Is the lower Minkowski content additive in any sense?

Question

Proposition 3.3.2 in the book The geometry of domains in space by S. Krantz and H. Parks states that if the sets $A$ and $B$ are separated by a positive distance, then $\mathcal{M}_*^K(A \cup B) = \mathcal{M}_*^K(A) +\mathcal{M}_*^K(B)$, where $\mathcal{M}_*^K$ denotes the the $K$-dimensional lower Minkowski content. The proof is left to the reader. Is it true, and if yes, how is it proved?

Answer 1

Notation follows LINK.

For $r$ small enough (less than half the separation between $A$ and $B$), we have $$ \frac{\mu(\{x\;:\;d(x,A \cup B) < r\})}{\alpha(n-m)r^{n-m}} =\frac{\mu(\{x\;:\;d(x,A) < r\})}{\alpha(n-m)r^{n-m}} +\frac{\mu(\{x\;:\;d(x,B) < r\})}{\alpha(n-m)r^{n-m}} $$ but then we take the liminf as $r \to 0^+$. If the liminf and limsup disagree, we cannot conclude that the liminf of the sum is the sum of the liminfs. So then what?

No, it need not be additive.

For our example, we take the $1$-dimensional lower Minkowski content in $\mathbb R^2$. For a set $A \subseteq \mathbb R^2$ $$ M^1_*(A) = \liminf_{r \to 0+}\frac{\mu(\{x: d(x,A)<r\})}{2r} \tag{1}$$ The denominator here is the one-dimensional measure of a $1$-ball of radius $r$; and $\mu$ is $2$-dimensional Lebesgue measure.

We will first consider the self-similar set $A$ in the plane: three maps are similarities with ratio $1/3$ centered at the vertices of the equilateral triangle with side $1$. We have chosen this so that its Hausdorff dimension (and Minkowski dimension) is $1$. Here is a picture:

Now consider the set $A_r := \{x : d(x,A) < r\}$ for various values of $r$, $0<r<1/2$:

Because of the self-similarity, when $r$ shrinks by factor $1/3$, the area of $A_r$ also shrinks by factor $1/3$. (The first and last of these pictures have $r=1/2$ and $r=1/6$ to illustrate this.) It follows (as Lalley first observed) that $$ \mu(A_r) = p(-\log_3 r) \cdot r,\qquad 0 < r < \frac{1}{2} $$ where $p$ is a function with period $1$. So $p(-\log_3 r)$ is something like this:

We can observe by computing two of these values accurately enough that function $p$ is not constant. So this is a case with limsup and liminf different in $(1)$.

Now for our set $B$ we take a copy of $A$, translated far away so there is no overlap, and expanded by a factor $q$. Then $$ \mu(B_r) = q^2 \mu(A_{r/q}) = q^2 p(-\log_3 r+\log_3 q) \cdot r,\qquad 0 < r < \frac{1}{2} $$ So that we translate the argument of funtion $p$ and multiply by a constant factor. Now we choose the factor $q$ so that, in the translation of $p$, the minimum of $p$ concides with the maximum of translated $p$:

When we do this, the liminf of the sum is not the sum of the liminf.