If $f\colon\Bbb R\to\Bbb R$ then $\operatorname{var}(f, [a,b]):=\sup \{\sum_{k=1}^n |f(x_k)-f(x_{k-1})| \}$, where supremum is taken over all finite sequences $(x_k)$ such that $a=x_0
Let $f:R \rightarrow R$ be $1$-periodic and $c \in R$.
Is it then $\operatorname{var}(f, [0,1])=\operatorname{var}(f, [c, c+1])$ ?