set $A_{\delta}$ must be finite. We will show that $\mu(\Omega \thicksim A_{\delta}) = 0$. Since $\Omega \thicksim A_{\delta}$ is open, we have $$\Omega \thicksim A_{\delta} = \bigcup_{x \in \Omega \thicksim A_{\delta}} B_{R_x}(x) .$$ Moreover, since there are no atoms in $\Omega \thicksim A_{\delta}$, for every $x$ in $\Omega \thicksim A_{\delta}$ we can choose $R_x$ in such a way that $$\mu(B_{R_x}(x)) = 0.$$ Furthermore, since $\Omega$ is separable, Lindelöf’s lemma yields $$\Omega \thicksim A_{\delta} = \bigcup_{x \in A} B_{R_x}(x),$$ where $A$ is a countable subset of $\Omega \thicksim A_{\delta} $, and $\mu( \Omega \thicksim A_{\delta} ) = 0$ follows.

To state the next lemma, given some space $F(\Omega)$ of functions on some set $\Omega$, denote by $$F_c (\Omega) :=\{\phi \in F(\Omega): \hbox{spt} \phi  \subset \subset \Omega \}$$ the subset of $F(\Omega)$ consisting of those functions whose support is a compact subset of $\Omega$.

\[aproksymacja\] Here $(X,d)$ is a locally compact metric space with two Radon measures $\mu$ and $\nu$, and $ \Omega \subset X$ is a precompact open set. Then for every $p$ and $r$ in $[1, \infty)$ the set $\hbox{Lip}_c (\Omega)$ is equidense both in $L^{r}(\Omega,\mu)$ and in $L^{p}(\Omega,\nu)$. This means that for every $\epsilon >0$ and every $f \in 