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Here the lower box dimension and the upper box dimension are exactly what Falconer talks about in his book "Fractal Geometry". I already know that the upper box dimension is finitely stable since $N_\delta(A\cup B)\le N_\delta (A)+N_\delta(B)$, where $N_\delta(A)$ is defined as usual, and Falconer gives an example to show the box dimension may not be countably stable. (We know that, however, the Hausdorff dimension is countably stable).

But now I want to find the example to show the lower box dimension may not even be finitely stable.It seems that we can use von Koch curve to illustrate this, but what is the explicit explanation for this? Since the lower box dimension is monotonic, we need to find set A and B and show that $\underline{\dim}_B(A\cup B)>\max\{\underline{\dim}_B(A),\underline{\dim}_B(B)\}$.

Martin Sleziak
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Bob brant
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    You need two sets, $A$ and $B$, with the property that $$\frac{\log(N_{\varepsilon}(A))}{\log(1/\varepsilon)} : : \text{ and } : : \frac{\log(N_{\varepsilon}(B))}{\log(1/\varepsilon)}$$ both oscillate as $\varepsilon \searrow 0$. You also need one to be small when the other is large. An explicit example showing how to construct one such set is here. Note that the Koch curve is far too regular, as is any self-similar set. – Mark McClure May 13 '17 at 10:48

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Recall that the standard middle-third Cantor set $C$ has lower box/Minkowski dimension $d=\log 2/\log 3$ (same as upper, same as Hausdorff). One can reduce this dimension by changing the construction so that at certain steps we throw out a lot more than middle-third. Let's say we remove middle $\lambda_k/3$ portion instead of $1/3$ at step $n=n_k$. The sequence $\{n_k\}$ will grow super-exponentially, for example $n_k = 2^{2^k}$.

The resulting set, call it $A$, can be covered by $N = 2^{n_k}$ intervals of length $$\epsilon = 3^{-n_k}\prod_{j=1}^k \lambda_j$$ We'll choose these so that $\log N /\log(1/\epsilon) \to d/2$. So, we need $$ \frac{n_k \log 2}{n_k\log 3 + \sum_{j=1}^k \log(1/\lambda_j)} \to \frac{d}{2}$$ which can be achieved by choosing $\lambda_k$ so that $\log(1/\lambda_k) = n_k \log 3$. So, $\lambda_k = 3^{-n_k}$. Note that the sum over $j$ is asymptotic to its largest term, due to super-exponential growth of $n_k$.

What happens if we try to cover this set by elements of a different scale? After $n_k+m$ steps, where $n_k+m < n_{k+1}$, there are $2^{n_k+m}$ intervals of length $$\epsilon = 3^{-n_k-m}\prod_{j=1}^k \lambda_j$$ so $$ \frac{\log N}{\log(1/\epsilon)} = \frac{(n_k+m) \log 2}{(n_k+m)\log 3 + \sum_{j=1}^k \log(1/\lambda_j)} \sim \frac{(n_k+m) \log 2}{(2n_k+m)\log 3} $$ When $m\ge n_k$, this quotient is at least $2d/3$.

The upshot is that the lower box dimension of $A$ is $d/2$ but we can only achieve that by covering it at the scales between $$ \epsilon_k = 3^{-n_k} \prod_{j=1}^k 3^{-n_j} \quad\text{and}\quad 3^{-n_k} \epsilon_k $$ which is a relatively narrow range of scales, considering that $\epsilon_{k+1} = 3^{-n_{k+1}} \prod_{j=1}^{k+1} 3^{-n_j}$ is much smaller than $3^{-n_k} \epsilon_k$. Outside of this range of "convenient" scales, the covering numbers yield the dimension of at least $2d/3$.

It remains to repeat this construction but using $\tilde n_k=2n_k$ instead. The resulting set, call it $B$, will again have lower box dimension $d/2$. But the union $A\cup B$ has lower box dimension at least $2d/3$, because the "convenient" scales for $A$ are not "convenient" for $B$.