Is the measure density condition a necessary condition for bounding the Sobolev norm $W^{n,p}(\Omega)$ by the extremal terms?

Let $$\Omega \subseteq \mathbb{R}^M$$ be a measurable subset of positive measure. R. A. Adams and J. Fournier in their article have proven that if $$\Omega$$ satisfies the so-called weak cone property, then the Sobolev space $$W^{n,p}(\Omega)$$ norm $$$$\lVert f \rVert_{W^{n,p}(\Omega)} = \sum_{\alpha \colon \lvert \alpha \rvert \le n } \lVert D^\alpha f \rVert_{ L^p(\Omega)}$$$$ can be bounded by only considering the extreme terms, i.e. that there exists $$C > 0$$ such that: $$$$\forall f \in W^{n,p}(\Omega) \qquad \lVert f \rVert_{W^{n,p}(\Omega)} \le C \left( \lVert f \rVert_{L^{p}(\Omega)} + \sum_{\alpha \colon \lvert \alpha \rvert = n } \lVert D^\alpha f \rVert_{ L^p(\Omega)}\right).$$$$

There are spaces for which this bound does not hold. Indeed, let $$I_k = (k - 2^{-k}, k + 2^{-k} )$$ for $$k \in \mathbb{N}$$ and let us consider a space $$\Omega = \bigcup_{k =1}^\infty I_k$$. We can notice that $$\lvert \Omega \rvert = 1$$. Let us consider a sequence of functions $$f_K$$ defined by the formulas: $$$$f_K(x) = \sum_{k=1}^K 2^k(x-k) 1_{I_k}.$$$$ Then we have that $$f_K \in W^{n,p}(\Omega)$$ for all $$n \in \mathbb{N}$$ and $$p \in [1, \infty)$$. Since $$\lvert f_K \rvert \le 1$$, this means that $$\lVert f_K \rVert_{L^p(\Omega)} \le 1$$. We also have that $$\lVert D f \rVert_{L^p(\Omega)} = K^{\frac{1}{p}}$$ and $$\lVert D^m f \rVert_{L^p(\Omega)} = 0$$ for $$m > 1$$.

Therefore, we see that the left-hand side of the previously mentioned bounded diverges to $$\infty$$, when $$K \to \infty$$. while the right-hand side (assuming the existence of appropriate $$C$$), remains bounded.

Set $$\Omega$$ has a rather property that it does not satisfy the measure density condition, i.e. that there exists a constant $$L>0$$ such that for all $$x \in \Omega$$ and all balls $$B(x, r)$$ with $$r \le 1$$ we have: $$$$\lvert \Omega \cap B(x,r) \rvert \ge Lr^M,$$$$ where in our case we have $$M=1$$.

Question:

I would like to know whether there are domains $$\Omega$$ such that they do not satisfy the measure density condition, but they do allow for the mentioned bounding of the Sobolev norm. In other words, whether the measure density condition is a necessary condition for the bounding to occur.

No, the measure density condition is not necessary I would say. Possibly there are more precise arguments, but I would argue as follows, in a nutshell: The desired inequality can be proven using Ehrling's lemma if we have the compact embedding $$W^{n,p}(\Omega) \hookrightarrow W^{n-1,p}(\Omega)$$ at our disposal, and this in turn follows from compactness of $$W^{1,p}(\Omega) \hookrightarrow L^p(\Omega)$$.
However, you can have the latter also for domains which do not submit to a measure density condition, so in particular for non-Sobolev-extension domains. The prime example would be that of domains with sufficiently mild outwards cusps, where one maybe does not have the full range of Sobolev embeddings, but still $$W^{1,p}(\Omega) \hookrightarrow L^q(\Omega)$$ for some $$q>p$$. This is sufficient to obtain a compact embedding into $$L^p(\Omega)$$. (Write $$u = w + v$$ such that $$w \in W^{1,p}_0(\Omega')$$ on a subset $$\Omega'$$, this space is compactly embedded into $$L^p(\Omega')$$, and by making $$\Omega\setminus \Omega'$$ small you can leverage the gap $$q>p$$ to make the $$L^p(\Omega)$$ norm of $$v$$ as small as desired.)