I know that every absolutely continuous functions are of bounded(finite) variations but converse need not be true. and the cantor function is well-known example of function of bounded variation which is not absolutely continuous. I want to know some other examples. Please help me! Thanks in advance!
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$\begingroup$ math.stackexchange.com/questions/4683/… $\endgroup$– Michael GreineckerCommented Jun 22, 2012 at 13:13
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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 EdgarCommented Jun 22, 2012 at 13:24
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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 MajerCommented Jun 22, 2012 at 14:05
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$\begingroup$ 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. $\endgroup$– RickyCommented Jun 22, 2012 at 14:08
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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)
