I am charged with the following
Assume $f:[a,b] \rightarrow \mathbb{R}^n$ is of bounded variation. Show that for all $0<h<b-a$
$\int_a^{b-h} |f(x+h) - f(x)| dx = h V_f(a,b)$
I am allowed the fact that $f$ having bounded variation implies that $|f'|$ is integrable. It seems like that should play a role but I'm having difficulty putting it together. Any help would be greatly appreciated.
The equality doesn't hold.
Take for $f$ the sine function on the interval $[0, 4\pi]$ and $h=2\pi$. You have
$$\int_a^{b-h} |f(x+h) - f(x)| dx =\int_0^{2\pi} \vert \sin(x+2\pi)-\sin x \vert \ dx=0$$ while $h V_f(0,4\pi)=16\pi$.