Hausdorff dimension is the unique critical exponent

If $s<d$ and ${\mathcal{H}}^s(A)$ were finite, choose a finite $t$ strictly between $s$ and $d$ (also possible for $d=\mathrm{\infty }$). Then ${\mathcal{H}}^t(A)=0$, contrary to $d$ being the infimum of the zero exponents. Hence ${\mathcal{H}}^s(A)=\mathrm{\infty }$.

If $d<s<\mathrm{\infty }$, the zero-exponent set is nonempty and contains $u<s$ by its infimum property. Exponent comparison gives ${\mathcal{H}}^s(A)=0$. Thus when $d=0$ all positive exponents vanish, and when $d=\mathrm{\infty }$ every finite exponent has infinite measure.

The first two steps place every finite-measure exponent at least $d$, and every exponent strictly greater than finite $d$ among the finite-measure exponents. Their infimum is $d$, also when the set is empty. Similarly the infinite-measure exponents lie at most $d$ and contain every nonnegative exponent strictly below $d$. Their supremum is $d$; for $d=0$ it is zero whether that set is empty or consists of zero. Finite positive measure at $s$ excludes $s<d$ and $s>d$, hence forces equality. Empty $A$ has $d=0$ and no infinite-measure exponent.