is Lipschitz with compact support, and also the measure $$\mu_n :=  \mu \llcorner ( \theta^{\frac{r}{r - p}} \mathbf{1}_{\theta \leq n} ).$$

Assuming (\[hypothesis\]) and recalling Lemma \[aproksymacja\], use the decomposition (\[decomposition\]) to obtain $$\label{ineq-rp}
\| \phi_n \|_{L^r(\Omega,\mu)} \leq C \| \phi_n \|_{L^p(\Omega,\nu)} = C \| \phi_n \|_{L^p(\Omega,  \mu \llcorner \theta + \sigma)} 
=  C \| \phi_n \|_{L^p(\Omega,  \mu \llcorner \theta )} .$$ However $$\label{eq-p}
\| \phi_n \|^p_{L^p(\Omega,  \mu \llcorner \theta )} =
\int_{\Omega} \left| \psi \right|^p \theta^{\frac{p}{r-p}} \mathbf{1}_{\theta \leq n} \theta d \mu = 
\int_{\Omega} \left| \psi \right|^p \theta^{\frac{r}{r-p}} \mathbf{1}_{\theta \leq n} d \mu =
\| \psi \|^p_{L^p(\Omega , \mu_n)} ,$$ and similarly $$\label{eq-r}
\| \psi \|_{L^r(\Omega , \mu_n)} = \| \phi_n \|_{L^r(\Omega, \mu)} .$$

Then use (\[eq-p\]) and (\[eq-r\]) in (\[ineq-rp\]) to infer that $$\| \psi \|_{L^r(\Omega , \mu_n)} \leq C \| \psi \|_{L^p(\Omega ,  \mu_n)}$$ for every $n$.\
Hence by Step 1 $$\mu_n = \sum_{i \in I_n} \mu_{n,i} \delta_{x_{n,i}}$$ for every $n$. Recall the definition of the measures $\mu_n$, and note that $\hbox{spt} \mu_n \subset \hbox{spt} \mu_{n+1}$, in particular $I_n \subset I_{n+1}$. Let $I=\bigcup_{n=1}^{\infty}I_n$ and define $x_i : =x_{n,i} \big|_{I_n}$; one can write $$\mu_n = \sum_{i \in I_n} \mu_{n,i} \delta_{x_i}.$$ Since $\mu_{n,i} =\mu_n (\{x_i\})\leq  \mu_{n+1} (\{x_i\})=\mu_{n+1,i} $, it 