Define $\rho:\mathbb{R}\to[0,\infty)$ by $\rho((a,b])=(b-a)^\alpha$ for some $\alpha>1$, need to show the outer measure $\mu^*$ induced by $\rho$ is trivial.
I think it is sufficient to show $\mu*([0,1])=0$, but I have no idea where to start. Could someone please give me a hint?
Cover $[0,1]$ by two intervals with length close to $1/2$. Take $\alpha=2$ for example. Then the sum of their lengths is close to $1/2$. If you now cover it by three intervals with length close to $1/3$, the sum is close to $3(1/3)^2=1/3$ and so on. Therefore the ínfimum defining the outer measure is zero. The same works for any $\alpha>1$. Basically we are saying that any $\alpha$- fractal measure is zero, just like the area or the volume of a segment is zero.