A integrable condition for the distance to boundary function Let $\Omega$ be an open bounded domain in $\mathbb{R}^{n}$. We denote
$d\left(x\right)$ the distance from $x$ to the boundary of $\Omega$,
that is 
$$
d\left(x\right):=\inf\left\{ \left\Vert x-y\right\Vert :y\in\partial\Omega\right\} .
$$
In the book of A.Kufner: Weighted Sobolev spaces, page 50, he claimed
that (without any explanation): there always exists $\varepsilon_{0}>0$
such that 
$$
\int_{\Omega}d^{-\varepsilon_{0}}\left(x\right)dx<\infty.
$$
I suspect this is not true. I think at least we need some assumption
on the regularity of $\partial\Omega$, for example $\partial\Omega$
is Lipschitz. 
My question: is the above statement is true for any
open bounded domain?
 A: No, that seems false. Let's work in $\mathbb R^2.$  Suppose $p>0.$ Let $S =(0,1)\times (0,1).$ Let $I=[a,b], 0<a<b<1.$ Consider a uniform partition of $I$ into $n$ subintervals. Above each partition point, put a closed vertical line segment of length $1/2.$ Let $\Omega$ be $S$ minus the union of these $n+1$ segments. We have removed a "comb" from $\Omega.$ You'll want to draw a picutre. By the way this $\Omega$ is just a first version. The final $\Omega$ will be fancier.
Now let $\Omega_0$ be the part of $\Omega$ that lies between these line segments (or "tines") and below the line $y=1/2.$ We have
$$\tag 1 \int_\Omega d((x,y),\partial \Omega )^{-p}\, dx\,dy > \int_{\Omega_0} d((x,y),\partial \Omega )^{-p}\, dx\,dy.$$
Consider the region between any two adjacent tines. This region will have the form $(c,d)\times (0,1/2).$ For any $(x,y)\in (c,d)\times (0,1/2),$ $d((x,y),\partial \Omega )\le x-c.$ Thus the integral over this region it at least
$$\int_0^{1/2}\int_c^d (x-c)^{-p}\, dx\, dy = \frac{1}{2}\frac{(d-c)^{1-p}}{1-p}= \frac{1}{2}\frac{(|I|/n)^{1-p}}{1-p}.$$
Here $|I|$ is the length of $I.$
Now we have $n$ such regions. Thus the integral on the right side of $(1)$ is at least
$$n^p\frac{|I|^{1-p}}{2(1-p)}.$$
This expression $\to \infty$ as $n\to \infty.$
That was just for one comb. Let's now think about removing a countable number of combs. So choose disjoint closed intervals $I_1,I_2,\dots \subset (0,1) $ travelling to $0.$ For each $I_k,$ we can choose a comb with $n_k$ tines as above in order to make the associated integral $>1.$ Our final $\Omega$ is then $S$ minus all of these combs. This $\Omega$ is open and connected, with the left side of $(1)$ equal to $\infty.$
