Content and measure have the same null sets

Suppose $s>0$ and ${\mathcal{H}}_{\mathrm{\infty }}^s(A)=0$. Given finite $\delta >0$ and $\epsilon >0$, take a cover with cost less than $\min (\epsilon ,{\delta }^s)$. Each diameter is strictly below $\delta$, so ${\mathcal{H}}_{\delta }^s(A)<\epsilon$. Therefore every scale value is zero.

When $s=0$, any nonempty member costs one. Content less than one forces the empty family, hence $A=\mathrm{\varnothing }$. All its scale values are zero. This deals with empty and singleton possibilities without division by the exponent.

If all finite-scale values vanish their supremum vanishes. Conversely, a zero supremum forces all those nonnegative values, and then the smaller unrestricted content, to vanish. These implications complete both equivalences.