Cantor staircase function is not BV if we redefine it to be zero outside the Cantor set The Cantor staircase function has bounded variation. Is it true that Cantor staircase function is _not_ BV _if we redefine it to be zero outside the Cantor set_?

Yes. Let $C$ denote the Cantor set, $c(x)$ the Cantor staircase function, and $f(x)$ your modified Cantor staircase function that is zero outside the Cantor set. There exist infinitely many points $x \in C$ for which $c(x)> \frac{1}{2}$, and thus $f(x)>\frac{1}{2}$. Between any two points $x, y \in C$, we can find $z \
otin C$ (and we have $f(z)=0$).

Now bounded variation on $[0,1]$ says there exists $M>0$ such that for any choice of values $0=x_0