Minkowski content

Remarks

The limit superior and limit inferior always exist in $[0, \infty]$. The same formulas with a real exponent $s$ in place of $m$ are also used, with a suitable constant.

Example: let $A \subseteq \R^2$ be a segment of length $L$ and $m = 1$. Then $A_r$ is a rectangle of area $2 r L$ with two half-discs of radius $r$, so $\mathcal{L}^2(A_r) = 2 r L + \pi r^2$. Since $\alpha(1) = 2$, the ratio is $L + \pi r / 2 \to L$, so $\mathcal{M}^1(A) = L$.

Minkowski content does not ignore countable sets. For $A = \Q \cap [0, 1]$ in $\R$ with $m = 1$, one has $A_r = (-r, 1 + r)$, so $\mathcal{M}^1(A) = 1$, although $A$ has Lebesgue measure $0$. Thus Minkowski content is not countably additive.

For a closed set $A \subseteq \R^n$ that is the image of a bounded subset of $\R^m$ under a Lipschitz map, $\mathcal{M}^m(A)$ exists and equals its normalized $m$-dimensional Hausdorff measure.