Show Hausdorff measure $H^d$ on $R^n$ is not $\sigma$-finite if d<n

Question

Suppose $H^d$ $(0<d<n)$ is $d$-dimensional($d$ is an integer) Hausdorff measure on $R^n$, then how to show $H^d$ is not a $\sigma$ finite measure? Is it $\sigma$ finite if $d$ is not an integer?

Answer 1

HINT:

If the $d$ (outer) measure of $A$ is finite then the $d'$ measure of $A$ is $0$ for all $d'>d$. So if $H^d$ were $\sigma$ finite then the $n$ measure of $\mathbb{R^n}$ would be zero. But we know it's not so.