Could someone provide some intuition behind the $n$-dimensional Minkowski Contentthe $n$-dimensional upper Minkowski Content of $\mathcal{A}$ as $$\mathfrak{M}^{*m} (\mathcal{A}) : = \lim_{\epsilon \to 0} \sup \frac{\mu(\mathcal{A})_\epsilon - \mu(\mathcal{A})}{\alpha_{n-m} \epsilon^{n-m}}$$ we define the $n$-dimensional lower Minkowski Content of $\mathcal{A}$ as $$\mathfrak{M}^{m}_* (\mathcal{A}) : = \lim_{\epsilon \to 0} \inf \frac{\mu(\mathcal{A})_\epsilon - \mu(\mathcal{A})}{\alpha_{n-m} \epsilon^{n-m}}$$

where $\alpha_{n-m}$ is a $n-m$ dimensional sphere and $\mu$ is the standard Lebesgue Measure.

Could someone please provide some intuition for this definition?

What does it mean to talk about the Minkowski content of a natural object, i.e. a sphere in euclidean space? What would be the Minkowski content for such an object?

Still looking for a clear answer from someone...

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    $\begingroup$ The $m^{th}$ Minkowski content $\mathfrak{M}(\mathcal{A})$ is something which when computed for regular enough $m$-dim object $\mathcal{A}$ embedded in $\mathbb{R}^n$, give you back the $m$-dim volume of $\mathcal{A}$. Just think of them as a way to generalize length/area/volume to complex geometric shapes. To appreciate this, just compute it for some simple geometric objects. $\mathfrak{M}^1$ for a line segment gives you its length. $\mathfrak{M}^2$ for a planar polygon gives you its surface area. $\mathfrak{M}^{n-1}$ for a $(n-1)$-dim sphere gives you its surface area and so on. $\endgroup$ – achille hui Dec 14 '13 at 6:01
  • $\begingroup$ @achillehui What then is meant by the Minkowski Content of a fractal object? $\endgroup$ – Anthony Peter Dec 14 '13 at 18:43

The Minkowski content is a rather simplistic way to define an $m$-dimensional measure of an object immersed in the euclidean space (however mathematically speaking Minkowski content is not a measure).

If $\mathcal A$ is a regular $m$-dimensional surface in $\mathbb R^n$ it is easy to understand that the Minkowski content (both lower and upper) gives the $m$-dimensional area of the surface. In fact the surface is locally close to its tangent plane, which is a $m$-dimensional affine space in $\mathbb R^n$. The fattening of the $m$-plane gives a cylinder whose section is a $(n-m)$-dimensional ball of radius $\epsilon$. So the volume of $(\mathcal A)_\epsilon$ is approximately given by the $m$-dimensional surface area of $\mathcal A$ times the volume of the section, which is $\alpha_{n-m} \epsilon^{n-m}$.

To make a formal proof I think that a $C^2$ manifold $\mathcal A$ with an upper bound on the curvature should be enough to get the result.

There is also a physical intuition behind the definition. Suppose you want to know how long is a rope. You can simply compute the volume occupied by the rope (either by its weight and density, or by immersing it in water) then you divide the volume by the area of the section (i.e. $\pi r^2$ if $2r$ is the diameter of the section) to find the length. You may notice that bending the rope could change the volume, and that the rope is not exactly the set of points with distance less than $r$ by the transversal axis. However if $r$ goes to $0$ you can imagine that this errors become negligible.

A similar experiment can be imagined to compute the surface area of a thick surface. Just compute the volume and divide by the thickness.

Actually in your definition you are subtracting the volume $\mu(\mathcal A)$ from the volume of the fattened object $\mathcal A$. This makes the definition a little bit different in the case when $\mu(\mathcal A)>0$. For example if you have an $n$-dimensional object (let say a set with regular boundary) what you get is half the measure of the boundary, not the measure of the whole object. This because $$ \mu((\mathcal A)_\epsilon) - \mu(\mathcal A) = \mu((\mathcal A)_\epsilon \setminus \mathcal A) = \mu((\partial \mathcal A)_\epsilon \setminus \mathcal A) \approx \frac 1 2 \mu((\partial \mathcal A)_\epsilon). $$ This also could be understand from a practical point of view. Imagine you want to measure the surface area of an irregular object. A possibility is to immerse the object in some dense paint and measure the mass of painting which remains attached to the object. This you expect to be proportional to the surface area of the object.

If the set $\mathcal A$ is not regular you can get very bad results. Consider for example $\mathcal A = \mathbb Q^3 \cap [0,1]^3$. Such a set is composed by a countable number of points, hence its Hausdorff dimension is $0$. Instead the Minkowski content (for any $m$) is the same of the cube $[0,1]^3$ since $(\mathcal A)_\epsilon \supset [0,1]^3$ for all $\epsilon>0$. This example shows that Minkowski content is not additive since $$\mathfrak M^m(\mathcal A) = \mathfrak M^m([0,1]^3\setminus \mathcal A) = \mathfrak M^m([0,1]^3)$$

However in applications it might be true that Minkowski content makes sense even at an $\epsilon$ level. For example consider the problem of computing the length of the coast of Great Britain (this is the example provided by Mandelbrot, leading to fractals). The problem, like this, does not make sense, because as you inspect the coast at smaller and smaller details, you always get a larger total length. However if you fix a scale $\epsilon$, the Minkowski content gives you an answer. For example if you are interested in the length of the coast to know how long a road following the coast should be, you should take $\epsilon$ to be approximately the distance of the road from the coast.

In general I expect that the Minkowski content should work also for fractal dimension. I'm not absolutely sure but I think that in many cases it should give the same result as Hausdorff measure, hence could be used to measure fractal dimension of objects. For example I expect that Minkowski content (both lower and upper) should correctly measure the dimension (and measure) of fractals like: Cantor-set, Sierpinksi carpet, Koch snowflake...

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  • $\begingroup$ What is fattening of the m-plane ? My English is poor, can't understand it. Could you give a picture to describe it ? Thanks. $\endgroup$ – lanse7pty Mar 28 '19 at 4:41

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