follows that for each $i$ the number $\mu_{n,i}$ is non decreasing with respect to $n$.\
Denote by $\mathcal{M}(\Omega)$ the set of measures on $\Omega$ endowed with the weak-$\ast$ topology. Let $\tilde{\mu}_n= \mu_n \llcorner (\theta^{-\frac{r}{r-p}}\mathbf{1}_{\{\theta>0\}})$, and observe that $\tilde{\mu}_n \rightarrow \mu \llcorner \mathbf{1}_{\{\theta>0\}}$ in $\mathcal{M}(\Omega)$. We claim that $$\tilde{\mu}_n \rightarrow  \mu .$$ To prove that, it suffices to show that $\mu(\{\theta=0\})=0$. Since $\mu$ is singular with respect to $\sigma$, there exist subsets $A$ and $B$ with $A \cap B =\emptyset$, such that for every measurable set $E$ we have $\mu(E)= \mu (A\cap E)$ and $\sigma(E)= \sigma (B\cap E)$. Therefore $$\int_{\Omega}\mathbf{1}_A \mathbf{1}_{\{\theta=0\}} d \nu =\int_{\Omega}\mathbf{1}_A \mathbf{1}_{\{\theta=0\}} \theta d \mu + \int_{\Omega}\mathbf{1}_A \mathbf{1}_{\{\theta=0\}} d \sigma=0,$$ hence $\nu (A\cap \{\theta=0\})=0$, and using (\[hypothesis\]) $$\| \mathbf{1}_{A\cap \{\theta=0\}} \|_{L^r(\Omega,\mu)} \leq C \| \mathbf{1}_{A\cap \{\theta=0\}} \|_{L^{p}(\Omega,\nu)}=0.$$ Thus $\mu ( \{\theta=0\})=\mu (A\cap \{\theta=0\})=0$, as required.

Now we continue with Lemma \[prod\]:

(Hajłasz-Leibniz) \[prod\] If $v \in M^p_1(X,\mu)$ and $\phi \in \hbox{Lip}(X)$, then $f = v \phi \in M^p_1(X,\mu)$. Moreover, $$g_f = g_v |\phi| + |v|\|\phi\|_{\hbox{Lip}}$$ is a Hajłasz gradient for $v \phi$.

The result follows from the string of inequalities $$\begin{aligned}
  |v(x)\phi(x) - v(y)\phi(y)| &\leq& |v(x)-v(y)|\ \min\{|\phi(x)|,|\phi(y)|\} + \max \{|v(x)|,|v(y)| \}\ |\phi(x)-\phi(y)|\\
  &\leq& \big( g_v(x)+g_v(y) \big)\  \min \{ 