Lipschitz Equivalents

Question

How can I show that this proposition. Can you help me, please.

In this set, give a set $X$ different from null and $c$ and $d$ measures so that $c$ and $d$ aren't Lipschitz equivalents. but $\tau_c= \tau_d$.

Answer 1

On $\Bbb R$, $d(x,y)=|x-y|$ and $d'(x,y)=\min(|x-y|,1)$ are equivalent metrics in that $\tau_d = \tau_{d'}$ but they are not Lipschitz equivalent.