  Consider a regular domain $\Omega \subset \mathbb{R}^N$ and let
$d(x)=\operatorname{dist}(x,\partial\Omega)$. Denote $L^{1,\infty}_a(\Omega)$
the space of functions from $L^{1,\infty}(\Omega)$ having absolutely continuous
quasinorms. This set is essentially smaller than $L^{1,\infty}(\Omega)$ but, at
the same time, essentially larger than a union of all $L^{1,q}(\Omega)$,
$q\in[1,\infty)$.
  A classical result of late 1980's states that for $p\in (1,\infty)$ and $m
\in \mathbb{N}$, $u$ belongs to the Sobolev space $W^{m,p}_0(\Omega)$ if and
only if $u/d^m\in L^p(\Omega)$ and $\left|\nabla^m u\right|\in L^p(\Omega)$.
During the consequent decades, several authors have spent considerable effort
in order to relax the characterizing condition. Recently, it was proved that
$u\in W^{m,p}_0(\Omega)$ if and only if $u/d^m\in L^1(\Omega)$ and
$\left|\nabla^m u\right|\in L^p(\Omega)$. In this paper we show that for
$N\geq1$ and $p\in(1,\infty)$ we have $u\in W^{1,p}_0(\Omega)$ if and only if
$u/d\in L^{1,\infty}_a(\Omega)$ and $\left|\nabla u\right|\in L^p(\Omega)$.
Moreover, we present a counterexample which demonstrates that after relaxing
the condition $u/d\in L^{1,\infty}_a(\Omega)$ to $u/d\in L^{1,\infty}(\Omega)$
the equivalence no longer holds.
