Content, spherical covers, and normalisation

Assume the Axiom of Countable Choice. The covering convention matters. For the diameter-power definition, replacing nonempty covering sets by their closures leaves diameters unchanged, hence gives the same infimum even at a fixed scale. Open enlargement gives the same limiting measure for positive exponents, with scale and cost slack. Finite-scale content should not be confused with the limiting measure: Content and measure have the same null sets establishes only their common null sets.

If ${\mathcal{S}}^s$ denotes the analogous limiting infimum restricted to open metric balls, then for $s>0$,

${\mathcal{H}}^s(A)\le {\mathcal{S}}^s(A)\le 2^s{\mathcal{H}}^s(A)\mathrm{.}$

For the second inequality, enclose each nonempty cover member of diameter $d_j$ in a ball about one of its points, of radius slightly greater than $d_j$. Its diameter is at most twice that radius. Choose positive enlargements with summable cost error, including when $d_j=0$; the covering scale is increased by a factor tending to two and still tends to zero. Infimisation, vanishing cost error, and the small-scale limit give the inequality. The first inequality is inclusion of cover families. The ball-only convention is called spherical Hausdorff measure; equality with arbitrary-cover measure is not a general convention equivalence (Falconer §1.2).

Recorded, not proved here. The sharper Euclidean identification for this unnormalised convention is

$c_n=\frac{2^n}{{\lambda }_n(B(0,1))}\mathrm{.}$

Fremlin 264H–I supplies the isodiametric argument and exact factor. The local theorem Euclidean Hausdorff measure is proportional to Lebesgue measure proves proportionality and elementary bounds only. No proof or example here uses the displayed exact identification for $n\ge 2$.