported in a Lebesgue null set. The Cantor function is of this form, and in fact every BV function, up to removing an absolutely continuous part. $\endgroup$ – Pietro Majer Jun 22 '12 at 14:05

$\begingroup$ When you are looking for a counterexample in analysis you should always have a look at the very beautiful "counterexamples in analysis" by Gelbaum and Olmsted (books.google.de/…). It is very likely it contains what you are looking for. $\endgroup$ – Ricky Jun 22 '12 at 14:08
A real valued function $f:[a,b]\rightarrow \mathbb {R} $ is of bounded variation iff it is the difference of two bounded monotonically increasing functions. This gives you many  in fact all real real valued  examples.
(See, eg., Rudins 'Real and Complex Analyis', Exercise 7.13)