Minkowski content of a Cantor-like fractal

Question

Let $K_0 = [0,1]$. Split $K_0$ into 4 intervals and remove the middle half. This gives $K_1 = [0,1/4] \cup [3/4, 1]$ and so on and set $K = \cap K_i$. I computed the upper and lower Minkowski content of K and found that they are different, but how could I prove this without evaluating them? Many thanks in advance!

Answer 1

The $s$-dimensional upper and lower Minkowski contents, up to some constant factor (same for both), are $$\limsup_{r\to 0}\ r^{s-1}|K_r| \quad \text{ and }\quad \liminf_{r\to 0}\ r^{s-1}|K_r| $$ where $|K_r|$ is the Lebesgue measure of the $r$-neighborhood of the set $K$. It suffices to exhibit two sequences $r_n\to 0$ and $r_n'\to 0$ along which $|K_r|$ follows two different patterns.

When $r = 4^{-n}$, the set $K_r$ is the $r$-neighborhood of $K_{n-1}$ (all gaps of size $2r$ and smaller are closed). Therefore, $K_r $ consists of $2^{n-1}$ intervals of size $4^{1-n}+2r$. This adds up to $$|K_r| = 2^{1-n} + 2^{-n} = 3\cdot 2^{-n},\qquad r = 4^{-n} \tag1$$

When $r = 2 \cdot 4^{-n}$, the set $K_r$ still consists of $2^{n-1}$ intervals of size $4^{1-n}+2r$. This adds up to $$|K_r| = 2^{1-n} + 2^{1-n} = 4\cdot 2^{-n},\qquad r = 2\cdot 4^{-n} \tag2$$

Since $s=1/2$, the values of $r^{s-1}|K_r|$ between (1) and (2) are different.

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Nothing special about $2$ here: $r = x\cdot 4^{-n}$ would work for $x\in (1,4)$. The idea is that in this range of scales the size of $r$-neighborhood grows linearly, which does not exactly match the fractional dimension of the set.